Foundations of Chemical and Biological Engineering I by Jonathan Verrett is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.
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This book seeks to integrate interactive open educational resources (OER) to support learning within and outside class in Chemical & Biological Engineering (CHBE) 220 – Founding Principles in Chemical and Biological Engineering I.
CHBE 220 introduces chemical engineering design and has students practice design through a course-long design project. This is a mandatory course in the CHBE degree program, which about 120 students take per year. Students work in teams of four to design and analyze various components of a chemical process. Process topics studied in the past include carbon capture from power plant emissions and ammonia production for use in fertilizers, to name a few.
The topics introduced in the lectures are constructed according to the timeline of the engineering design process, as shown below.
First, we would like to thank the UBC Vancouver students who supported this project through the Open Educational Resources Fund administered by the Centre for Teaching, Learning and Technology at UBC Vancouver. The support made this project a possibility.
We would also like to thank UBC Library staff, particularly Erin Fields and Stephanie Savage, for all the help and support they’ve offered us while writing this book, specifically around resource curation and licensing.
Finally, we would like to acknowledge the team at BCcampus for the useful guides they created on publishing open textbooks open educational resource content for educators and students. As well as the support they offer through the access to Pressbooks.
BCcampus Open Education began in 2012 as the B.C. Open Textbook Project with the goal of making post-secondary education in British Columbia more accessible by reducing student costs through the use of openly licensed textbooks and other OER. BCcampus supports the post-secondary institutions of British Columbia as they adapt and evolve their teaching and learning practices to enable powerful learning opportunities for the students of B.C. BCcampus Open Education is funded by the British Columbia Ministry of Advanced Education, Skills & Training, and the Hewlett Foundation.
Open textbooks are open educational resources (OER) created and shared in ways so that more people have access to them. This is a different model than traditionally copyrighted materials. OER are defined as teaching, learning, and research resources that reside in the public domain or have been released under an intellectual property license that permits their free use and re-purposing by others."Open Educational Resources," Hewlett Foundation, https://hewlett.org/strategy/open-educational-resources/ (accessed September 27, 2018). Our open textbooks are openly licensed using a Creative Commons license and are offered in various e-book formats free of charge, or as printed books that are available at cost. For more information about this project, see the BCcampus Open Education website.This section is adopted from David W. Ball, Jessie A. Key, Introduction to Chemistry - 1st Canadian Edition, 2014
Pressbook can be exported in various types of files. However, the codes to show mathematical equations are not compiled in some of the file types. We recommend viewing this textbook online because it shows the best format for displaying mathematical equations.
This book is divided into 8 chapters. Once you select a chapter, there will be a short introduction text along with the chapter’s learning objectives. The learning objects can help guide your learning as you go through each chapter. Here is an example:
In this chapter, we will introduce some of the basics in the design aspect of chemical and biological engineering. We will focus on basic engineering and general engineering economics. This chapter also introduces the components and characteristics of 4 key process diagrams.
Learning Objectives
By the end of this chapter, you should be able to:
Define the components of a design basis memorandum
Apply basic measures of economic merit to perform initial estimates of project economic feasibility
Recognize the 4 key types of flow sheets
Create input-output diagrams
Construct block flow diagrams (BFDs) for chemical processes
Define common unit operations and use these when constructing process flow sheets
Recognize the need to apply basic material balances
Build process flow diagrams (PFDs)
Understand piping and instrumentation diagrams (P&IDs)
We’ve also included terms you should take note of or keep in mind while you’re going through the chapter:
As you’re going through this chapter, here are some important terms for you to take note of:
Each chapter has sections focusing on more specific topics within the chapter. Each section in the chapter starts off with the learning objectives that are specific to that section.
Learning Objectives
By the end of this chapter, you should be able to:
In each section, we’ve included 2 types of questions. The first question type is an “Example” (purple text box). These examples are used as an educational guide to demonstrate how problems in this section are approached. You can use these examples to learn how questions are solved step by step. The second question type is an “Exercise” (blue text box). These exercises are made for you to test your knowledge. Try solving the question before looking at the solution to see how much you know and what concepts you should review.
Example
Say we want to build a plant that produces 360,000 tonnes per year of a vinyl chloride monomer () to be used in the production of polyvinyl chloride (PVC). One reaction pathway is shown below.
A DBM for this process might consist of the following items:
For contractual issues:
- High level description of vinyl chloride and uses
- Ethane and chlorine feeds to be used
- Reactor type being used
- Plant capacity of 360,000 tonnes/yr of vinyl chloride, located in Houston, Texas
- Operating composition purity, temperature, and pressure
- Operating philosphy
- Gantt chart showing project timeline
For site-specific issues:
- Temperature, pressure, and composition of feed (inside battery limits)
- Cooling water used at and 2 atm and steam at and 3 atm (outside battery limits)
- Emission predictions and requirements
- Atmospheric pressure, little wind and rainfall, maximum and minimum temperatures at location
- Environmental and safety issues at locations
- Control and electrical requirements of the plant
For design-specific issues:
- Equipment, instrument, and line numbering
- Units used in calculation (imperial or SI)
- Standards
- Certain technical equipment have certain standards (pressure vessels, heat exchangers, electrical work)
Exercise
Consider a process where toluene is fed into a process at 1000 kg/hr and hydrogen is fed at 82 kg/hr. The following reaction takes place in the process:
The reaction achieves 75% conversion and the products of the process are separated into 2 streams. The first product stream is benzene, exiting at 821 kg/hr. The second product stream is a mixed gas stream (including the unreacted reactants and by-products) exiting at 261 kg/hr. Draw an input-output diagram for this process.
Solution
In some sections, we have put in extra useful information (such as resources or finding data). We’ve used grey text boxes for these information sources.
Finding this data can be very challenging. Companies typically want to keep pricing information secret (for competition reasons). The best way I have found to search for chemical pricing data is as follows:
This process is tedious and often takes some time, even for people experienced in this process. So don’t get discouraged when looking for this data and feel free to ask questions (through the discussion board, in tutorials or office hours, etc.).
This book has a chapter review section at the end of each chapter that consists of important equations discussed in the chapter as well as a set of flashcards you can use to gauge how well you understand the concepts discussed in the chapters. Note that you don’t need to memorize these terms, but it is important to understand what the terms mean and the context in which they appear.
Kinetic Energy | |
Potential Energy | |
First Law of Thermodynamics | |
Flow Work | |
Steady-state Open System Energy Balance | |
Enthalpy | |
Heat Capacity (closed system) | |
Internal Energy (closed system) | |
Heat Capacity (open system) | |
Enthalpy (open system) | |
Heat of Reaction Method | |
Heat of Formation Method |
An interactive or media element has been excluded from this version of the text. You can view it online here:
https://pressbooks.bccampus.ca/chbe220/?p=2227
A section consisting of extra practice problems is included at the end of each chapter. The purpose of this section is to give you more practice problems to work on as well as get more confident with the course material. The exercises consist of 3 question types: multiple-choice questions, short answer questions, and long answer questions. This allows you to practice solving different question types and timing yourself. The solutions to the questions are posted below the question in a different text box. We would highly recommend solving the problem before looking over the solution.
Exercise (Multiple Choice)
Consider a reactor in which an endothermic reaction takes place and 15.8 kJ are absorbed. A mixer is used in the reactor and provides 6.3 kJ of work on the system. What is the overall heat that needs to be provided to the system?
a) -22.1 kJ
b) 9.5 kJ
c) 22.1 kJ
d) -9.5 kJ
b) 9.5 kJ
Performing an energy balance on the system, we know that . Since this is an endothermic reaction and work is done on the system, both the enthalpy of the reaction and the work are positive:
\begin{align*}
Q&= \Delta H -W \\
&= 15.8 kJ – 6.3 kJ\\
&=9.5kJ
\end{align*}
Exercise (Short Answer)
You have a feed stream of methanol at 1 atm and . You’d like to bring it to a temperature of so that it enters a reactor as a gas. The boiling point of methanol is at 1 atm. What process path (with enthalpy changes) can you take to achieve this change and calculate the overall enthalpy?
The following steps can be taken to calculate the overall change in enthalpy of going from liquid methanol at 1 atm and to gas methanol at 1 atm and .
The enthalpy change of the process will be the sum of steps 1-3.
Exercise (Long Answer)
Consider the combustion of ethylene:
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g)
The enthalpy of combustion can be determined using the enthalpies of the following reactions:
Determine the enthalpy of the combustion of ethylene from reactions 1, 2, and 3 using Hess’s Law at standard temperature and pressure.
Step 1: Determine the combination of reactions that will give us the desired combustion reaction.
We know that we need C2H4 in the reactants, so we multiply reaction 1 by -1. We notice that by adding the other 2 reactions to reaction 1 (multiplied by -1), we get our desired reaction:
Step 2: Calculate the heat of combustion from the reaction enthalpies.
The appendix introduces some sites and software for finding useful data or perform certain types of calculations. Each session contains a guide to walk through the site or software, and some can be found with a worked example when the site or software is used.
Finding this data can be very challenging. Companies typically want to keep pricing information secret (for competition reasons). The best way I have found to search for chemical pricing data is as follows:
This process is tedious and often takes some time, even for people experienced in this process. So don’t get discouraged when looking for this data and feel free to ask questions (through the discussion board, in tutorials or office hours, etc.).
The “Glossary” session in the appendix summarizes the important terms in this course arranged in alphabetical order. You can find a short definition of each term and a link to the section that the term is introduced.
In this chapter, we will introduce some of the basics in the design aspect of chemical and biological engineering. We will focus on basic engineering and general engineering economics. This chapter also introduces the components and characteristics of 4 key process diagrams.
Learning Objectives
By the end of this chapter, you should be able to:
Define the components of a design basis memorandum
Apply basic measures of economic merit to perform initial estimates of project economic feasibility
Recognize the 4 key types of flow sheets
Create input-output diagrams
Construct block flow diagrams (BFDs) for chemical processes
Define common unit operations and use these when constructing process flow sheets
Recognize the need to apply basic material balances
Build process flow diagrams (PFDs)
Understand piping and instrumentation diagrams (P&IDs)
As you’re going through this chapter, here are some important terms for you to take note of:
Learning Objectives
By the end of this section, you should be able to:
Define the components of a design basis memorandum
Chemical plant construction can be split into four broad stages which come one after another:
In this course, we will only focus on basic engineering.
Basic Design Package: This document is provided to clients by engineers and may consist of the following:
A thorough basic engineering design package can be found on the website of Startail Management and Consulting Services.
Design Basis Memorandum (DBM): This document contains key information and data related to the project deliverables.
A design basis memorandum is typically 10-20 pages and mainly consists of the following information:
The discussion of what is inside and outside battery limits is usually established early in the design process to give a clear indication of what engineers are responsible for in their design.
Example: DBM Content – Production of Vinyl Chloride
Say we want to build a plant that produces 360,000 tonnes per year of a vinyl chloride monomer () to be used in the production of polyvinyl chloride (PVC). One reaction pathway is shown below.
A DBM for this process might consist of the following items:
For contractual issues:
- High level description of vinyl chloride and uses
- Ethane and chlorine feeds to be used
- Reactor type being used
- Plant capacity of 360,000 tonnes/yr of vinyl chloride, located in Houston, Texas
- Operating composition purity, temperature, and pressure
- Operating philosphy
- Gantt chart showing project timeline
For site-specific issues:
- Temperature, pressure, and composition of feed (inside battery limits)
- Cooling water used at and 2 atm and steam at and 3 atm (outside battery limits)
- Emission predictions and requirements
- Atmospheric pressure, little wind and rainfall, maximum and minimum temperatures at location
- Environmental and safety issues at locations
- Control and electrical requirements of the plant
For design-specific issues:
- Equipment, instrument, and line numbering
- Units used in calculation (imperial or SI)
- Standards
- Certain technical equipment have certain standards (pressure vessels, heat exchangers, electrical work)
Exercise: Classifying Inside & Outside Battery Limits
Consider a process where benzene is fed into a process for oxidization and purification in a plant. The process also consists of using 2 utility streams (cold water and steam) used in heat exchangers in the plant, as well as an onsite wastewater treatment operation. Your company has been hired to do basic engineering for the oxidation and purification process. Classify the oxidation, purification, wastewater treatment, and utility supply as either inside battery limit or outside battery limit.
Note that in other cases the outside batter limit elements above could be included inside the battery limits depending on what the client asks the engineering design firm to do. However with our example above we had clearly stated the design firm was only focusing on the oxidation and purification steps.
An interactive or media element has been excluded from this version of the text. You can view it online here:
https://pressbooks.bccampus.ca/chbe220/?p=1756
[1] CII – CII Glossary. [Online]. Available: https://www.construction-institute.org/resources/knowledgebase/about-the-knowledge-base/glossary/cii-glossary. [Accessed: 29-May-2020].
Learning Objectives
By the end of this section, you should be able to:
Apply basic measures of economic merit to perform initial estimates of project economic feasibility
As with any venture, we want to ensure a chemical plant is profitable to operate. In order to get an initial sense of this in a project, we may perform basic economic analysis. This analysis relies on four basic figured of merit and we will see how these are used to perform basic economic analysis on a process.
The four basic figures of merit are:
1- Gross Economic Potential (GEP): generally in dollars/yr
GEP has to be positive, otherwise, our product is worth less than our feed and we will definitely be losing money accounting for the operating costs.
2- Net Economic Potential (NEP): generally in dollars/yr
NEP again has to be positive, otherwise, we cannot pay for the process to operate.
3- Capital Costs (CC): generally in dollars
Cost of design plant and construction
4- Return on Investment (ROI): generally in %yr
ROI also has to be positive, generally most processes require a certain minimum ROI, typically above 10%. ROI represents the percentage of capital costs recovered each year. The higher the ROI is, the more favorable the process is.
One thing that is essential to find our first figure of merit, GEP, is chemical pricing data. We will look at how to find this data now as it is so essential for evaluating processes. For now, we will not focus on how to assess production and capital costs but will revisit these later in the course.
Finding this data can be very challenging. Companies typically want to keep pricing information secret (for competition reasons). The best way I have found to search for chemical pricing data is as follows:
This process is tedious and often takes some time, even for people experienced in this process. So don’t get discouraged when looking for this data and feel free to ask questions (through the discussion board, in tutorials or office hours, etc.).
Example: Pricing of Vinyl Chloride
From this article, we find that vinyl chloride was priced at 870-895 USD/tonne by CFR India, 840-870 USD/tonne by CFR GCC (Gulf Coast in the southern US) and 860-870 USD/tonne CFD East Med.[2] This range of values is dependent on multiple factors, and as you see depend on geography. We may choose a value depending on where our plant is located. Since our vinyl chloride plant is in the Gulf Coast, we might choose 855 USD/tonne as a rough average for that specific range.
Example: Vinyl Choride Production Economic Analysis
Our client wants to produce 360,000 tonnes/yr (800 million lbs/yr) of vinyl chloride monomer. We decide to use the following reaction pathway, which contains two reaction steps as follows:
We gather the following information on the process:
Name | Formula | Molecular Weight (g/mol) | Price (USD/tonne) |
---|---|---|---|
Ethene | 28.05 | 733 | |
Chlorine | 70.91 | 240 | |
Dichloroethane | 98.96 | 300 | |
Hydrogen Chloride | 36.46 | 140 | |
Vinyl Chloride | 62.50 | 750 |
Assume as well that our production costs are of VC produced, and assume our VC production plant costs to build. Note these values usually depend on the size of the plant, location, and other factors.
Calculate or find all four economic figures of merit.
We know that one of the product streams will be 360,000 tonnes/yr of vinyl chloride. Hydrogen chloride is produced in a 1:1 molar ratio with vinyl chloride:
Similarly for ethane and chlorine, we get:
1 – Gross Economic Potential (GEP) in USD/yr
GEP = Value of Products – Value of Feeds
2 – Net Economic Potential (NEP) in USD/yr
NEP = GEP – production costs
Using our assumed production costs of of VC produced.
3 – Capital Costs (CC) in USD
Total cost of plant design and construction
This will be our VC production plant capital building cost of .
4 – Return on Investment (ROI) in %/yr
ROI = NEP/CC*100%
ROI=22 \frac{\%}{yr}
ROI is generally written simply in % and this is assumed to be %/yr. The higher the ROI the better.
[1] CII - CII Glossary. [Online]. Available: https://www.construction-institute.org/resources/knowledgebase/about-the-knowledge-base/glossary/cii-glossary. [Accessed: 29-May-2020].
[2] Ho, Z. X. (2019). PVC. ICIS Chemical Business, 296(9), 36. [Online].Available: http://ezproxy.library.ubc.ca/login?url=https://search.proquest.com/docview/2292001228?accountid=14656 [Accessed: 06-June-2020].
Learning Objectives
By the end of this section, you should be able to:
Recognize the 4 key types of flow sheets
There are 4 key types of flow sheet and they are shown below in increasing complexity:
Image from Con-struct / CC BY-SA
Learning Objectives
By the end of this section, you should be able to:
Create input-output diagrams
These diagrams are considered to be the simplest of process flow sheets. In input-output diagrams, the entire process is represented by 1 block only. They usually only reference the main process streams in and out of a process.
Example: Ammonia Production Input-output Diagram
Consider an ammonia production process where 3 main streams of raw materials enter the process:
The process produces 2 main product streams:
We can put together the following input-output diagram for this process:
Exercise: Benzene Production Input-output Diagram
Consider a process where toluene is fed into a process at 1000 kg/hr and hydrogen is fed at 82 kg/hr. The following reaction takes place in the process:
The reaction achieves 75% conversion and the products of the process are separated into 2 streams. The first product stream is benzene, exiting at 821 kg/hr. The second product stream is a mixed gas stream (including the unreacted reactants and by-products) exiting at 261 kg/hr. Draw an input-output diagram for this process.
Exercise: Production of Vinyl Chloride
Our client wants to produce 360,000 tonnes/yr (800 million lbs/yr) of vinyl chloride monomer. We decide to use the following reaction pathway, which contains two reaction steps as follows:
We gather the following information on the process:
Name | Formula | Molecular Weight (g/mol) | Price (USD/tonne) |
---|---|---|---|
Ethene | 28.05 | 733 | |
Chlorine | 70.91 | 240 | |
Dichloroethane | 98.96 | 300 | |
Hydrogen Chloride | 36.46 | 140 | |
Vinyl Chloride | 62.50 | 750 |
Assume as well that our production costs are of VC produced, and assume our VC production plant costs to build. Note these values usually depend on the size of the plant, location, and other factors.
Create an input-output diagram of the process.
We know that one of the product streams will be 360,000 tonnes/yr of vinyl chloride. Hydrogen chloride is produced in a 1:1 molar ratio with vinyl chloride:
Similarly for ethane and chlorine, we get:
From this and the information in the table, we get the following input-output diagram:
Learning Objectives
By the end of this section, you should be able to:
Construct block flow diagrams (BFDs) for chemical processes
We typically break down chemical and biological processes (or any other process) into units representing equipment. These are called unit operations and are drawn as circles or squares:
Lines with arrows represent the flow of materials in streams to and from units. These lines have only 1 arrowhead:
Materials in each stream are labeled above or beside the stream:
Exercise: BFD Components
What are some errors in the following flowsheet? How would you change the flowsheet to address these errors?
Solution
We can correct the BFD to following:
Learning Objectives
By the end of this section, you should be able to:
Define common unit operations and use these when constructing process flow sheets
Recognize the need to apply basic material balances
We will go over four common unit types below, many other units are variants of these basic types of units.
Processes are often represented as interconnected process units consisting of mixers, reactors, splitters, and separators among other unit operations.
Examples of other common unit operations:
Example
Recall the vinyl chloride production process from the last chapter:
\begin{align*}
C_{2}H_{4} + Cl_{2} → C_{2}H_{4}Cl_{2}&\;\;\;\; (1)\\
C_{2}H_{4}Cl_{2} → C_{2}H_{3}Cl + HCl&\;\;\;\; (2)
\end{align*}
We now consider an additional reaction in the vinyl chloride production process. Most vinyl chloride manufacturers use the combination of direct chlorination and oxychlorination, which is called the balanced process .
\begin{align*}
C_{2}H_{4} + Cl_{2} → C_{2}H_{4}Cl_{2}&\;\;\;\; (1)\;\;\;\; \\
C_{2}H_{4} + 2HCl + \frac{1}{2}O_{2} → C_{2}H_{4}Cl_{2} + H_{2}O&\;\;\;\; (2) \\
C_{2}H_{4}Cl_{2} → C_{2}H_{3}Cl + HCl&\;\;\;\; (3)
\end{align*}
- Direct Chlorination (last lecture)
- Oxychlorination
- Ethylene dichloride pyrolysis (last lecture)
What is the overall reaction and what advantage does this have over what we saw last chapter?
To get the overall reaction, we add all the reactions and cancel any compounds that appear in both the reactants and products of the overall reaction. In the separation process, HCl is recycled back to the oxychlorination reactor if it is not depleted to ensure no HCl is released into the environment. Reaction (1) is balanced with the rest of the process in a way to provide the source of for reaction (3). This process gives us only 1 by-product: water. Using the balanced process gives us water as a by-product as opposed to hydrogen chloride.
This process is carried out in multiple unit operations. To achieve the maximum efficiency and avoid releasing hazardous chemicals, hydrogen chloride produced in reaction 3 is recycled back to the oxychlorination reactor and used reaction (2), as well as unreacted EDC(ethylene dichloride, ) in reaction 3 is recycled to react again.
- Ethylene enters the process and is split into 2 streams: 1 stream goes to the oxychlorination reactor an the other stream goes to the direct chlorination reactor
- Air and/or oxygen are fed into the oxychlorination reactor and chlorine is fed into the direct chlorination reactor
- EDC produced by both the oxychlorination and direct chlorination processes, and is fed into a purification unit to separate from the other components in the system
- EDC is then fed into the EDC pyrolysis reactor, followed by the VCM(vinyl chloride monomer, ) purification unit, which separates HCl, VCM, and EDC into 3 separate streams
Compound | Boiling Point () |
---|---|
-103.7 | |
-85.05 | |
-34.04 | |
-13.4 | |
83.47 | |
114 |
As depicted above, we have light products (the components that are more volatile than our desired product, EDC), heavy products (the components that are less volatile than our desired product), and a stream of EDC. Let’s look at separating the components after direct chlorination:
For simplicity, say we only have chlorine (), ethylene dichloride (EDC, ), and trichloroethane (TCE, ) produced from a side reaction as products of the direct chlorination step. In reality, there may be other components in the stream, but we will stick to these three as the main components for this analysis. How many streams will come out after the EDC purification processes, and what are the components of each stream? How do we select the temperature for the purification process?
Compound Boiling Point () -34.04 83.47 114
From the boiling point temperatures, we can deduce that the light product will be (boils at a lower temperature) and the heavy product will be (TCE) (boils at a higher temperature).
We can consider the separation of EDC from TCE and chlorine using 2 distillation columns. In this course, we’ll assume that distillation columns separate a single feed stream into two exit streams. There are more complex column arrangements with a greater number of entering and exit streams, but we won’t go over these in this course. Distillation columns usually operate with a range of temperatures inside them, and we’ll see why this is later in this course. For now, we’ll just assign them an average temperature.
The first column separates EDC and TCE from chlorine by operating at a temperature that is lower than EDC’s boiling temperature, but higher than ‘s boiling temperature (let’s say on average ):
The second column separates EDC from TCE by operating at a temperature that is lower than TCE’s boiling temperature, but higher than EDC’s boiling temperature (let’s say, ):
Exercise
Consider the following vinyl chloride separation process:
The following data is available:
Compound | Boiling Point () |
---|---|
-85.05 | |
(VC) | -13.4 |
(EDC) | 83.47 |
How would you set up distillation columns for the separation of vinyl chloride (VC) from hydrogen chloride and EDC? Provide a temperature estimate could these columns be run at for separation.
Solution
Step 1: Recognize that an easy way to separate 3 compounds is using 2 different distillation columns by separating out EDC in the first column and then separating vinyl chloride from in the second column.
Step 2: For the first column, we separate EDC as a liquid, while and vinyl chloride exit the column as gases. We choose a temperature that is lower than the boiling point temperature of EDC, but higher than the boiling point temperature of vinyl chloride. We have a large temperature range that meets these criteria, but we choose since it is an easy temperature to work with (as it is close to ambient temperatures generally).
Step 3: For the second column, we separate VC as a liquid and HCl as a gas. We choose a temperature that is lower than the boiling point temperature of VC, but higher than the boiling point temperature of HCl. For example, we may choose .Note: this is a very cold working temperature for a distillation column and we might try to find ways to change this by manipulating pressure, but we will explore this later in this course and you will learn more through other courses.
General material balance rules:
We won’t cover material balances in-depth in this course. Instead, I will rely on you applying the concepts you are learning in your Material and Energy Balances course.
Learning Objectives
By the end of this section, you should be able to:
Build process flow diagrams (PFDs)
Process flow diagrams usually include:
This is a typical way to label a piece of equipment using its position in the process. This is not the only way to label equipment, but a very common way. We will stick with this in this class and you will likely see it very commonly.
P-101A
Label | Equipment Description |
---|---|
C | Compressor or Turbine |
E | Heat Exchanger |
P | Pump |
R | Reactor |
T | Tower |
TK | Storage Tank |
V | Vessel |
Generally, rather than using numbers, utility streams will be labeled using a certain descriptor. These streams are labeled in this way as the same type of stream will generally have the same conditions, and so this saves repeatedly specifying these streams with numbers.
Label | Utility Description |
---|---|
lps | Low-Pressure Steam (3-5 bar) |
mps | Medium-Pressure Steam (10-15 bar) |
hps | High-Pressure Steam (40-50 bar) |
htm | Heat Transfer Media (organic) |
cw | Cooling Water |
rw | Refrigerated Water |
rb | Refrigerated Brine |
cs | Chemical Waste Water |
ss | Sanitary Waste Water |
el | Electric Heat |
ng | Natural Gas |
fg | Fuel Gas |
fo | Fuel Oil |
The legend should include everything to allow someone to read the PFD with no prior knowledge of the project.
Certain essential information must be included in stream summary tables, this may vary slightly depending on the industry, what we will stick within this class is the following:
Other information may or may not be included as well:
Stream Number | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|
Temperature () | 25 | 25 | 90 | 90 | 95 | 250 |
Pressure (atm) | 1.5 | 1.5 | 1.5 | 1.5 | 30 | 20 |
Vapor Fraction | 1.0 | 1.0 | 0.3 | 0.6 | 0.5 | 1.0 |
Mass Flow (kg/hr) | 50,000 | 105,000 | 149,000 | 276,000 | 276,000 | 276,000 |
Molar Flow (kmol/hr) | 1,500 | 1,500 | 1,500 | 2,600 | 1,750 | 5,000 |
Component Molar Flow (kmol/hr) | ||||||
Ethylene | 1,500 | 0 | 0 | 0 | 0 | 0 |
Chlorine | 0 | 1,500 | 0 | 0 | 0 | 0 |
1,2-dichloroethane | 0 | 0 | 1,500 | 0 | 0 | 0 |
Vinyl Chloride | 0 | 0 | 0 | 2,600 | 0 | 2,500 |
Hydrogen Chloride | 0 | 0 | 0 | 0 | 1,750 | 2,500 |
The equipment summary table provides information on equipment specifications. These are essentially a short list of key details used for the designing or costing of equipment. Examples of what might be specified in this table for different types of equipment are shown below. As you can see, these can be quite detailed, and you’ll learn more about how to specify these parameters throughout your degree.
Equipment | Required Specification |
---|---|
Vessels | Height; diameter; pressure; temperature; materials of construction |
Pumps | Flow; driver type; suction and discharge pressures; temperature; shaft power; materials of construction |
Towers | Height; diameter; orientation; pressure; temperature; number and type of trays; height and type of packing; material of construction |
Compressors | Inlet flow; driver type; suction and discharge pressures; temperature; shaft power; material of construction |
Heat Exchangers | Type; area; duty; number of shell and tube passes; operating temperatures; pressures; pressure drops; material of construction |
Fired Heaters | Type; tube pressure; tube temperature; duty; radiant heat transfer area; convective heat transfer area; material of construction |
Exercise
Consider the following diagram:
If we wanted to turn this into a PFD, what would need to be added or modified?
Solution
Learning Objectives
By the end of this section, you should be able to:
Understand piping and instrumentation diagrams (P&IDs)
P&IDs are based off of PFDs but include more specific information
P&IDs do not include:
A P&ID often depicts one unit operation in thorough detail.
Exercise: P&ID Components
Consider the following diagram:
Image from Jim Cahill / CC BY-ND 2.0
What are some components that need to be added or adjusted to turn this into a P&ID?
You will learn more about control symbols as you go through your degree. However, below we have a brief introduction to some important symbols.
Instrument connections
- Capillary: fluid flow control
- Pneumatic: air or gas flow control
- Electrical: electrical currents controls
Instrument locations
- Located in plant: generally means the instrument is near what it is measuring (local), it does not go to the control room
- Front panel in control room: Control rooms used to have instrument panels and would have manual controls on the front panel facing the operators, these have been replaced largely with digital (or virtual) screens which the operators can switch between different views. You could think of the main view in these digital screens as the front panel, with the most important controls or indicators shown.
- Back panel in control room: place for instruments that are perhaps less important or do not need to be looked at frequently. In our class we won’t distinguish between these, but this is common notation you may see in P&IDs
Control panel rooms have changed quite a bit throughout the years. Below are examples of older and newer control panels.
Image from XXBuilderXx / CC BY-SA
Image from PEO ACWA / CC BY 2.0
For the control symbology identification with the ‘XYY’ format:
First Letter (X) | Second or Third Letter (Y) | |
---|---|---|
A | Analysis (Composition/Concentration) | Alarm |
C | – | Control |
F | Flowrate | – |
H | – | High |
I | – | Indicate |
L | Level | Light or Low |
P | Pressure | – |
T | Temperature | Transmit |
Exercise: P&ID Control
What information can you derive from the following control symbol?
This is a flow indicator and controller, with the instrument located on the front panel of the control room with an electrical connection.
Here is a thorough table of control symbology:
First Letter (X) | Second or Third Letter (Y) | |
---|---|---|
A | Analysis | Alarm |
B | Burner Flame | – |
C | Conductivity | Control |
D | Density or Specific Gravity | – |
E | Voltage | Element |
F | Flowrate | – |
H | Hand (Manually Initiated) | High |
I | Current | Indicate |
J | Power | – |
K | Time or Time Schedule | Control Station |
L | Level | Light or Low |
M | Moisture or Humidity | Middle or Intermediate |
O | – | Orifice |
P | Pressure or Vacuum | Point |
Q | Quantity or Event | – |
R | Radioactivity or Ratio | Record or Print |
S | Speed or Frequency | Switch |
T | Temperature | Transmit |
V | Viscosity | Valve, Damper, or Louver |
W | Weight | Well |
Y | – | Relay or Compute |
Z | Position | Drive |
Exercise
A simple system consists of two feed streams (stream 1: toluene; stream 2: benzene) that enter a heater separately and combine into 1 stream to exit. The heater uses high-pressure steam as the utility. Build a simple process flow diagram of the process described above.
Solution:
Gross Economic Potential (GEP) | GEP = Value of Products – Value of Feeds |
Net Economic Potential (NEP) | NEP = GEP – production costs |
Capital Costs (CC) | CC = Total cost of plant design and construction |
Return of Investments (ROI) | ROI = NEP/CC*100% |
An interactive or media element has been excluded from this version of the text. You can view it online here:
https://pressbooks.bccampus.ca/chbe220/?p=1801
Exercise: Design Basis Memorandum
A client for a vinyl chloride production plant is looking to find information in the design basis memorandum you’ve provided on how long it will take to complete the plant construction. Where can the client look for this information in the DBM?
a) In the contractual issues section, under project scope
b) In the plant site-specific issues section, under site environmental conditions
c) In the design-specific issues section, under standards
d) In the contractual issues section, under deliverables and schedule
d) In the contractual issues section, under deliverables and schedule
The deliverables and schedule section of the DBM will contain a Gantt chart which can give the client information on
Exercise: Process Diagrams
You are upgrading the control system on a reactor and heat exchanger in the plant. You’d like to present these updates to a group of engineers and operators in your company. What would be the best type of diagram to present in this scenario?
a) P&ID
b) BFD
c) PFD
d) Input-Output
c) P&ID
For a technically-literate stakeholder group such as engineers and operators, it would be beneficial to present a process diagram with enough detail about the process and what streams and unit operations are involved. This requirement eliminates choices b and d. Given that this update focuses on the control system of the units, a PFD would not include this information. Therefore, a P&ID would be the best option for this scenario.
Exercise: Equipment Labeling
What information does the label TK-203B give you?
a) Storage Tank 2 in section 3B of the plant
b) Storage Tank 3 in section 2B of the plant
c) Storage Tank 2 in section 3 of the plant with other duplicate storage tanks
d) Storage Tank 3 in section 2 of the plant with other duplication storage tanks
d) Storage Tank 3 in section 2 of the plant with other duplication storage tanks
Exercise: Unit Operations
What sequence of unit operations can we use to mix the ingredients to make cookie dough, then divide the dough into equal portions, followed by baking to make a batch of cookies?
Mixer > Splitter > Reactor.
A mixer combines multiple process streams (separate ingredients) into one stream (one cookie dough). We can use a splitter to separate the dough into equal amounts. A splitter is different than a separater that a splitter divides the stream into streams with the same composition, while a separtor separates the stream into streams with different compositions. Finally, a reactor (oven) can be used to bake the cookies.
Exercise: Process Economics
Consider a company that has developed a process that produces cinnamon-scented wax melts used in spas, hotels, and homes for fragrancing. This company prides itself on making everything from the essential oil to the soy wax and beeswax in-house.
What is the gross economic potential of this process?
Recall that the gross economic potential (GEP) is:
GEP = Value of Products – Value of Feeds ($/yr)
The feed value can be calculated as follows:
The product value can be calculated as follows:
Therefore, the GEP will be:
Exercise: Process Flow Diagrams
What is missing from the following process flow diagram (PFD)?
The appropriate equipment labeling along with stream 7 content label (“HCl” in this case) are missing from this PFD.
Exercises: Block Flow Diagram
Carbon capture is a process that is gaining a lot of attention due to its ability to mitigate the environmental effects in the oil and gas industry (along with other industries). One proposed carbon capture method is through the use of MDEA (N-methyl diethanolamine) absorption. The following steps describe the process:
Draw a block flow diagram (BFD) for this process
*Flue gas: a gas stream exiting from an upstream process that usually consists of carbon dioxide, nitrogen, and oxygen (only nitrogen and carbon dioxide are considered for this process).
In a chemical process, the rate of production is limited by the rate of reaction in the reactors. This chapter introduces methods to predict the rate of reaction and concentration of components. The reactions are analyzed using rate law, reaction equilibrium and reaction mechanisms. By learning this chapter, we can gain knowledge to select the optimal conditions for maximum production rate and model the productivity of our process.
Learning Objectives
By the end of this chapter, you should be able to:
Calculate the reaction rate for the reaction and rate of formation of compounds
Calculate the reaction extent
Determine reaction order and rate constant from kinetic data
Analyze equilibrium reaction equations
Explain the effect of temperature on reaction rates
Analyze elementary reactions, reaction molecularity
Select the rate-determining step given the consecutive reactions and overall rate law of the reaction
Derive the rate law of a reaction using steady-state approximations to pre-equilibria assumptions
Understand kinetic and thermodynamic control
As you are going through this chapter, some important terms for you to take note of:
Learning Objectives
By the end of this section, you should be able to:
Define Reaction rate and reaction extent
Calculate the reaction rate for the reaction, the rate of formation of compounds and the reaction extent
A reaction rate shows the rates of production of a chemical species. It can also show the rate of consumption of a species; for example, a reactant. In general though, we want an overall common reaction rate to describe changes in a chemical system.
Let’s look at an example. Say we have reaction represented as:
For this system reaction rate can be expressed as follows:
The reaction rate is represented by the letter “r” or the Greek letter upsilon “”
NOTE: I will stick with r as upsilon looks like “” which we will use to represent the stoichiometric coefficient
Above we have written the reaction rate as if a substance with a coefficient of 1 was reacting (or being produced). This is the typical form of an overall reaction rate describing a reaction.
Rearranging the above equation, we can find the rate of production/consumption for any species based on this overall reaction rate, note that stoichiometric coefficients are positive for products and negative for reactants:
The general equation for calculating the reaction rate:
General notation: J is used to denote any compound involved in the reaction.
Reaction rate at a given time can also be found from the graph of concentration of components in a system vs. time:
Exercise: Calculating the Reaction Rate
If we have the reaction
and we measure that the rate of formation of NO is 1.6 mmol/(L·s), what are the overall reaction rate, and the rate of formation of and ?
Step 1: Determine the overall reaction rate from the rate of formation for NO.
\begin{align*}
r & = \frac{1}{\nu_{j}} \frac{d[NO]}{dt} \\
& = \frac{1}{2} \left( 1.6 \frac{mmol}{L·s}\right)\\
& = 0.8\frac{mmol}{L·s}
\end{align*}
Step 2: Use the reaction rate to determine rate of formation for the other compounds
NOTE: rate of formation is positive for products and negative for reactants
\begin{align*}
\frac{d[Br_{2}]}{dt}& = r \\
& = 0.8\frac{mmol}{L·s}
\end{align*}
\begin{align*}
\frac{d[NOBr]}{dt}& = -2r \\
& = -1.6\frac{mmol}{L·s}
\end{align*}
Reaction rates can be given in a variety of units over time. In this class we will just explore molarity and partial pressure, although other forms exist.
Molarity – molar concentration – expressed in units of (eg. )
Partial pressure – the pressure produced by one gaseous component if it occupies the whole system volume at the same temperature, commonly used for gasses – units of ( eg. )
We use the extent of reaction () to describe the change in an amount of a reacting speicies J.
where:
We can get a relationship between the reaction extent and the rate of reaction when the system volume is constant:
where:
= volume
Learning Objectives
By the end of this section, you should be able to:
Define reaction order and overall order
Determine the reaction order and the rate constant from kinetic data
The dependence of the rate of reaction on the reactant concentrations can often be expressed as a direct proportionality, in which the concentrations may be raised to be the zeroth, first, or second power. The exponent is known as the order of the reaction with respect to that substance.
The overall order of a reaction is the sum of the orders with respect to the sum of the exponents
For example:
The reaction is:
In the more complicated example for the reaction previously given between hydrogen and bromine:
The reaction is:
One of the simplest methods for determining the rate law is the isolation method.
Say we have a reaction with two reactants, A and B. We put a large amount of B in our reactor compared to A; so much that the concentration barely changes.
If the true rate law is:
What we would observe is , where
With B in great excess: where , reaction appears first order.
With A in great excess: where , reaction appears second order.
is read as k prime effective
It is easier to analyse these individual effective rate laws than the more complex combined rate law when both concentrations change significantly.
The method of initial rates is commonly used in conjunction with the isolation methods to determine reaction order.
Exercise: Method of Initial Rates
Say we have a reaction with two species, A and B. We put in the same amount of B into the reactor each time. The following is our initial rates are observed with different concentrations of A. What is the reaction order in A?
[A] (mol/L) | Initial rate(mol/L·s) |
1 | |
2 |
As the concentration of A doubles, the reaction rate quadruples
So the reaction is second order in A.
If we take the logarithm of the general equation for reaction rate, we can linearize the equation:
\begin{align*}
r_{0} & = k_{eff}[A_{0}]^a \\
log(r_{0})& = log(k_{eff}) + a*log([A_{0}])\\
y & = intercept + slope * x
\end{align*}
Plot vs , where the slope is the reaction order in A; and y-intercept equals to
Exercise: Reaction Order and Rate Constant
Sulfuryl chloride () decomposes to and by the following reaction:
Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.
Experiment | Initial rate(M/s) | |
1 | 0.0050 | |
2 | 0.0075 | |
3 | 0.0100 | |
4 | 0.0125 |
Step 1: determine the reaction order with respect to sulfuryl chloride
Comparing Experiments 1 and 3, for example, shows that doubling the concentration doubles the reaction rate , which means that the reaction rate is proportional to .
The reaction is first order with respect to .
Step 2: calculate
We have . We can calculate the rate constant () using data from any experiment in the table.
\begin{align*}
rate & = k_{r}[SO_{2}Cl_{2}] \\
1.10×10^{-7} M/s& = k_{r}×0.0050M \\
2.2×10^{-5}s^{-1} & = k_{r}
\end{align*}
Method 2
The following graph is produced when plotting vs
From the line of best fit, we can see the slope is 1 (meaning a first order relationship) and
Therefore the reaction is first order in , .
[1] Chemistry LibreTexts. 2020. The Rate Law. [online] Available at: <https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Rate_Laws/The_Rate_Law> [Accessed 23 April 2020].
[2] Chemistry LibreTexts. 2020. 14.4: The Change Of Concentration With Time (Integrated Rate Laws). [online] Available at: <https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.4%3A_The_Change_of_Concentration_with_Time_(Integrated_Rate_Laws)> [Accessed 23 April 2020].
Learning Objectives
By the end of this section, you should be able to:
Determine the reaction order and rate constant from kinetic data using the linearized form of integrated rate laws
Rate laws are differential equations that can be integrated to find how the concentrations of reactants and products change with time. We can imagine a fairly complex system with multiple reactions described by
The concentration of all components can be described by equations such as We need to integrate these to find concentrations at a given time.
Any of these rate laws can be solved numerically (you’ll learn about this in CHBE 230), but some simpler cases can also be solved analytically.
A zeroth-order reaction is one whose rate is independent of concentration; Say we have a reaction:
It’s differential rate law would be represented as:
Integrating from t=0, when the system has a concentration of A as , to some time t, when the system has a concentration represented by
or |
Using a similar approach to the zeroth order reactions:
or |
Example: Half-life for First-order Reactions
Based on the equation we derived, what is an expression for the time taken for half of A to be consumed (half-life)?
\begin{align*}
k_{r}* t_{1/2} & = -ln \frac{[A]}{[A]_{0}} \\
& = -ln \frac{\frac{1}{2}[A]_{0}}{[A]_{0}} \\
& =-ln(\frac{1}{2}) \\
& = ln(2)
\end{align*}
Note that this half-life does not depend on [A].
Practice: use , you should get the same result.
Solution
\begin{align*}
[A] & = [A]_{0}*e^{-k_{r}t_{1/2}} \\
\frac{[A]}{[A]_{0}}& = e^{-k_{r}t_{1/2}} \\
\frac{1}{2} & = e^{-k_{r}t_{1/2}} \\
ln(\frac{1}{2}) & = -k_{r}*t_{1/2}
\end{align*}
or |
Using this to find half life, we find that:
[1] Chemistry LibreTexts. 2020. 12.4 Integrated Rate Laws. [online] Available at: <https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chemistry_(OpenSTAX)/12%3A_Kinetics/12.4%3A_Integrated_Rate_Laws#:~:text=We%20measure%20values%20for%20the,different%20concentrations%20of%20the%20reactants.&text=Integrated%20rate%20laws%20are%20determined,various%20times%20during%20a%20reaction.> [Accessed 24 April 2020].
Learning Objectives
By the end of this section, you should be able to:
Analyze equilibrium reaction equations
When the concentrations of reactants and products become constant, the system is said to have reached a point of equilibrium. The equilibrium state is dynamic. At equilibrium, the reactions are not stopped, but rather the rates of opposing reactions have become equal such that concentrations of chemical species do not change.
Laboratory kinetic studies may be performed on reactions far from equilibrium where products are in low concentration and the reverse reactions are not important (this is what we considered previously).
Closer to equilibrium however we must take these reverse reactions into account.
Unpack the notation:
Consider a system with the following reaction:
Let’s say the system is composed of 2 first-order reactions:
Forward reaction: , where the reaction rate is expressed by:
Reverse reaction: , where the reaction rate is expressed by:
Net rate of change in considering both reactions is:
If the initial concentration of A is and no B is present, then at all times: . Using this in the equation above:
\begin{align*}
\frac{d[A]}{dt} &=k’_{r}[B]-k_{r}[A]\\
& =k’_{r}([A]_{0}-[A])-k_{r}[A]\\
& =k’_{r}[A]_{0}-(k_{r}+k’_{r})[A]
\end{align*}
This is what is known as an ordinary differential equation, as is both a differential and a separate term. You may not know how to solve these equations and do not need to know this for this course. However, if we were to solve this ordinary differential equation, we would get the following:
Using the expression for above, , taking arbitrary values of and , the following graph is produced:
As you can see from the graph, the concentrations change less over time. After a long time, they should reach a state of equilibrium. The length of time to reach equilibrium (or close to it) will depend on the reaction rates.
As , we approach equilibrium and we get the following:
and
The equilibrium constant in this example is defined as:
At equilibrium, the rates of forward and reverse reactions should be the same (meaning no change in overall concentrations), so it makes sense that when we rearrange this we get:
Consider another system with the following reaction:
Forward reaction: , where the forward reaction rate is expressed by:
Reverse reaction: , where the reverse reaction rate is expressed by:
Here, in equilibrium we have:
The equilibrium constant is defined as dimensionless, so here is where the terms make a difference. The equilibrium constant can be espressed as:
Recall that and here will have different units, so the term ensures that these become dimensionless when one is divided by the other.
Exercise: Equilibrium Constant
1. For the balanced chemical reaction below, write an equation for the equilibrium constant(K).
2. Consider the following reaction system:
From initial rate and equilibrium experiments we know the following: , , , .
What are the values of the equilibrium constant() and reverse reaction constant()?
, at equilibrium
1.
2. Start with expressing K in forms of concentrations and :
\begin{align*}
K & =\frac{[B]_{eq}/c^\theta*[C]_{eq}/c^\theta}{[A]_{eq}/c^\theta}\\
& = (\frac{[B][C]}{[A]})_{eq}\frac{1}{c^\theta}\\
& = \frac{k_{r}}{k’_{r}*c^\theta}
\end{align*}
Calculate K from the expression with concentrations:
\begin{align*}
K & = (\frac{[B][C]}{[A]})_{eq}\frac{1}{c^\theta}\\
& = (\frac{0.1\frac{mol}{L}*0.2\frac{mol}{L}}{0.5\frac{mol}{L}})_{eq}\frac{1}{c^\theta}\\
& = 0.04
\end{align*}
Subsitude the K value into the expression with :
\begin{align*}
K & = \frac{k_{r}}{k’_{r}*c^\theta}\\
k’_{r}& = \frac{k_{r}}{K*c^\theta}\\
& = \frac{0.1min^{-1}}{0.04*1\frac{mol}{L}}\\
& = 2.5 \frac{L}{mol*min}
\end{align*}
Back substitute into the rate equations to check this:
Forward and reverse reaction rates are the same at , as we would expect at equilibrium.
[1] Chemistry LibreTexts. 2020. Characteristics of the Equilibrium State. [online] Available at: <https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Principles_of_Chemical_Equilibria/Principles_of_Chemical_Equilibrium/Characteristics_Of_The_Equilibrium_State> [Accessed 24 April, 2020].
Learning Objectives
By the end of this chapter, you should be able to:
Explain the effect of temperature on reaction rates
The Arrhenius Equation describes the relationship between the reaction rate and temperature.
Where
The Arrhenius Equation is in linear form if we graph as y and as x:
Where
Once is known for a reaction, the rate of reaction can be predicted at a given temperature if it is known at another temperature:
and
Subtract the first from the second to get:
This simplifies to:
Exercise: Reaction Rate Change with Changing Temperature
Say a certain biological reaction has an activation energy of 50 kJ/mol, how much faster will this reaction be at body temperature (37°C) vs room temperature (25°C)?
Note R = 8.314
So with this 12°C temperature increase, the reaction rate would increase 2.18 times. A rule of thumb is that reaction rates double every 10°C, but we know this depends on the activation energy.
Manipulating the previous Arrhenius equation, we can isolate by taking the exponent of each side to get:
If Ea is 0 (all collisions result in a reaction) then:
Meaning the rate constant when each collision is successful is equal to the frequency factor.
When , then only a subset of reactions are successful represented by
Catalysts speed up the reactions usually by decreasing the activation energy. Sometimes catalysts can affect frequency factors as well, but the effect of frequency factor is minor in comparison to the decrease of activation energy. This is demonstrated in the example question after the activation energy session.
Image from Wikimedia Commons
Exercise: Effect of Temperature on Equilibrium
1. True or False: A reaction is in a state of equilibrium when the equilibrium constant K is equal to 0.
2. Which of the following does not affect the concentration of each component at equilibrium for a reaction system?
a. Adding products
b. Increasing temperature
c. Using a catalyst
d. Decreasing volume
Hint: An understanding of Le Châtelier’s principle can be helpful for this question, but you can also analyze this question using the definition of
1. False– because for each specific reaction, the equilibrium constant only depends on temperature. The reaction is at equilibrium when the value of is equal to (notice instead of , because we don’t know if the system is at equilibrium when we calculate this). Also experimentally, we can say that the system is at equilibrium when the concentrations of each component are not changing.
2. Answer: c
Using a catalyst decreases the energy barrier needed for the reaction to proceed (activation energy), but does not affect the equilibrium constant.
Both a and d will cause shifting of equilibrium due to Le Chatelier’s principle. For a, the reaction will shift to the reverse, meaning that the reverse reaction will happen more often so more reactants will be formed and more products will be consumed. For d, the system will shift to the side that produces a larger number of gaseous molecules to increase the volume. There may be a case where gas volume stays the same no matter with products or reactants, but this is not known.
If you analyze a and d using , the equilibrium constant will not change, but the concentration of the components will change, and the system will approach the same equilibrium constant after the change.
b – The equilibrium constant will change when temperature changes, shifting reactant and product concentrations.
The activation energy() represents the energy needed for the reactants to collide and get to a state such that their bonds can rearrange to form the products.
Graph (a) illustrates the general situation in which the products have lower potential energy than the reactants, so the overall reaction releases energy; that is, it is energetically downhill, and ΔE < 0.
In contrast, graph (b) illustrates the case in which the products have higher potential energy than the reactants, so the overall reaction requires an input of energy; that is, it is energetically uphill, and ΔE > 0.
Although the energy changes that result from a reaction can be positive, negative, or even zero, in most cases an energy barrier must be overcome before a reaction can occur. This means that the activation energy is almost always positive.
Image from Chemistry Libretexts / CC BY-NC-SA 3.0
Example: Catalyst Affecting Activation Energy
Enzymes are biological catalysts. Suppose we have a reaction with an activation energy of 80 kJ and a frequency factor of 1 without catalysts at 298K, calculate the reaction rate constant without catalyst and compare the rate constant (by taking the ratio of k) if the enzyme:
a) Decreases the activation energy to 8kJ.
b) Increases the frequency factor to 10.
Solution
without catalyst:
\begin{align*}
k(uncatalyzed)& =Ae^{(-\frac{E_{a}}{RT})}\\
& = 1×e^{(-\frac{80×10^3J}{8.314J/(molK)298K})}\\
& = 9.48×10^{-15}
\end{align*}
a)
\begin{align*}
\frac{k'(catalyzed)}{k(uncatalyzed)} & =\frac{Ae^{(-\frac{E_{a}’}{RT})}}{Ae^{(-\frac{E_{a}}{RT})}}\\
& = e^{\frac{-E_{a}’+E_{a}}{RT}}\\
& = e^{\frac{(-8+80)×10^3J/mol}{8.314J/(molK)298K}}\\
& = 4.2×10^{12}
\end{align*}
b)
\begin{align*}
\frac{k'(catalyzed)}{k(uncatalyzed)} & =\frac{A’e^{(-\frac{E_{a}}{RT})}}{Ae^{(-\frac{E_{a}}{RT})}}\\
& = \frac{A’}{A}\\
& = 10
\end{align*}
A decrease in has a drastic effect on reaction rate.
Exercise: Activation Energy Calculation
The equation for the decomposition of to and is second order in :
Data for the reaction rate as a function of temperature are listed in the following table. Calculate Ea for the reaction and the rate constant at 700 K.
T(K) | k |
592 | 522 |
603 | 755 |
627 | 1700 |
652 | 4020 |
656 | 5030 |
Step 1: Plot the Arrhenius Equation in the linear form:
Step 2: Calculate the activation energy ($E_{a}$) and the frequency factor ($A$) from the slope and intercept\begin{align*}
E_{a} & = -slope × R\\
& = -(-13572 K)×8.314 J/(molK)\\
& = 112838 J/mol\\
& = 112.8 kJ/mol
\end{align*}\begin{align*}
A & = e^{y-intercept}\\
& = e^{29.145}\\
& = 4.545×10^{12} M^{-1}s^{-1}
\end{align*}Note that the units of A are the same as kr, but we tend to then remove the units when we take logarithm as we usually consider products of logarithmic. The important thing to remember is if we use A to solve for kr (as we do below), we need to include the relevant units.Step 3: Calculate the rate constant () using the , and given temperature
\begin{align*}
k_{r} & =Ae^{(-\frac{E_{a}}{RT})} \\
& = 4.545×10^{12}e^{(-\frac{112838 J/mol}{8.314 J/(molK)×700 K})} \\
& = 17265 M^{-1}s^{-1}\\
& = 1.73 × 10^{4} M^{-1}s^{-1}
\end{align*}
[1] Chemistry LibreTexts. 2020. Characteristics of the Equilibrium State. [online] Available at: <https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Principles_of_Chemical_Equilibria/Principles_of_Chemical_Equilibrium/Characteristics_Of_The_Equilibrium_State> [Accessed 24 April, 2020].
[2] Chemistry LibreTexts. 2019. 4.6 Activation Energy and Rate. [online] Available at: <https://chem.libretexts.org/Courses/Bellarmine_University/BU%3A_Chem_104_(Christianson)/Phase_2%3A_Chemical_Kinetics/4%3A_Kinetics%3A_How_Fast_Reactions_Go/4.6%3A_Activation_Energy_and_Rate> [Accessed 24 April, 2020].
Learning Objectives
By the end of this section, you should be able to:
Analyze elementary reactions and reaction molecularity
The sequence of events that occur at the molecular level during a reaction is the mechanism of the reaction. It describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism. Each step of the reaction is called an elementary reaction. The sum of the elementary reactions in the mechanism must give the balanced chemical equation for the overall reaction.
The decomposition of ozone, for example, appears to follow a mechanism with two steps:
Elementary reactions add up to the overall reaction:
With the elementary reactions, we do not specify phases, and the formula represents specific processes occurring with individual molecules or atoms.
Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called intermediates.
The molecularity is the number of molecules or atoms coming together to react in an elementary reaction.
Example: isomerization of cyclopropane to propene
Example: decomposition of cyclobutane
Example: Hydrogen attacking a bromine molecule
Reaction Order vs. Molecularity
Reaction order – An empirical quantity obtained from experimentally determined rate law
Molecularity – Number of molecules coming together in an elementary reaction proposed as an individual step in a mechanism
The order of the elementary reaction is the same as its molecularity. In contrast, the rate law cannot be determined from the balanced chemical equation for the overall reaction (unless it is a single step mechanism and is therefore also an elementary step).
(can be several products):
(can be several products):
(can be several products):
Generally these depend on concentration of reactants. However, these rate laws must be tested against experimental data.
Some reactions proceed through the formation of an intermediate (I). One example of this with irreversible reactions is:
The intermediate() is present in the reaction steps but does not appear in the overall reaction:
Example: Express Rate of Formation for Reactive Compounds
Given the consecutive elementary reaction:
The rate of change for [A] is:
What are the equations for the rates of change for the other two species, and , in terms of their concentrations and the reaction rate constants?
Solution
the intermediate I is produced by the reaction that converts A to I and consumed by the reaction that produces P:
P is produced by the reaction that converts I to P:
Taking 3 sets of arbitary values for and , we can plot the following graphs of concentration vs.time:
[1] Chemistry LibreTexts. 2020. 14.6 Reaction Mechanisms. [online] Available at: <https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.6%3A_Reaction_Mechanisms#:~:text=A%20balanced%20chemical%20reaction%20does,step%20is%20an%20elementary%20reaction.> [Accessed 24 April, 2020].
[2] OpenStax Chemistry 2016. 12.6 Reaction Mechanisms. [online] Available at: <https://opentextbc.ca/chemistry/chapter/12-6-reaction-mechanisms> [Accessed 27 April, 2020].
Learning Objectives
By the end of this section, you should be able to:
Select the rate-determining step given the consecutive reactions and overall rate law of the reaction
Derive the rate law of a reaction given the consecutive reactions
Derive the rate law under pre-equilibria conditions
This approximation assumes intermediate, I, is in a low constant concentration,ie:
Generally the approximation is more accurate when ( is much greater than ). The bigger the difference, the more accurate the assumption.
much greater than means that the reaction converts intermediate to product much faster than converting reactant to intermediate, then all the intermediate molecules will quickly be consumed after being produced. The intermediate molecule will stay at a relatively low concentration.
As we saw before with first order reactions: leads to
Now: so
For product: if we solved this, we would find:
Detailed proof: This is not required for the course, but might be useful for understanding.
For reactant:
is a function of time. A function’s derivative is a multiple of itself, the function would be in an exponential form (will learn more of this in MATH 256 next term)
, to find the constant, we know that [A] is equal to at time 0. Substitute t=0; the exponential term is equal to 1. We are left with . Therefore:
For intermediate:
so (use the second equivalence of the equation, no calculus needed)
For product:
\begin{align*}
\frac{d[P]}{dt} & = k_{a}[A] \\
\frac{d[P]}{dt} & = k_{a}*[A]_{0}*e^{-k_{a}t}\\
[P] & = \int k_{a}*[A]_{0}*e^{-k_{a}t}\\
[P] & = -[A]_{0}*e^{-k_{a}t} + constant
\end{align*}
To find the constant, we know that there is no product in the system, initially. At t=0, [P]=0. Subsitute t=0, the exponential term is equal to 1, so we are left with , so , Therefore:
Taking arbitrary values of and , we can plot the concentration vs. time for all the components:
Exercise: Using Steady-state Approximation to Find Rate Law Expressions
To derive the overall rate law for this reaction:
Assume the reaction follows the following three-step mechanism:
The final expression of the rate law should only contain and the rate constants of the elementary steps.
Step 1: Find the intermediates and express their rate of formation. Equate their rate of formation to 0 due to steady-state approximation.
Step 2: Express the overall rate law. The easiest way is to express because it is a product that has stoichiometric coefficient of 1.
Step 3: From the expression of , we can see that to find the overall rate law, we need to find as an expression of the concentrations of reactants or products (no intermediates) by manipulating (1) and (2).
From (1):
Substitute into (2):
\begin{align*}
k_{f}[N_{2}O_{5}]-k_{b}[NO_{2}][NO_{3}]-k_{2}[NO_{3}][NO_{2}]-k_{2}[NO_{3}][NO_{2}] & = 0\\
k_{b}[NO_{2}][NO_{3}]+2k_{2}[NO_{3}][NO_{2}] & = k_{f}[N_{2}O_{5}]\\
[NO_{3}][NO_{2}] & =\frac{k_{f}[N_{2}O_{5}]}{k_{b}+2k_{2}}
\end{align*}
Step 4: We can substitute the expression for we just found into the expression for , which is the same as the overall rate law.
Rate determining step is the step that determines the overall rate of reaction in a series of reactions. The slowest forward reaction step is referred to as the rate-determining step, as it limits the rate of the entire reaction. Note that this must be a forward reaction step, a reverse reaction in an equilibrium reaction cannot be the rate-determining step.
An analogy that illustrates this concept is an hourglass having two different sized openings. The rate of the sand falling to the bottom-most chamber is determined by the smaller of the two openings. Similarly, the rate law of the overall reaction is determined from its rate-determining slowest step.
Image from Introduction to Chemistry, 1st Canadian Edition/ CC BY 4.0
Say we have the following forward and reverse reaction:
Consider a case where an equilibrium is established due to :
The equilibrium constant, expressed using the forward and reverse reaction constants ( and ) would remain the same over time
The reaction of would decrease over time. Due to Le Châtelier’s principle, the equilibrium reaction will shift to produce more
Because , the reaction will shift concentrations to achieve equilibrium again far faster than the forward reaction () depletes
Therefore, we can approximate with only the equilibrium part of the reaction
Review: From lecture 4: reaction equilibrium:
The forward reaction: , reaction rate
The reverse reaction: , reaction rate
At equilibrium: \begin{align*}
r & =r’\\
k_{a}[A][B] & = k_{a}'[I]\\
\frac{k_{a}}{k_{a}’}& = \frac{[I]}{[A][B]}
\end{align*}
So
Using the equation for based on this equilibrium, we can develop an equation for the rate of formation of :
Where . The final step of is rate determining, whereas the steps prior control the amount of intermediate.
Recall from Lecture 4:
For each step of the reaction, we can use the Arrhenius equation to substitute , where is the frequency factor and is the activation energy
So we can express the overall rate constant of the reaction above as:
Although each rate constant may increase with temperature, this may not be true of the overall rate constant .
The effective activation energy is: and this may be positive or negative depending on of individual steps.
Exercise: Using Pre-Equilibria to Find the Rate Law Expression
The two-step mechanism below has been proposed for a reaction between nitrogen monoxide and molecular chlorine:
Use this mechanism to derive the equation and predicted rate law for the overall reaction.
The rate law should be expressed in terms of , , and the rate constants of the elementary reactions shown above.
Step 1: The overall reaction is the sum of the 2 elementary reactions:
Step 2: Write each reaction separately and express the rate law for each reaction (including the reverse reaction in the equilibrium):
\begin{align*}
NO_{(g)}+Cl_{2(g)}→NOCl_{2(g)},\;\;\; & r_{1}=k_{a}[NO][Cl_{2}]\\
NOCl_{2(g)}→NO_{(g)}+Cl_{2(g)},\;\;\; & r_{-1}=k_{a}'[NOCl_{2}]\\
NOCl_{2}+NO→2NOCl,\;\;\; & r_{2}=k_{b}[NOCl_{2}][NO]
\end{align*}
Step 3: Use the equilibrium reaction to express the concentration of the intermediate ():
Step 4: Express the overall reaction rate using the step 2 elementary reaction, then subsitute in :
\begin{align*}
overall\;rate&=\frac{1}{2}\frac{d[NOCl]}{dt}\\
&=\frac{2k_{b}[NO][NOCl_{2}]}{2}\\
&=k_{b}[NO][NOCl_{2}]\\
&=\frac{k_{b}k_{a}}{k_{a}’}[NO]^2[Cl_{2}]
\end{align*}
[1] Chemistry LibreTexts. 2020. Steady State Approximation. [online] Available at: <https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Reaction_Mechanisms/Steady-State_Approximation> [Accessed 04 May, 2020].
[2] Introduction to Chemistry. 1st Canadian Edition. 2014. Reaction Mechanisms. [online] Available at: <https://opentextbc.ca/introductorychemistry/chapter/reaction-mechanisms-2/> [Accessed 04 May, 2020].
[3] OpenStax Chemistry 2e. 2019. 12.6 Reaction Mechanisms. [online] Available at: <https://openstax.org/books/chemistry-2e/pages/12-6-reaction-mechanisms> [Accessed 05 May, 2020].
Learning Objectives
By the end of this section, you should be able to:
Understand kinetic and thermodynamic control
Reactants can sometimes give rise to a variety of products.
Consider the nitration of nitrobenzene:
The relative portion of the products before reaching equilibrium is given by the ratio of the rates of production.
where
where
Here, before equilibrium:
This is called kinetic control, and it is dictated by reaction rates.
As opposed to thermodynamic control, which is dictated by reaction equilibrium (after a long time):
Say we have the system:
If
Then at any time before the equilibrium reaction start severely affecting product concentration, the reaction simplifies to:
The reaction is kinetically controlled: the amount of products depends on the rates of reaction.
Proof to show:
\begin{align*}
r_{P1}=k_{r1}[A][B]&=\frac{d[P1]}{dt}\\
r_{P2}=k_{r2}[A][B]&=\frac{d[P2]}{dt}
\end{align*}
Say that both , start at a concentration of 0. We can express the change in concentration for and at any time before [A][B] reaches 0. Note that once [A][B] reaches 0, the equilibrium reaction starts to dominate as we no longer have forward reactions that consume A and B to produce P1 and P2.
If
Then at any given time, this reaction simplifies to:
The reaction is thermodynamically controlled : the amount of products depends on the equilibrium state.
Proof to show: :
At equilibrium:
Forward reaction rate:
Reverse reaction rate:
At equilibrium, the forward and reverse reaction rates are equal:
\begin{align*}
k_{e2}[P1] & = k_{e1}[P2]\\
\frac{[P_{1}]}{[P_{2}]} & = \frac{k_{e1}}{k_{e2}}
\end{align*}
Reaction rate | |
Extent of reaction | |
Relating extent of reaction with reaction rate | |
Rate law (general form) | |
Zeroth-order rate law | |
First-order rate law | |
Second-order rate law | |
Equilibrium constant | |
Arrhenius equation | Linear form: Simplification for temperature dependency calculations:
|
Unimolecular rate law | |
Biomolecular rate law | |
Kinetic control | If : |
Thermodynamic control | If : |
An interactive or media element has been excluded from this version of the text. You can view it online here:
https://pressbooks.bccampus.ca/chbe220/?p=1427
Exercise: Stoichiometry
Which of the following statement is true about the reaction ?
a) the stoichiometric coefficient for A is 1
b) the stoichiometric coefficient for B is -1
c) the reaction extent can be positive, negative
d) the stoichiometric coefficient for P is -1
b) the stoichiometric coefficient for B is -1
The stoichiometric coefficient is negative for reactants and positive for products. Since the reaction only goes forward, the reaction extent is always positive (or 0 if it doesn’t happen)
Exercise: Stoichiometric Expression
Which of the following statement is true about the equilibrium reaction ?
a) the stoichiometric coefficient for A is 1
b) the stoichiometric coefficient for P is -1
c) the reaction extent can be positive or negative
d) the equilibrium constant K can be positive or negative
c) the reaction extent can be positive, negative
The stoichiometric coefficient is negative for reactants and positive for products. Since it is an equilibrium reaction, if the reverse reaction happens more frequently than the forward reaction, the reaction extent can be negative. The equilibrium constant is . Since the concentration of any compound must be positive, the equilibrium constant K must be positive as well.
Exercise: Reaction Order
The following data shows the concentration of reactive species [A] in the system at points of time.
Time (s) | [A] (mol/L) |
0 | 2.000 |
0.100 | 0.286 |
0.200 | 0.154 |
0.300 | 0.105 |
What is the reaction order of A, determined by the integrated rate law?
a) zeroth-order
b) first-order
c) second-order
d) none of the above
c) second-order
Zeroth order: , would be linear if graph against
First order: , would be linear if graph against
Second order: , would be linear if graph against So we plot against , against , and against and see which one is linear:
We can see that only against is linear, so the reaction is second order in A.
Exercise: Reaction Rate Law Units
.
If the concentration of , and are measured in units of mol/L, what is the unit of the reaction rate constant ?
a)
b)
c)
d)
c)
Based on the unit that the reactive species are measured, we can know that the unit for is . So we can equate the units on both sides of the rate law:
\begin{align*}
\frac{mol}{L·s}&=k_r×\frac{mol}{L}×\frac{mol}{L}\\
\frac{mol}{L·s}×\frac{L}{mol}×\frac{L}{mol}&=k_r\\
\frac{L}{mol·s}&=k_r
\end{align*}
Exercies: Reaction Rates
We have the following reaction happening in a fixed reaction chamber:
The partial pressure of was measured at each time point and shown in the table below.
Time (ms) | (Pa) |
0 | 1670 |
100 | 1300 |
200 | 1012 |
300 | 789 |
400 | 614 |
500 | 478 |
a) Calculate the initial rate of reaction
b) Calculate the initial rate of change in partial pressure for every compound involved in the reaction
*Hint: for small time intervals, we can take , where J is any species involved in the reaction
a) \begin{align*}
r&=\frac{1}{v_j}\frac{d[J]}{dt}=\frac{1}{v_j}\frac{\Delta[J]}{\Delta t}\\
&=\frac{1}{-2}\frac{\Delta[NH_{3}]}{\Delta t}\\
&=\frac{1}{-2}\frac{(1300-1670)Pa}{(100-0)ms×\frac{1s}{1000ms}}\\
&=1850 Pa/s
\end{align*}
b) Because is it a "fixed reaction chamber", which means the total volumn of the gas chamber stays constant, so we can do the following balance using the definition of reaction rate:
Exercise: Reaction Order and Rate Constant
At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.
Experimental data for the reaction at 300°C and four initial concentrations of NO2 are listed in the following table:
Experiment | (M) | Initial rate(M/s) |
1 | 0.015 | |
2 | 0.010 | |
3 | 0.0080 | |
4 | 0.0050 |
Step 1: Determine the reaction order with respect to nitrogen dioxide
For example, comparing Experiments 2 and 4 shows that doubling the concentration quadruples the reaction rate
, which means that the reaction rate is proportional to .
Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to .
Therefore, the reaction is second order with respect to nitrogen dioxide.
Step 2: calculate
We have . We can calculate the rate constant () using data from any experiment in the table.
\begin{align*}
rate & = k_{r}[NO_{2}]^2 \\
5.40×10^{-5} M/s& = k_{r}×(0.010M)^2\\
0.54M^{-1}s^{-1} & = k_{r}
\end{align*}
Method 2
The following graph is produced when plotting vs
From the line of best fit, we can see ,
Therefore the reaction is first order in , .
Exercise: Rate Constant and Arrhenius Equation
Suppose a reaction has an activation energy of 85 kJ/mol. If the reaction has a reaction rate constant of at 0°C, then what is the reaction rate constant at 25°C?
We can use the rearranged Arrhenius Equation to relate the reaction rate constant to temperature. Note that the temperature should be expressed in K.
\begin{align*}
ln(\frac{k_{r2}}{k_{r1}})&=\frac{E_{a}}{R}\big(\frac{1}{T_{1}}-\frac{1}{T_{2}}\big)\\
ln\bigg(\frac{k_{r2}}{70s^{-1}}\bigg)&=\frac{85×10^3J/mol}{8.314J/(mol·K)}\bigg(\frac{1}{(0+273.15)K}-\frac{1}{(25+273.15)K}\bigg)\\
ln\bigg(\frac{k_{r2}}{70s^{-1}}\bigg)&=3.138\\
\frac{k_{r2}}{70s^{-1}}&=e^{3.138}\\
k_{r2}&=e^{3.138}×70s^{-1}\\
k_{r2}&=1615s^{-1}
\end{align*}
Exercise: Using Steady-state Approximation to Find the Rate Law Expression
Suppose we have a system with the reaction:
The elementary reactions are as follows:
Using the steady state assumption, we can prove that
This will also allow us to determine the values of and in terms of the elementary reaction rate constants to
Step 1: Start by expressing the rate of change of the intermediates and , and set them equal to 0:
Step 2: Express based on the elementary reactions:
We can see that to get the espression in the question statement, we need to replace and and simplify.
Step 3: Adding equations (1) and (2) together, we can see that 3 of the terms cancel, allowing us to solve for :
\begin{align*}
2k_{1}[Br_{2}]-2k_{5}[Br]^2 & = 0\\
2k_{1}[Br_{2}] & = 2k_{5}[Br]^2\\
(\frac{k_{1}}{k_{5}})^{1/2}[Br_{2}]^{1/2} & =[Br]
\end{align*}
Step 4: Next, we can determine using equation (2) by substituting what we derived for :
\begin{align*}
k_{2}[Br][H_{2}]-k_{3}[H][Br_{2}]-k_{4}[H][H\!Br] & = 0\\
(k_{3}[Br_{2}]+k_{4}[H\!Br])[H] & = k_{2}[Br][H_{2}]\\
[H] & =\frac{k_{2}[Br][H_{2}]}{k_{3}[Br_{2}]+k_{4}[H\!Br]}\\
[H] & = \frac{k_{2}(\frac{k_{1}}{k_{5}})^{1/2}[Br_{2}]^{1/2}[H_{2}]}{k_{3}[Br_{2}]+k_{4}[H\!Br]}
\end{align*}
Step 5: Substitute and into , then simplify
Therefore, the values of and are:
Reaction Mechanism Exercise 1
For the overall reaction
The step-wise elementary reactions are:
Express reaction rate of the overall reaction using the steady-state approximation. The expression should only include , , and the rate constants of the elementary steps.
Step 1: Steady-state approximation states that the rate of generation and consumption of intermediates sum up to 0. The intermediates in the reaction are and
\begin{align*}
&\frac{d[N_2O_2]}{dt}=k_1[NO]^2-k_{-1}[N_2O_2]-k_2[N_2O_2][H_2]=0\;\;\;\;\;(1)\\
&\frac{d[N_2O]}{dt}=k_2[N_2O_2][H_2]-k_3[N_2O][H_2]=0\;\;\;\;\;(2)
\end{align*}
Step 2: Express the rate of the overall reaction. In this reaction, the rate of the overall reaction can be expressed as the rate of generation of :
Step 3: Use equations (1) and (2) to replace the intermediate terms in the expression for r:
From (2):
Then we can use equation (1) to replace with an expression with and :
\begin{align*}
k_1[NO]^2-k_{-1}[N_2O_2]-k_2[N_2O_2][H_2]&=0\\
k_1[NO]^2&=k_{-1}[N_2O_2]+k_2[N_2O_2][H_2]\\
k_1[NO]^2&=[N_2O_2]×(k_{-1}+k_2[H_2])\\
\frac{k_1[NO]^2}{k_{-1}+k_2[H_2]}&=[N_2O_2]
\end{align*}
Step 4: Substitute the result of step 3 into the expression for r:
Reaction Mechanism Exercise 1 - with Pre-equilibria
For the overall reaction
The step-wise elementary reactions are:
Express reaction rate of the overall reaction using the steady-state approximation and the pre-equilibrium assumption. The expression should only include , , and the rate constants of the elementary steps.
Step 1: Set the sum of the rate of generation and consumption of intermediates to 0:
\begin{align*}
&\frac{d[N_2O_2]}{dt}=k_1[NO]^2-k_{-1}[N_2O_2]-k_2[N_2O_2][H_2]=0\;\;\;\;\;(1)\\
&\frac{d[N_2O]}{dt}=k_2[N_2O_2][H_2]-k_3[N_2O][H_2]=0\;\;\;\;\;(2)
\end{align*}
Step 2: Express the rate of the overall reaction:
Step 3: Replace the intermediate terms:
From (2):
Under pre-equilibria assumptions, we assume the rate of the equilibrium reaction is much faster than the reaction that consumes the product of the equilibrium reaction. So we can express using the definition of equilibrium that the rate of forward and reverse reactions in step 1 are the same:
\begin{align*}
k_1[NO]^2&=k_{-1}[N_2O_2]\\
\frac{k_1[NO]^2}{k_{-1}}&=[N_2O_2]
\end{align*}
Step 4: Substitute the result of step 3 into the expression for r:
Why are the answers different?
The pre-equilibria assumption states that the rate of the equilibrium reaction is much faster than the reaction that consumes the product of the equilibrium reaction, which means . Therefore, the term is negligible comparing to the term in the denominator under pre-equilibria assumption.
Reaction Mechanism Exercise 2
For the overall reaction to produce formaldehyde:
The step-wise elementary reactions are:
Express reaction rate of the overall reaction using the steady-state approximation. The expression can involve the concentration of the reactants of the overall reaction, , and the rate constants of the elementary steps. Then state how to replace in the expression (you do not have to calculate it).
Step 1: Set the sum of the rate of generation and consumption of intermediates to 0. Notice how the coefficients affect the rate law:
\begin{align*}
&\frac{d[H]}{dt}=2k_1[H_2]-2k_{-1}[H]^2-k_2[H][CO]-k_3[H][HCO]=0\;\;\;\;\;(1)\\
&\frac{d[HCO]}{dt}=k_2[H][CO]-k_3[H][HCO]=0\;\;\;\;\;(2)
\end{align*}
Step 2: Express the rate of the overall reaction:
Step 3: Replace the intermediate terms:
From (2):
We can get an expression of by solving for it in equation (1), also using the relationship from (2) that :
\begin{align*}
2k_1[H_2]-2k_{-1}[H]^2-k_2[H][CO]-k_3[H][HCO]&=0\\
2k_1[H_2]-2k_{-1}[H]^2-2k_2[H][CO]&=0\\
\end{align*}
This is a quadratic equation about . We can solve it using the quadratic formula:
where , ,
Reaction Mechanism Exercise 2 - with Pre-equilibria
For the overall reaction to produce formaldehyde:
The step-wise elementary reactions are:
Express reaction rate of the overall reaction using the steady-state approximation and the pre-equilibrium assumption. The expression should only include , , and the rate constants of the elementary steps.
Step 1: Set the sum of the rate of generation and consumption of intermediates to 0:
\begin{align*}
&\frac{d[H]}{dt}=2k_1[H_2]-2k_{-1}[H]^2-k_2[H][CO]-k_3[H][HCO]=0\;\;\;\;\;(1)\\
&\frac{d[HCO]}{dt}=k_2[H][CO]-k_3[H][HCO]=0\;\;\;\;\;(2)
\end{align*}
Step 2: Express the rate of the overall reaction:
Step 3: Replace the intermediate terms:
From (2):
Using the pre-equilibrium assumptions to replace [H]:
\begin{align*}
k_1[H_2]&=k_{-1}[H]^2\\
\frac{k_1}{k_{-1}}&=[H]^2\\
\Big(\frac{k_1}{k_{-1}}\Big)[H_2]^{\frac{1}{2}}&=[H]
\end{align*}
Step 4: Substitute the result of step 3 into the expression for r:
Reaction Mechanism Exercise 3
For the reaction of ozone depletion:
The step-wise elementary reactions are:
Express reaction rate of oxygen gas generation using the steady-state approximation. The expression should include , , and the rate constants of the elementary steps.
Step 1: Set the sum of the rate of generation and consumption of intermediates to 0:
Step 2: Express from the elementary reactions. Notice that is generated in multiple elementary steps:
Step 3: Replace using equation (1)
\begin{align*}
k_1[O_3]-k_{-1}[O_2][O]-k_2[O][O_3]&=0\\
k_1[O_3]&=k_{-1}[O_2][O]+k_2[O][O_3]\\
k_1[O_3]&=[O](k_{-1}[O_2]+k_2[O_3])\\
\frac{k_1[O_3]}{k_{-1}[O_2]+k_2[O_3]}&=[O]
\end{align*}
Step 4: Sub the expression for into
\begin{align*}
\frac{d[O_2]}{dt}&=k_1[O_3]-k_{-1}[O_2][O]+2k_{2}[O][O_{3}]\\
\frac{d[O_2]}{dt}&=k_1[O_3]-\frac{k_{-1}k_1[O_3][O_2]}{k_{-1}[O_2]+k_2[O_3]}+\frac{2k_{2}k_1[O_3]^2}{k_{-1}[O_2]+k_2[O_3]}
\end{align*}
Reaction Mechanism Exercise 3 - with Pre-equilibria
For the reaction of ozone depletion:
The step-wise elementary reactions are:
Express reaction rate of oxygen gas generation using the steady-state approximation and the pre-equilibrium assumption. The expression should include , , and the rate constants of the elementary steps.
Step 1: Set the sum of the rate of generation and consumption of intermediates to 0:
Step 2: Express :
Step 3: Using the pre-equilibrium assumptions to replace [O]:
\begin{align*}
k_1[O_3]&=k_{-1}[O_2][O]\\
\frac{k_1[O_3]}{k_{-1}[O_2]}&=[O]
\end{align*}
Step 4: Sub the expression for into
\begin{align*}
\frac{d[O_2]}{dt}&=k_1[O_3]-k_{-1}[O_2][O]+2k_{2}[O][O_{3}]\\
&=k_1[O_3]-k_{-1}[O_2]\frac{k_1[O_3]}{k_{-1}[O_2]}+\frac{2k_2k_1[O_3]^2}{k_{-1}[O_2]}\\
&=k_1[O_3]-k_1[O_3]+\frac{2k_2k_1[O_3]^2}{k_{-1}[O_2]}\\
&=\frac{2k_2k_1[O_3]^2}{k_{-1}[O_2]}\\
\end{align*}
[1] Chemistry LibreTexts. 2020. 14.4: The Change Of Concentration With Time (Integrated Rate Laws). [online] Available at: <https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.4%3A_The_Change_of_Concentration_with_Time_(Integrated_Rate_Laws)> [Accessed 23 April 2020].
Phase equilibrium can be used to control the concentration of components. It is also a concept widely used in various types of separation processes. In this chapter, we will learn to collect information from existing diagrams, including variations of phase diagrams for pure components and diagrams summarizing vapour-liquid equilibrium conditions for multicomponent systems. We can use the information to find the phase(s) of components in our processing equipment and the mole fractions of components in each phase.
Learning Objectives
By the end of this chapter, you should be able to:
Interpret phase diagrams for relevant thermodynamic information (e.g. composition, temperature, pressure…)
Apply the definition of intensive and extensive properties to process variables
Describe the relationship between variables of state for gases
Describe gas behaviour using the ideal gas law and compressibility factor for equations of state
Estimate the vapour pressure of pure compounds at a given temperature using Antoine’s equation
Apply properties of ideal mixtures to describe vapour-liquid separations
Read Txy and Pxy diagrams for binary mixtures
Find bubbles points and dew points on the diagrams
Calculate the mole fraction of substances in the vapour and liquid phases
As you’re going through this chapter, here are some important terms for you to take note of:
Learning Objectives
By the end of this section, you should be able to:
Calculate pressure using its definition
Understand the difference between absolute & gauge pressure
Physical properties are properties that can be measured without changing the molecular structure of the substance.
The three main physical properties we will introduce in this lecture are pressure, temperature, and volume:
Pressure – the amount of force exerted per area in a system
Temperature – a measure of the average kinetic energy of a system
Volume – the space occupied by a system
To see more about how these are related for gases, there is an interesting gas property simulator available online: https://phet.colorado.edu/en/simulation/gas-properties
When a gas molecule elastically collides with the wall of a container, it exerts a force on the wall. These forces are the source of pressure in a gas. In a sample of gas in a container, the randomness of the molecular motion causes the number of collisions to fluctuate in a given time. However, because a huge number of molecules collide with the wall in a short time, the number of collisions on the scales of time and space we measure fluctuates by only a tiny, usually unobservable fraction from the average.
As the number of molecules increases, the number of collisions, and thus the pressure, increases. If the average velocity of the molecules is higher, each collision exerts a larger force on the wall, therefore the gas pressure is higher.
Image obtained from OpenStax University Physics Volume 2 / CC BY 4.0
General notation:
M – any units of mass (g, kg, lbs…)
L – any units of length (m, cm, in…)
t – any units of time (s, min, hr…)
Common units for pressure:
[S.I.] (i.e. Pascal)
[cgs]
[American]
Other common pressure units:
(millimetres of mercury column at 0 °C) = (i.e. pound force per square inch)
Exercise: Pressure Calculation
What is the force exerted from a column of air on us in mass per square metre?
Take atmosphere pressure =
We can use Newton’s second law to find mass, with being the acceleration of gravity. Since the mass we are looking for is “per square metre”, we can also divide the force by area(in ), which gives us pressure on the left side of the equation.
A manometer can be used to determine gas pressures. The manometer is essentially a U-shaped tube containing some kind of fluid with known density, and one side is connected to the region of interest while the reference pressure is applied to the other. The difference in liquid level represents the applied pressure.
Manometer fluid – mercury was used originally (hence mmHg), but we now have a wide variety of fluids of various densities.
Manometer types:
Right side image obtained from Apply Science Concepts to Trades Applications / CC BY 4.0
The difference in heights of the liquid columns correlates to the difference in the pressure of the gases. The pressure of the gas in the vessel is calculated by:
Exercise: Manometer Calculation
Compressed air has been used to transmit and store power since the 1870s. Cities such as Paris, Dresden and Buenos Aires had compressed air lines to transmit power to homes and businesses. The technology is now being applied to energy storage in former salt mines underground. One such plant in McIntosh, Alabama stores gas at 1100 psi. If a sealed end manometer with a fluid with a density of 3450 was used to measure this pressure, what would the height of the fluid column be? (take )
Manipulate h from the equation:
\begin{align*}
h& = \frac{P-P_{0}}{\rho g}\\
h& = \frac{1100psi×\frac{101325Pa}{14.696psi}}{3450\frac{kg}{m^3}×9.8\frac{m}{s^2}}\\
&=\frac{7.58×10^6Pa}{3.381×10^4\frac{kg}{m^2s^2}} \\
& = 2.242×10^2 \frac{Pa}{\frac{kg}{m^2s^2}}×\frac{1\frac{kg}{ms^2}}{1Pa}\\
& = 2.24×10^2m
\end{align*}
height = or 224 m, very high pressure, likely difficult to measure with a manometer.
Gauge vs. Absolute Pressure
The absolute pressure is the actual pressure at the point of interest. The absolute pressure is 0 in a vacuum and cannot be negative.
Gauge pressure is defined to be the difference between absolute pressure and atmospheric pressure:
when gauge pressure is 0, absolute pressure = atmospheric pressure
Most measuring devices measure gauge pressure. Many measuring devices use a flexible membrane between a chamber of known pressure (the outside chamber, which is connected to the atmosphere) and a vessel with the pressure we want to measure. If there is a difference in pressure on both sides of the membrane, the membrane will expand or contract. Through this, we can relate the membrane’s position to the gauge pressure of the vessel.
The units often used on the devices include:
= pounds per square inch
= inches of mercury
Exercise: Reading Absolute and Gauge Pressures
1. If the gauge in the image reads 5 psi at the orange mark and the atmospheric pressure is 14.7 psi, which of the following are the gauge and absolute pressures?
A – gauge = -5 psi, absolute = 9.7 psi
B – gauge = 5 psi, absolute = 9.7 psi
C – gauge = -5 psi, absolute = 19.7 psi
D – gauge = 5 psi, absolute = 19.7 psi
2. If the gauge in the image reads 10 inHg at the purple mark and the atmospheric pressure is 29.9 inHg, which of the following are the gauge and absolute pressures?
A – gauge = -10 inHg, absolute = 39.9 inHg
B – gauge = 10 inHg, absolute = 19.9 inHg
C – gauge = -10 inHg, absolute = 19.9 inHg
D – gauge = 10 inHg, absolute = 39.9 inHg
1. Answer: D
The arrow is on the “pressure” side, which indicates that the gauge pressure is positive. The absolute pressure is the sum of absolute pressure and gauge pressure.
\begin{align*}
absolute\;pressure& =gauge\;pressure+atmospheric\;pressure\\
& = 5psi+14.7psi\\
&=19.7 psi
\end{align*}
2. Answer: C
The arrow is on the “vacuum” side, which indicates that the gauge pressure is negative. The absolute pressure is the sum of absolute pressure and gauge pressure.
\begin{align*}
absolute\;pressure& =gauge\;pressure+atmospheric\;pressure\\
& =-10inH\!g+29.9inH\!g\\
&=19.9inH\!g
\end{align*}
[1] OpenStax University Physics Volume 2. 2016. 2.2 Pressure, Temperature, and RMS Speed. [online]<https://openstax.org/books/university-physics-volume-2/pages/2-2-pressure-temperature-and-rms-speed> [Accessed 11 May 2020].
[2] Line D – Organizational Skills Competency D-2: Apply Science Concepts to Trades Applications. 2015. Manometer. [online] <http://solr.bccampus.ca:8001/bcc/file/3654f43b-8e2b-43be-99ce-acd1b2e4d809/1/D-2.pdf> [Accessed 11 May 2020].
Learning Objectives
By the end of this section, you should be able to:
Understand the physical properties described by the temperature
Calculate conversions between scales of temperature
Temperature is operationally defined as the quantity of what we measure with a thermometer. The average kinetic energy of a molecule is directly proportional to its absolute temperature. Differences in temperature maintain heat transfer through different systems. Heat transfer is the movement of energy from one place or material to another as a result of a difference in temperature.
Assigned scales are based on physical measurements:
Celsius (°C) – based on water freezing (0 °C) and boiling (100 °C) at atmospheric pressure. Absolute zero at -273.15°C.
Fahrenheit (°F) – based on water freezing (32 °F) and boiling (212 °F) at atmospheric pressure. Absolute zero at -459.67°F.
There are also other absolute scales with 0 being absolute zero:
Kelvin (K) – Same size of a degree as Celsius
Rankine (°R) – Same size of a degree as Fahrenheit
Celsius to Kelvin:
Fahrenheit to Rankine:
Kelvin to Rankine:
Celsius to Fahrenheit:
Exercise: Temperature Conversions
Some solar plants use molten salt as a heat transfer agent that can be stored efficiently. This allows the plant to operate after the sun has set by extracting energy from the molten salt they have stored during the day. The temperature difference is important to know as it will dictate the energy obtainable from the salt. If the molten salt is heated to 1000°F on a given day and if the environment is at 25°C what is the temperature difference in Kelvin?
First, convert the two temperatures to the same units. Because the question asks for the temperature difference in Kelvin, it is more convenient to compare on a Celsius scale because the sizes of 1 degree are the same for Celcius/Kelvin.
Step 1: Convert the temperature in Fahrenheit to Celsius:
Step 2: Calculate the difference in temperature in Celsius, which is equal to the difference in Kelvin (you can prove this to yourself if desired if you convert both temperatures above from Celsius to Kelvin).
We can use Newton’s second law to find mass, with being the acceleration of gravity. Since the mass we are looking for is “per square metre”, we can also divide the force by area(in ), which gives us pressure on the left side of the equation.
[1] OpenStax University Physics Volume 2. 2016. 1.1 Temperature and Thermal Equilibrium. [online]<https://openstax.org/books/university-physics-volume-2/pages/1-1-temperature-and-thermal-equilibrium> [Accessed 13 May 2020].
[2] OpenStax University Physics Volume 2. 2016. 2.2 Pressure, Temperature, and RMS Speed. [online]<https://openstax.org/books/university-physics-volume-2/pages/2-2-pressure-temperature-and-rms-speed> [Accessed 13 May 2020].
Learning Objectives
By the end of this section, you should be able to:
Apply the definitions of intensive and extensive properties to process variables
Extensive Variables – depend on the size of a system
Intensive Variables – do not depend on the size of a system
Are temperature, pressure, and volume extensive or intensive?
Temperature: intensive
Pressure: intensive
Volume: extensive. Instead we may use molar volume if we want to express volume as an intensive property. Molar volume is defined by the volume occupied by 1 mole of the substance inside the system: .
Learning Objectives
By the end of this section, you should be able to:
Interpret phase diagram for relevant thermodynamic information (e.g. composition, temperature, pressure…)
The phase of a substance and separation between molecules are related to the intermolecular forces in a substance. Therefore, a source of energy needs to be supplied when the substance transitions to a higher energy state to separate the molecules in the substance.
Something similar you may recall from a previous chemistry course (say CHEM 154): Bond-breaking requires energy; Bond making releases energy. Similarly, a transition to a lower energy state releases energy.
image obtained from General Chemistry: Principles, Patterns, and Applications/CC BY-NC-SA
The phase of a given substance depends on pressure and temperature. Thus, plots of pressure versus temperature showing the phase in each region provide considerable insight into the thermal properties of substances. Such a P-T graph is called a phase diagram.
In the phase diagram, we can observe that the boiling point curve (the line that distinguishes liquid and vapour phases) ends at a point, which is called critical point. Above the critical point, liquid and gas can no longer be distinguished and are classified as a supercritical fluid.
Vapour phase is a gaseous species below its critical temperature. On the other hand, a gaseous phase describes a gaseous species above its critical temperature but below its critical pressure.
Triple point is defined to be the set of pressure and temperature that solid, liquid and vapour are in equilibrium together.
Reduced properties are defined as variables (pressure, temperature) scaled by their values at the critical point. Reduced properties can be useful in coming up with general equations describing different substances.
= critical temperature
= critical pressure
Critical Pressure =
Critical Temperature =
Triple point with ice = 611 Pa
Triple point with ice = 273.16 K
Ice is the most common crystal structure of ice, but ice has a total of 17 known crystal forms. The latest form was discovered in 2014.
Image obtained from Chemistry libretexts / CC BY-NC-SA 3.0
At low temperatures, solid is the stable phase. At moderate temperatures and high pressure, liquid is the stable phase, and at high temperature and low pressure, gas/vapor is the stable phase. Lines separate these phases, representing sets of pressure/temperature for phase changes.
The melting curve or fusion curve of ice/water is very special. It has a negative slope due to the fact that when ice melts, the molar volume decreases. At higher pressure, ice actually melts at a lower temperature. When we skate, skate applies a very high pressure to the ice, so the ice below the skate melts without increasing temperature. The liquid formed between the skate and ice acts as a lubricant so that the skater moves gracefully across the ice.
A thermodynamic system includes anything whose thermodynamic properties are of interest. It is embedded in its surroundings or environment; it can exchange heat with, and do work on, its environment through a boundary, which is the imagined wall that separates the system and the environment.
In reality, the immediate surroundings of the system are interacting with it directly and therefore have a much stronger influence on its behavior and properties. For our thermodynamic piston system, the gas inside the cylinder is the thermodynamic system; the cylinder, piston, and air outside form the surroundings of the system. The boundary then consists of the inner surfaces of the cylinder and piston.
Image obtained from OpenStax University Physics Volume 2 / CC BY 4.0
The path from A to E below shows a process in a thermodynamic piston system containing pure water. The water vapor (point A) in the system is first compressed under the same temperature till a certain point at liquid state (point C). Then, the water is heated under the same pressure till a certain point at vapour state (point E).
During the process, we can see that the state of water passes through the boiling curve twice (points B and D). The curve indicates a transition of state between liquid and vapor. During a boiling (point D) or condensation (point B) process, the substance does not change in temperature or pressure. Since this phase diagram does not contain information on molar volume, points B and D could be a mixture of liquid and vapor at any ratio, including pure liquid or vapor at the boiling temperature and pressure.
We have seen phase diagrams with pressure and temperature on the axes from the previous chapters; this was one projection of a phase diagram. Instead, we could also have other intensive variables on the axes. As a result, a 3-dimensional phase diagram can be produced with pressure, temperature, and molar volume on separate axises. Note that we could also use other intensive variables such as enthalpy or internal energy per mole, etc.
Image obtained from Wikimedia Commons/ CC BY-SA 3.0
A link to a 3-D model of phase diagram to help to visualize: http://biomodel.uah.es/Jmol/plots/phase-diagrams/
A phase diagram can be produced with the projection on any of the planes spanned by two of the axises in the 3-dimensional phase diagram above. For example, if we project it onto the Pressure-Volume (P-V) plane we get the image below. Note that volume here appears to be written as an extrinsic variable, but should rather be a specific volume (say in ).
Image obtained from College Physics / CC BY 4.0
The red lines represent the substance at constant temperatures (isotherms). The isotherms have steeper slopes in the liquid region and gentler slopes in the vapour region. This agrees with what we observed before that the volume of gases experiences more significant changes due to pressure changes, but the volume of liquids is hardly affected by pressure.
The segments of isotherms in the vapour-liquid equilibrium region are horizontal, which agrees with our previous observation that the pressure is constant during the boiling process for a pure substance at a given temperauture. This diagram provides information that allows us to calculate the ratio of vapour to liquid under vapour-liquid equilibrium conditions based on molar volume, which is not represented on the P-T diagram because it collapses to a point. The vapour pressure is represented by the y-coordinate of the horizontal segment. Note each vapour pressure will thus correspond to specific to a temperature (given by the isotherm).
Image obtained from Chapter 2: Pure substances /CC BY-NC-SA 3.0 US
If we project the 3-dimensional phase diagram onto the temperature-volume (T-V) plane, the phase diagram will be similar to the one shown above. Each curve on the diagram represents processes with constant pressure (isobars). Similar to the P-V diagram, the pressure curve is horizontal in the region of vapour-liquid equilibrium. The vapour pressure is constant for each curve while the boiling temperature can be found as the y-coordinates on the horizontal segments. These correspond directly to the temperature and vapour pressure lines we would see on the P-V diagram.
[1] OpenStax University Physics Volume 2. 2016. 1.5 Phase Changes. [online] <https://openstax.org/books/university-physics-volume-2/pages/1-5-phase-changes> [Accessed 14 May 2020].
[2] Chemistry Libretext. 2020. Phase Diagrams. [online] <https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Phase_Transitions/Phase_Diagrams> [Accessed 14 May 2020].
[3] OpenStax University Physics Volume 2. 2016. 3.1 Thermodynamic Systems. [online] <https://openstax.org/books/university-physics-volume-2/pages/3-1-thermodynamic-systems> [Accessed 15 May 2020].
Learning Objectives
By the end of this section, you should be able to:
Calculate degrees of freedom using Gibb’s Phase Rule
Gibb’s phase rule is used to determine the number of intensive variables required to fully specify the thermodynamic properties of a system (thermodynamic degrees of freedom).
Once this number of intensive variables is specified, we can find all the thermodynamic properties of the substance (eg. other intensive variables such as pressure, temperature, enthalpy, internal energy, etc.) at that state using previous experimental data from a table or chart.
= degrees of freedom (number of intensive variables that needs to be specified)
= number of independent chemical species
= number of phases in a system at equilibrium
= number of independent equilibrium chemical reactions among the species
One way of understanding this equation is thinking that we want to find a point (0 dimension) on the phase diagram, as this represents a single set of thermodynamic properties. If we have a single component system, the phase diagram is similar to the ones we have seen so far in this lecture. Specifying one intensive property reduces the dimension by 1. Adding a component adds another axis to the graph (to represent the composition of the mixture), which adds one dimension. Each equilibrium provides an equation that relates the composition of substances in the system, which allows us to secure a position on the composition axis.
For example, if we have a pure substance in the system and we know that the state is in the liquid region(2 dimension), specifying pressure ensures the point of the state we are looking for is on a horizontal line (1 dimension). If we then also specify the temperature, it ensures the point is on a vertical line. There is only one point that satisfies both requirements of the temperature and pressure. We have reduced the dimension to 0.
In another example, if we know the state is on one of the lines indicating state transitions (eg. boiling – being vapour-liquid equilibrium), we start at 1 dimension, then we just need to specify one of the x, y coordinates (temperature or pressure) to reduce the position of the state to a point.
Triple point is already a point (0 dimension). No other variables need to be specified.
Exercise: Gibb’s Phase Rule
In a system containing pure water, how many degrees of freedom are there in the system at:
a) somewhere in the liquid phase shown by area C.
b) The triple point
Because we only have water in the system, , and there is no equilibrium chemical reaction in the system (as there is nothing to react with and no reaction specified), so .
a) At area C, we only have a liquid phase, so
This area represents pure water as a liquid. To fix this and know its thermodynamic properties we would need to know 2 intensive variables, say temperature and pressure. Knowing these two then rather than having an area, we would have a specific point.
b) At the triple point, pure vapour, liquid, and solid water exist at the same time, so
Specifying the water is at its triple point is equivalent to specifying the pressure and temperature, because the triple point only exists at a certain set of pressure and temperature.
Learning Objectives
By the end of this section, you should be able to:
Describe the relationship between variables of state for gases
Describe gas behaviour using ideal gas law and compressibility factor using equations of state
Relates pressure (), volume (), temperature () and the number of moles () of an ideal gas species using the ideal gas constant ():
Can also relate pressure, molar volume () and temperature:
The ideal gas law is an approximation that works well under some conditions:
It is known experimentally that for gases at low density (such that their molecules occupy a negligible fraction of the total volume) and at temperatures well above the boiling point, these proportionalities hold to a good approximation.
There are different ways to estimate how well the ideal gas approximation applies to a system. We will not go over these estimates explicitly in this course. Generally for this course, unless otherwise noted, we will assume the ideal gas law applies. However, in future thermodynamic courses, you will see when the ideal gas approximation may not be appropriate and will see other methods of relating gas properties in non-ideal scenarios.
If two or more gases are mixed, they will come to a thermal equilibrium as a result of collisions between molecules. When the gases have the same temperature, their molecules have the same average kinetic energy. Thus, each gas obeys the ideal gas law separately and exerts the same pressure on the walls of a container that it would if it were alone.
Therefore, in a mixture of gases, the total pressure is the sum of partial pressures of the component gases, assuming ideal gas behavior and no chemical reactions between the components.
Using Dalton’s law, we can calculate the partial pressure of a gas component, which is defined by the pressure that individual component in the gas mixture exerts on the wall if it were alone.
where
the mole fraction of the gas component
Amgat’s Law is analogous to Dalton’s Law, but is applied to the volume of the gas. The partial volume that each gas occupies will add up to the total system volume.
|
These partial volumes are proportional to the molar fraction of each gas in a system.
A word of caution that standard conditions may not always be the same in different industries or applications. This is shown below with two examples of standard conditions for gases from IUPAC (the International Union of Pure and Applied Chemistry) and NIST (the National Institute of Standards and Technology).
IUPAC | NIST | |
---|---|---|
Standard Temperature | 273.15K or 0°C | 20°C |
Standard Pressure | kPa or 1 bar | 1 atm or 101.325kPa |
Using the IUPAC numbers we can calculate a standard volume from the ideal gas law. This is the amount of volume a mole of gas would occupy at these standard conditions:
In some applications, gas flows may be expressed in standard cubic metres (SCM) or feet (SCF). A flow of 18.2 SCMH means 18.2 at standard conditions (whatever T/P these are defined to be).
We can calculate the temperature, pressure, and volume by comparing fluids at standard and non-standard conditions using the fact that is constant. We usually use it to compare the same system under different conditions, which means the number of gas molecules (n) may be constant when we perform the calculation.
Exercise: Natural Gas Standard Flow
Natural gas (mostly that we might use to heat our homes) is normally measured with a correction to a certain temperature and pressure (to ensure you are always getting the same number of gas molecules). If it is a very cold winter day in Quebec City, where natural gas is entering your home at -30°C at a pressure of 2 bar and a flow of 100 L per hour. What would the standard flow (in L/hour) be if our standard conditions were 0°C and 1 bar (the IUPAC conditions)?
The number of moles of gas (n) is the same in both conditions. So we can cancel n from both sides and rearrange the equation to isolate . Remember that T in the equation needs to be converted to an absolute scale.
In thermodynamics, and generally in this course:
means that the partial pressure is a function of temperature
The brackets are not for multiplication, but are for showing what variables the variable of interest is a function of. This is confusing, but a common notation in thermodynamics. Think of f(x) you have seen in math, this means f is a function of x, rather than f is multiplied by x.
Some other examples are:
means the density of gasses is a function of Temperature and Pressure
means density of liquids is a function of Temperature
In this course, we will consider solids and liquids incompressible, meaning their density does not change significantly with changes in pressure. We will consider liquid density to be a weak function of temperature.
Density values can be found in:
http://resolve.library.ubc.ca/cgi-bin/catsearch?bid=5888251 – 2008 edition
http://resolve.library.ubc.ca/cgi-bin/catsearch?bid=9292999 – 2019 edition
http://tinyurl.com/yxv3r8vy – library ebook link
http://resolve.library.ubc.ca/cgi-bin/catsearch?bid=5888251 – 2008 edition
http://resolve.library.ubc.ca/cgi-bin/catsearch?bid=9292999 – 2019 edition
Review: Assumptions of the ideal gas law
When rearranging the ideal gas law, we get that for ideal gases:
For a real gas, we adjust the ideal gas law by the compressibility factor(Z):
We can generalize the value of Z based on the reduced temperature and reduced pressure, which can be obtained from charts or correlations:
where and are the temperatrue at the critical point.
The charts to find compressibility factor Z can be found in the “module” tab from CHBE 220 course on canvas.
Exercise: Compressibility Factor
If our substance is propane at P = 20 bar and T = 100 °C, with critical parameters: = 4.23 MPa, = 97°C, find the compressibility factor .
First, we calculate the reduced temperature and pressure:
Unit conversions used: , :
Find the lines representing on the x-axis, and on the curves with solid lines. The compressibility factor is the y-coordinate of the intersection.
[1] OpenStax University Physics Volume 2. 2016. 2.1 Molecular Model of an Ideal Gas. [online] <https://openstax.org/books/university-physics-volume-2/pages/2-1-molecular-model-of-an-ideal-gas> [Accessed 14 May 2020].
[2] OpenStax University Physics Volume 2. 2016. 2.2 Pressure, Temperature and RMS Speed. [online] <https://openstax.org/books/university-physics-volume-2/pages/2-2-pressure-temperature-and-rms-speed> [Accessed 14 May 2020].
Learning Objectives
By the end of this section, you should be able to:
Estimate the vapour pressure of pure compounds at a given temperature using Antoine’s equation
Vapour pressure() is defined as the pressure at which a gas coexists with its solid or liquid phase. Vapour pressure is created by faster molecules that break away from the liquid or solid and enter the gas phase. The vapour pressure of a substance depends on both the substance and its temperature—an increase in temperature increases the vapour pressure.
One empirical method to estimate is the Antoine equation:
where
A, B and C are constants depending on the type of substance
NOTE: Sometimes different sources of data can use different units for temperature, or used (base e) instead of (base 10).
Finding Antoine Equation Parameters on NIST:
Example: Estimating Vapour Pressure for Water
Temperature(K) | A | B | C |
---|---|---|---|
379-573 | 3.55959 | 643.748 | -198.043 |
273-303 | 5.40221 | 1838.675 | -31.737 |
304-333 | 5.20389 | 1733.926 | -39.485 |
data obtained from NIST
Based on the data above, what is the vapour pressure of water at 100°C and 200°C?
For T=100°C: Since the temperature is given in the Kelvin scale on the chart, we convert 100°C to 373K, which is closest to the range in the first row of data, so we will use that range.
\begin{align*}
log_{10}(p*) &=A-\frac{B}{T+C}\\
p*& = 10^{A-\frac{B}{T+C}}\\
&= 10^{3.55959-\frac{643.748}{373-198.043}}\\
& = 0.76 bar
\end{align*}
For T=200°C: We first convert the temperature to 473K, which is in the range for the first row of data.
\begin{align*}
log_{10}(p*) &=A-\frac{B}{T+C}\\
p*& = 10^{A-\frac{B}{T+C}}\\
&= 10^{3.55959-\frac{643.748}{473-198.043}}\\
& = 16.53 bar
\end{align*}
Consider the calculated result for vapour pressure at 100 °C, compare it with your knowledge of vapour pressure of water. Does it seem accurate?
Temperature(K) | calculated (bar) | actual(bar) | % difference |
---|---|---|---|
373 | 0.758798 | 1.013 | 25.09403 |
473 | 16.53187 | 15.55 | 6.314304 |
[1] OpenStax University Physics Volume 2. 2016. 13.5 Phase Changes. [online] <https://openstax.org/books/college-physics/pages/13-5-phase-changes> [Accessed 15 May 2020].
Learning Objectives
By the end of this section, you should be able to:
Apply properties of ideal mixtures to describe vapour-liquid separations using Raoult’s law and Henry’s law
An ideal mixture contains substances with similar molecular interactions (which usually means similar structures or functional groups). e.g. benzene and toluene, with both being non-polar and of a similar molecular size
Image from Darkness3560 via Wikimedia Commons/CC0
A combination of two substances is called a binary mixture. For a binary mixture, there is no longer a set temperature and pressure for a boiling point. Instead, there is a range of temperature or pressure values where the mixture is in vapour-liquid equilibrium. These are bounded by two points called the bubble point and dew point. We’ll explore these concepts more thoroughly when looking at diagrams for these binary mixture systems.
In a closed container, vapour-liquid equilibrium can be achieved by the same number of molecules vapourizing and condensing at any given period of time (meaning no net change in the system). The temperature and pressure have to be kept constant if we are truly at equilibrium (again, no net change in any intensive variable).
We usually denote:
: overall temperature
: overall pressure
: the mole fraction of substance in the liquid phase
: the mole fraction of substance in the vapour phase
Raoult’s law is used when all components are in relatively significant quantities or they are chemically very similar (generally in terms of intermolecular interactions).
means that the vapour pressure is a function of temperature, which can be calculated using the Antoine Equation (there are other equations, but in this class we will stick to the Antoine equation).
Exercise: Estimating Vapour Pressure for Ethanol
At 298 K, the vapour pressure of water is 3,138 Pa and the vapour pressure of ethanol is 7,817 Pa. Suppose we have a mixture of only water and ethanol in a closed container, at 298 K, what is the mole fraction of ethanol in the vapour phase () at equilibrium if the liquid is 50 mol% water and 50 mol% ethanol?
Step 1: First note here ethanol and water both have very polar OH groups, so we will consider their intermolecular interactions similar and apply Raoult’s law. Rearrange the Raoult’s law to isolate :
\begin{align*}
y_{i}×P & =x_{i}×p^*_{i}(T)\\
y_{i} & = \frac{x_{i}×p^*_{i}(T)}{P}
\end{align*}
Here we know both , which was given to be 0.5 (50 mol%) for ethanol, and for ethanol, again given in the question prompt. But now we need to find pressure ().
Step 2: To find we’ll first apply the ideal gas law to the gas phase in this system. We know that for ideal gases the partial pressures () add up to the total system pressure (). We can then use Raoult’s law to express partial pressures in terms of liquid mole fractions and vapour pressures (all of which we know in this case):
\begin{align*}
P & =p_{water}+p_{ethanol}\\
& = x_{water}×p^*_{water}+x_{ethanol}×p^*_{ethanol}\\
& = 0.5×3138Pa + 0.5×7817Pa \\
& = 5477.5 Pa
\end{align*}
Step 3: Subsitute $P$ into the expression for :
\begin{align*}
y_{ethanol} & = \frac{x_{ethanol}×p^*_{ethanol}(298K)}{P} \\
& =\frac{0.5×7817Pa}{5477.5Pa} \\
& = 0.71
\end{align*}
Finally we could ask ourselves, “does this answer make sense?”. If we look up the normal boiling point temperature of water (100 °C ) and ethanol (78 °C), we see ethanol is more volatile (it boils at a lower temperature). This tells us to expect a greater mole fraction of it in the vapour phase than the liquid phase, which matches what we find with our calculations.
Henry’s law is used to describe the concentration of a gas (or very volatile component) dissolved in a liquid. This is typically used to describe components in a mixture that are near or above their critical temperature (think back to the difference between vapours () and gases ().
Henry’s constant, usually given for a solvent in a given range of temperature.
The Henry’s law constant can be given in different units, such as , , , which will change the corresponding unit for . can be used to represent the solubility of the gas in per mole / per volume / per mass of solvent.
Henry’s law is generally more valid when is close to 0 (very dilute solutions).
Exercise: Calculating Partial Pressure Using Henry’s Law
The concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa at 20 °C is . Use Henry’s law to determine the solubility of oxygen when its partial pressure is 20.7 kPa at the same temperature. Note that this approximates the pressure of oxygen in the earth’s atmosphere.
Step 1: Calculate Henry’s law constant from data with oxygen pressure at 101.3 kPa
\begin{align*}
p_{i} &=\frac{x_{i}}{H_{i}(T)}\\
H_{i}(T) & =\frac{x_{i}}{p_{i}}\\
& =\frac{1.38×10^{−3}\frac{mol}{L}}{101.3kPa}\\
&= 1.36×10^{−5} \frac{mol}{L·k\!Pa}
\end{align*}
Step 2: Apply Henry’s law another time for an oxygen partial pressure at 20.7 kPa:
\begin{align*}
p_{i} &=\frac{x_{i}}{H_{i}(T)}\\
x_{i} & = p_{i}×H_{i}(T)\\
& =20.7kPa×1.36×10^{−5} \frac{mol}{L·k\!Pa}\\
&= 2.82×10^{−4} \frac{mol}{L}
\end{align*}
[1] OpenStax Chemistry 2e. 2019 Feb. 11.3 Solubility. [online] <https://openstax.org/books/chemistry-atoms-first-2e/pages/11-3-solubility> [Assessed 18 May 2020]
Learning Objectives
By the end of this section, you should be able to:
Read Txy diagrams for binary mixtures
Find bubble points and dew points on a Txy diagram
Calculate the mole fraction of substances in the vapour and liquid states
The following diagram is the Txy diagram (at a constant pressure of 1 atm) for a mixture of benzene and toluene. The y-axis (ordinate) reads the temperature value, and the x-axis (abscissa) corresponds to the mole fraction of benzene. We can use the x-axis to find the mole fraction of benzene in the liquid and vapour phase.
Because this is a binary mixture, we can calculate the mole fraction of toluene as in the liquid phase and in the vapour phase.
The two curves separate the digram into 3 regions:
Liquid region: when the mixture of both components are in liquid phase, this is the lowest temperature region. This makes sense as liquid is a lower energy state than a gas, so we would expect it at lower temperatures.
Vapour region: when the mixture of both components are in vapour phase, this is the highest temperature region. Again, the higher the temperature, the more energy, and thus more likely to be a gas.
Vapour-liquid equilibrium region: when the mixture splits into a liquid and a vapour, this is the region in the middle
Given a mole fraction of benzene and temperature of this system, we can locate which of the three regions our system would be in. Note we have to ensure we are at the same pressure as the graph states when using these graphs.
Two particular points we will analyze are when the mole fractions of benzene in the system are 1 or 0:
mole fractions of benzene = 1: , “mixture” is pure benzene. The system vapourizes/condenses at a contant temperature, which is the normal boiling point of benzene at 1 atm (~349 K in this case).
mole fractions of benzene = 0: , “mixture” is pure toluene. The system vapourizes/condenses at a contant temperature, which is the normal boiling point of toluene at 1 atm (~383 K in this case).
Let’s take the following example to demonstrate how to read a Txy diagram: suppose we start with a mixture of 0.5 mole fraction of liquid benzene in the system at 80°C and 1 atm.
Shown above is the point we can locate on the Txy diagram. We can see that the point is completely in the liquid region, which means the mixture is completely in the liquid phase.
We then increase the temperature (keeping the pressure constant as defined by the diagram) until we hit the first curve on the boundary of the liquid region:
This point is called the bubble point of this mixture. At this point, we see the first bubble of vapour being formed in the liquid. The composition of this first vapour can be found at the point on the other curve at the same temperature:
Calculate the Composition of Vapour at the Bubble Point
We can also use the phase change data to calculate the composition of the first vapour. This is useful when the phase diagram is not provided:
The phase change data can be obtained from the NIST website:
substance | A | B | C |
benzene | 4.60362 | 1701.073 | 20.806 |
toluene | 4.07827 | 1343.943 | -53.773 |
Solution
We can also use the phase change data to calculate the composition of the first vapour. This is useful when the phase diagram is not provided:
The phase change data can be obtained from the NIST website:
Using Raoult’s law, we can calculate the molar composition of the vapour:
where
We are looking for the molar fraction of components in the vapour phase: and .
P is the pressure of the system, in this case, it is kept constant at 1 atm=1.013bar
At the bubble point, there is a very small amount of vapour being formed, therefore we assume the change in the composition of the liquid is negligible. We can take .
can be calculated using the Antoine Equation:
However, we don’t know the T for the bubble point (we are assuming the diagram is not given):
When we don’t know the bubble point temperature, we use a trial and error approach: We guess a temperature and use this to find for the substances involved. We then calculate the for both substances using Raoult’s law and the in the liquid phase. If we are at the right temperature (the bubble point temperature) this should result in the sum of being equal to 1 ().
A safe place to start is taking the average of the boiling points of the two substances, but the starting point does not affect the result. It may just affect the number of iterations needed before finding the actual bubble point temperature.
From the NIST website, we can find that the boiling point of benzene at 1 atm ≈ 80°C and the boiling point of toluene at 1 atm ≈ 110°C, so we can take for our initial guess:
Step 1: Calculate for both substances using the Antoine Equation:
\begin{align*}
log_{10}(p*) &=A-\frac{B}{T+C}\\
p*& = 10^{A-\frac{B}{T+C}}\\
p*_{benzene}&= 10^{4.60362-\frac{1701.073}{368+20.806}}= 1.692 bar \\
p*_{toluene}&= 10^{4.07827-\frac{1343.943}{368-53.773}}=0.633 bar
\end{align*}
Step 2: Calculate the for both substances using Raoult’s law:
\begin{align*}
y_{i}×P & =x_{i}×p_{i}^*(T)\\
y_{benzene}& =\frac{x_{benzene}×p_{benzene}^*(T)}{P}\\
&= \frac{0.50×1.692bar}{1.013bar}\\
& = 0.835
\end{align*}
\begin{align*}
y_{toluene}& =\frac{x_{toluene}×p_{toluene}^*(T)}{P}\\
&= \frac{(1-0.50)×0.633bar}{1.013bar}\\
& = 0.312
\end{align*}
Step 3: Check our result by adding up the and adjust the guess of temperature accordingly.
So the guess for should be decreased for the next guess.
After several trials, we can prove that is the bubble point temperature:
Step 1: Calculate for both substances using the Antoine Equation
\begin{align*}
log_{10}(p*) &=A-\frac{B}{T+C}\\
p^*& = 10^{A-\frac{B}{T+C}}\\
p^*_{benzene}&= 10^{4.60362-\frac{1701.073}{363+20.806}}= 1.484 bar\\
p^*_{toluene}&=10^{4.07827-\frac{1343.943}{363-53.773}}=0.633 bar=0.540bar
\end{align*}
Step 2: Calculate the for both substances using Raoult’s law:
\begin{align*}
y_{i}×P & =x_{i}×p_{i}^*(T)\\
y_{benzene}& =\frac{x_{benzene}×p_{benzene}^*(T)}{P}\\
&= \frac{0.50×1.484bar}{1.013bar}= 0.732
\end{align*}
\begin{align*}
y_{toluene}& =\frac{x_{toluene}×p_{toluene}^*(T)}{P}\\
&= \frac{(1-0.50)×0.540bar}{1.013bar}= 0.267
\end{align*}
Step 3: Check our result:
Because there are rounding errors involved in the calculations, we can say that T=363 K is our bubble point temperature (at least to +/- 1 K).
Therefore, the molar fractions of the substances for the first vapour at the bubble point are and . Note that this matches what we see on the diagram.
The composition of the vapour at the bubble point can also be solved with the solver add-in in excel:
How to load “Solver Add-in” in Excel:
Step 1: Click “file” in the header and open the “Options” menu:
Step 2: In the Excel Options menu, select “Solver Add-in” in the “Add-ins” section, then click “OK”.
Step 3: Then you can find the “Solver” in the “Analyze” section under the “Data” menu.
Solve for the composition of the vapour at the bubble point with the solver add-in in excel:
Step 1: To find the composition of the vapour at the bubble point, we first input all the know information for the calculation, including:
Step 2: Compute the vapour pressure of both substances using the temperature we guessed:
Step 3: Compute for both substances using:
Step 4: Compute the by adding the for both substance. With our initial guess, the we get should be close, but not equal to 1.
Step 5: Open the “Solver”, indicate that we want to solve for the temperature that makes our .
Step 6: Click “OK” if the window says the solver found a solution. If the solver cannot find a solution, it is maybe because our initial guess is too far away from the actual bubble point temperature, if so, we can change the temperature and try to solve it again.
Step 7: After the window closes automatically, the value for temperature and all cells affected by temperature are changed to the corresponding values calculated for the bubble point temperature found.
Therefore, according to the result generated by solver:
Bubble point temperature = 363.04K
The molar fractions of the substances for the first vapour at the bubble point are and . Note this matches what we see on the diagram and our result from the trial-and-error approach. Note, that you could also use excel without solver to make guess and check faster, we would simply guess temperatures and see the effect on . This then uses excel to do the calculations each time we update the guess, making solving much faster. However, solver automates this process of guessing temperatures.
If we keep increasing the temperature, the system will be in the vapour-liquid equilibrium region. We locate the point using the total mole fraction of benzene and the temperature of the system, so the x-coordinate is 0.5 no matter what phase the system is at.
After we find the point, we draw a horizontal line (representing that the temperature of the liquid and vapour are equal under vapour-liquid equilibrium conditions), the line will hit both curves. The x-coordinates of the points where the horizontal line hits the curves are the mole fraction of benzene, with the one on the boundary of the liquid region as the mole fraction in the liquid phase and the other one as the mole fraction in the vapour phase.
At vapour-liquid equilibrium, there is always more benzene in the vapour phase compared to benzene in the liquid phase. That is because benzene is the more “volatile” compound compared to toluene, which means it has weaker intermolecular interactions, therefore it’s more likely to vapourize. A more volatile compound has a lower boiling point.
Exercise: Calculate the Mole Fraction at Vapour-Liquid Equilibrium
Calculate the molar fraction of the total liquid component () in the system at the point shown above in the vapour-liquid region. Use values for the mole fraction of benzene in the vapour and liquid phases as read off the graph to find this.
We know that the total benzene in the system is from the sum of the benzene molecules from the liquid and vapour phase, which is equal to 0.5 (as we start with a half/half mixture):
We can replace the expression of , starting from because the are only liquid phase and vapour phase present in the system:
\begin{align*}
n_{vapour}+n_{liquid} &= n_{total}\\
\frac{n_{vapour}}{n_{total}}+\frac{n_{liquid}}{n_{total}}&=1\\
\frac{n_{vapour}}{n_{total}}&=1-\frac{n_{liquid}}{n_{total}}
\end{align*}
Substitute into the in the equation:
From reading the x-coordinates of the points on the curves, we can get: $x_{benzene}≈0.39,y_{benzene}≈0.62$
\begin{align*}
0.39×\frac{n_{liquid}}{n_{total}}+0.62×(1-\frac{n_{liquid}}{n_{total}}) & =0.5\\
(0.39-0.62)\frac{n_{liquid}}{n_{total}}& = 0.5-0.62\\
\frac{n_{liquid}}{n_{total}}&= \frac{0.5-0.62}{0.39-0.62}\\
& = 0.52
\end{align*}
Note this means we have roughly half vapour and half liquid. This is shown by our overall system point (red dot) being roughly between the two lines. Within the vapour-liquid region, the closer this point is to a certain line on the diagram the more of that phase will be in the system. Think back to our bubble point, where our system was almost entirely liquid (and very close to the liquid line).
If we then keep heating up the system under the same pressure, the system will hit the second curve at a certain point. We call this point the dew point. At this point, the last drop of liquid turns into vapour. The composition of the last drop of liquid can be found on the curve on the boundry of the liquid region at the same temperature.
In the Txy graph, the line connecting all the dew points (saturated vapour line) is always higher than the line connecting all the bubble points (saturated liquid line). Whether the slopes of the curves are positive or negative, the saturated vapour line will still be above the saturated liquid line because the system is more likely to be in the vapour phase when the temperature is higher.
Exercise: Dew Point Mole Fraction
Taking a similar approach to the bubble point and using Raoult’s law, calculate the and for the last drop of liquid at the dew point using the Antoine equation constants below.
Substance | A | B | C |
benzene | 4.60362 | 1701.073 | 20.806 |
toluene | 4.07827 | 1343.943 | -53.773 |
Raoult’s law states:
We know that P = 1.013 bar for the system is kept constant, then we just need to calculate the at the temperature of the dew point, then substitute this into the Raoult’s law equation to calculate .
To find using the Antoine equation, we must first guess a temperature. Similar to the bubble point calculation, a safe place to start is the average of the boiling points of the pure substances T=368K.
Step 1: Calculate the for both substances:
\begin{align*}
log_{10}(p*) &=A-\frac{B}{T+C}\\
p*& = 10^{A-\frac{B}{T+C}}\\
p*_{benzene}&= 10^{4.60362-\frac{1701.073}{368+20.806}}= 1.692 bar \\
p*_{toluene}&= 10^{4.07827-\frac{1343.943}{368-53.773}}=0.633 bar
\end{align*}
Step 2: Substitute into the Raoult’s law equation to calculate for both substances:
Use , because when there is only very little amount of liquid left in the system, we can approximate the mole fraction of benzene in the vapour to be equal to the total mole fraction of benzene in the system.
\begin{align*}
y_{i}×P & =x_{i}×p_{i}^*(T)\\
x_{benzene}& =\frac{y_{benzene}×P}{p^*_{benzene}}\\
&= \frac{0.5×1.013bar}{1.692bar}\\
& = 0.299
\end{align*}
\begin{align*}
x_{toluene}& =\frac{y_{toulene}×P}{p^*_{toluene}}\\
&= \frac{0.5×1.013bar}{0.633bar}\\
& = 0.800
\end{align*}
Step 3: Calculate the sum of the liquid mole fractions, and adjust the temperature guess accordingly:
The next guess for T should be higher so that the substances tend to condense less.
After several trials, we can show that is the bubble point temperature:
Step 1: Calculate the for both substances:
\begin{align*}
y_{i}×P & =x_{i}×p_{i}^*(T)\\
x_{benzene}& =\frac{y_{benzene}×P}{p^*_{benzene}}\\
&= \frac{0.5×1.013bar}{1.837bar}\\
& = 0.276
\end{align*}
\begin{align*}
x_{toluene}& =\frac{y_{toulene}×P}{p^*_{toluene}}\\
&= \frac{0.5×1.013bar}{0.695bar}\\
& = 0.725
\end{align*}
Step 3: Calculate the sum of the liquid mole fractions, and adjust the temperature guess accordingly:
Because there are rounding errors involved in the calcuations, we can say that T=371.2K is our bubble point temperature.
Therefore the molar fractions of ths substances in the last drop of liquid at the dew point are and .
If we keep increasing the temperature above the dew point, the system will be in the vapour phase only. The mole fraction of benzene will still be 0.5 since it is the ratio of total benzene molecules and total molecules in the system, regardless of the state.
Learning Objectives
By the end of this section, you should be able to:
Read Pxy diagrams for binary mixtures
Find bubble points and dew points on a Pxy diagram
Calculate the mole fraction of substances in the vapour and liquid states
The following diagram is the Pxy diagram for a benzene-toluene mixture at a constant temperature of 373K. They y-coordinates show the pressure of the mixture and the x-coordinates show the mole fraction of benzene in liquid and/or vapour phases.
Higher pressure forces molecules to come closer together, which causes the molecules to be more likely to condense from vapour to liquid. From top to bottom, the regions are liquid region, vapour-liquid equilibrium region, vapour region, which is the opposite from the Txy diagram:
Suppose we have a mixture with 0.5 mole fraction of benzene at a pressure of 0.5 bar, and we slowly increase the pressure while keeping the temperature constant at 373K.
We can locate our starting point using the pressure and mole fraction of benzene:
The point is fully in the vapour region, which means the mixture of benzene and toluene is in the vapour phase at a pressure of 0.5 bar and temperature of 373K.
As we keep increasing the pressure, the point will hit the first curve on the boundary of the vapour region. This point is the dew point of the system. At the dew point, the first drop of liquid is being formed. The composition of the first liquid can be found using the x-coordinate of the point on the other curve where the pressure is the same.
If we keep increasing the pressure while keeping the temperature constant, the point will move up into the vapour-liquid equilibrium region. Similarly to the Txy diagram, we can read the mole fraction of benzene by drawing a horizontal line through the point and reading the x-coordinate where the line hits the two curves.
As the pressure goes up, the system will hit the bubble point, where the last drop of vapour condenses into a liquid.
Notice that the line connecting all the bubble points at different mole fractions appears to be linear. We can show why this occurs using Raoult’s law:
Starting with the total pressure in the system being the sum of individual partial pressures, we replace each partial pressure with the liquid mole fraction and vapour pressure (using Raoult’s law):
\begin{align*}
P=p_{1}+p_{2} & =x_{1}×p^*_{1}(T)+x_{2}×p^*_{2}(T)\\
& =x_{1}×p^*_{1}(T)+(1-x_{1})×p^*_{2}(T)\\
&= x_{1}×[p^*_{1}(T)-p^*_{2}(T)]+p^*_{2}(T)
\end{align*}
This equation is in a linear form, where is the slope and is the y-intercept. Because the Pxy diagrams assume constant temperature, the vapour pressures are constant.
When we raise the pressure of the system over the dew point, the system will stay fully in the liquid region. There will be only liquid in the system and the composition of the liquid will be equal to the overall system composition.
Pressure definition | |
Calculating pressure using a manometer | |
Converting gauge/absolute pressure | |
Temperature conversions | Celsius to Kelvin: Fahrenheit to Rankine: Kelvin to Rankine:
Celsius to Fahrenheit: |
Reduced properties | |
Gibb’s Phase Rule | |
Ideal Gas Law | |
Dalton’s Law | |
Amgat’s Law | |
Relating ideal gas properties | |
Compressibility Factor | |
Antoine Equation | |
Raoult’s Law | |
Henry’s Law |
An interactive or media element has been excluded from this version of the text. You can view it online here:
https://pressbooks.bccampus.ca/chbe220/?p=1753
Exercise: Phase Change – Carbon Dioxide
Identify the phase changes that carbon dioxide will undergo as its temperature is increased from −100 °C while holding its pressure constant at 1,000 kPa. At what approximate temperatures do these phase changes occur? The phase diagram of carbon dioxide is shown below.
Image obtained from OpenStax Chemistry/ CC BY 4.0
Note: This diagram is missing the “vapour” phase. Usually, we call the gaseous phase under the liquid region “vapor” because it is able to condense to a liquid, while the “gas” phase is in a temperature that is too high to condense. Sometimes the terms are not strictly distinguished.
The phase change from solid to liquid at ~-55°C; from liquid to gas at ~-40°C.
To represent holding pressure constant while increasing temperature, we can draw a line moving horizontally to the right on the phase diagram.
A phase change occurs every time when the horizontal line hits a curve on the diagram.
Exercise: Gibb’s Phase Rule
How many degrees of freedom exist for a vapour-liquid (V-L) mixture of acetone () and methyl ethyl ketone ()? Assume no chemical reaction happens.
vapour-liquid mixture:
mixture of acetone and methyl ethyl ketone:
no chemical reactions:
$$D\!F=2+c-\pi-r=2+2-2-0=2$$
Exercise: Ideal Gas Law
Calculate the density (mass per cubic metre) of dry air (a) under IUPAC standard conditions and (b) in a hot air balloon at a temperature of 120°C under the same pressure. Assuming ideal gas behaviour.
Take average molar mass =29.0 mol/L for dry air.
a) To find the mass per cubic metre, we need the number of moles contained in a cubic metre under the IUPAC standard temperature and pressure. Calculate n by rearranging the ideal gas law:
\begin{align*}
PV & = nRT \\
n & =\frac{PV}{RT}\\
n & =\frac{1×10^5Pa×1m^3}{8.314\frac{J}{molK}×273.15K}\\
n & = 44.1mol
\end{align*}
The mass of air is equal to the number of moles contained in a cubic metre multiplied by molar mass:
$$m=n×MW=44.1mol×29.0g/mol=1278.9g=1.28kg$$
The density of air is the mass divide by volume:
$$\rho=\frac{m}{V}=\frac{1.28kg}{1m^3}=1.28kg/m^3$$
b) Here, we also have the same volume (1 ) but different temperature. We can once again find the number of moles this will hold.
\begin{align*}
PV & = nRT \\
n & =\frac{PV}{RT}\\
n & =\frac{1×10^5Pa×1m^3}{8.314\frac{J}{molK}×(273.15+120)K}\\
n & = 30.6mol
\end{align*}
We do a similar calculation as before to find the mass of the gas in the cubic metre
$$m=n×MW=30.6mol×29.0g/mol=887.4g=0.89kg$$
We can then use this to find density under this new condition:
$$\rho=\frac{m}{V}=\frac{0.89kg}{1m^3}=0.89kg/m^3$$
This is less dense, which makes sense as the temperature is higher, so we would expect less gas molecules in the same volume at a similar pressure.
One other way to solve this is to observe that the density is proportional to the number of moles, which is inversely proportional to the temperature:
\begin{align*}
\frac{\rho_{2}}{\rho_{1}} & = \frac{T_{1}}{T_{2}} \\
\rho_{2} & =\frac{T_{1}}{T_{2}}\rho_{1}\\
\rho_{2} & =\frac{273.15K}{(120+273.15)K}×1.28kg/m^3\\
\rho_{2} & = 0.89 kg/m^3
\end{align*}
Exercise: Manometer in a Pressure Alarm
a) The pressure of the gas in the sealed tube is higher than atmospheric pressure.
b) The tube leaks or depressurizes.
Closing the circuit requires the liquid level to rise up to the level of the red arrows. When the pressure in the gas tube decreases, the liquid level on the left side will increase to balance the pressure.
Exercise: Measurement Range of a Pressure Gauge
Suppose we use a flexible membrane to measure the pressure in a sealed container. The membrane is attached to the atmosphere on one side (take) and a sealed container of gas at the other. The membrane has an area of 0.005 and will be permanently deformed when experiencing a net force larger than 300N.
a) It is known that the absolute pressure in the gas container is 89kPa at 273K temperature. What is the range of pressure in the container that can be measured without permanently deforming the membrane?
b) Assuming the volume of the container is constant, what is the corresponding temperature range that allows the pressure in the container to be measured?
a) The net force experienced by the membrane is caused by the difference of the pressure in the gas container and atmospheric pressure (gauge pressure). Using the definition of pressure , we can calculate the range of gauge pressure we can measure with the membrane. The net force can be exerted in both directions, which means the gauge pressure may be positive to negative:
$$P=\frac{F}{A}=\frac{300N}{0.005m^2}=60000Pa$$
So the range of gauge pressure we can measure is: $$-60000Pa<P<60000Pa$$
The range of absolute pressure we can measure is:
\begin{align*}
1.0×10^5Pa-60000Pa<&P<1.0×10^5Pa+60000Pa\\
0.4×10^5Pa<&P<1.6×10^5Pa
\end{align*}
b) We can calculate the corresponding temperatures by rearranging the ideal gas law:
$$n=\frac{P_1V_1}{RT_1}=\frac{P_2V_2}{RT_2}$$
Because R and the volume of the container are constant, we can simplify the equation:
$$\frac{P_1}{T_1}=\frac{P_2}{T_2}$$
And we can solve for the temperature range by substituting in the range of absolute pressure and the known set of temperature and pressure:
\begin{align*}
\frac{1.6×10^5Pa}{T_{high}}&=\frac{89×10^3Pa}{273K}\\
T_{high}&=\frac{1.6×10^5Pa×273K}{89×10^3Pa}\\
&=491K
\end{align*}
\begin{align*}
\frac{0.4×10^5Pa}{T_{low}}&=\frac{89×10^3Pa}{273K}\\
T_{low}&=\frac{0.4×10^5Pa×273K}{89×10^3Pa}\\
&=123K
\end{align*}
So the temperature range to measure the pressure in the container is $$123K<T<491K$$
Exercise: Pxy Diagram
The following graph shows the Pxy diagram for a mixture of n-hexane and n-heptane at a constant temperature of 373K.
Using the diagram above, answer the following questions:
a) What is the vapour pressure of pure n-hexane at T=373K?
b) What is the phase(s) in the system when the mole fraction of n-hexane in the system is 0.3 at a pressure of 2 bar?
c) What is the dew point pressure for a mixture with 30% n-heptane?
d) At vapour-liquid equilibrium, if the mole fraction of n-hexane in the liquid phase is 0.3, what is the mole fraction of n-hexane in the vapour phase?
e) At vapour-liquid equilibrium, if the mole fraction of n-hexane in the vapour phase is 0.3, what is the mole fraction of n-hexane in the liquid phase?
(Since the answers are read from the graph, each question will allow a range of correct answers)
a) 2.45 bar
Find the y-coordinate for the point when the mole fraction of n-hexane is 1 and the two lines meet at a point.
b)the system only has a liquid phase
c) 1.75 bar
Because the x-axis represents the mole fraction of n-hexane, we need to convert 30% n-heptane to mole fraction of n-hexane=0.7:
d)
e)
Exercise: Txy Diagram
The following graph shows the Txy diagram for a mixture of n-hexane and n-heptane at a constant pressure of 1.013 bar.
Using the diagram above, answer the following questions:
a) What is the boiling temperature for pure n-heptane at P=1.013 bar?
b) What is the bubble point temperature when the mole fraction of n-hexane is 0.6?
c) If the mole fraction of n-hexane is 0.4, at T=360K, what is the mole fraction of n-hexane in liquid and vapour phase?
d) Say we use the following process to purify the n-hexane in the stream:
(Because the answer is read from the graph, each question will allow a range of correct answers)
a)372K
Find the y-coordinate for the point when the mole fraction of n-hexane is 0 and the two lines meet at a point.
b) 351K
c) ,
d) mole fraction of n-hexane in the purified stream=0.85
[1] OpenStax University Physics Volume 2. 2016. 2.1 Molecular Model of an Ideal Gas. [online] <https://openstax.org/books/university-physics-volume-2/pages/2-1-molecular-model-of-an-ideal-gas> [Accessed 14 May 2020].
In this chapter, we will introduce energy balances and relate them to chemical processes. Energy balances are very important in chemical and biological engineering since we often deal with large process equipment that either requires a lot of energy or releases a lot of energy. Knowing how much energy is going into our system and coming out of it is crucial. As chemical and biological engineers, we also deal with reactive systems, which are highly dependant on the energy of the system.
Learning Objectives
By the end of this chapter, you should be able to:
Identify relevant terms for energy balances for open and closed systems
Use thermodynamic data tables to identify enthalpy, internal energy, and other thermodynamic properties using system temperatures and pressures
Solve energy balance problems using thermodynamic data
Evaluate the cost of utilities in processes
Characterize energy changes in a system due to changes in temperature
Analyze energy balances on processes involving phase changes
Explain heats of reaction as well as endothermic and exothermic reactions
Determine the standard heat of reaction given other heats of reaction or heats of formation (Hess’s Law)
Analyze energy balances involving reactive systems
As you’re going through this chapter, here are some important terms for you to take note of:
Learning Objectives
By the end of this section, you should be able to:
Identify relevant terms for energy balances for open and closed systems
Use thermodynamic data tables to identify enthalpy, internal energy, and other thermodynamic properties using system temperatures and pressures
Solve energy balance problems using thermodynamic data
Systems are typically divided into three main categories depending on how the system interacts with its surroundings:
Kinetic Energy is energy associated with motion, which can be described as translational or rotational energy.
where:
is mass
is velocity relative to a reference. Generally we refer to earth’s surface as stationary.
Potential energy can be described as energy present due to position in a field, such as gravitational position or magnetic position
Usually for chemical processes, we consider the potential energy change due to the gravitational position of the process equipments.
where:
is mass
is the gravitational acceleration (approximately 9.8 )
is the height about the point of reference
Internal energy can be described as all other energy present in a system, including motion, and molecular interaction.
Closed systems are defined as systems with no mass transfer across the system’s boundaries. All the energy forms described above are applicable to closed systems.
Exercise: Energy Balance Sign Conventions
Consider a system that consists of a stirred tank reactor where an exothermic reaction is taking place, where an external motor is mixing the contents in the reactors. What are the signs of and for this system?
Energy in closed systems follows the Law of Conservation of Energy
In terms of general energy:
The initial energy in the system can be defined as:
The final energy in the system can be defined as:
The energy transfer of the system can be defined as:
This yields the following closed system energy balance, defined as
the First Law of Thermodynamics:
Assumptions made by the First Law of Thermodynamics
- If no acceleration exists in the system, the change in the kinetic energy term will be 0, and can be ommitted from the balance
- If no change in height (or other fields) exist in the system, the change in the potential energy term will be 0, and can be ommitted from the balance
- Internal energy depends on chemical composition, state (solid, liquid, or gas) and temperature; Pressure’s effect is negligible.
- If the system has the same temperature as its surroundings or is adiabatic, the heat term will be 0, and can be ommitted from the balance
- If there are no moving parts, electrical currents, or radiation in the system, the work term will be 0 and can be omitted from the balance
Open systems are defined as systems where both mass and energy cross the system’s boundaries.
Two types of work are typically observed in these systems:
For stream ‘j’ in a system:
Rearranging the energy terms, we get:
Example: Energy Flow in a System
Consider the following system:
There are 2 streams with energy entering the system (streams 1 and 2), and 2 streams with energy exiting the system (streams 3 and 4).
For this system:
Recall the three forms of energy:
Each energy flow term can be further separated into:
Specific Property “” : This denotes an intensive property obtained by dividing an extensive property by a total amount of flow rate (can be total number of moles or total mass)
Combining all these terms:
Recall the work terms expansion:
where flow work is dependant on system pressure and volume
Now we have:
Because usually appear together in the energy balances, we define them to be "enthalpy" ():
where is the internal energy and is the flow work
The following terms are defined:
Finally, an open system steady-state energy balance is defined:
Exercise: Heat for an Ideal Gas
Prior to entering a furnace, air is heated from to and the change in specific enthalpy for the whole heating process is 3640 J/mol. The flow rate of air at the outlet of the heater is and the air pressure at this point is 150 kPa absolute.
Calculate the heat needed for the process in kW. Assume the ideal gas behavior and that kinetic and potential energy changes from the heater inlet to the outlet are negligible.
Step 1: Calculate the molar flowrate using the ideal gas law.
\begin{align*}
\dot{n} & = \frac{1.5m^{3}}{min}*\frac{273K}{150+273K}*\frac{150kPa}{101.3kPa}*\frac{1mol}{22.4L}*\frac{10^{3}L}{1m^{3}}\\
\dot{n} & =64.0\frac{mol}{min}
\end{align*}
Step 2: Calculate the heat using the specific enthalpy. Since the potential and kinetic energy changes are zero, the following calculations are made
\begin{align*}
\dot{Q} & =\Delta\dot{H}= \dot{n}\Delta\hat{H}\\
\dot{Q} & =\frac{64.0 mol}{min}*\frac{1min}{60s}*\frac{3640J}{mol}*\frac{kW}{10^{3} J/s} \\
\dot{Q} & =3.88kW
\end{align*}
Reference State: a substance at some pressure, temperature, and state of aggregation (solid, liquid, gas; pure or mixture).
It is much easier to estimate the energy of a system as a change from a reference state rather than the absolute energy.
Exercise: Cooling in a Heat Exchanger
Water is used to cool a liquid in a heat exchanger. Water enters the heat exchanger at and exits at . Using the table below, find the change in enthalpy of water in its liquid state.
Entry # | ||
1 | 5 | 21.02 |
2 | 10 | 42.02 |
3 | 100 | 419.17 |
Step 1: Determine which reference state you are going to use. In this case, we are using as the reference state.
Step 2: Find the change in enthalpy by taking the difference of the system's specific enthalpies at different temperatures.
\begin{align*}
\Delta\hat{H} & = \hat{H}_{3} - \hat{H}_{2}\\
\Delta\hat{H} & = (419.17 - 42.02) kJ/kg\\
\Delta\hat{H} & = 377.15 kJ/kg
\end{align*}
Since water is a commonly used resource in processes for heating and cooling, detailed information on its state properties at different temperatures and pressures is available.
For example, the steam table below provides , , and at multiple sets of temperature and pressure:
0.01 | 0.000612 | 0.001000 | 205.9912 | 0.00 | 2374.92 | 2374.92 | 0.00 | 2500.92 | 2500.92 | 0.0000 | 9.1555 | 9.1555 |
5 | 0.000873 | 0.001000 | 147.0113 | 21.02 | 2360.76 | 2381.78 | 21.02 | 2489.04 | 2510.06 | 0.0763 | 8.9485 | 9.0248 |
10 | 0.001228 | 0.001000 | 106.3032 | 42.02 | 2346.63 | 2388.65 | 42.02 | 2477.19 | 2519.21 | 0.1511 | 8.7487 | 8.8998 |
15 | 0.001706 | 0.001001 | 77.8755 | 62.98 | 2332.51 | 2395.49 | 62.98 | 2465.35 | 2528.33 | 0.2245 | 8.5558 | 8.7803 |
A. Select 'Water' from the 'Please select the species of interest:' drop-down menu
B. Choose the steam table units you'd like to work with, in step 2.
C. In Step 3, choose what kind of data you're looking to obtain. For an isothermal system, select 'Isothermal properties'. For a constant pressure system, select 'Isobaric properties'.
D. Select the desired standard state convention. This course will most likely only use the 'Default for fluid' convention.
Exercise: Steam Tables
Superheated steam at 40 bar absolute and flowing at a rate of is sent to an adiabatic turbine which expands to 5 bar. The turbine outputs . The expanded steam is then sent to a heat exchanger where isobaric heating occurs, resulting in the stream being reheated to its initial temperature. Assume no changes in kinetic energy. Write the energy balance for the turbine and determine the outlet stream temperature.
Learning Objectives
By the end of this section, you should be able to:
Evaluate the cost of utilities in processes
Characterize energy changes in a system due to changes in temperature
Analyze energy balances on processes involving phase changes
The figure below is a shell-and-tube heat exchange:
Image from Turbojet /CC BY-SA 4.0
Utilities: Utilities are services at a site such as water, electricity, and gas.
Utility | Inlet | Outlet | P () | Cost (dollars/GJ) | Cost (dollars/hr) | Cost dollars/yr @ 8,000 hrs |
High-pressure steam | 250 | 249 | 575 | 5.66 | 16.98 | 135,840 |
Electricity | N/A | N/A | N/A | 18.72 | 56.16 | 449,280 |
The high-pressure steam is a much cheaper option here
Phase changes take place when a compound or mixture of compounds undergoes a change in their state of matter. The enthalpy changes for phase changes occur at a constant temperature and pressure and are determined experimentally.
Parameter | Phase Transition |
---|---|
Liquid → Gas | Heat of Vapourization () |
Solid → Liquid | Heat of Fusion () |
Gas → Liquid | (-) Heat of Vapourization () |
Liquid → Solid | (-) Heat of Fusion () |
Note that heat of vapourization is also referred to as latent heat
Heat capacities are physical properties that describe how much heat is needed to increase the temperature of a compound for a unit temperature per unit mass of a compound.
For a closed system where the only changing system variable is temperature:
is a function of temperature :
Notation: means that the volume is kept constant during the process
This expression can be rearranged and integrated to obtain the following:
|
If the is constant:
For an open system, the heat capacity is defined under constant pressure conditions:
|
If the is constant:
Example: Phase Changes and Heat Capacity in Energy Balances
Compound | Formula | Boiling Point at 1 atm () | Heat of Vapourization at 1 atm () |
EDC | 84 | 35 | |
HCl | -85 | 16 | |
VC | -13 | 21 |
The mixture exits a reactor at and has the following properties:
Compound | Formula | Flow to Separate (tonne/hr) | Flow to Separate (tonne-mol/hr) | (gas mole fraction) |
EDC | 47.5008 | 0.48 | 0.25 | |
HCl | 26.2512 | 0.72 | 0.375 | |
VC | 45 | 0.72 | 0.375 |
Recall Raoult's Law:
Compound Formula Flow to Separate (tonne/hr) Flow to Separate (tonne-mol/hr) (gas mole fraction) (liquid mole fraction) EDC 47.5008 0.48 0.25 0.94 HCl 26.2512 0.72 0.375 0.005 VC 45 0.72 0.375 0.053
Suppose we operate the distillation column at the dew point of the system to separate benzene. For this system, the dew-point temperature is , therefore we must use a heat exchanger to bring the mixture from to
How much energy is removed to cool this stream from from to ?
Compound | Formula | Flow to Separate (tonne/hr) | Flow to Separate (tonne-mol/hr) | Cp (J/mol-K) |
EDC | 47.5008 | 0.48 | 29 | |
HCl | 26.2512 | 0.72 | 29 | |
VC | 45 | 0.72 | 29 |
There is no exchange of energy in the form of work in a heat exchanger. Therefore, the total heat removed is the sum of the change in enthalpy for each species:
Exercise: Energy Balance for a Heat Exchanger
Consider an equimolar binary mixture of n-hexane and n-heptane at a constant pressure of 1 atm flowing at 1 kmol/h. This mixture is originally at and needs to be cooled to for a process in order to ensure vapor-liquid equilibrium is satisfied. The process uses a heat exchanger to achieve this cooling. The heat capacities for both compounds can be described by the following expression:
where is in J/mol-K and the constants A, B, C, and D are listed below:
Compound | A | B | C | D |
n-Hexane | -4.413 | 0.528 | -3.119E-04 | 6.494E-8 |
n-Heptane | -5.146 | 0.6762 | -3.651E-04 | 7.658E-08 |
How much heat needs to be removed by the heat exchanger to reach the required temperature for the process?
Step 1: Calculate the change in enthalpy for each compound using the heat capacities taking 1 kmol/hr as the molar flow.
where and
For n-hexane:
Similarly for n-heptane:
Step 2: Sum up the enthalpy changes for the components.
\begin{align*}
\dot{Q} &= \Sigma\Delta\dot{H}\\
\dot{Q} & = (929.5+1212)\frac{kJ}{h}\\
\dot{Q} &= 2142\frac{kJ}{h}
\end{align*}
Option | Utility | Inlet T () | Outlet T () | P | Cost (dollars/GJ) |
A | Cooling Water | 20 | 25 | N/A | 0.378 |
B | Refrigerated Water | 5 | 15 | N/A | 4.77 |
C | Low T Refrigerant | -20 | -5 | N/A | 8.49 |
How much will cooling this stream cost?
\begin{align*}
Cost \Big(\frac{dollars}{h}\Big) &= \dot{Q}*cost\Big(\frac{dollars}{GJ}\Big)\\
& = 8.56\frac{GJ}{h}*0.378\frac{dollars}{GJ}\\
& = 3.24\frac{dollars}{h}
\end{align*}
Recall that and are state properties. These properties depend on the state and not the path to that state.
It is easier to calculate the enthalpy change by changing one variable at a time through a hypothetical process path.
1 - Calculate change in enthalpy by only changing the temperature at the same state
2 - Calculate change in enthalpy by only changing the pressure at the same state (in the course, we usually neglect the effect of pressure on enthalpy change)
3 - Calculate change in enthalpy by only changing the phase
To obtain the desired change in enthalpy, add each enthalpy change where only 1 state property changes:
Example: EDC Process Path
Consider the enthalpy change of transforming from vapour at and 3 atm to liquid at and 1 atm:
In this process, there are 3 variables changing:
The following steps are taken in the process path and added together to get the overall change:
is the enthalpy change from going from 3 atm to 1 atm at a constant temperature of 200°C in the vapour phase ()
is the enthalpy change from going from 200°C to 84°C at a constant pressure of 1 atm in the vapour phase ()
is the enthalpy change for a phase change, going from the vapour phase () to the liquid phase () at a constant pressure and temperature of 1 atm and 84°C
is the enthalpy change from going from 84°C to 25°C at a constant pressure of 1 atm in the liquid phase ()
When we add these changes together, we can get the overall change in enthalpy:
Exercise: Process Path
What process paths can be taken to calculate the change in enthalpy for acetone going from in the liquid phase to in the vapour phase?
Step 1: Bring the acetone to the boiling point temperature from without changing the phase using the in the liquid phase.
Step 2: Use the latent heat (or heat of vaporization) to calculate the enthalpy of changing the phase from liquid to vapour.
Step 3: Bring the acetone to from the boiling point temperature using the in the vapour phase.
Learning Objectives
By the end of this section, you should be able to:
Explain heats of reaction as well as endothermic and exothermic reactions
Determine the standard heat of reaction given other heats of reaction or heats of formation (Hess’s Law)
Analyze energy balances involving reactive systems
Bonds can be formed and broken. Breaking bonds takes energy and forming bonds releases energy.
Example: Chemical Bonds
Consider the simple reaction of the formation of water from hydrogen and oxygen:
Image from すじにくシチュー / CC0
The formation of bonds takes energy while the breakage of bonds releases energy
In this reaction:
2 H-H bonds are broken
1 O-O bonds are broken
4 H-O bonds are formed
Heat of Reaction : the stoichiometric enthalpy difference when reactants react completely to form products at a specified constant temperature and pressure.
In exothermic reactions,
In endothermic reactions,
Heats of reaction are directly proportional to the amount of reactants or products in a reaction.
Example: Formation of Nitrogen
Consider the formation of nitrogen dioxide
The formation of 1 mole of yields an enthalpy change of 33.2 kJ
The formation of 2 moles of yields an enthalpy change of 66.2 kJ
Heat of reactions for a reaction in the forward direction is *equal to the negative heat of reaction* for the backward reaction.
Consider the formation of hydrogen chloride (or the corresponding decomposition):
Exercise: Heat of Reaction
Consider the oxidation of ammonia taking place in an isothermal and constant pressure reactor:
where the heat of reaction is
The ammonia is fed into the reactor at and the oxygen enters at . Assuming the limiting reactant is completely consumed, what is the enthalpy change for this reaction?
Step 1: Find the limiting reactant by finding the smallest reactant extent if all of a reactant is consumed.
being the limiting reagent yields the smaller reaction extent, therefore is the limiting reagent. The reaction extent is .
Step 2: Multiply the reaction extent by the to obtain
Some reactions may be difficult to reproduce in a laboratory setting. Instead, for analyzing a certain overall reaction, we may break the reaction down in multiple steps:
For example, say we want to find the standard heat of reaction for the following hypothetical reaction:
If this reaction is difficult to carry out in a laboratory, we might break it up into 2 reactions, that are easier to carry out and where we can determine specified standard heats of reaction:
Reaction 1: with
Reaction 2: with
We can combine reactions 1 and 2 to obtain our desired reaction by subtracting reaction 2 from reaction 1:
We obtain our desired reaction:
Therefore, the heat of reaction of the desired reaction will be:
Exercise: Hess’s Law
Consider the formation of chlorine trifluoride from chlorine monofluoride and fluorine:
Data on the following reactions are available:
Number | Reaction | (kJ) |
---|---|---|
1 | -49.4 | |
2 | 214.0 | |
3 | 236.2 |
What is the for the desired reaction?
Step 1: Determine what reaction will give us in the reactants.
Since is needed as a reactant, we can multiply reaction 2 by to obtain:
Step 2: Determine what reaction will give us in the reactants.
Since is needed as a reactant, we can multiply reaction 1 by to obtain:
Step 3: Determine what reaction will give us as a product.
Since is needed as a product, we can multiply reaction 3 by -1 to obtain:
Step 4: Add the reactions up to ensure that the desired reaction is obtained:
Step 5: Add up all the manipulated heat of reactions to obtain the desired
Now let’s try applying the heat of reaction to determine energy use in a process, where reactants and products are coming in and exiting at a given temperature. The following process path is taken for the heat of reaction method, where the reference state is at :
Generally, refers to standard state conditions at and 1 atm, but always double-check whether a different standard state condition is used.
In this process path, the enthalpy calculations are done in three steps:
The calculated enthalpy changes for each step are then added to obtain .
Example: Heat of Reaction Method for Energy Use
Consider a propane combustion chamber, where of propane are feed to the chamber at and air is fed at ( and ). The products stream exits at and consists of , , , and . How much heatis released by this combustion chamber? Assumming atmoshpere pressure, so water boils at 100°C.
The following information is provided:
The values listed under each compound are specific enthalpies in kJ/mol
Temperature () | ||||
25 | 0 | 0 | 0 | 0 |
100 | 2.24 | 2.19 | 2.90 | 2.54 |
300 | 8.47 | 8.12 | 11.58 | 9.57 |
1000 | 32.47 | 30.56 | 48.60 | 37.69 |
1 – Solve material balances as much as possible
This example already has the material balances solved
2 – Choose reference states for energy calculations. Like we say in the previous figure we want to calculate the changes in energy associated with bringing our reactants to our reference state, calculating the energy of the reaction at the reference state, and then the energy associated with bringing the products to their final state. We will choose reference states based on the information we have to try to make these calculations as easy as possible.
For reacting species:
Since is given, we will assume this is at $25^{\circ}C$ and 1 atm (as there is nothing indicating otherwise), and we will use this as our reference state.For non-reacting species:
We can use any convenient temperature and pressure as a reference state (inlet temperature, outlet temperature, temperature in enthalpy table). With this example, the enthalpy table given uses a reference state of $25^{\circ}C$, and we will assume everything is at 1 atm (as no other values for pressure are given).
3 – Calculate the extent of reaction for all reactions (in this case we just have the one reaction)
4 – Prepare an inlet-out enthalpy table (this will show what enthalpies we need to calculate associated with energy changes in the reactants or products)
Substance | (mol/s) | (kJ/mol) | (mol/s) | (kJ/mol) |
100 | – | – | ||
600 | 100 | |||
2256 | 2256 | |||
– | – | 300 | ||
– | – | 400 |
5 – Calculate all the enthalpies. For , the specific enthalpies at different temperatures are given as the difference in enthalpy from the reference state (25°C, 1atm, which is the same as what we choose as reference state):
6 – For , the specific enthalpy for water is calculated. From the reference state of 25°C to the stream temperature of 1000°C, water is first heated to 100°C, which is its natural boiling point at 1atm, then vaporized, then heated in vapor phase to 1000°C. The enthalpy change of each process is calculated separately and added together to get .
Note: The in the formulas to calculate is given in kelvin. This doesn’t make a difference when is given as a number, as the scales for one degree of Celcius and kelvin are the same, but the temperature must be converted to kelvin when is higher power of is used in calculation.
\begin{align*}
\hat{H}_{7}& = \int^{100^{\circ}C}_{25^{\circ}C} C_{P,l}dT + \Delta \hat{H}_{vap}(100^{\circ}C) + \int^{1000^{\circ}C}_{100^{\circ}C} C_{P,v}dT\\& = \int^{100^{\circ}C}_{25^{\circ}C} 75.4*10^{-3}dT + \Delta \hat{H}_{vap}(100^{\circ}C) \\& \;\;\;\;+\int^{1273K}_{373K} (33.46*10^{-3}+0.688*10^{-5}*T + 0.7604*10^{-8}*T^{2} – 3.593*10^{-12}*T^{3})dT\\& = 75.4*10^{-3}*(100-25)+ 40.66 \\& \;\;\;\; +(33.46*10^{-3}*T+\frac{1}{2}*0.688*10^{-5}*T^2 + \frac{1}{3}*0.7604*10^{-8}*T^{3} – \frac{1}{4}*3.593*10^{-12}*T^{4})\bigg\vert^{1273K}_{373K}\\& = (5.65 + 40.66 + 35.1)kJ/mol\\& = 81.46 kJ/mol
\end{align*}Therefore, we have calculated all the specific enthalpy for the reactants and products:
Substance | (mol/s) | (kJ/mol) | (mol/s) | (kJ/mol) |
100 | 0 | – | – | |
600 | 8.47 | 100 | 32.47 | |
2256 | 8.12 | 2256 | 30.56 | |
– | – | 300 | 48.60 | |
– | – | 400 | 81.46 |
8 – Finally, solve the energy balance
, , and are assumed negligible for this system (as no information is provided on these)
The combustion chamber releases of heat to the environment
Formation Reaction: a reaction in which the compound is formed from its elemental constituents as they would normally occur in nature (eg. rather than ).
For elemental consistituents, the energy of formation is = 0, since they would be forming themselves:
Standard specific heat of formation : the enthalpy change associated with forming 1 mole of the compound of interest at standard temperature () and pressure (1 atm)
Compound | Reaction | (kJ/mol) |
---|---|---|
Water | -285.83 | |
Methane | -74.8936 | |
Ethane | -83.82 | |
Propane | -104.68 | |
Benzene | 82.88 | |
Toluene | 50.17 | |
Oxygen | 0 | |
Nitrogen | 0 |
We can use these heats of formation and Hess’s law to find the heat of reaction for a given reaction:
Example: Heat of Formation
Consider the combustion of ethane:
where
To calculate the enthalpy using the heat of formation method, the following steps are taken:
The enthalpy of reaction using the heat of formation method is very close to the enthalpy of reaction (combustion).
One source of heat of formation data is the National Institute of Standards and Technology (NIST) Webbook NIST
To find the heat of formation data at this site:
1 – Look up the compound of interest
2 – Heat of formation data will be under “condensed phase thermochemistry data” (for liquids or solids) or “gas phase thermochemistry data” (for gases).
Heats of formation can also be found in Appendix E of “Introductory Chemical Engineering Thermodynamics” by J. Richard Elliot and Carl T. Lira.
Example: Heat of Formation Method for Energy Use
Let’s consider the propane combustion chamber problem we analyzed before with the heat of reaction method, and let’s see if we can get a similar answer using the heat of formation method.
1 – Solve the material balance as much as possible
This example already has the material balances solved
2 – Choose reference states for energy calculations
For reacting species: elemental species that make up reacting species at standard conditions; we will choose at 1 atm with , , and .
For non-reacting species (same as ): Use any convenient temperature (inlet temperature, outlet temperature, temperature in enthalpy table) Here, and 1 atm works because of our enthalpy table values
3 – Prepare an inlet-outlet enthalpy table
Substance | (mol/s) | (kJ/mol) | (mol/s) | (kJ/mol) |
100 | – | – | ||
600 | 100 | |||
2256 | 2256 | |||
– | – | 300 | ||
– | – | 400 |
4 – Calculate all the enthalpies
Substance | (kJ/mol) |
---|---|
-103.8 | |
0 | |
0 | |
-393.51 | |
-241.835 |
Here, you assume that water forms as vapor directly in the reaction. Therefore, there is no need to account for heat of vaporization.
Substance | (mol/s) | (kJ/mol) | (mol/s) | (kJ/mol) |
100 | -103.8 | – | – | |
600 | 8.47 | 100 | 32.47 | |
2256 | 8.12 | 2256 | 30.56 | |
– | – | 300 | -344.9 | |
– | – | 400 | -204.1 |
5 – Calculate $\Delta\dot{H}$ for the reactor
6 – Finally, solve the energy balance
Exercise: Heat of Formation
Consider a reactor where the following reaction reaction takes place :
The reactor uses an 0.8 kW impeller (mixer for liquids) that mixes the reactor contents. The reaction occurs at standard temperature and pressure. A feed consisting of 100 mol/s and 300 mol/s of water and , respectively, enters the reactor. The reaction goes to completion and the products stream consists of 200 mol/s of and 100 mol/s of .
The following heats of formation are available:
Number | Reaction | (kJ/mol) |
---|---|---|
1 | 33.2 | |
2 | -285.8 | |
3 | -207.4 | |
4 | 90.2 |
How much heat must be removed from the reactor in one day for it to remain at standard temperature?
Solution
Step 1: Determine what combination of the given reactions will give the desired reaction (Hess’s Law)
For this desired reaction:
\begin{align*}
& -3*Reaction1\\
& -1*Reaction2\\
& +2*Reaction3\\
& +1*Reaction4
\end{align*}
Summing all the reactions above yields:
Step 2: Add all the individual reaction enthalpies:
Step 3: Calculate the overall change in enthalpy using the extent of the reaction:
\begin{align*}
\xi &= \frac{|(\dot{n}_{i})_{out}-(\dot{n}_{i})_{in}|}{|(\dot{\nu}_{i})|}hat\\& = \frac{|(\dot{n}_{H_{2}O})_{out}-(\dot{n}_{H_{2}O})_{in}|}{|(\dot{\nu}_{H_{2}O})|} \\& =\frac{|0\frac{mol}{s} – 100 \frac{mol}{s}|}{|-1|}\\ &= 100 mol/s
\end{align*}
Step 4: Calculate how much work is done on the reactor by the impeller in one day:
Step 5: Perform an energy balance on the reactor
Step 1: Determine what combination of the given reactions will give the desired reaction (Hess’s Law)
For this desired reaction:
\begin{align*}
& -3*Reaction1\\
& -1*Reaction2\\
& +2*Reaction3\\
& +1*Reaction4
\end{align*}
Summing all the reactions above yields:
Step 2: Add all the individual reaction enthalpies:
Step 3: Calculate the overall change in enthalpy using the extent of the reaction:
\begin{align*}
\xi &= \frac{|(\dot{n}_{i})_{out}-(\dot{n}_{i})_{in}|}{|(\dot{\nu}_{i})|}hat\\& = \frac{|(\dot{n}_{H_{2}O})_{out}-(\dot{n}_{H_{2}O})_{in}|}{|(\dot{\nu}_{H_{2}O})|} \\& =\frac{|0\frac{mol}{s} – 100 \frac{mol}{s}|}{|-1|}\\ &= 100 mol/s
\end{align*}
Step 4: Calculate how much work is done on the reactor by the impeller in one day:
Step 5: Perform an energy balance on the reactor
where assuming there are no phase changes
If a phase change occurs, an additional heat term (i.e. heat of vapourization) will be added to the term in both methods
where assuming there are no phase changes
For the same system, equating for both methods, we can see that the terms cancel. We obtain the following:
Overall, the final for both methods will be the same. This means we can choose either method based on which is easiest to use with the information we have available.
Exercise: Comparing Both Methods
Ethane and oxygen are fed into a furnace at 100 kmol/hr and 500 kmol/hr, respectively. Ethane goes through complete combustion in the furnace. Oxygen, carbon dioxide, and water exit the furnace at 150 kmol/hr, 200 kmol/hr, and 300 kmol/hr, respectively.
The following data on heats of formations is available:
Compound | (kJ/mol) |
Ethane | -83.82 |
Oxygen | 0 |
Carbon Dioxide | -393.51 |
Water (liquid) | -285.83 |
Assume the reaction takes place in the furnace at standard temperature and pressure. What is the for the desired reaction? Calculate the change in enthalpy using both the heat of reaction and heat of formation method
Heat of Reaction Method
Step 1: Determine the extent of the reaction:
\begin{align*}
\xi &= \frac{|(\dot{n}_{i})_{out}-(\dot{n}_{i})_{in}|}{|(\dot{\nu}_{i})|}hat\\& = \frac{|(\dot{n}_{C_{2}H_{6}})_{out}-(\dot{n}_{C_{2}H_{6}})_{in}|}{|(\dot{\nu}_{C_{2}H_{6}})|} \\& =\frac{|0\frac{kmol}{hr} - 100 \frac{kmol}{hr}|}{|-1|}\\ &= 100 kmol/hr
\end{align*}
Step 2: Multiply the extent of the reaction by the heat of reaction for combusion
Heat of Formation Method
Step 1: Set up the heat of formation formula
Step 2: Plug in the enthalpies of formation and calculate the change in enthalpy
Compound | (kmol/hr) | (kmol/hr) |
---|---|---|
Ethane | 100 | 0 |
Oxygen | 500 | 150 |
Carbon Dioxide | 0 | 200 |
Water | 0 | 300 |
Notice that the methods aren't exactly equal. This is due to slight differences in available enthalpy data and assumptions made regarding temperature
Kinetic Energy | ||
Potential Energy | ||
Flow Work | ||
Steady-state Open System Energy Balance | ||
Enthalpy |
| |
Heat Capacity (closed system) | ||
Internal Energy (closed system) | | |
Heat Capacity (open system) | ||
Enthalpy (open system) | ||
Heat of Reaction Method | ||
Heat of Formation Method |
An interactive or media element has been excluded from this version of the text. You can view it online here:
https://pressbooks.bccampus.ca/chbe220/?p=1783
Exercise: Endothermic Energy Balance
Consider a reactor in which an endothermic reaction takes place and 15.8 kJ are absorbed. A mixer is used in the reactor and provides 6.3 kJ of work on the system. What is the overall heat that needs to be provided to the system?
a) -22.1 kJ
b) 9.5 kJ
c) 22.1 kJ
d) -9.5 kJ
b) 9.5 kJ
Performing an energy balance on the system, we know that . Since this is an endothermic reaction and work is done on the system, both the enthalpy of the reaction and the work are positive:
\begin{align*}
Q&= \Delta H -W \\
&= 15.8 kJ – 6.3 kJ\\
&=9.5kJ
\end{align*}
Exercise: Heat Capacity
A heat exchanger heats methanol from to at a pressure of 1 atm. The methanol is entering and exiting the heat exchanger at a molar flow rate of 500 mol/s. The following data is available:
Cp (liquid) [J/molK] | 80 |
Cp (vapour) [J/molK] | 40 |
Heat of Vapourization [kJ/mol] | 40 |
Normal Boiling Point Temperature [latex]^{\circ}C[/latex] | 65 |
What is the rate of change in enthalpy for the heat exchange?
a) 22.3 MW
b) 2.3 MW
c) 2.3 kW
d) 22.3 kW
a) 22.3 MW
Recall that . We can expand this expression in order to include phase changes:
Plugging in our values, we get:
Exercise: Utilities
We have the following utility options for heating toluene from to .
Utility | Inlet T() | Outlet T() | Cost ($/GJ) |
Cooling Water | 20 | 25 | 0.378 |
Refrigerated Water | 5 | 15 | 4.77 |
Low-pressure Steam | 125 | 124 | 4.54 |
Medium-pressure Steam | 175 | 174 | 4.77 |
Which utility would be the best option for this scenario?
a) Cooling water
b) Refrigerated water
c) Low-pressure steam
d) medium-pressure steam
c) Low-pressure steam
A minimum of of difference in temperature is required for efficient heat exchange.
The cooling and refrigerated water are not hot enough to heat the toluene. Both the low and medium pressure steam are hot enough to heat the toluene to , but low-pressure steam is the cheaper option.
Exercise: Hess’s Law
Consider the following desired reaction:
Express the enthalpy change of the desired reaction using the following reactions’ enthalpies:
Using Hess’s Law we can multiply reaction 1 by -3/2, reaction 2 by 1/2, and reaction 3 by -1 to obtain our desired reaction:
\begin{align*}
3NO_{2}(g) → 3NO(g) + \frac{3}{2}O_{2}(g) \;\;\;\;\;&-3/2\Delta H_{1}\\
N_{2}(g) + \frac{5}{2}O_{2}(g) + H_{2}O(l) → 2HNO_{3}(aq)\;\;\;\;\;&1/2\Delta H_{2}\\
2NO(g) → N_{2}(g) + O_{2}(g)\;\;\;\;\;&-\Delta H_{3}
\end{align*}
Therefore, our change in enthalpy can be represented by:
Exercise: Heat of Formation
Consider the combustion of acetylene. Find the enthalpy of combustion using the heats of formation at :
2C2H2(g) + 5O2(g) ⟶ 4CO2(g) + 2H2O(g)
Compound | (kJ/mol) |
C2H2(g) | 226.7 |
O2(g) | 0 |
CO2(g) | -393.5 |
H2O(g) | -241.8 |
Recall that
Using the data provided and the reaction’s stoichiometry:
Exercise: Process Paths
You have a feed stream of methanol at 1 atm and . You’d like to bring it to a temperature of so that it enters a reactor as a gas. The boiling point of methanol is at 1 atm. What process path (with enthalpy changes) can you take to achieve this change and calculate the overall enthalpy?
The following steps can be taken to calculate the overall change in enthalpy of going from liquid methanol at 1 atm and to gas methanol at 1 atm and .
The enthalpy change of the process will be the sum of steps 1-3.
Exercise: Energy Balance on Heat Exchanger
A heat exchanger uses superheated steam to heat up a stream of pure benzene at 1 MPa.
The heat capacity of benzene can be represented using the following expression:
where T is in K and is given in J/mol K.
A | B | C | D | |
Benzene | -6.211 | 5.65 x 10-1 | -3.141 x 10-4 | 0 |
The enthalpy change for steam can be found using data from the steam table:
300°C, 0.1 MPa | 350°C, 0.1 MPa | |
(kJ/kg) | 3074.5 | 3175.8 |
The benzene’s boiling point temperature is 80.1°C at 1 atm, and its heat of vaporization is 30.765 kJ/mol.
Does the benzene vapourize in the heat exchanger? What is the final temperature of the benzene exiting the heat exchanger? Assume there is no energy loss due to heat transfer.
Because there are no other forms of energy transfer other than the heat transfer between the streams, the energy balance simplifies to:
The enthalpy change for steam in the energy balance can be expanded to:
\begin{align*}
\Delta\dot{H}_{benzene}&=\int_{T_i}^{T_f}C_pdT\\
&=\int_{T_i}^{T_f}A+BT+CT^2+DT^3 dT\\
&=AT+\frac{1}{2}BT^2+\frac{1}{3}CT^3+\frac{1}{4}DT^4\;\Bigg|_{T_i}^{T_f}\\
&= A(T_f-T_i)+\frac{1}{2}B(T_f-T_i)^2+\frac{1}{3}C(T_f-T_i)^3+\frac{1}{4}D(T_f-T_i)^4
\end{align*}
Step 1: Input given information that will be used in the calculation. This includes the given data to calculate enthalpy change, the molar flow rate of both substances, molecular weights, and the initial temperature of the benzene stream.
Step 2: Calculate the molar enthalpy change. For water, this is calculated by subtracting the molar enthalpy values calculated by subtracting the initial enthalpy from the final. We also convert the units by multiplying the water’s mass enthalpies by the molecular weights at the final and initial temperatures to keep it consistant with the benzene stream.
For benzene, we need to provide a guess of the final temperature to substitute into the equation. For example, we can guess 400 K.
Step 3: Calculate the total enthalpy change for both substances by multiplying the molar flow rate by the molar enthalpy change.
Step 4: Use a cell to represent the sum of enthalpy change. If we have guessed the correct final temperature, the value of this cell should be 0. We will use Solver to let excel perform the guess-and-check process for us.
Step 5: Select “Solver” from the “Data” menu in the header, and indicate that we want to solve for the final temperature of benzene that makes our :
Step 6: Click “OK” if Solver finds a solution. If the solver cannot find a solution, it is maybe because our initial guess is too far away from the actual final temperature, if so, we can change the temperature and try to solve it again.
Step 7: After the window closes, the value of the final temperature of benzene and all cells affected by that temperature are changed to the corresponding values calculated with the actual final temperature.
Therefore, the final temperature of the benzene stream is , which is equal to . This temperature is lower than the boiling temperature of benzene at 1 atm, so the stream will not start to vapourize.
Exercise: Heat of Combustion of Ethylene
Consider the combustion of ethylene:
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g)
The enthalpy of combustion can be determined using the enthalpies of the following reactions:
Determine the enthalpy of the combustion of ethylene from reactions 1, 2, and 3 using Hess’s Law at standard temperature and pressure.
Step 1: Determine the combination of reactions that will give us the desired combustion reaction.
We know that we need C2H4 in the reactants, so we multiply reaction 1 by -1. We notice that by adding the other 2 reactions to reaction 1 (multiplied by -1), we get our desired reaction:
Step 2: Calculate the heat of combustion from the reaction enthalpies.
Exercise: Enthalpy of Formation of Ethylene Combustion
Consider the combustion of ethylene:
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g)
Using the enthalpies of formation listed below, determine the enthalpy of combustion for ethylene at standard temperature and pressure. Compare the enthalpy of combustion to the one found in the previous exercise found using Hess’s Law.
Compound | (kJ/mol) |
C2H2(g) | 226.7 |
C2H4(g) | 52.51 |
O2(g) | 0 |
CO2(g) | -393.5 |
H2O(g) | -241.8 |
H2(g) | 0 |
Step 1: Calculate the enthalpy of reaction for each of the reactions using the enthalpy of formation of the reactants and products.
Step 2: Compare the calculated enthalpy of combustion using heats of formation to the enthalpy of combusion using Hess’s Law from the previous exercise.
They are both equal to -1323.11 kJ!
In this chapter, we will introduce unsteady-state operations, which are very common in chemical processing. Since unsteady-state balance systems are transient, we will explore deriving material balances and determining their initial conditions. We will also look at process control strategies, which are used to measure and manipulate processes in plants.
Learning Objectives
By the end of this chapter, you should be able to:
Recognize unsteady-state balance systems.
Derive material balance equations.
Provide initial conditions for well-mixed separable transient single-unit processes.
Solve separable transient balances to find a property of a system at any given time.
Provide initial conditions for well-mixed non-separable transient single-unit processes.
Set-up separable transient balances.
Apply control strategies to control common process variables: flow, liquid pressure, liquid level, and temperature of endothermic process streams.
As you’re going through this chapter, here are some important terms for you to take note of:
Learning Objectives
By the end of this section, you should be able to:
Recognize unsteady-state balance systems
Derive material balance equations
Unsteady-State: in unsteady-state processes, mass and energy in the system are changing with time (not constant).
Understanding and modeling these systems allow us to:
Example: Draining Tank
Consider the following draining tank:
This tank demonstrates a very simple unsteady-state system. Taking the tank as our system of interest, as the water drains from the tank, the system’s overall mass changes. The height of the liquid in the tank also changes as the liquid drains out of the tank.
To characterize systems (generally process units), we can do a balance on any quantity (total mass, energy, moles, etc.). For this class, we will focus on mass, energy, and mole balances, but we could do these balances on any quantity.
- Input: Enters through the system boundaries ()
- Output: Exits through the system boundaries ()
- Generation: Produced in the system ()
- Consumption: Consumed in the system ()
- Accumulation: Build-up or overall loss in the system ()
Note that the generation and consumption terms are often associated with chemical reactions
Exercise: Balance Terms
Consider a reactor with continuous inlet and exit streams. A reaction occurs in the reactor that is first order:
The inlet stream only contains species A and the reaction goes to completion in the reactor. The reactor is initially filled entirely with A. What terms in the following expression are considered (i.e. non-zero) for this system when assessing the moles of species A as the reaction starts occurring?
Since the reaction goes to completion in the reactor, there is no A exiting in the outlet stream. Therefore, will be omitted from the expression. This reaction is a non-reversible reaction, so no A is generated in the system ( ). All the other terms are present, A is fed in ( ), A is consumed or reacted ( ) and the amount of A in the reactor will change with time as B is produced ( ).
Because we start with a reactor filled with A initially, the moles of A decreases in the reactor during start-up, and approaches a steady-state, as shown in the image below. The change in moles of A in the reactor represents the ACC term in the balance.
Transient balances are any balances that where some quantities involved in the balance are functions with time. If we have a continuous process, we can analyze what happens to a given quantity for a given change in time . As an example, let’s take the mass of a given species (we’ll call species A) to be our quantity of interest:
Where:
- , are the mass flows in and out of the system of species A
- , are the rates of generation and consumption of A in the system
- is the total mass of A in the system at a given time
If we want to turn this into a differential balance, we will isolate and let it approach to 0 (just like in calculus classes):
We can mutiply on both sides, and take the integral of each term to calculate the change of each term over time:
For steady-state processes, the term goes to 0, as there would be no change in the total mass of A in the system over time. For unsteady-state processes, this differential equation must be solved using boundary conditions.
An example of a boundary condition is
This equations says that the mass of A in the system at time (t = 0) is equal to . This mass can be equal to 0, meaning that there is no A in our system at time zero. It can also be equal to a specific value, meaning that our system starts with an initial mass of A.
Learning Objectives
By the end of this section, you should be able to:
Provide initial conditions for well-mixed separable transient single-unit processes.
Solve separable transient balances to find a property of a system at any given time.
Separable differential equations are differential equations where the variables can be isolated to one side of the equation. Take the following differential equations:
1 –
This equation is separable because you can completely isolate the x and y variables as follows:
2 –
This equation is non-separable because you cannot completely isolate the x and y variables:
These non-separable equations we will discuss later.
Example: Chemical Reactor
Consider a “continuous stirred-tank reactor” (CSTR). CSTRs are reactors with continuous feed and exit streams and some kind of mixer.
Say we know the following information about this CSTR:
We want to write a balance for the total (overall) mass in the system under transient conditions.
We start off by writing out the overall balance:
Mass is not being consumed or generated, just changed from one substance to another. This means the and terms are negligible. We get:
The units for the , , and terms are kg/s. Simplifying the balance, we get:
Since the densities are all constant, we can cancel them out:
Using separation of variables (from calculus), we can integrate both sides. We want to find a given value about our system at a specific final time (), starting from an initial time (say ). We know the initial volume and want to find the final volume .
Let’s try substituting in some numbers to this equation. Say the rate of flow in is 5 L/min, and the flow out is 6 L/min with 300 L initially in the tank. How much water remains in the tank after 1 hour? Using the formula we derived, we can find this:
. What is the height of the tank at 10 minutes?
Step 1: Determine what terms in the general balance are zero or negligible.
Since there are no feed streams and no reactions (no mass generated or consumed), the , , and terms can be omitted from the balance.
Step 2: Write out the mass balance for species A in the system.
Step 3: Expand the mass terms.
Recall that , so we can use that to express both terms in the mass balance:
Since the density is constant, the density terms on both sides cancel each other out
The volume terms can be further expanded. Recall that the volume of a cylinder or . In differential terms, . Since the area is constant in this case, this simplifies to: .
Step 4: Solve the integral from t = 0 to t = 10 mins
Exercise: Transient Mass Balance
Suppose we have a tank with an outlet at the bottom as shown below. Water is the only thing in the tank (we'll call this species A). The water flows out of the tank at a rate of , where h is the height of water in m at any specific time (note that expression like this for outlet flow come from conservation of energy in a tank with gravity driving flow out of the tank). The tank has a cross-sectional area of and the outlet pipe has a cross-sectional area . If the initial height of the water in the tank is 1 m, using a transient mass balance, calculate the height of water in the tank after 5 mins.
Step 1: Determine what terms in the general balance are zero or negligible.
Since there is no inlet stream and no chemical reactions in the tank, there is no , or terms:
Step 2: Write out the mass balance for species A in the system.
Step 3: Express each term using the given quantities:
Replace by for both terms:
Because are multiplied by both sides and we assume it is constant throughout the system, we can cancel in this step:
The volume of fluid flowing out of a pipe at any instant can be calculated using , where u is the instantaneous velocity of the fluid and A is the cross-sectional area of the pipe. We can use this to express the outlet flow. For the accumulation term, it is not beneficial to do this because we don't know the expression for .
For the accumulation term, we can espress dV as because we know that is constant, and this leaves only one variable (h) for us to solve for in the mass balance:
We can replace using given in the question. T
Step 4: Solve the integral to find :
Separate the equation by moving all terms involving h to one side:
Replace the variables by the given information: , , , .
\begin{align*}
\int_{0}^{t=300}-\frac{10×10^{-4}}{1} dt&= \int_{h_{0}=1}^{h_{f}}\frac{1}{\sqrt{19.6}}×h^{-\frac{1}{2}}dh \\
-10×10^{-4}× t\;\bigg|_{0}^{t=300}& = \frac{2}{\sqrt{19.6}}×h^{\frac{1}{2}}\;\bigg|_{h_{0}=1}^{h_{f}}\\
-10×10^{-4}×(300-0)& = \frac{2}{\sqrt{19.6}}×h_{f}^{\frac{1}{2}}-\frac{2}{\sqrt{19.6}}×(1)^{\frac{1}{2}}\\
\bigg(\frac{-10×10^{-4}×(300-0)+\frac{2}{\sqrt{19.6}}×(1)^{\frac{1}{2}}}{\frac{2}{\sqrt{19.6}}}\bigg)^2&=h_{f} \\
h_{f}&=0.11m
\end{align*}
Learning Objectives
By the end of this section, you should be able to:
Provide initial conditions for well-mixed non-separable transient single-unit processes.
Set-up separable transient balances.
Non-separable differential equations are differential equations where the variables cannot be isolated. These equations cannot be easily solved and require numerical or analytical methods that will be taught in future courses. You will learn some methods to solve non-separable equations in CHBE 230 (numerical methods) and MATH 256 (differential equations).
You cannot isolate the x and y variables in this differential equation, making it a non-separable ordinary differential equation. Without separation you cannot use the integration techniques you learned in your calculus courses to solve this equation for y as a function of x.
In this class, we will only formulate non-separable differential equations without solving them. We will use general balance equations to come up with differential equations and identify the initial conditions, but we will not find the solutions.
Example: Chemical Reactor
Consider a “continuous stirred-tank reactor” (CSTR). CSTRs are reactors with continuous feed and exit streams and some kind of mixer. Let’s say the following first-order reaction takes place in our CSTR:
Say the units of the rate of consumption of A are .
Say we know the following information about this CSTR:
We want to write a balance for the moles of A in the system under transient conditions.
We start off by writing out the general balance for the moles of A for a CSTR:
For this example, we can assume that the reaction is irreversible, indiceted by the one-way arrow in the reaction. Therefore, there will be no $GEN$ term in the mole balance for A. We get the following terms:
– this is the reaction rate in terms of species A
– since the volume V is constant
The units for all these terms are mol/s. Simplifying the balance, we get:
with initial conditions: when
For this course, formulating this differential equation and listing initial conditions without solving it is sufficient. You will learn how to solve these different equations in future classes.
Exercise: Transient Mole Balance
Consider a CSTR where carbonic acid irreversibly decomposes to form water and carbon dioxide as follows:
There are two streams entering the CSTR that both contain carbonic acid. The first stream has a concentration of and a volumetric flowrate of , while the second stream has a concentration of and a volumetric flowrate of . The volume of the reactor is constant V. The concentration of the exit stream is at 0 minutes. If the reaction is first order with the following rate of reaction:
What is the transient mole balance on carbonic acid? Set up the ordinary differential equation and initial conditions for this system without solving the differential equation.
Step 1: Determine what terms are present in the mole balance.
Step 2: Write out the mole balance with the derived terms and determine the initial conditions.
The intial conditions are
Note you can see from this equation that cannot be isolated. When we try to put all terms containing to the right side by dividing it, it induced terms into the first and second term on the left. and thus we cannot use separation of variables. You will learn to solve these equations in your ordinary differential equations class.
Learning Objectives
By the end of this section, you should be able to:
Apply control strategies to control common process variables: flow, liquid pressure, liquid level, and temperature of endothermic process streams.
Some common terms used in process control:
Exercise: Control System Variables
You are driving a car and are approaching a hill. Assume there are three variables in this system:
Classify each of these variables as process/controlled variable, manipulated variable, or disturbance variable.
The gas pedal will be the manipulated variable since you have direct control on the gas pedal. The hill will be the disturbance variable, because you cannot control the hill’s steepness and how this will affect the system. Finally, the car’s speed is the process or controlled variable, since it is the system’s output variable.
1 – Obstruction Meter
An obstruction meter measures the change in pressure $\Delta P$ and correlates it to flow.
2 – Rotational or Turbine Flowmeter
A rotational or turbine flowmeter measures the speed of the rotation of the turbine and correlates it to flow.
For the flow control loop depicted above:
The transmitter is generally placed sufficiently upstream of the valve so that the flow reading is typically not affected by the valve.
Many level measurement devices are available (pressure, weight, thermal), but float and sonic/radar level measures are the most common.
1 – Float
2 – Sonic/Radar
A sonic or radar device measures the level of the tank by measuring the time for wave to reach the liquid-gas interphase and reflect back to the device, and correlating the time to distance. The sonic or radar device can be installed from the top or bottom of the tank.
A typical level control loop is depicted:
For the level control loop depicted above:
Note that we could also manipulate the inlet valve to affect the tank level, but typically in these systems, the outlet valve is used to control the level. We will stick with this convention in this class.
1 – Manometer
Manometers measure difference in fluid height and correlate it to pressure.
2 – Differential Pressure Cells
A differential pressure cell measures the deflection of the membrane and correlates it to change in pressure $\Delta P$.
A typical liquid pressure control loop is depicted:
For the level control loop depicted above:
For a centrifugal pump (least expensive and most common for liquids), the line can be closed without immediate damage to the pump. Other pumps (positive displacement pumps) will use other strategies.
1 – Expansion-based
Image from Pearson Scott Foresman / Public domain
Expansion-based thermometers measure fluid expansion and correlate it to temperature.
2 – Electrical
Image from Nanite / CC0
Thermocouples or resistance temperature devices measure changes in electrical properties and correlate them to temperature. Electrical temperature measurements are the most commonly used inplants as switching combination of metals can adopt to a wide range on temperature.
A typical slow-response endothermic temperature control loop is depicted:
For the temperature control loop depicted above:
A slow-response endothermic temperature control loop is classified as slow because it takes a longer time to see a temperature change. The response of the system is dependant on the amount of steam that condenses in the heat exchanger. When the outflow of condensate is manipulated, the vapour-liquid equilibrium point is shifted, which affects the rate of condensation/vaporization. The equilibrium shift speed causes a delay in the response time of temperature control.
A typical critical endothermic temperature control loop is depicted:
For the temperature control loop depicted above:
A critical endothermic temperature control loop is faster than a slow-response endothermic control loop because the temperature of the process fluid is directly changed. This eliminates the lag that would be present from manipulating the steam temperature.
Example: Endothermic Temperature Control Loops
Consider a heat exchanger where an endothermic temperature control loop is present in the system (which means the process stream absorbs heat from the utilities). There is a (from the set-point) increase in the outlet temperature of the process fluid. Which endothermic control loop will decrease the temperature back to its original set-point faster and why?
For a slow-response endothermic temperature control loop :
When the temperature change is detected, the controller will close the valve slightly to allow for less condensate to flow out of the heat exchanger. This will allow less steam in the heat exchanger to provide heat through condensation, and eventually cause the outlet process fluid temperature to decrease to the set-point temperature
For a critical endothermic temperature control loop :
When the temperature change is detected, the controller will open the valve of the cold process fluid. This will directly add cold process fluid to the outlet process fluid and directly decrease the temperature back to its set-point, without any delay.
Therefore, a critical endothermic temperature control loop is going to decrease the temperature back to its original set-point faster.
Exercise: Heat Exchanger Control
Consider the following heat exchanger system. We’d like to control the outlet cold temperature .
Suggest 2 different control loops that we can implement to this system.
The first control strategy measures the process variable () and adjusts the outlet hot stream’s temperature () by manipulating the outlet flowrate of the hot fluid. This control strategy is known as the slow-response endothermic control loop.
The second control strategy measures the process variable () and adjusts the inlet cold stream’s temperature () by adding more cold process fluid to the outlet cold stream. This control strategy is known as the critical endothermic control loop.
General balance | |
General mass balance |
An interactive or media element has been excluded from this version of the text. You can view it online here:
https://pressbooks.bccampus.ca/chbe220/?p=1804
Exercise: Separable Differential Equations
Which of the following is not a separable differential equation?
a)
b)
c)
d)
d)
If you try to seperate this equation, you get: . There is no way to isolate the T and t variables further.
How to separate the other choices:
a) \begin{align*}
\frac{t^{2}}{C_{A}}& = 4*C_{A}\frac{dC_{A}}{dt}\\
t^{2}dt&= 4*C_{A}^2\,dC_{A}
\end{align*}
b) \begin{align*}
\frac{t^{2}}{T} &= 3*T \frac{dT}{dt}\\
t^{2}dt &= 3*T^2dT
\end{align*}
c)\begin{align*}
\frac{dM_{A}}{dt} + e^{M_{A} + t} &= 0\\
\frac{dM_{A}}{dt} + e^{M_{A}}· e^{t} &= 0\\
e^{M_{A}}· e^{t} &= -\frac{dM_{A}}{dt} \\
e^{t}dt &=-\frac{1}{e^{M_{A}}}dM_{A}
\end{align*}
Exercise: Process Control Variables
Consider a mixing tank with 2 inlet streams and 1 outlet stream. Stream 1 is feeding the tank with a solution with a concentration of species A. We can change the flowrate of stream 1 using a valve. Stream 2 is feeding the tank with a solution with a concentration of species B, but we do not have control over stream 2’s flowrate or concentration (species B concentration fluctuates and is decided by an upstream process). We measure the concentration of the outlet stream and adjust our system accordingly. Which of the following variables is the disturbance variable?
a) Stream 1 flowrate
b) Stream 2 flowrate
c) Outlet stream flowrate
d) Outlet stream concentration
b) Stream 2 flowrate
The flowrate of stream 2 is the best answer here because it is an input variable that is affecting our controlled variable (outlet stream concentration), but we do not have control over it. We must adjust our manipulated variable (stream 1 flowrate) to achieve our desired outlet flowrate or concentration.
Exercise: Unsteady-state Systems
Which of the following is not a steady-state system?
a) CSTR (continuous stirred-tank reactor) with constant flow in and out of all species
b) A filtration system that has fluid entering and exiting the system at constant volumetric flowrates
c) A bathtub filling up with water at the same rate as the drain empties the water
d) Heat exchanger in a process that maintains the process fluid at 50°C
b) A filtration system that has fluid entering and exiting the system at the same rate
This system will accumulate a build-up of the filtered product. Although the flowrates entering and exiting the system are constant, the concentration will change over time due to the filter present.
Exercise: Methods to Decrease Reactor Temperature
We have the following reactor with an endothermic reaction taking place inside the reactor. There is a heating jacket outside of the reactor that acts in a similar manner to a heat exchanger with steam as the utility stream. If the reactor temperature is currently lower than the desired temperature that is required to achieve the optimum yield from the reaction, how should we manipulate both valves to increase the temperature in the reactor?
a) V-101 and V-102 both open
b) V-101 and V-102 both close
c) V-101 opens; V-102 closes
d) V-101 closes; V-102 opens
d) V-101 closes; V-102 opens
Since the reaction is endothermic, we want to slow down the reaction rate inside the reactor to increase the temperature, which can be achieved by closing V-101 to slow down the inlet flow of the reactant.
The heating jacket releases more heat into the reactor when there is more steam and less condensate in the jacket, due to most of the heat released comes from the condensation process. This can be achieved by opening V-102 to let out some of the condensates in the heating jacket.
Exercise: General Balance for Semi-Batch Reactor
Consider a semi-batch reactor that is initially filled with species A prior to the startup of the reactor. Species B is added to the reactor slowly until a stoichiometrically balanced (species A and B) amount is reached. There is a continuous outlet stream in the reactor. Species B reacts with species A with a conversion of 75% at the end of the process. The reaction is reversible.
What terms in the general balance (IN – OUT + GEN – CON = ACC) are considered for species A in this system?
GEN – OUT – CON = ACC
Exercise: Level Control Loop
Consider the following control loop. The number indicated by the reader is the level of water (height of the water inside the tank), not the height from the top of the tank to the surface of the water. The exit valve is initially closed.
What action will the level control take on the valve and why?
The level controller will signal the valve to open to drain some of the water in the tank. The setpoint height in the tank is less than or equal to 12 m. Given that the tank water level is at 15 m, we must drain some of the water in the tank to bring the water level back to an acceptable range.
Exercise: Temperature Boundary Conditions
The change in temperature with respect to time for a system can be described by the following function, where the temperature(T) is in K and time(t) is in mins:
The temperature of the system is initially at 270 K. Solve for the time when the temperature change is equal to 95% of the temperature change t = 0 minutes to t →
\begin{align*}
T(t) &= \int 20e^{-0.2t}dt\\
T(t) &= -100e^{-0.2t} + C
\end{align*}
Our boundary condition is T(t=0) = 270 K. We can use this to solve for the constant in th temperature function.
\begin{align*}
270 &= -100e^{-0.2*0} + C\\
270 &= -100 + C\\
C &= 370 K
\end{align*}
Find the system temperature when time goes to infinity:
\begin{align*}
T(t=\infty) &= -100*0 + 370 K\\
T(t=\infty) &= 370 K
\end{align*}
Our temperature is going to be 95% of temperature change plus :
\begin{align*}
T_{2} &= 0.95(370 K – 270 K) + 270 K\\
T_{2} &= 365 K
\end{align*}
Now, to find the time, we plug in our temperature :
\begin{align*}
365 &= -100e^{-0.2t_{2}} + 370\\
-5 &= -100e^{-0.2t_{2}}\\
0.05 &= e^{-0.2t_{2}}\\
t_{2} &= \frac{ln(0.05)}{-0.2} = 15 mins
\end{align*}
Exercise: Mass Balance in a Filtration System
Suppose we have the settling tank system above to separate solid particles from a waste stream. Given the following information:
We know that we need to stop the process and clean the tank when there is 500 kg of solids accumulated in the tank. If the tank is initially empty, what is the maximum time before we must clean the tank up?
We write each term in the mass balance in terms of the slurry and simplify the mass balance:
Since the tank is initially empty, we know that the when t=0 hr, so we can solve the mass balance to find t when .
\begin{align*}
\int_{t_i=0hr}^{t_f}3\frac{kg}{hr}dt-\int_{t_i=0hr}^{t_f}2.775\frac{kg}{hr}dt&=\int_{M_i=0kg}^{M_f=500kg}dM\\
3\frac{kg}{hr}×(t_f-0)hr-2.775\frac{kg}{hr}×(t_f-0)hr&=(500-0)kg\\
t_f×0.225\frac{kg}{hr}&=500kg\\
t_f &=2222hr (\cong 93days)
\end{align*}
Exercise: Mass Balance in a Filtration System
For the same settling tank system, suppose we have the following information instead:
Express each term in the mass balance for the mass of the solids inside the tank. Construct the balance equation to solve for the time when 500kg of solids accumulated in the tank. You do not need to solve the equation.
Let M be the total mass of solids in the tank. We write each term in the mass balance in terms of the slurry and simplify the mass balance:
Since the tank is initially empty, we know that the when t=0hr, so we can solve the mass balance to find t when .
\begin{align*}
\int_{t_i=0hr}^{t_f}3\frac{kg}{hr}dt-\int_{t_i=0hr}^{t_f}(2.5+0.1M)\frac{kg}{hr}dt&=\int_{M_i=0kg}^{M_f=500kg}dM
\end{align*}
Because M is also a function of t, this equation is not separable, see we don’t need to solve it in this course.
Exercise: Batch Reactor with Unsteady-state Energy Balance
A batch reactor is used for a reaction that requires a high temperature prior to adding a catalyst to the system. A heated jacket is used to heat the batch reactor. The heat jacket provides heat to the batch reactor at a rate of . The batch reactor vessel is losing energy at a rate described by the following function, where energy is in and time is in mins:
The accumulation of heat in the system (batch reactor) can be described by , where and the mass of the reactor is . The reactor is initially at 300 K.
a) Write out the overall energy balance, with each of the terms in the IN-OUT+GEN-CON=ACC formula.
b) At what time will the reactor reach a temperature of 350 K?
a) For each of the energy balance terms, we have:
b) Once again, we solve the energy balance:
\begin{align*}
45\frac{kJ}{min} – (-45e^{-0.2t}+45)\frac{kJ}{min} &= C_{P}*m \frac{dT}{dt}\\
45\frac{kJ}{min} + (45e^{-0.2t} – 45)\frac{kJ}{min} &= 0.38 \frac{kJ}{kg*K}* 10 kg * \frac{dT}{dt}\\
45e^{-0.2t}\frac{kJ}{min}&= 0.38\frac{kJ}{kg*K}*10 kg*\frac{dT}{dt}\\
45e^{-0.2t}\frac{kJ}{min}&= 3.8\frac{kJ}{K}*\frac{dT}{dt}
\end{align*}
We know that the temperature of the batch reactor is initially at 300 K. Therefore, our first boundary condition is (t1=0 mins, 300 K). Our second boundary condition will be (t2, 350 K). We can now separate our differential equation and take the integral of both sides:
\begin{align*}
\int_{0 mins}^{t_{2} mins} 45e^{-0.2t}\frac{kJ}{min}*dt &= \int_{300 K}^{350 K} 3.8 \frac{kJ}{K}* dT\\
(\frac{-45}{0.2}e^{-0.2t_{2}} + \frac{45}{0.2}e^{-0.2*0})\frac{kJ}{min} &= 3.8\frac{kJ}{K}*(350 K – 300 K)\\
\frac{45}{0.2}(1-e^{-0.2*t_{2}}) \frac{kJ}{min}&= 190 kJ\\
225(1-e^{-0.2*t_{2}}) \frac{kJ}{min}&= 190 kJ\\
(1-e^{-0.2*t_{2}}) &= \frac{190 kJ}{225 \frac{kJ}{min}}\\
(1-e^{-0.2*t_{2}})&= 0.84\\
e^{-0.2*t_{2}}&= 0.16\\
-0.2*t_{2}&= ln(0.16)\\
t_{2} &= \frac{ln(0.16)}{-0.2}\\
t_{2}&= 9.2 minutes
\end{align*}
Cost is an important factor that stakeholders in engineering projects consider when presented with proposals. However, finding chemical, equipment, and labour pricing is not always an easy and straight forwards task. Therefore, it is essential for us to be able to come up with a valid estimation using historical data and existing formulas. In this chapter, we are going to introduce some methods to estimate the cost of building and operating plants.
Learning Objectives
By the end of this chapter, you should be able to:
Understand The elements that contribute to the capital cost for a plant
Estimate the purchased equipment price based on baseline data using the effect of capacity and time
Estimate the total capital investment (TCI) and fixed capital investment(FCI) based on the lang factor
Understand the components of manufacturing costs
Estimate the cost of operating labour
Calculate the cost of manufacturing based on a few known process costs
As you’re going through this chapter, here are some important terms for you to take note of:
Learning Objectives
By the end of this section, you should be able to:
Understand The elements that contribute to the capital cost for a plant
Total Capital Investment (TCI) is the amount of money (A.K.A. capital) needed for a project to proceed. TCI is the total amount of funds needed to build and start a chemical plant, including the testing and troubleshooting stages before the actual production. It does not include operating expenses or maintenance during the production process. This is further divided into 2 types of capital:
Fixed capital investment is the total price to purchase equipment, get it installed and tested.
FCI is made up of direct costs and indirect costs. These are described below and common direct and indirect cost categories are listed in the table below.
Table 1: Examples of Direct and Indirect Costs that Contribute to FCI
Directs costs | Indirect costs |
---|---|
1. purchased equipment (PE) cost | 1. Engineering and supervision |
2. PE installation | 2.Construction expenses |
3. Instrumentation and controls | 3. Contractor’s fee |
4. Piping | 4. Contingency |
5. Electrical equipment and materials | |
6. Buildings | |
7. Yard improvements | |
8. Service facilities | |
9. Land |
The Working Capital is the initial money required to start the plant. You can think of this like a float (initial money in a cash box) when selling something like samosas or baked goods in CHBE. It is money needed for supplies or people that is not recovered until after the plant closes. Since this is generally a long-time in the future, it is considered a capital investment rather than a manufacturing cost.
WC includes capital invested in:
Typically the WC is 10-20% of the TCI for most chemical plants
Learning Objectives
By the end of this section, you should be able to:
Estimate the purchased equipment price based on baseline data using the effect of capacity and time
PE is the purchased price of equipment from a vendor (someone selling the equipment). It is one of the major factors in the TCI direct costs.
It includes the cost to build the equipment but does not include the cost associated with transportation of that equipment to the site, and installation, etc.
Many times we may estimate PE costs based on costs of PE from previous projects. We will usually use factors to adjust the cost for capacity or changes in prices over time.
There are a variety of ways to adjust PE pricing to account for different equipment capacity (bigger or smaller equipment). The most common, simple relationship, and the one we’ll use in this class, is shown below:
Where
– desired equipment cost
– base cost for known equipment
– desired capacity
– base capacity
– cost exponent (given for each type of equipment)
We can rearrange this equation and plot it in a linear form as well (many times you may see log-based plots comparing equipment capacity and cost):
Where
If you are curious, we show how this equation is derived below:
Since and are known values for equipment, we can treat them as constants and take use to represent , and take the ln of both sides of the equation.
\begin{align*}
\frac{C_{a}}{C_{b}}&=\Big(\frac{A_{a}}{A_{b}}\Big)^n\\
C_{a} & = \frac{C_{b}}{A_{b}^n}×A_{a}^n\\
C_{a} & = K × A_{a}^n \\
ln(C_{a}) &= ln(K) + n ln(A_{a}) \\
\end{align*}
Table 1: Examples of Cost Exponents for Process Equipment
Equipment Type | Range of Correlation | Capacity Units | Cost Exponent (n) |
---|---|---|---|
Air compressor, multiple stages | 1 -1500 | 0.85 | |
Shell and tube heat exchanger stainless steel | 1.9 – 1860 | 0.60 | |
Horizontal tank carbon steel | 0.5-74 | 0.30 | |
centrifugal pump stainless steel | 1-70 | 0.67 | |
Crystalizer | 0.2-3.8 | 0.47 |
Note: heat exchanger capacities are measured by the area of heat exchange, thus the unit is .
Analogous exponent values are shown due to copyright considerations. You can find the updated values in Analysis, Synthesis and Design of Chemical Processes, fifth edition, Section 2, Chapter 7.2, Table 7.3.
A general rule for cost exponents is called the six-tenths rule. This is a generalization that for many processes, the exponent will be close to six-tenth’s (0.60). You can see this applies better to certain equipment more than others based on the table above.
Exercise: Using Capacity Correlation to Estimate Purchased Equipment Cost
If a 0.8 crystalizer costs $35,000, how much does a piece of similar equipment with a capacity of 3.0 ? Compare the results using the given cost exponent and the six-tenth rule.
Using n=0.48 given by the cost exponent:
\begin{align*}
\frac{C_{a}}{C_{b}}&=\Big(\frac{A_{a}}{A_{b}}\Big)^n\\
C_{a} & = C_{b}×\Big(\frac{A_{a}}{A_{b}}\Big)^n\\
C_{a} & = $35000 × \Big(\frac{3.0}{0.8}\Big)^{0.47}\\
C_{a} & ≈ $65,100 \\
\end{align*}
Using n=0.60 given by the six-tenth rule:
\begin{align*}
\frac{C_{a}}{C_{b}}&=\Big(\frac{A_{a}}{A_{b}}\Big)^n\\
C_{a} & = C_{b}×\Big(\frac{A_{a}}{A_{b}}\Big)^n\\
C_{a} & = $35000 × \Big(\frac{3.0}{0.8}\Big)^{0.60}\\
C_{a} & ≈ $77,400 \\
\end{align*}
This shows the difference these factors can make, so we want to try to use factors that are as accurate as possible (or be clear about our uncertainties).
As you may know, the value of money changes with time due to factors such as inflation. Cost indexes give us an estimate of what a common item costs at different points in time.
where
– known cost of equipment/plant at a known (past) time
– cost of equipment/plant at a time of interest
– the cost index at a known time
– the cost index at a time of interest
Table 2: Cost indexes for 1996-2020
Year | Marshall and Swift Equipment cost index | Chemical Engineering Plant Cost Index |
---|---|---|
1996 | 211.7 | 154.7 |
2000 | 264.8 | 169.3 |
2004 | 301.7 | 186.3 |
2008 | 339.2 | 212.8 |
2012 | 391.5 | 227.8 |
2016 | 418.9 | 237.8 |
2020 | 458.3 | 258.8 |
Analogous time indexes are shown due to copyright considerations. You can find the updated values in Analysis, Synthesis and Design of Chemical Processes, fifth edition, Section 2, Chapter 7.2.2, Table 7.4.
Marshall and Swift Equipment cost index and Chemical Engineering Plant Cost Index are two common types of cost indexes that we will use for this course.
Note the Marshall and Swift is used for equipment, whereas the plant cost index is for an overall plant cost including installation, piping, etc.
Exercise: Estimation of Equipment Price Over Time
If a stainless steel shell-and-tube heat exchanger costs $18,000 in 1996, what is the cost of a shell-and-tube heat exchanger with the same capacity in 2020?
Marshall and Swift Equipment cost index is used because we are considering the cost for a heat exchanger, which is a single piece of equipment.
\begin{align*}
C_{2} &=C_{1}\Big(\frac{I_{2}}{I_{1}}\Big)\\
C_{2} & = $18000×\Big(\frac{458.3}{211.7}\Big)\\
C_{2} & ≈ $39,000
\end{align*}
Exercise: Estimation of Plant Price Over Time
If a plant producing 200,000 tons of ammonia per year costs $25,000,000 to build in 2000, what would the cost of such a plant be in 2020?
Since we are interested in the cost for the whole plant, we can use the Chemical Engineering Plant Cost Index.
\begin{align*}
C_{2} & = C_{1}\Big(\frac{I_{2}}{I_{1}}\Big)\\
C_{2} & = $25,000,000×\Big(\frac{258.8}{169.3}\Big)\\
C_{2} & ≈ $38,000,000
\end{align*}
Exercise: Combining the Effect of Time and Capacity for Equipment Cost
If a 150kW multiple-stage air compressor costs $80,000 in 2004, how much does the same type compressor with 500kW compacity cost in 2020?
We use Marshall and Swift Equipment cost index to solve for effect of time and cost exponent to solve for the effect of capacity. Both factors are multiplied so the order does not matter.
\begin{align*}
C_{2} &=C_{1}×\Big(\frac{A_{a}}{A_{b}}\Big)^n×\Big(\frac{I_{2}}{I_{1}}\Big)\\
C_{2} & = $80,000×\Big(\frac{500}{150}\Big)^{0.85}×\Big(\frac{458.3}{301.7}\Big)\\
C_{2} & ≈$340,000
\end{align*}
[1] Joseph A. Shaeiwitz; Debangsu Bhattacharyya; Wallace B. Whiting; Richard C. Bailie; Richard Turton. Analysis, Synthesis and Design of Chemical Processes, fifth edition. [online]<https://gw2jh3xr2c.search.serialssolutions.com/?sid=sersol&SS_jc=TC0002267093&title=Analysis%2C%20synthesis%20and%20design%20of%20chemical%20processes> [Accessed 11 June, 2020].
Learning Objectives
By the end of this section, you should be able to:
Estimate the total capital investment (TCI) and fixed capital investment(FCI) based on the lang factor
Lang factor relates purchased equipment cost to total capital investment (TCI) and fixed capital investment(FCI). This factor represents the cost of building a major expansion onto an existing chemical plant, including other costs such as piping, control, etc.
where
– total plant cost
– lang factor
– sum of the cost of all purchased equipment
Lang factors for common plant types are given below. Note that fluid processing means plants processing liquids and/or gasses.
Table 1: Lang Factors for Common Plant Types
type of plant | FCI | TCI |
---|---|---|
solid processing plant | 3.8 | 4.5 |
solid-fluid processing plant | 4.3 | 4.8 |
fluid processing plant | 5.0 | 5.8 |
Exercise: Estimating FCI Using the Lang Factor
Suppose we want to build a sulfuric acid production plant. Say the reaction mechanism for the process is as follows:
\begin{align*}
&S_{(s)} + O_{2(g)} → SO_{2(g)}\\
&SO_{2(g)} + \frac{1}{2}O_{2(g)} ⇌ SO_{3(g)}\\
&SO_{3(g)} + H_{2}SO_{4(l)} → H_{2}S_{2}O_{7(l)}\\
&H_{2}S_{2}O_{7(l)} +H_{2}O_{(l)} → 2H_{2}SO_{4(l)}
\end{align*}
The purchased price for all equipment is shown below:
Equipment categories | Puchased price |
pre-treatment | $3,000,000 |
reactors | $8,000,000 |
absorption tower (for separation) | $5,000,000 |
tanks | $2,000,000 |
pumps and compressors | $4,000,000 |
Estimate the total capital investment for the plant.
From the reaction mechanisms, we see solid, liquid, and gasous phases in the process streams, so we select the lang factor to calculate TCI for solid-fluid processing plant ().
\begin{align*}
TCI &=F_{lang}\sum_{i=1}^nC_{p,i}\\
TCI &=4.9×($3,000,000+$8,000,000+$5,000,000\\
&\;\;\;\;\;+$2,000,000+$4,000,000)\\
TCI & = $107,800,000
\end{align*}
Return on investment is the ratio of the net profit and total capital investment (). The net profit can be calculated by the total revenue minus cost of manufacturing (). It provides a measure of how favourable the investment is.
Typically ranges from 10%-20%. An over 30% for a plant is considered very desirable.
Revenue is the total profit we generated from selling the product without deducting the cost of production:
is the total cost of operating the plant, including raw materials, utilities, labours, etc. We are going to learn about how to estimate later.
Exercise: Estimating ROI
Suppose we want to build a solid-fluid processing plant that produces soda ash (), and it also produces and as profitable by-products. The annual production rates and market prices for our products are shown below:
product | production rate (tonne/year) | price per tonne |
120,000 | $220 | |
95,000 | $145 | |
90,000 | $150 |
Suppose the cost for the total purchased equipment is 18,000,000 dollars and the annual cost of manufacturing is 60,000,000 (which includes the cost of raw materials). Assuming all the products are sold, what is the return on investment for our plant? Based on this estimate would it be economically feasible?
Step 1: Calculate the total revenue generated by all types of products:
\begin{align*}
\text{revenue} &=\sum_{i} \text{unit price of product } i × \text{sales}\\
&=120,000×$220+95,000×$145+90,000×$150\\
& = $53,675,000
\end{align*}
Step 2: Estimate using the lang factor:
We select the lang factor to calculate TCI for solid-fluid processing plants ( ) as stated in the question.
Step 3: Calculate using the values we calculated in the previous steps:
\begin{align*}
ROI &=\frac{(revenue-COM)}{TCI}×100\text{%}\\
&=\frac{($53,675,000-$60,000,000)}{$88,200,000}×100\text{%}\\
& = -7.17 \text{%}
\end{align*}
The project is not feasible because the return on investment is negative, which means we are deficient even if we assume we sold all of the products.
Chemical Engineering Magazine; Accessible from UBC library at http://tinyurl.com/ybaf56kt
G. P. Towler and R. K. Sinnott, Chemical Engineering Design: Principles, Practice, and Economics of Plant and Process Design. (2nd ed.) Waltham, MA; Oxford; Butterworth-Heinemann, 2013; Accessible from UBC library at http://tinyurl.com/y9oubo7n
[1] Joseph A. Shaeiwitz; Debangsu Bhattacharyya; Wallace B. Whiting; Richard C. Bailie; Richard Turton. Analysis, Synthesis and Design of Chemical Processes, fifth edition. [online]<https://gw2jh3xr2c.search.serialssolutions.com/?sid=sersol&SS_jc=TC0002267093&title=Analysis%2C%20synthesis%20and%20design%20of%20chemical%20processes> [Accessed 11 June, 2020].
Learning Objectives
By the end of this section, you should be able to:
Understand the components of manufacturing costs
Estimate the cost of operating labour
Typically divided into three main elements:
– Number of operators needed in the plant at a given time to ensure it runs smoothly.
The number of operators () needed per shift can be estimated from a correlation by Alkhayat and Gerrard which is based on data from five major chemical companies.
where
– number of operating labour workers at a given time
– process steps involving handling solids (transport, partical size control, particulate removal)
– number of process steps not invloving handling solids, including compression, heating, cooling, mixing, separation and reaction, but not including pumps and vessels
We will take the convention of rounding up to the nearest integer to ensure there is enough labour to operate the plant, although in practice there may be ways to have part-time staff etc.
The number of operators is rounded up to the nearest integer when the calculation does not result in a whole number.
The number of total shifts in a year and the number of shifts per operator can be calculated as follows:
On average an operator works five eight-hour shifts per week for 49 weeks per year (3 weeks off for sick leave and vacation deducted from 52 total weeks in a year).
A plant generally requires operators 365 days a year including maintenance and shutdowns. One day is split into 3 eight-hour shifts.
The number of operators needed to fill a position that needs one operator at a given time is .
To find the total cost of operating labour, we can use an average salary for a plant operator. The average salary for a plant operator can be found from an agency such as Statistics Canada.
Suppose each shift requires 4 operators on the site (), we can calculate the total cost of operating labour:
The number should be rounded up if it does not result in a whole number.
Say an operator receives a salary of 70,000 dollars per year, the total cost of operating labour is .
Exercise: Calculating the Cost of Operating Labour
A typical catalytic hydrodesulfurization process used in oil refineries to convert organic sulfur compounds in various process streams into hydrogen sulfide gas is shown below. All streams in the process only involve liquid and gaseous phases.
Image obtained from mbeychok via Wikimedia commons / CC BY-SA 3.0
Given the information:
Calculate the total cost of operating labour per year.
Step 1: Calculate the number of operating labour ():
From the labels, the process does not involve handling solids(). And we can count that we have a total of 9 process steps that handle fluids without accounting for the number of pumps and vessels. This is shown in the diagram below:
So . Substituting this into the formula for the number of operators:
\begin{align*}
N_{OL}&=(6.29+31.7×(0)^2+0.23×9)^{0.5}\\
& = 2.89\\
& ≈ 3
\end{align*}
Therefore, each shift needs 3 operators.
Step 2: Calculate the number of operators needed to fill a position that needs one operator at a time:
Step 3: Calculate the total number of operators needs to be hired for the plant:
\begin{align*}
\text{# of operator} & =N_{OL}×\frac{\text{# of total shifts in a year}}{\text{# of shifts per operator}} \\
& =3×4.5\\
& = 13.5 \\
& ≈ 14
\end{align*}
Step 4: Calculate the total cost of operating labours:
\begin{align*}
C_{OL} & =\text{# of operators}×\text{salary} \\
& =14×$80,000\\
& = $1,120,000
\end{align*}
[1] Alkhayat, W. A., and A. M. Gerrard, Estimating Manning Levels for Process Plants, AACE Transactions, I.2.1–I.2.4, 1984.
Learning Objectives
By the end of this section, you should be able to:
Calculate the cost of manufacturing based on a few known process costs
Direct Costs of Manufacturing are costs that are directly impacted by the manufacturing rate. To assess the impacts of some of the main elements of the direct costs, we will use multiplication factors based on certain known costs.
Direct manufacturing costs | Symbol or Range of multiplying factors | Avg. factor used |
---|---|---|
1. Raw Materials | ||
2. Utilities | ||
3. Operating Labour | ||
4. Waste Treatment | ||
5. Direct Supervisory and clerical labour | ||
6. Maintenance and repairs | ||
7. Operating supplies | ||
8. Laboratory charges | ||
9. Patents and Royalties | ||
Total direct costs of manufacturing |
Fixed Costs of Manufacturing are costs that are not directly impacted by the manufacturing rate. To assess the impacts of some of the main elements of the fixed costs, we will use multiplication factors based on certain known costs.
Fixed manufacturing costs | Range of multiplication factors | Avg. factor used |
---|---|---|
1. Depreciation – costs associated with buildings and plant equipment. Note that this is not an actual expense, but just accounted for in terms of tax purposes | 0.1 FCI (Note there are a variety of ways depreciation can be calculated, this is a very crude approximation) | |
2. Local taxes and insurance | ||
3.Plant overhead costs (referring to the costs related to running the business but not directly related to the process of making the product) | ||
Total fixed costs of manufacturing |
Governments will not let companies write off all expenses to build a plant immediately. Instead, they are written off over time, in what is known as depreciation.
Depreciation is calculated based on the purchase value of equipment minus the salvage value and evaluated over a number of years specified. This does not include land costs, since land generally retains its value, whereas equipment and buildings tend to lose their value over time.
Straight-line depreciation is the simplest calculation of depreciation:
Exercise: Estimate Depreciation Value
The following information is the purchased cost for categories of equipment at a small waste treatment facility.
Equipment categories | Puchased price |
---|---|
pre-treatment | $2,000,000 |
reactors | $5,500,000 |
tanks | $4,000,000 |
pumps and compressors | $3,500,000 |
sensor and control devices | $1,000,000 |
If the life of all of the equipment shown in the table above is 8 years, and the total salvage value of all the equipment is $89,000. Calculate the depreciation per year using the straight-line method.
\begin{align*}
Depreciation &=\frac{\text{purchased value } (FCIL)-\text{salvage value}}{\text{life of equipment } (n)}\\
& = \frac{($2,000,000+$5,500,000+$4,000,000+$3,500,000+$1,000,000)-$89,000}{8}\\
& ≈$ 2,000,000
\end{align*}
Exercise: Estimating Depreciation Using FCI
Calculate the yearly straight-line depreciation based on the following information about a hydrodesulphurization plant:
Item | Value (dollars or in units specified) |
Fixed Capital Investment | $87,000,000 |
land cost | $1,000,000 |
salvage value | $3,200,000 |
life of equipment | 12 years |
The purchased equipment cost can be calculated using the Fixed Capital Investment excluding the land cost:
\begin{align*}
Depreciation &=\frac{\text{purchased value } (FCIL)-\text{salvage value}}{\text{life of equipment } (n)}\\
& = \frac{($87,000,000-$1,000,000)-$3,200,000}{12}\\
& =$6,900,000
\end{align*}
General expenses are overhead costs related to carrying out business functions. To assess the impacts of some of the main elements of general expenses we will use multiplication factors based on certain known costs.
General expenses | Range of multiplication factors | Avg. factor used |
---|---|---|
1. Administrative costs | ||
2. Product distribution and selling costs | ||
3. Research and development | ||
Total general expenses |
Summing together the Direct Costs of Manufacturing, Fixed Costs of Manufacturing, and General Expenses, we can get a general formula to estimate the total cost of manufacturing.
Direct Cost of Manufacturing:
Fixed Cost of Manufacturing:
General Expenses:
Cost of Manufacturing:
Depreciation is not included in most of the calculations we perform – depreciation is not a real cost that we need to pay, but it is part of the money we lose for tax purposes.
Exercise: Estimating the Cost of Manufacturing
Suppose the following information is obtained from a plant producing 100,000 tonnes of ethylene oxide annually.
Item | Value (dollars) |
Fixed Capital Investment | $46,000,000 |
Raw Material Cost | $27,600,000 |
Waste Treatment Cost | $5,000,000 |
Utility Cost | $800,000 |
Direct Labour Cost | $840,000 |
a. What is the cost of manufacturing?
b. Calculate the percentage of each element (Direct Cost of Manufacturing, Fixed Cost of Manufacturing, and General Expenses) as a portion of the Cost of Manufacturing.
Do not include depreciation in the calculations above.
a. Cost of Manufacturing (COM):
\begin{align*}
&\;\;\;\;0.180FCI+1.23(C_{RM}+C_{WT}+C_{UT})+2.73C_{OL}\\
& = 0.180×46,000,000+1.23×(27,600,000+5,000,000+800,000)+2.73×840,000\\
& ≈ 51,700,000
\end{align*}
b.
Direct Cost of Manufacturing
\begin{align*}
& \;\;\;\;C_{RM}+C_{WT}+C_{UT}+1.33C_{OL}+0.03COM +0.069FCI\\
& =27,600,000+5,000,000+800,000+1.33×840,000+0.03×51,700,000+0.069×46,000,000\\
& ≈ 39,200,000
\end{align*}
\begin{align*}
\frac{\text{Direct Cost of Manufacturing}}{\text{Cost of Manufacturing}}×100\text{%}&=\frac{39,200,000}{51,700,000}×100\text{%}\\
&≈75.8\text{%}
\end{align*}
Fixed Cost of Manufacturing
\begin{align*}
& \;\;\;\;0.708C_{OL}+0.068FCI\\
& =0.708×840,000+0.068×46,000,000\\
& ≈ 3,700,000
\end{align*}
\begin{align*}
\frac{\text{Fixed Cost of Manufacturing}}{\text{Cost of Manufacturing}}×100\text{%}&=\frac{3,700,000}{51,700,000}×100\text{%}\\
&≈7.2\text{%}
\end{align*}
General Expenses
\begin{align*}
&\;\;\;\; 0.177C_{OL}+0.009FCI+0.16COM\\
& =0.177×840,000+0.009×46,000,000+0.16×51,700,000\\
& ≈ 8,800,000
\end{align*}
\begin{align*}
\frac{\text{General Expenses}}{\text{Cost of Manufacturing}}×100\text{%}&=\frac{8,800,000}{51,700,000}×100\text{%}\\
&≈17.0\text{%}
\end{align*}
Total Capital Investment | |
Effect of capacity on equipment prices | |
Effect of time on equipment prices | |
Preliminary Estimation of Purchased Equipment Costs | |
Lang Factor | |
Return on Investment | |
Revenue | |
Cost of manufacturing | \begin{align*} &cost\; of\; manufacturing\\ &=direct\;cost+fixed\;cost\\ &\;\;+general\;expenses \end{align*} |
Number of operating labour | |
Cost of operating labour | \begin{align*} &\text{# of operators}\\ &=N_{OL}×\frac{\text{# of shifts in a year}}{\text{# of shifts per operator}} \end{align*}$$C_{OL}=\text{# of operators}×salary$$ |
Total direct cost of manufacturing | \begin{align*} &\text{total direct cost of manufacturing}\\ &=C_{RM}+C_{WT}+C_{UT}+1.33C_{OL}+0.03COM+0.069FCI \end{align*} |
Total fixed cost of manufacturing | \begin{align*} &\text{total fixed cost of manufacturing}\\ &=0.708C_{OL}+0.068FCI+depreciation(0.1FCI) \end{align*} |
Depreciation | \begin{align*} &\text{depreciation}\\ &=\frac{\text{purchased value}-\text{salvage value}}{\text{life of equipment }(n)} \end{align*} |
General expenses | \begin{align*} &\text{general expenses}\\ &=0.177C_{OL}+0.009FCI+0.16COM \end{align*} |
Cost of manufacturing | \begin{align*} &\text{cost of manufacturing}\\ &=0.180FCI+1.23(C_{RM}+C_{WT}+C_{UT})+2.73C_{OL}+depreciation(0.1FCI) \end{align*} |
An interactive or media element has been excluded from this version of the text. You can view it online here:
https://pressbooks.bccampus.ca/chbe220/?p=2401
Exercise: Equipment Prices with the Effect of Capacity and Time
Equipment Type | Range of Correlation | Capacity Units | Cost Exponent (n) |
---|---|---|---|
Air compressor, multiple stages | 1 -1500 | 0.85 | |
Shell and tube heat exchanger stainless steel | 1.9 – 1860 | 0.60 | |
Horizontal tank carbon steel | 0.5-74 | 0.30 | |
centrifugal pump stainless steel | 1-70 | 0.67 | |
Crystalizer | 0.2-3.8 | 0.47 |
Year | Marshall and Swift Equipment cost index | Chemical Engineering Plant Cost Index |
---|---|---|
1996 | 211.7 | 154.7 |
2000 | 264.8 | 169.3 |
2004 | 301.7 | 186.3 |
2008 | 339.2 | 212.8 |
2012 | 391.5 | 227.8 |
2016 | 418.9 | 237.8 |
2020 | 458.3 | 258.8 |
In 2008, a carbon steel horizontal tank with a capacity of 26.5 gallons costs $6700, then which of the following is the closest estimation to the cost of the same type of tank with a capacity of 50 cost in 2020?
a)$9800
b)$10,100
c)$7,500
d)$8,000
b) $10,100
We combine the effect of capacity and time by multiplying the two factors. For time, we use the Marshall and Swift Equipment cost index because we are interested in the cost of a piece of equipment.
\begin{align*}
C_{2} &=C_{1}×\Big(\frac{A_{a}}{A_{b}}\Big)^n×\Big(\frac{I_{2}}{I_{1}}\Big)\\
C_{2} & = $6700×\Big(\frac{50}{26.5}\Big)^{0.30}×\Big(\frac{458.3}{339.2}\Big)\\
C_{2} & ≈ $10,100
\end{align*}
Exercise: Preliminary Estimation of Purchased Equipment Costs
Using the preliminary estimation of purchased equipment costs, estimate the price of a centrifugal compressor with the power of 400kW.
The value of a, b and n are shown as follows:
Equipment | Units for Size, S | S lower | S upper | a | b | n |
centrifugal compressors | driver power, kW | 75 | 30,000 | 490,000 | 16,800 | 0.6 |
a) $1,100,000
b) $7,210,000
c) $490,000
d) the size of the equipment is out of the range for effective estimation
a) $1,100,000
The size of the compressor is 400kW, which is within the range of S lower and S upper, so we can eliminate choice d).
Then we can use the formula:
\begin{align*}
C_e&=a+bS^n\\
&=490,000+16,800×(400)^{0.6}\\
&≈ $1,100,000
\end{align*}
Exercise: Calculating Depreciation
Equipment A | Equipment B | |
Purchased Value | $80,000 | $100,000 |
Salvage Value | $30,000 | $40,000 |
Life of Equipment | 5 years | 10 years |
Suppose we have 2 choices for a piece of equipment that serves the same purpose in our process, and we want to choose the equipment with the smallest depreciation, which one should we choose?
a) equipment A
b) equipment B
c) they have the same depreciation value
b) equipment B
\begin{align*}
&\text{Depreciation of equipment A}\\
&=\frac{\text{purchased value}-\text{salvage value}}{\text{life of equipment }(n)}\\
&=\frac{$80,000-$30,000}{5years}\\
&=$10000/year
\end{align*}
\begin{align*}
&\text{Depreciation of equipment B}\\
&=\frac{\text{purchased value}-\text{salvage value}}{\text{life of equipment }(n)}\\
&=\frac{$100,000-$40,000}{10years}\\
&=$6000/year
\end{align*}
Exercise: Estimating Project Cost Using the Lang Factor
Shown above is the block flow diagram of a carbon-capture process. The cost of equipment for each step is listed as follows:
Equipment | Price |
Absorber | $80,000 |
Heat Exchanger | $65,000 |
Stripper | $50,000 |
5-stage Compression | $120,000 |
Separator | $30,000 |
Mixer | $30,000 |
Type of Plant | FCI | TCI |
solid processing plant | 3.9 | 4.6 |
solid-fluid processing plant | 4.1 | 4.9 |
fluid processing plant | 4.8 | 5.7 |
All of the compounds are in liquid or vapour states. Estimate the TCI of the project.
Since all of the compounds are in liquid or vapour states, we take
\begin{align*}
C_{TM}&=F_{lang} \sum_{i=1}^{n}C_{p,i}\\
&=5.7×($80,000+$65,000+$50,000+$120,000\\
&\;\;\;\;\;\;+$30,000+$30,000)\\
&=$2,137,500
\end{align*}
Exercise: Calculating the number of Operating Labour
Shown below is a PFD of a catalytic reforming plant used in oil refineries to produce a high-octane gasoline component. Calculate the number of operators needed per shift for the process.
Image obtained from User: mbeychok via Wikimedia Commons / CC BY-SA 3.0
Note: Naphtha is a flammable liquid hydrocarbon mixture
From the labels, the process does not involve handling solids(). And we can count that we have a total of 12 process steps that handle fluids without accounting for the number of pumps and vessels, shown in the diagram below:
So . Substitute this into the formula:
\begin{align*}
N_{OL}&=(6.29+31.7×(0)^2+0.23×12)^{0.5}\\
& = 3.008\\
& ≈ 4
\end{align*}
The number of operating labour is rounded up to the nearest integer.
Exercise: Return on Investment
Suppose the estimation of the key manufacturing costs for our process are listed below:
$1,650,000 | |
$2,000,000 | |
$600,000 | |
$280,000 | |
$320,000 | |
$500,000 |
a) Calculate the cost of manufacturing (do not include depreciation)
b) If the market price of the product is $500/tonne, how many tonnes of the product do we need to sell per year to get an ROI of 10%?
a)
\begin{align*}
COM&=0.180FCI+1.23(C_{RM}+C_{WT}+C_{UT})+2.73C_{OL}\\
&=0.180×$1,650,000+1.23×($600,000+$500,000+$280,000)\\
&\;\;\;\;\;\;+2.73×$320,000\\
&=$2,868,000
\end{align*}
b)
\begin{align*}
ROI&=\frac{revenue-COM}{TCI}×100\text{%}\\
ROI×TCI&=revenue-COM\\
revenue&=ROI×TCI+COM\\
revenue&=10\text{%}×$2,000,000+$2,868,000\\
revenue&=$3,068,000
\end{align*}
\begin{align*}
revenue&=unit\;price\;of\;product×sales\\
sales&=\frac{revenue}{unit\;price\;of\;product}\\
sales&=\frac{$3,068,000}{$500/tonnes}\\
sales&=6136\;tonnes
\end{align*}
In this chapter, we will be introducing the basic concepts of green engineering. There is a growing demand for green engineering and environmental considerations in today’s world. Chemical and biological engineers play a key role in determining the effect engineering processes will have on the environment, which is why it is crucial for us to adopt green engineering analyses and practices in our work. This chapter provides basic concepts and definitions of green engineering and introduces methods that we can use to predict the effect our product’s life cycle will have on the environment.
Learning Objectives
By the end of the chapter, you should be able to:
Characterize green engineering developments
Assess the physical properties of substances in order to determine their environmental fate if released
Understand the steps involved in life cycle assessment (LCA)
Interpret LCA results critically
As you’re going through this chapter, here are some important terms for you to take note of:
Learning Objectives
By the end of this section, you should be able to:
Characterize green engineering developments
Green engineering is the design, commercialization, and use of processes and products in a way that reduces pollution, promotes sustainability, and minimizes risks to human health and the environment without sacrificing economic viability and efficiency..
• Clean fuels
• Sustainable or better regenerative development
• Waste reduction
• Not having to deal with byproducts, or finding a value use for byproducts
Within green engineering, we can think about ways to improve processes and generally see that some of these may be more effective than others. Examples of a few means of process improvement listed in order of most effective to least effective follow.
Concepts in green engineering processes generally translate to economic, environmental, and social benefits. Processes meeting these three aims will generally thrive in the long term.
Environmental regulations generally determine how companies must act as a minimum. The standards may vary in each country. These will dictate:
• Human safety, ensuring people aren’t exposed to harmful substances
• Pollution control measures that must be reported
• Emission limits
Over time, limits generally get more restrictive → means that the most efficient processes will be the ones that succeed in the longer term.
As an example, environmental restrictions of chemicals in Canada can be found on the site of Health Canada from the Government of Canada.
The waste management hierarchy shows certain methods of process improvements and lists them from least to most effective. The most effective process improvements generally require more detailed engineering as they are more complex. However, they also can create much more efficient processes.
Learning Objectives
By the end of this section, you should be able to:
Assess the physical properties of substances in order to determine their environmental fate if released
The environmental fate of contaminants describes where substances will go when released into the environment.
We may use our knowledge of chemical structures to predict the physical properties of the chemical substances.
Taking common chemical groups from a molecule, we can use the interactions from these groups to predict the properties the molecule will have based on parameters regressed from known molecules.
This is an example of how physical chemistry simulation and modeling can be used to predict important properties.
These properties can inform us of how a substance will act when dispersed in the environment.
Example: Group Contribution Method
The solubility of a substance is an important property for us to predict the way it migrates in the environment after release. Consider a simple situation where we have the following 3 compounds being released into a body of water:
Ethanol- a fuel commonly produced from biomass, also used as an industrial solvent for organic chemicals.
Nonane – a flammable liquid produced from petroleum.
1-nonanol – a compound commonly found in oils of citrus fruits and industrially used in manufacturing of artificial lemon oil.
The chemical structures of the 3 compounds are shown below:
The -OH functional group is polar, which helps a compound to interact with and dissolve in water (a polar solvent). The long chain of alkanes are nonpolar and will decrease a compound’s solubility in water.
Using the group contribution analysis principle, ethanol will have the highest solubility in water due to a significant portion of its structure being polar (the -OH functional group). 1-nonanol will be less soluble owing to the larger carbon chain as opposed to ethanol. Nonane will be the least soluble as there are no polar groups in the molecule. There are group contribution correlations that allow us to calculate estimates of physical properties (solubility, boiling point, etc.) based on chemical structure, however, will not cover these methods in this course.
Experimental data for solubility of each compound is shown below, you’ll see this supports our rough analysis using the group contribution principles above:
Ethanol is miscible in water, which means it can dissolve with water of any proportions.
1-nonanol has a solubility of 0.13g/L in water.
Nonane has a solubility of 0.00002g/L in water.
We can conclude that if the three substances are released in a water body, ethanol will migrate much faster and will likely be more difficult to collect for treatment.
We can also use software to calculate physical properties. One such software is from the US EPA and is called EPI Suite – Estimation Program Interface Suite. It is only available to be run on windows and can be downloaded from https://www.epa.gov/tsca-screening-tools/epi-suitetm-estimation-program-interface.
2. Click “Calculate” to calculate the physical properties.
3. The calculated properties will be shown in the pop-up window or on the bottom of the EPI suite window. We can scroll between tabs to see subsets of results.
Example: Hazardous Materials & Contamination
Naphthalene, toluene, and methylene chloride are common hazardous materials used in the metal production industry. The following data is gathered from the EPI Suite software:
Property | naphthalene | toluene | methylene chloride |
---|---|---|---|
Boiling point (°C) | 217.9 | 110.6 | 40 |
of octanol/water partition coefficient | 3.30 | 2.73 | 1.25 |
Water solubility(mg/L) | 31 | 526 | |
Henry’s law coefficient in water (atm·m3/mol) | |||
of soil adsorption coefficient | 3.189 | 2.369 | 1.337 |
Say a mixture of these three substances is accidentally released into the soil with some groundwater. Based on the physical property data:
a. Which substance is the most likely to be transported in the groundwater?
b. Which substance is the most likely to contaminate the soil?
Solution
a. Methylene chloride is the most likely to be transported in the water because of its much higher solubility compared to the other two substances.
b. Naphthalene is the most likely to contaminate the soil because it has the highest soil adsorption coefficient.
What is the octanol/water partition coefficient ()?
The octanol/water partition coefficient() can be measured by dissolving a substance in a container with n-octanol(a common non-polar solvent) and water(a common polar solvent). N-octanol is not soluble in water, so water and n-octanol will form separate phases. This allows us to measure the distribution of the third substance between water (a polar phase) and n-octanol (a non-polar phase).
is the ratio of the concentration of a substance dissolved in n-octanol and in water. The higher the , the less likely it is for the substance to dissolve in water. is used in predicting the distribution of the substance when released into the environment if it is exposed to different solvents (such as water or oils).
We often express the partition coefficient in the form of .
In general, the process of dissolving one substance in another is called absorption
Exercise: Using the Property of to Select a Pollution Control Method
We will continue with the previous example with naphthalene, toluene, and methylene chloride. Their properties predicted by EPI Suite are shown below:
Property | naphthalene | toluene | methylene chloride |
---|---|---|---|
Boiling point (°C) | 217.9 | 110.6 | 40 |
of octanol/water partition coefficient | 3.30 | 2.73 | 1.25 |
Water solubility(mg/L) | 31 | 526 | |
Henry’s law coefficient in water (atm·m3/mol) | |||
of soil adsorption coefficient | 3.189 | 2.369 | 1.337 |
a. Say each of these substances is present in a wastewater stream at a steel manufacturing plant in equal concentrations. We are to try to remove these substances by contacting them with a non-polar solvent that has similar properties to n-octanol (and leaving this for sufficient time). Which substance will have the lowest concentration in the wastewater stream after this contact?
b. Suppose the steel-manufacturing plant produces of wastewater per day containing naphthalene at a concentration of . This wastewater is sent to a wastewater treatment plant, where the wastewater produced in one day is gathered and put in contract with 5000L of a non-polar solvent that has similar properties to n-octanol. What is the concentration of naphthalene in the water after this treatment? You can assume there is sufficient contact time for the solvent to absorb the naphthalene.
a. Naphthalene will end up with the lowest concentration in the wastewater because it has the highest . This means out of the three substances naphthalene will have the largest proportion absorbed by the non-polar solvent.
b. Step 1: Calculate the total moles of naphthalene:
Step 2: Calculate the moles of naphthalene still dissolved in the water after the absorption process:
We can set the number of moles of naphthalene remaining in the water to be x mol, then the number of moles of naphthalene entering the non-polar solvent can be expressed as (1000-x)mol.
\begin{align*}
\frac{\text{concentration of the substance in nonpolar solvent}}{\text{concentration of the substance in water}}& = K_{ow} \\
\frac{\frac{1000-x}{5000}}{\frac{x}{1×10^{6}}}& = 10^{3.30} \\
\end{align*}
We can solve the equation using excel solver or a graphing calculator to find: .
Step 3: Calculate the final concentration of naphthalene:
Learning Objectives
By the end of this section, you should be able to:
Understand the steps involved in life cycle assessment (LCA)
Interpret LCA results critically
Life Cycle Assessment (LCA) is a technique used to quantify the environmental impact of a product from raw material acquisition through end of life disposition (cradle-to-grave)
Product assembly image from Sammutawe / CC BY-SA
There are four stages of a life cycle assessment:
Life Cycle Assessments are based on international standards set by the International Organization for Standardization (ISO 14040).
Each of the steps are set by different ISOs:
- Step 1: ISO 14041
- Step 2: ISO 14041
- Step 3: ISO 14042
- Step 4: ISO 14043
Exercise: Biofuels Product Scope
Consider the following system boundary diagram for the production of biofuels. Which scope does this representation cover (cradle-to-grave, cradle-to-gate, or gate-to-gate)?
This diagram represents a cradle-to-gate scope because it shows the raw materials used for the production of biofuels, the transportation of raw materials, as well as initial processing.
Define the goal:
Define the scope:
*Functional Units: a measure of the function of the studied system. It provides a reference to which the inputs and outputs can be related and enables a comparison of two essentially different systems.
Examples of functional units:
- For a paint system (the paint as well as layers of coating materials applied to a surface), the functional unit may be defined as the unit surface protected for 10 years
- For a printer, the funcional unit may be defined as the number of printed pages of an acceptable print quality
- For a power generation system, the functional unit may be defined as 1kWh of electricity
Example: Reusable Shopping Bags
Let’s say you own a local grocery store in Kitsilano. Your customers start requesting that you sell reusable shopping bags to minimize waste. You think its a great idea, but want to make sure that the reusable shopping bags you sell have the lowest environmental impact. Since you are an engineer and learned about LCA in CHBE 220, you decide to perform a streamlined LCA.
What is your goal?
Determine which shopping bag has the lowest environmental impact
After doing a search of what is available on the market, you determine that your options for shopping bags are as follows:
What are your system boundaries and process?
To simplify your LCA, you make the following assumptions:
What is the functional unit for this system?
Let’s take our functional unit to be the amount of shopping bags consumed by a household to carry 70 grocery items (~7 items per bag per trip) home from the supermarket each week for 52 weeks. Note you’ll see that each bag type in the table below has relative capacities, and this is used to account for some bags having the ability to carry more or fewer items than others while comparing to single-use plastic bags as a basis.
Bag Type | Single-Use Plastic | Single-Use Paper | Reusable Plastic | Reusable Cotton |
---|---|---|---|---|
Material | HDPE | Unbleached Kraft paper | Polypropylene | Cotton |
Mass per bag | 7g | 42.6g | 95g | 85g |
Relative Capacity | 1 | 0.9 | 1.1 | 1.1 |
Bags per year | 520 | 578 | 4.55 | 4.55 |
Mass bags per year | 3640g | 24622.8g | 432.25g | 386.75g |
The life cycle inventory step is a highly data-intensive step that is based on detailed material and energy balance analyses over the life cycle of a product. The aim of this step is to account quantitatively for the components which make up the life cycle for our products.
Advantages:
Challenges:
A life-cycle inventory can be summarized by block flow diagrams. The following diagram is one example for a typical manufacturing process for lithium-ion battery cells.[1]
Example: Reusable Shopping Bags Life-Cycle Inventory
Once again, you’d like to continue your LCA on the four options of shopping bags that you can offer in your grocery store. You gather the following information for quantities involved in each step of the lifecycle of these bags. Note most of these are measured in terms of a mass (eg grams, g) or mass per distance for travel (eg. tonne-kilometers, tkm).
Single-Use Plastic Bag | |
---|---|
Polyethylene, HDPE, granulate | 3640g |
Stretch blow moulding | 3640g |
Transport, 100km manufacturing to customer | 0.364tkm |
Transport, municipal waste collection, 10km customer to landfill | 0.0364tkm |
Disposal, polyethylene, 0.4% water, to sanitary landfill | 3640g |
Single-Use Paper Bag | |
---|---|
Kraft paper, unbleached, at plant | 24622.8g |
Production of paper bags | 24622.8g |
Transport, 100km manufacturing to customer | 2.4623tkm |
Transport, municipal waste collection, 10km customer to landfill/recycling | 0.24623tkm |
Disposal, packaging paper, to sanitary landfill | 12311g |
Recycling paper | 12311g |
Reusable Plastic Bag | |
---|---|
Polyethylene, granulate | 432.25g |
Extrusion, plastic film | 432.25g |
Transport, 100km manufacturing to customer | 0.043225tkm |
Transport, municipal waste collection, 10km customer to landfill | 0.0043225tkm |
Disposal, polyethylene, to sanitary landfill | 432.25g |
Reusable Cotton Bag | |
---|---|
Textile, woven cotton, at plant | 386.75 |
Transport, 100km manufacturing to customer | 0.03867tkm |
Transport, municipal waste collection, 10km customer to landfill | 0.003867tkm |
Disposal, inert material, to sanitary landfill | 386.75g |
Impact Assessment Results: impact assessment converts the inventory into impact categories or endpoints, detailing human health and environmental impact.
*ADP: abiotic depletion potential – depletion of non-living resources such as minerals, clay, fossil fuels, etc.
*AP: acidification potential – the potential of the battery releasing harmful acids into the environment
*EcoTP: ecological toxicity potential – the potential of the battery being toxic to surrounding ecology
*EP: eutrophication potential – nutrient addition that leads to the (over)growth of aquatic plant species; can lead to things like algal blooms and food chain disruptions
*GWP: global warming potential – the potential of warming caused by the release of greenhouse gases as caused by the battery
*HTP: human toxicity potential – the potential the battery process has for releasing chemicals that are toxic (dosage is taken into account) into the environmental
*OCH: occupational cancer hazard – the potential the battery process has for exposing people that work on this battery to carcinogens
*ODP: ozone depletion potential – the potential the battery process has for releasing chemicals that degrade the ozone layer
*OnCH: occupational non-cancer hazard – the potential the battery process has for exposing people that work on this battery to other diseases, injuries, or other health problems
*POP: photochemical oxidation potential – the potential the battery process has for releasing chemicals that react in the sunlight and release harmful products into the environment
*EOL: end-of-life; note that the negative values for the EOL bar signify recovery of the material at the end of battery’s life, which reduced the life-cycle impact.
Life cycle interpretation: findings of the inventory analysis or impact assessment are evaluated in relation to the goal and scope of the study to reach conclusions and recommendations.
In this stage we aim to:
Example: Reusable Shopping Bags LCA Results
Consider the following chart that you put together after conducting an impact assessment on the 4 options for shopping bags. Note in this case that a lower score will mean less environmental impact (ie. lower score is better).
After contacting suppliers, you find that the cost per year for each of the shopping bags is as follows:
Some questions you might ask:
- Which bag would you promote given the impact information?
- Which bag would you promote given the cost information?
- Which bag would customers most likely lean towards given the cost?
- Are there any assumptions you would change?
- Did your final result differ from your initial guess?
Given both the cost information and the environmental assessment, you might decide to go with the reusable plastic shopping bags, or offer these alongside re-useable cotton bags with some information to customers.
Exercise: LCA Results on Different Lithium-Ion Batteries
A life-cycle assessment is conducted on two different types of vehicle batteries: electric vehicle (EV) batteries and plug-in hybrid electric vehicle (PHEV) batteries. Each of these has different lithium-ion chemistries. The EV batteries have 3 different chemistries (, , and ), while the PHEV batteries have 2 different chemistries ( and ).[1]
Legend for the charts:
*ADP: abiotic depletion potential
*AP: acidification potential
*EcoTP: ecological toxicity potential
*EP: eutrophication potential
*GWP: global warming potential
*HTP: human toxicity potential
*OCH: occupational cancer hazard
*ODP: ozone depletion potential
*OnCH: occupational non-cancer hazard
*POP: photochemical oxidation potential
EV Batteries:
PHEV Batteries:
Your client is asking you to make a recommendation on which vehicle type (EV or PHEV) they should manufacture based solely on human toxicity potential through product use for batteries. Which vehicle type do you recommend based on the information you have and why?
Step 1: Determine where the human toxicity potential (HTP) is located on the graph.
Step 2: Compare the data of both EV and PHEV batteries side by side. Note that the ‘product use’ factor is the light blue/teal bar
EV:
PHEV:
Step 3: Determine the approximate percentages for ‘product use’ for each of the factors and the batteries.
EV:
Health Factor | |||
---|---|---|---|
HTP | 80% | 80% | 80% |
PHEV:
Health Factor | ||
---|---|---|
HTP | 90% | 90% |
Step 4: Determine which battery type is the best choice based on your client’s specifications.
Since the percentages are consistent among battery chemistries, our choice will be between the EV battery and the PHEV battery. The EV has a smaller impact on human toxicity potential through product use when compared to PHEV. Therefore, we recommend an EV battery.
[1]”Application of LifeCycle Assessment to Nanoscale Technology: Lithium-ion Batteries for Electric Vehicles”, United States Environmental Protection Agency, 2013.
An interactive or media element has been excluded from this version of the text. You can view it online here:
https://pressbooks.bccampus.ca/chbe220/?p=1979
Exercise: Waste Management Hierarchy
British artist Damien Hirst has a series of artwork of well-preserved animal corpses suspended in formaldehyde to express the concepts of life and the inevitable death at the end of it. Recent research shows that the artwork has a risk of leaking formaldehyde, a toxic and carcinogenic compound, during the exhibition.
Based on the concept of waste management hierarchy, what is the most effective way to prevent the leak of formaldehyde? Rank the following actions from the most to least effective?
a) Perform regular maintenance to make sure the container is sealed
b) Place activated carbon around the container to absorb the leaked formaldehyde
c) Use another method to preserve the animals that do not involve the use of toxic substances
d) Provide personal protective equipment for visitors
e) Warn people about the risk before visiting but do not implement any safety measurements
c, a, b, d, e
c – the production of formaldehyde is eliminated from the source
a – the production of formaldehyde fume is eliminated, but toxic material is still used, and there is still a risk of the leak of toxic materials
b – activated carbon will reduce the effect but not totally eliminate it
d – reduced the health effect to the visitors, but formaldehyde is released into the environment
e – formaldehyde is directly released into the environment
Exercise: Life Cycle Analysis
The graph above shows the life cycle analysis of paper cups. Use the graph to answer the following questions:
a) If we want to improve the life cycle of paper by focusing on reducing inorganics causing respiratory effects, which stage of the life cycle should we focus on?
b) If we want to compare the total mass of freshwater eutrophication substances with terrestrial eutrophication substances, can we get the information using the graph? Why?
a) production stage
Find the part of the graph investigating “respiratory effects, inorganics”:
The relative lengths of the bars represent the contribution from each life stage. We can see that the bar representing the production stage is much longer than all the other stages. This means that the production stage is emitting a lot more inorganics that result in negative respiratory effects. Therefore, focusing on improving the production stage will be the most effective in terms of eliminating inorganics causing respiratory effects.
For b), it is not possible to compare the total mass of types of hazardous substance emissions using this diagram. In this diagram, each category is scaled to 100% without stating the total overall impact.
Exercise: Environmentally Hazardous Chemicals
DEHP(2-ethylhexyl phthalate), , and lead are 3 types of environmentally hazardous chemicals released by the soft plastic that we use in daily life. The data is calculated using the EPI Suite software:
Property | DEHP | lead | |
---|---|---|---|
Boiling point (°C) | 384 | 223.4 | 485.37 |
of octanol/water partition coefficient | 7.60 | 5.51 | 0.73 |
Water solubility(mg/L) | 0.27 | 1.133 | 9581 |
Henry’s law coefficient in water (atm·m3/mol) | |||
of soil adsorption coefficient | 5.078 | 3.422 | 1.121 |
Say a mixture of these three substances is accidentally released into the soil with some groundwater. Based on the physical property data:
a) Which substance is the most likely to be transported in the groundwater?
b) Which substance is the most likely to contaminate the soil?
c) Say each of these substances is present in a wastewater stream at a steel manufacturing plant in equal concentrations. We are to try to remove these substances by contacting them with a non-polar solvent that has similar properties to n-octanol (and leaving this for sufficient time). Which substance will have the lowest concentration in the wastewater stream after this contact?
a) Lead is the most likely to be transported into the water due to its high solubility, especially when comparing it to the other two substances.
b) DEHP is the most likely to contaminate the soil because it has the highest soil adsorption coefficient.
c) DEHP will have the lowest concentration after the removal process because it has the highest value.
This chapter aims to allow us to gain awareness and appreciation of safety in chemical processes. The chapter introduces some common terminologies in safety measures and process hazard analysis techniques. By the end of this chapter, we should be able to use the process hazard analysis techniques to analyze potential hazards in simple process units and prioritize process safety in our design.
Learning Objectives
By the end of this chapter, you should be able to:
List and describe basic process safety terminology
Appreciate the need for process safety
Prioritize process safety changes based on inherently safer design strategies
List and describe process hazard analysis techniques
Apply the following process hazard analysis methods: what-if analysis, checklist / checklist combined with what-if analysis, fault-tree analysis
Interpret the following process hazard analysis methods: failure modes and effect analysis (FMEA)
As you are going through this chapter, some important terms for you to take note of:
Learning Objectives
By the end of this section, you should be able to:
List and describe process hazard analysis techniques
Apply the following process hazard analysis methods: what-if analysis, checklist/checklist combined with what-if analysis, fault-tree analysis
Interpret the following process hazard analysis methods: failure modes and effect analysis (FMEA)
PHA – Process Hazard Analysis
PHAs are separated into two major categories:
Generally, PHA should be reviewed and to some extent and redone at least every 5 years
Generally, we want diversity- Engineering and process operations expertise,
design, operations, maintenance…
One team member with process experience
One team member knowledgeable in PHA
methodology
Observe layout, verify information, obtain mental picture
Key contacts at facility for background, questions, etc
The whole team should participate
Information Required for PHA
e.g. American Society of Mechanical Engineers (ASME)
What-if analysis is a scenario PHA method as it investigates specific problems and traces what would happen if they occur.
Typical what-if questions:
What if the inlet stream is blocked?
What if the pipe leaks?
What if the temperature transmitter stops working?
What if the valve fails open / fails closed?
Suppose we focus on the analysis of a reboiler to heat a recycled portion of the bottoms stream coming from a distillation column before it is sent back to the column:
Example: Reboiler What-if Analysis
For the same reboiler system as describes previously and shown below, what is the most important consequences if the temperature transmitter stops working and keeps reporting a low temperature? What is a recommendation we can make to mitigate the most important consequence identified?
Solution
Identify the direct consequences:
If the temperature transmitter keeps reporting a low temperature, the valve will always be open to allow more steam to flow into the reboiler trying to heat up the process steam. This will cause the process steam to overheat, potentially leading to pressure build-up in the process stream.
Recommendations:
A pressure relief valve should be installed on the process stream in case of pressure build-up.
Exercise: Distillation Column What-if Analysis
For the distillation column shown above, perform what-if analyses for the following questions:
a) What if stream 2 is blocked?
b) What if V-102 fails open?
If stream 2 is blocked:
If the stream is blocked, the outflow of the vapor component will be stopped. This will cause the pressure to build up in the distillation column.
A pressure relief valve should be installed in the distillation column to relieve any pressure build-up.
If V-102 fails open:
If V-102 fails open, it will cause free flow of the bottoms out of the distillation column. None of the bottoms will be vapourized by the reboiler and go back to the distillation column. This will affect the vapour-liquid equilibrium in the distillation column, decreasing the efficiency of separation.
Though low separation efficiency may cause operational troubles, there is no significant immediate safety hazard to the distillation unit if V-102 fails open. No specific recommendation is required.
Checklist analysis is a non-scenario PHA method as it suggests common safety issues to assess, thus it is not scenario specific.
A sample checklist can be seen below with items taken from a process safety management (PSM) auditing checklist from the U.S. Occupation Safety and Health Administration (OSHA). This sample is a session from the Auditing Checklist from OSHA under US department of Labour. You can click on the link to see the full checklist.
Checklist analysis typically is broken into three steps:
Checklists can be used at any point in the plant design and operation process, from initial design to operation to decommissioning (permanently shutting down a plant).
Checklists can be combined with what-if technique to enhance that technique. With this combined approach, during PHA you would go through a checklist first to help identify hazards and then apply what-if analysis (this would then be considered a scenario method).
Below are common symbols used in fault-tree analysis logic diagrams:
Fault-tree Analysis
Calculate the frequencies of all the intermediate events for the given fault tree:
Solution
We start with the intermediate event that has known frequencies for all of its input events and work our way up.
Step 1: Calculate the frequency of “tube rupture due to over-pressurization”:
The intermediate event is connected to the input events by an “and” gate, so we multiply the frequencies.
Step 2: Calculate the frequency of “leaks between process fluids”:
The intermediate event is connected to the input events by an “or” gate, so we add the frequencies.
Step 3: Calculate the frequency of “heat exchanger fails to isolate and contaminates other equipment”:
The intermediate event is connected to the input events by an “and” gate, so we multiply the frequencies.
So in general, the resulting frequencies are:
Failure Modes and Effect Analysis (FMEA)
Sample failure mode of an equipment:
An interactive or media element has been excluded from this version of the text. You can view it online here:
https://pressbooks.bccampus.ca/chbe220/?p=2317
Exercise: Safety Action
During a yearly safety assessment, you find that a pressure relief valve in a heat exchanger is rusted and may not open properly in case of a pressure build-up. What is the best course of action?
a) Report to the project manager and immediately replace the pressure relief valve
b) Operate the process at a lower pressure to reduce the risk of over-pressurizing
c) Ignore it because the probability of over-pressurizing is very low
a) Report to the project manager and immediately replace the pressure relief valve
Any safety hazard should be reported and addressed as soon as possible.
For b), Operating the process at a lower pressure does not completely eliminate risks and may reduce process efficiency. In the long run, this will cost more than a new pressure relief valve.
Exercise: OSHA Incident Rate for Exposure to Hazardous Materials
Suppose that a total of 12 operators have been exposed to leaking hazardous materials in a chemical plant. The total working hours that might expose the workers to the risk are 870,000 hours. Calculate the OSHA incident rate for exposure to hazardous materials.
\begin{align*}
\text{incident rate} & = \frac{\text{number of incident exposure}×200,000}{\text{number of working hours}}\\
& =\frac{12×200,000}{870,000}\\
& = 2.76
\end{align*}
Exercise: Pump Kickback Loop
Sometimes a kickback loop is added on a pump to prevent damage to the pump when the outlet stream is blocked:
Compare the effect between the kickback loop is present or not when we close the valve (or when the stream after the valve is blocked), explain how the kickback loop prevents damage.
If no kickback loop present, the pump keeps doing work on the liquid before the valve, but the stream is not able to move forward. As a result, the stream will be heated up. If the stream vapourizes, the vapour may tear up the metal blade in the pump.
When there is a backflush cycle, the stream will circulate in the kickback loop, which increases the amount of liquid pumped and dissipates the heat.
Exercise: HCl Production What-if Analysis
Shown below is the simplified process of HCl production
The following reaction happens in the reactor at a temperature of 900°C:
In the absorber, HCl gas is dissolved in a water stream ( of solution = -74.8 kJ/mol), producing aqueous HCl solution
Perform what-if analysis regarding the following scenarios:
a) What if the cooling water stream in the reactor is blocked? State the direct consequences and recommendations.
b) If the cooling water stream in the absorber is blocked, how would the concentration of the product be affected?
a) If the cooling water stream is blocked, the temperature in the reactor will rise up due to the exothermic reaction. The gas in the reactor will expand, which leads to pressure increase and potential explosion.
Regular maintenance should be performed and a pressure relief valve should be installed in the reactor.
b) The absorber heats up as the process of HCl dissolving is exothermic. As the temperature increases, the solubility of HCl decreases, which leads to a lower concentration of HCl in the product.
Exercise: Glycerol Purification
We use a distillation column to purify a stream of glycerol with methanol impurities:
Boiling Point (°C) at 1 atm | |
Glycerol | 290 |
Methanol | 65 |
Base on the given information, answer the following questions:
a) What is the major component in stream 1 & 2?
b) Draw a control loop that regulates the rate of separation by controlling the level of the liquid in the column.
c) If the level in the column is too high, is the valve in the control loop open or closed?
d) Is there any safety hazard if valve V-102 fails closed?
a) Stream 1 is majorly methanol and Stream 2 is majorly glycerol.
Since methanol has a much lower boiling point than glycerol, methanol will mostly be in the vapour phase and glycerol will mostly be in the liquid phase when an appropriate temperature is selected for the distillation column.
b)
c) The valve in the control loop will be open. Because the valve is on the outlet of the liquid stream, opening the valve increases the flow of the liquid stream out of the column, decreasing the liquid level in the column.
d) There is no safety hazard, but it leads to ineffective separation.
When V-102 fails open all the liquid will be sent back into the column as vapour, which shifts the vapour-liquid equilibrium in the column.
Exercise: Nitric Acid Production
The following diagram shows the major process steps for the production of nitric acid.
Air and ammonia are first sent to a boiler where the reaction produces nitric oxide, water vapor, and heat. The process stream is then cooled through a heat exchanger, where nitric oxide oxidizes into nitric acid as temperature decreases. The first stream used in the first heat exchanger is also used as a utility stream to absorb excess heat in the boiler and convert it into energy when going through a turbine. The process stream is then sent to an absorber where it is mixed with a stream of water to produce liquid nitrate acid. The liquid acid is sent to further purification processes while the vapour produced in the absorber is sent to waste treatment.
a) What are the direct consequences if stream 5 is blocked? Analyze the effect on the process in the boiler and the first heat exchanger.
b) Using the description in the next paragraph, build a fault-tree analysis logic diagram on the risk of severe overpressure that could lead to an explosion in the absorber, and calculate the frequencies of all events in the diagram. Assuming all events with given frequencies are basic events.
Severe overpressure that could lead to an explosion happens when both system over maximum pressure and pressure relief valve fails happen at the same time. System over maximum pressure happens when the cooling water inlet is blocked (frequency = 0.015) or process vapour outlet stream is blocked (frequency = 0.008). Pressure relief valve fails happens when any one of the following circumstances happens: valve leaking (frequency=0.023), valve calibrated at a wrong pressure (frequency=0.005), or contaminants causing the valve to stick (frequency = 0.056).
a) The process stream in the heat exchanger will not be sufficiently cooled down, so the conversion to nitric acid will be lower, decreasing the concentration of the product in the absorber.
There will not be enough flow to absorb the heat in the boiler, which will lead to a temperature increase, pressure build-up and potential explosion of the gas in the boiler.
b)
Rana is a fourth-year chemical and biological engineering student at UBC. She takes interest in biomedical engineering, process control, machine learning, bioprocessing, and food processing. Outside of school and writing open-source textbooks, Rana spends most of her time leading the Biochemical Innovation Team (as part of the UBC Biomedical Engineering Student Design Team) and cooking. She hopes to pursue graduate school once she graduates in May 2021.
Rosie Qiao is a third-year student currently studying in the Chemical & Biological Engineering program at UBC. She takes interest in bioprocessing and medical-related technologies and materials. Outside of school, she is also interested in various forms of drawings, including watercolour, acrylics, and process flow diagrams. She is looking forward to gaining more experience and understanding of chemical and biological processing through further study at UBC.
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In the PDF files for this book, there are a few symbols that do not display properly and are replaced with a question mark (?). This is a known issue, and we are working on getting it fixed. Below is a list of these characters and their locations:
Term | Definition | Section |
“and” gate | The event only happens when all of the input events occur. When we compute the frequency of the event, we multiply the frequencies of all input events. Because all frequencies are smaller than 1, the resulting frequency will be smaller than any of the input frequencies. | Hazard Identification |
“or” gate | The event happens when any of the input events occur. When we compute the frequency of the event, we add the frequencies of all input events. The resulting frequency will be larger than any of the input event frequencies. | Hazard Identification |
Abiotic depletion potential (ADP) | Consumption of nonliving resources, such as fossil fuels, minerals, clay, and peat | Life Cycle Assessment |
Absolute pressure | The absolute pressure is the actual pressure at the point of interest. The absolute pressure is 0 in a vacuum and cannot be negative. | Pressure Definition; Absolute & Gauge Pressure |
Absolute temperature | Absolute temperature is temperature measured with reference to the absolute zero. | Temperature |
Accident | The occurrence of a sequence of events that produce unintended injury, death or property damage | An Introduction to Process Safety |
Acidification potential (AP) | The precursor compounds of acid rain. | Life Cycle Assessment |
Activation energy | The activation energy (Ea) represents the energy needed for the reactants to collide and get to a state such that their bonds can rearrange to form the products. | Arrhenius Equation |
Amgat’s Law | Amgat’s is analogous to Dalton’s law but is applied to the volume of the gas. The partial volume that each gas occupies will add up to the total system volume. | Ideal Gas Properties |
Antoine equation | log10(p∗)=A−B/(T+C), Used to estimate the vapour pressure of a substance at a certain temperature | Estimating Vapour Pressure |
Arrhenius equation | The Arrhenius Equation describes the relationship between the reaction rate and temperature. | Arrhenius Equation |
Basic design package | This document is provided to clients by engineers and may consist of the following: process definition, process control strategy, equipment design, selection and sizing, plant layout, safety review, environment assessment, economic analysis. | Project Scoping |
Basic engineering | The scale of the plant, the draft of the process flow diagram (PFD, we will introduce this later in the course), cost estimation, preliminary schedule, etc. | Project Scoping |
Basic event | A basic initiating fault (component failure) | Hazard Identification |
Bimolecular reaction | Two molecules collide, interact, and undergo some kind of change. It can be two of the same or different molecules. | Reaction Mechanisms – Elementary Reactions |
Binary mixture | A combination of two substances is called a binary mixture. | Multicomponent Equilibrium |
Block flow diagram (BFD) | Flowsheets that break down chemical and biological processes (or any other process) into units representing equipment (unit operations). | Block Flow Diagrams |
Boundary | The boundary is what separates the system from the surroundings. | Introduction to Energy Balances |
Boundary conditions | Boundary conditions are solutions to specific ordinary differential equations at specified conditions. | Introduction to Unsteady-state Operations |
Bubble point | The point (pressure or temperature) at which the first bubble (gas) in a multicomponent system appears. | Multicomponent Equilibrium |
Capillary connection | A fluid flow control connection. | Piping and Instrumentation Diagrams (P&IDs) |
Capital costs (CC) | Cost of designing and construction of the plant. | Basic Economic Analysis |
Catalysts | Catalysts speed up the reactions usually by decreasing the activation energy. Sometimes catalysts can affect frequency factors as well, but the effect of frequency factor is minor in comparison to the decrease of activation energy. | Arrhenius Equation |
Checklist analysis | Using a checklist to suggest common safety issues to assess. It can be combined with what-if technique to enhance that technique. | Hazard Identification |
Chemical engineering plant cost index | A common type of cost index that is used to estimate the overall plant cost (including installation, piping, etc.) over time. | Purchased Equipment (PE) Cost |
Closed systems | Energy, but no mass transfer between system and surroundings | Introduction to Energy Balances |
Consecutive elementary reactions | A set of reactions that proceed through the formation of one or more intermediate(s). | Reaction Mechanisms – Elementary Reactions |
Consumption | Mass and/or energy that is consumed in the system. | Introduction to Unsteady-state Operations |
Continuous stirred tank reactor (CSTR) | CSTRs are reactors with continuous feed and exit streams and some kind of mixer. | Separable Differential Equations |
Controlled variable | The output process variable we compare to the set-point. | Process Control |
Cost of manufacturing (COM) | The total cost associated with making the product. Divided into direct manufacturing costs, fixed manufacturing costs, and general expenses. | Costs of Manufacturing |
Cost of operating labour | The total cost of paying the salary of the operators. | Cost of Operating Labour |
Cradle-to-gate | The scope of life cycle assessment that includes all aspects of a product from raw material extraction to production, but does not include the product’s use or disposal. | Life Cycle Assessment |
Cradle-to-grave | The scope of life cycle assessment that includes all aspects of a product, starting from raw material extraction to the product’s end of life (use and disposal). | Life Cycle Assessment |
Critical point | In a phase diagram, we can observe that the boiling point curve (the line that distinguishes liquid and vapour phases) ends at a point, which is called a critical point. | Phase Diagram |
Dalton’s Law | The total pressure is the sum of partial pressures of the component gases, assuming ideal gas behavior and no chemical reactions between the components. | Ideal Gas Properties |
Degrees of freedom | The number of intensive variables that needs to be specified. | Gibb’s Phase Rule |
Depreciation | The expenses required to build a plant that are written off over time. Calculated based on the purchase value of equipment minus the salvage value and evaluated over a number of years specified. | Costs of Manufacturing |
Design basis memorandum (DBM) | This document contains key information and data related to the project deliverables. | Project Scoping |
Dew point | The point (pressure or temperature) at which the first dewdrop (liquid) in a multicomponent system appears. | Multicomponent Equilibrium |
Differential pressure cell | A differential pressure cell measures the deflection of the membrane and correlates it to change in pressure. | Process Control |
Direct costs in FCI | Costs related to tangible things or products such as process equipment. | Plant Capital Cost |
Direct manufacturing costs | Factors that vary directly with the rate of production. e.g. raw materials, plant operators | Costs of Manufacturing |
Disturbance variable | A variable that we have no control over in the process but affects the material or heat flow of the process. | Process Control |
Ecological toxicity potential (EcoTP) | The potential of biological, chemical or physical toxicity to ecosystems | Life Cycle Assessment |
Electrical connection | An electrical currents control connection. | Piping and Instrumentation Diagrams (P&IDs) |
Electrical temperature measure | Thermocouples or resistance temperature devices measure changes in electrical properties and correlate them to temperature. Electrical temperature measurements are the most commonly used in plants as switching combination of metals can adapt to a wide range of temperatures. | Process Control |
Elementary reactions | The stepwise changes are collectively called the reaction mechanism. Each step of the reaction is called an elementary reaction. The sum of the elementary reactions in the mechanism must give the balanced chemical equation for the overall reaction. | Reaction Mechanisms – Elementary Reactions |
Elimination | Remove the need for harmful materials or choosing other process pathways to eliminate the production of harmful products. | An Introduction to Process Safety |
End-of-life (EOL) | The EOL bar may be negative in life cycle assessment to signify recovery of the material at the end of the product’s life, which reduced the life-cycle impact. | Life Cycle Assessment |
Endothermic | If more energy is absorbed in breaking bonds than released in forming bonds, then the reaction is endothermic. | Reactive Energy Balances |
Endothermic control loop | A temperature control loop for a system that is absorbing thermal energy. | Process Control |
Environmental fate | Where substances will go when released into the environment. | Environmental Fate of Contaminants |
Equilibrium constants | The product of the activities of compounds at equilibrium raised to the stoichiometric coefficient. | Reaction Rate Law |
Eutrophication potential (EP) | The leak of nutrition sources into aquatic systems | Life Cycle Assessment |
Exothermic | If more energy is released in forming bonds than absorbed in breaking bonds, then the reaction is exothermic | Reactive Energy Balances |
Expansion-based temperature measure | Expansion-based thermometers measure fluid expansion and correlate it to temperature. | Process Control |
Extensive variables | Variables that depend on the size of a system. | Intensive & Extensive Variables |
Extent of hazards | The strength or magnitude of the hazard (e.g the area affected by a flood, the strength of an earthquake described by the Richter Scale) | Hazard Identification |
Failure modes and effect analysis (FMEA) | Tabulation of plant equipment, failure modes (how the equipment can fail), and the effects of these failures. Identify failure modes then analyze the result of these failures individually. | Hazard Identification |
FAR | Fatal Accident rate: Fatalities per 1000 employee over their working life (10E8 hours of exposure) | An Introduction to Process Safety |
Fatality rate | The fatality rate per person per year | An Introduction to Process Safety |
Fault-tree analysis | Identifies relevant events and potential failure pathways leading to one particular incident. Uses logic diagrams to express relations between initiating events and an incident. It can be implemented for PHA analysis at any stage of a process from design to the end of life | Hazard Identification |
First Law of Thermodynamics | The First Law of Thermodynamics is defined as: E(system,final) – E(system,initial) = E(system,transferred)When we expand the energy terms for a closed system, we get: ΔU + ΔEk + ΔEp = Q + W | Introduction to Energy Balances |
Fixed capital investment (FCI) | Fixed capital investment is the total price to purchase equipment, get it installed and tested. | Plant Capital Cost |
Fixed manufacturing costs | Factors not affected by the level of production, but related to manufacturing. e.g. taxes and insurance | Cost of Operating Labour |
Float measure | A device that measures the length of the line to float, which translates the line length to the liquid level in the tank. | Process Control |
Flow work | Flow work is work done on process fluid (inlet minus outlet). For the work flow in, the surroundings do work on the system, therefore it is positive. For the work flow out, the system does work on the surroundings, therefore it is negative. | Introduction to Energy Balances |
Formation reaction | A reaction in which the compound is formed from its elemental constituents as they would normally occur in nature. | Reactive Energy Balances |
Frequency factor | The frequency factor describes the frequency that molecules collide with the correct orientation to possibly make a chemical reaction. | Arrhenius Equation |
Functional unit | A measure of the function of the studied system. It provides a reference to which the inputs and outputs can be related and enables a comparison of two essentially different systems. | Multicomponent Equilibrium |
Gate-to-gate | The scope of life cycle assessment that includes the manufacturing/production and processing aspects of a product only. | Life Cycle Assessment |
Gauge pressure | Gauge pressure is defined to be the difference between the absolute pressure and atmospheric pressure. | Pressure Definition; Absolute & Gauge Pressure |
General expenses | Management and administrative expenses not related to manufacturing. | Plant Capital Cost |
Generation | Mass and/or energy that is produced in the system. | Introduction to Unsteady-state Operations |
Gibb’s Phase Rule | Gibb’s phase rule is used to determine the number of intensive variables required to fully specify the thermodynamic properties of a system (thermodynamic degrees of freedom). | Gibb’s Phase Rule |
Global warming potential (GWP) | Contribution to global warming caused by the release of gas. | Life Cycle Assessment |
Goal definition and scoping | Identify the LCA’s purpose, the products of the study, and determine the study boundaries. This can include things like what is included and what is not included in the study. | Life Cycle Assessment |
Green engineering | The design, commercialization, and use of processes and products in a way that reduces pollution, promotes sustainability and minimizes risks to human health and the environment without sacrificing economic viability and efficiency. | Green Engineering |
Gross economic potential (GEP) | GEP = Value of Products – Value of Feeds | Basic Economic Analysis |
Group contribution method | Taking common chemical groups from a molecule, we can use the interactions from these groups to predict the properties the molecule will have based on parameters regressed from known molecules. | Environmental Fate of Contaminants |
Half-life | The time it takes for half of the species to be consumed. | Integrated Rate Laws |
Hazard | A chemical or physical condition that has the potential to cause an accident (example of significant chemical plant hazard: flammable, explosive, reactive & toxic hazards). | Hazard Identification |
Heat | Heat is the energy flow due to temperature difference | Introduction to Energy Balances |
Heat capacity | Heat capacities are physical properties that describe how much heat is needed to increase the temperature of a compound for a unit temperature per unit mass of a compound | Phase Change and Heat Capacity |
Heat of formation | The heat of a reaction in which the compound is formed from its elemental constituents as they would normally occur in nature. The change in enthalpy of a reaction can be calculated by summing up the heats of formations of the reactants and the products multiplied by the reaction coefficients independently. | Reactive Energy Balances |
Heat of fusion | The enthalpy change for phase change from solid to liquid occurring at constant temperature and pressure. | Phase Change and Heat Capacity |
Heat of reaction | The stoichiometric enthalpy difference when reactants react completely to form products at a specified constant temperature and pressure. | Reactive Energy Balances |
Heat of vapourization | The enthalpy change for phase change from liquid to gas occurring at constant temperature and pressure. | Phase Change and Heat Capacity |
Heat transfer | Heat transfer is the movement of energy from one place or material to another as a result of a difference in temperature. | Introduction to Energy Balances |
Henry’s Law | pi=yi×P=xi×Hi(T) used to describe the concentration of a gas (or very volatile component) dissolved in a liquid. | Multicomponent Equilibrium |
Hess’s Law | The total enthalpy change of an overall reaction will be the sum of the enthalpies of the individual reactions that sum up to the total overall reactions. | Reactive Energy Balances |
Human toxicity potential (HTP) | A measure of the potential harm caused by chemicals released into the environment | Life Cycle Assessment |
Ideal Gas Law | Relates pressure (P), volume (V), temperature (T) and the number of moles (n) of an ideal gas species using the ideal gas constant (R): PV=nRT | Ideal Gas Properties |
Ideal mixture | An ideal mixture contains substances with similar molecular interactions (which usually means similar structures or functional groups). | Ideal Gas Properties |
Impact analysis | Assess the impacts on human health and the environment. | Life Cycle Assessment |
Incident | Loss of control of material of energy (e.g. leak in a pipe). | An Introduction to Process Safety |
Incompressible | The density does not change significantly with changes in pressure. | Ideal Gas Properties |
Indirect costs in FCI | Costs related to intangible items, these won’t stay at the end of building the plant, for example, supervision and project management expenses. | Plant Capital Cost |
Inherently safer design | what we don't have in the process we don't need to control."}">Permanently and inseparably reduce or eliminate process hazards -> what we don’t have in the process we don’t need to control. | An Introduction to Process Safety |
Input | Mass and/or energy that enters through the system boundaries. | Input-Output Diagrams |
Input-output diagram | These diagrams are considered to be the simplest of process flow sheets. In input-output diagrams, the entire process is represented by 1 block only. They usually only reference the main process streams in and out of a process. | Input-Output Diagrams |
Inside battery limit | Equipment and components that are associated with the main process feed streams, such as piping on process fluid streams. | Introduction to Chemical Processes and Process Diagrams |
Integrated rate law | Rate laws are differential equations that can be integrated to find how the concentrations of reactants and products change with time. | Integrated Rate Laws |
Intensive variables | Variables that do not depend on the size of a system. | Intensive & Extensive Variables |
Intermediate event | Occurs as a result of events at a lower level acting through logic gates. | Hazard Identification |
Intermediates | Species that are produced in one step and consumed in a subsequent step are called intermediates. | Reaction Mechanisms – Elementary Reactions |
Intermolecular forces | Intermolecular forces are forces that take place between individual molecules. | Ideal Gas Properties |
Internal energy | Internal energy can be described as all other energy present in a system, including motion, and molecular interaction. | Introduction to Energy Balances |
Isolated systems | No energy or mass transfer between system and surroundings, energy may change form within the system | Introduction to Energy Balances |
Isolation method | A method used for determining the rate constant in a reaction rate with multiple species. This method takes one of the species as a constant by adding a high concentration to the system and performing a concentration over time analysis on the species with a small concentration. | Reaction Order |
Kinetic control (of a set of reactions) | Controlling the products resulting from a reactant or reactants through reaction rates. | Kinetic & Thermodynamic Control |
Kinetic energy | The energy associated with motion, which can be described as translational or rotational energy. | Introduction to Energy Balances |
Lang Factor | Relates purchased equipment cost to total capital investment (TCI) and fixed capital investment(FCI). This factor represents the cost of building a major expansion onto an existing chemical plant, including other costs such as piping, control, etc. | Lang Factor and Return on Investment |
Life Cycle Assessment (LCA) | A technique used to quantify the environmental impact of a product from raw material acquisition through end of life disposition (cradle-to-grave). | Life Cycle Assessment |
Life cycle inventory | Quantifies the energy and raw material inputs and environmental releases associated with each life cycle phase. | Life Cycle Assessment |
Life of equipment | The amount of time the equipment is in use. | Costs of Manufacturing |
Linearize the equation (for reaction rate laws) | Plot log(r0) vs log([A0]), where the slope is the reaction order in A; and y-intercept equals to log(keff). | Reaction Order |
Loss prevention | Prevention of injury to people, damage to the environment, loss of equipment, inventory, production, reputation or economic loss. | An Introduction to Process Safety |
Manipulated variable | A variable that we can control in the process and directly affects the output of the process. | Process Control |
Manometer | a U-shaped tube containing some kind of fluid with known density, and one side is connected to the region of interest while the reference pressure is applied to the other. The difference in liquid level represents the applied pressure. | Process Control |
Marshall and Swift equipment cost index | A common type of cost index that is used to estimate the price of equipment over time. | Purchased Equipment (PE) Cost |
Method of initial rates | This method measures the concentration of a species over time (the species with the lower concentration) and takes the logarithm of the rate law to find the order of the reactant from the slope. | Reaction Order |
Mixer | Has multiple feed streams and mixes these streams into one output stream | Unit Operations and Material Balances |
Molar volume | The molar volume is the volume divided by the moles. | Ideal Gas Properties |
Molarity | The molar concentration is expressed in units of mole per (volume*time). | Definitions of Reaction Rate and Extent of Reactions |
Mole fraction | The mole fraction is the moles of the component over the total moles in the container. | Multicomponent Equilibrium |
Molecularity | The molecularity is the number of molecules or atoms coming together to react in an elementary reaction. | Reaction Mechanisms – Elementary Reactions |
Net economic potential (NEP) | NEP = GEP – Production Costs (utilities, labour, maintenance) | Basic Economic Analysis |
Non-scenario based methods | Look at a process in general; effectiveness of the output depends largely on the expertise of the PHA team members, such as checklist analysis. | Hazard Identification |
Nonseparable differential equation | Non-separable differential equations are differential equations where the variables cannot be isolated. These equations cannot be easily solved and require numerical or analytical methods that will be taught in future courses. | Unsteady-state Operations and Process Control |
Number of operating labour | The number of operators needed in the plant at a given time (per shift) to ensure it runs smoothly. | Cost of Operating Labour |
Obstruction meter | An obstruction meter measures the change in pressure and correlates it to flow. | Process Control |
Occupational cancer hazard (OCH) | Release of substances that can potentially cause cancer, such as certain chemicals, dust, radiation. | Life Cycle Assessment |
Occupational non-cancer hazard (OnCH) | Cause of potential diseases, injuries, or other health problems. | Life Cycle Assessment |
Octanol/water partition coefficient | The ratio of the concentration of a substance dissolved in n-octanol and in water. | Environmental Fate of Contaminants |
Open systems | Energy and mass transfer between systems and surroundings, typically use a dot with quantities that change over time in these open systems to denote the flow rate of energy or mass. | Introduction to Energy Balances |
Operating history | Process changes, failures, past incidents. | An Introduction to Process Safety |
OSHA | Occupational Safety & Health Administration | An Introduction to Process Safety |
Outcome / Consequence | The physical manifestation of an accident. | An Introduction to Process Safety |
Output | Mass and/or energy that exits through the system boundaries. | Input-Output Diagrams |
Outside battery limit | Equipment, components, utilities, and other facilities acting on the utilities or waste treatment (not included in inside battery limits) that are associated with the process, but not related to the current design scope. | Project Scoping |
Overall order | The overall order of a reaction is the sum of the orders of all substances involved. | Reaction Order |
Partial pressure | The pressure produced by one gaseous component if it occupies the whole system volume at the same temperature, commonly used for gasses. | Multicomponent Equilibrium |
Phase change | Phase changes take place when a compound or mixture of compounds undergoes a change in their state of matter (i.e. gas to liquid). | Reactive Energy Balances |
Phase diagram | Plots of pressure versus temperature showing the phase in each region provide considerable insight into the thermal properties of substances. | Phase Diagram |
Phase transitions | The phase of a substance and separation between molecules are related to the intermolecular forces in a substance. Therefore, a source of energy needs to be supplied when the substance transitions to a higher energy state to separate the molecules in the substance. | Phase Diagram |
Photochemical oxidation potential (POP) | Releasing chemicals that react in the sunlight and release harmful products into the environment | Life Cycle Assessment |
Physical properties | Physical properties are properties that can be measured without changing the molecular structure of the substance (volume, temperature, and pressure). | Phase Equilibrium |
Piping and instrumentation diagram (P&ID) | P&IDs are based on PFDs but include more specific information. P&IDs do not include: operating conditions (T, P) Stream flows Equipment locations Supports, structures, and foundations Pipe routings (lengths, fittings such as elbows, but connections above a certain size are generally shown). A P&ID often depicts one unit operation in thorough detail. | Piping and Instrumentation Diagrams (P&IDs) |
Pneumatic connection | An air or gas flow control connection. | Piping and Instrumentation Diagrams (P&IDs) |
Potential energy | Energy present due to position in a field, such as gravitational position or magnetic position. Usually, for chemical processes, we consider the potential energy change due to the gravitational position of the process equipment. | Introduction to Energy Balances |
Pre-equilibria | The pre-equilibrium approximation assumes that the reactants and intermediates of a multi-step reaction exist in dynamic equilibrium. | Steady-State Approximation |
Pressure | The amount of force exerted per area in a system. | Pressure Definition; Absolute & Gauge Pressure |
Process flow diagram (PFD) | More detailed chemical process flowsheets. Look at the section for a more detailed definition. | Process Flow Diagrams (PFDs) |
Process path | A real or hypothetical path that a system goes through from an initial to a final equilibrium state. Process paths usually progress in steps that change on state property at a time | Phase Change and Heat Capacity |
Process variable | The variable in the system or process that we desire to control. | Introduction to Unsteady-state Operations |
Purchased equipment cost | The purchased price of equipment from a vendor (someone selling the equipment). It is one of the major factors in the TCI direct costs. | Purchased Equipment (PE) Cost |
Purchased value | The cost of buying a piece of equipment. | Purchased Equipment (PE) Cost |
Pxy diagram | Multicomponent diagrams with the pressure on the y-axis and the liquid and vapour mole fractions on the x-axis. | Pxy Diagram |
Raoult’s Law | pi=yi×P=xi×pi(T) used when all components are in relatively significant quantities or they are chemically very similar | Multicomponent Equilibrium |
Rate constant | The rate constant is a constant in the rate law expression that independent of species’ concentrations, but generally dependent on temperature. | Integrated Rate Laws |
Rate-determining step | The rate-determining step is the step that determines the overall rate of reaction in a series of reactions. | Reaction Order |
Reaction mechanism | The sequence of events that occur at the molecular level during a reaction is the mechanism of the reaction. | Reaction Mechanisms – Elementary Reactions |
Reaction order in a substance | The dependence of the rate of reaction on the reactant concentrations can often be expressed as a direct proportionality, in which the concentrations may be raised to be the zeroth, first, or second power. The exponent is known as the order of the reaction with respect to that substance. | Reaction Order |
Reaction rate | A reaction rate shows the rates of production of a chemical species. It can also show the rate of consumption of a species; for example, a reactant. | Reaction Rate Law |
Reaction rate law | The relationship between the rate of reaction and the concentration of reactants. | Reaction Rate Law |
Reactor | A reactor typically consists of a vessel in which a reaction (or multiple reactions) takes place. | Unit Operations and Material Balances |
Recycling | Putting energy or materials back into use. | Concepts and Definitions |
Reduced properties | Reduced properties are defined as variables (pressure, temperature) scaled by their values at the critical point. Reduced properties can be useful in coming up with general equations describing different substances. | Ideal Gas Properties |
Reference state | A state of a substance at some pressure, temperature, and state of aggregation (solid, liquid, gas; pure, or mixture) used as a reference for calculation purposes. | Introduction to Energy Balances |
Results interpretation | Evaluate opportunities to reduce energy, material inputs, or environmental impacts at each stage of the product life-cycle. | Life Cycle Assessment |
Return on investments (ROI) | The ratio of net profit and total capital investment. ROI=NEP/CC*100% | Lang Factor and Return on Investment |
Revenue | The total profit we generated from selling the product without deducting the cost of production. | Lang Factor and Return on Investment |
Risk | The probability of an accident occurring and the consequence of it | An Introduction to Process Safety |
Rotational or turbine flowmeter | A rotational or turbine flowmeter measures the speed of the rotation of the turbine and correlates it to flow. | Process Control |
Safety | The strategy of accident prevention | An Introduction to Process Safety |
Safety triangle | The safety triangle shows three essential aspects to achieve workplace safety: Technical – characterization and control of hazards Management – policies and procedures for safety Cultural – motivating people to do things safely | An Introduction to Process Safety |
Salvage value | The value we can recover from a piece of equipment’s end of life. | Costs of Manufacturing |
Saturated liquid | Liquid that is about to vapourize. It will vapourize with any little amount of decrease in pressure or increase in temperature. | Txy Diagram |
Saturated vapour | Vapour that is about to condense. It will condense to a liquid with any little amount of increase in pressure or decrease in temperature. | Txy Diagram |
Scenario | A sequence of events causing the hazard to result in an accident | Hazard Identification |
Scenario-based methods | Used to predict accident scenarios; a lot more detailed and specific, such as what-if analysis | Hazard Identification |
Separable differential equation | Separable differential equations are differential equations where the variables can be isolated to one side of the equation. | Separable Differential Equations |
Separator | A separator is similar to a splitter but has the ability to split streams into different exiting compositions | Unit Operations and Material Balances |
Set-point | This is a target or desired value of a variable in the system (such as flowrate, temperature, etc.) | Process Control |
Shaft work | Shaft work is work done on process fluid by a moving part, such as a pump, rotor, or a stirrer. | Introduction to Energy Balances |
Solubility | The ability of a substance to dissolve. | Environmental Fate of Contaminants |
Sonic / Radar | A sonic or radar device measures the level of the tank by measuring the time for a wave to reach the liquid-gas interphase and reflect back to the device and correlating the time to distance. The sonic or radar device can be installed from the top or bottom of the tank. | Process Control |
Specific property | This denotes an intensive property obtained by dividing an extensive property by a total amount of flow rate (can be the total number of moles or total mass). | Introduction to Energy Balances |
Splitter | A splitter will split a feed stream into multiple streams with the same composition. The exiting streams are not necessarily equal in terms of flow rates, but will always have the same composition | Unit Operations and Material Balances |
Standard specific heat of formation | The enthalpy change associated with forming 1 mole of the compound of interest at standard temperature 25°C and pressure (1 atm). | Reactive Energy Balances |
Steady-state approximation | This approximation assumes that the intermediate concentration is very low so that it does not change with time. | Steady-State Approximation |
Steady-state open system energy balance | An energy balance performed on a steady-state (unchanging) open system (mass and energy are transferred between the system and the surroundings). Q̇ + Ẇ = Σout(Ėj) – Σin(Ėj) | Introduction to Energy Balances |
Steam table | Detailed information on water’s state properties at different temperatures and pressures available in a tabulated form | Introduction to Energy Balances |
Stoichiometric coefficient | Stoichiometric coefficients are the numerical coefficients associated with each reactant or product. | Definitions of Reaction Rate and Extent of Reactions |
Substitution | Using less harmful materials, or optimizing the process to reduce the formation rate of harmful by-products. | Concepts and Definitions |
Supercritical fluid | Above the critical point, liquid and gas can no longer be distinguished and are classified as a supercritical fluid. | Phase Diagram |
Surroundings / Environment | The environment that is ‘outside’ of the system’s boundaries. | Introduction to Energy Balances |
Temperature | A measure of the average kinetic energy of a system. | Temperature |
Thermodynamic control (of a set of reactions) | Controlling the products resulting from a reactant or reactants through reaction equilibrium | Kinetic & Thermodynamic Control |
Thermodynamic system | A thermodynamic system includes anything whose thermodynamic properties are of interest. It is embedded in its surroundings or environment; it can exchange heat with and do work on its environment through a boundary. | Kinetic & Thermodynamic Control |
Total capital investment (TCI) | The amount of money (aka. capital) needed for a project to proceed. TCI is the total amount of funds needed to build and start a chemical plant, including the testing and troubleshooting stages before the actual production. It does not include operating expenses or maintenance during the production process. | Basic Economic Analysis |
Transient balances | Transient balances are any balances that where some quantities involved in the balance are functions with time. | Introduction to Unsteady-state Operations |
Triple point | The triple point is defined to be the set of pressure and temperature that solid, liquid and vapour are in equilibrium together. | Phase Diagram |
Txy diagram | Multicomponent diagrams with the temperature on the y-axis and the liquid and vapour mole fractions on the x-axis. | Txy Diagram |
Undeveloped event | Undeveloped event due to lack of information or significance. Treated like a basic event in the analysis, but could have been developed into another fault tree. | Process Safety |
Unimolecular reaction | Only one molecule is involved, examples might include an isomerization or decomposition | Reaction Mechanisms – Elementary Reactions |
Unit operation | Major chemical and biological processing equipment. They are usually drawn as circles or squares in BFDs. | Unit Operations and Material Balances |
Unsteady-state | In unsteady-state processes, mass and energy in the system are changing with time (not constant). | Introduction to Unsteady-state Operations |
Utilities | Utilities are services at a site such as water, electricity, and gas. | Phase Change and Heat Capacity |
Vapour pressure | Vapour pressure is defined as the pressure at which a gas coexists with its solid or liquid phase. | Estimating Vapour Pressure |
Vapour temperature | Vapour temperature is defined as the temperature at which a gas coexists with its solid or liquid phase. | Estimating Vapour Pressure |
Vapour-liquid equilibrium | The same number of molecules vapourizing and condensing at any given period of time (meaning no net change in the system). | Multicomponent Equilibrium |
Volatility | How likely a compound will evaporate. | Multicomponent Equilibrium |
Volume | The space occupied by a system. | Ideal Gas Properties |
Waste management hierarchy | A hierarchy describing actions for managing waste that goes from most desirable to least desirable. | Concepts and Definitions |
What-if analysis | What-if analysis is a scenario PHA method as it investigates specific problems and traces what would happen if they occur. Results in a list of potential consequences and suggestions of mitigation methods (if needed). | Hazard Identification |
Work | Work is energy resulting from driving forces (not temperature) such as force, torque, or voltage. | Introduction to Energy Balances |
Working capital (WC) | The initial money required to start the plant. It is the money needed for supplies or people that is not recovered until after the plant closes. Since this is generally a long-time in the future, it is considered a capital investment rather than a manufacturing cost. | Plant Capital Cost |
Zeroth/first/second-order rate law | The overall reaction orders for rate laws. Each of these rate laws has distinct graphical characteristics. | Reaction Order |
Step 1: Click “file” in the header and open the “Options” menu:
Step 2: In the Excel Options menu, select “Solver Add-in” in the “Add-ins” section, then click “OK”.
Step 3: Then you can find the “Solver” in the “Analyze” section under the “Data” menu.
Sample Question
A heat exchanger uses superheated steam to heat up a stream of pure acetic acid.
The following data can be used for calculating the enthalpy change of the acetic acid stream:
$$C_{p}=A+BT+CT^2+DT^3$$
where T is in K and is given in J/mol K.
A | B | C | D | |
acetic acid |
The enthalpy change for superheated steam can be found using data from the steam table:
300°C, 1 MPa | 250°C, 1 MPa | |
(J/mol) | 54990 | 53035 |
The boiling point for acetic acid at 1 atm is 117.9°C. Does the acetic acid boil in the heat exchanger? If not, calculate the final temperature of the acetic acid stream. Assume there is no energy loss due to heat transfer.
Because there are no other forms of energy transfer other than THE heat transfer between the streams, the energy balance simplifies to:
$$\Delta\dot{H}_{water}+\Delta\dot{H}_{acetic \;acid}=0$$
Based on the type of data given to calculate enthalpy change, the term in the energy balance can be expanded to:
$$\Delta\dot{H}_{water}=\dot{H}_f-\dot{H}_i$$
\begin{align*}
\Delta\dot{H}_{acetic \;acid}&=\int_{T_i}^{T_f}C_pdT\\
&=\int_{T_i}^{T_f}A+BT+CT^2+DT^3 dT\\
&=AT+\frac{1}{2}BT^2+\frac{1}{3}CT^3+\frac{1}{4}DT^4\;\Bigg|_{T_i}^{T_f}\\
&= A(T_f-T_i)+\frac{1}{2}B(T_f-T_i)^2+\frac{1}{3}C(T_f-T_i)^3+\frac{1}{4}D(T_f-T_i)^4
\end{align*}
Step 1: Input given information that will be used in the calculation. This includes the given data to calculate enthalpy change, the molar flow rate of both substances, and the initial temperature of acetic acid.
Step 2: Calculate the molar enthalpy change. For water, this is calculated by subtracting the molar enthalpy values given at the final and initial temperatures. For acetic acid, we need to provide a guess of the final temperature to substitute into the equation.
Step 3: Calculate the total enthalpy change for both substances by multiplying the molar flow rate by the molar enthalpy change.
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Step 4: Use a cell to represent the sum of enthalpy change. If we have guessed the correct final temperature, the value of this cell should be 0. We will use solver to let excel perform the guess-and-check process for us.
Step 5: Select “Solver” from the “Data” menu in the header, and indicate that we want to solve for the final temperature of acetic acid that makes our :
Step 6: Click “OK” if solver finds a solution. If the solver cannot find a solution, it is maybe because our initial guess is too far away from the actual final temperature, if so, we can change the temperature and try to solve it again.
Step 7: After the window closes, the value of the final temperature of acetic acid and all cells affected by temperature are changed to the corresponding values calculated with the actual final temperature.
Therefore, the final temperature of the acetic acid stream is , which is equal to . This temperature is lower than the boiling temperature of acetic acid at 1 atm, so the stream will not start to boil.
Step 1: After you open a new notebook, click the “File” session in the header. Note that you will only be able to turn on iterations when there is a notebook open.
Step 2: Select the “Options” appearing at the bottom of the menu.
Step 3: Check the “Enable Iterative Calculations” in the “Formulas” session and click “OK”. This will allow excel to perform looped calculations when every variable is dependent on previous variables in the loop, for example, calculating flowrates when a recycle stream is present.
Sample Question
Grain extract is produced by the process below. Ethanol is synthesized by yeast in the reactor. In this idealistic process, the yeast converts 2kg of grain into 1 kg of ethanol and 1 kg of water. A perfectly efficient yeast reactor (efficiency = 1) would convert all the grain entering the reactor. A reactor with efficiency =0.5 would convert half of the grain entering the reactor and so on. Below are some additional specifications:
• The feed is 100 kg/min, 20 wt% grain, 80 wt% water
• The reactor efficiency is 0.25
• The grain in the slurry retains 1kg of ethanol-water solution for every 10kg of grain
• 5 wt% of the slurry is purged.
• The mass ratio of water and ethanol in the slurry is the same as in the reactor effluent
Step 1: Set up the spreadsheet using the given process specifications and components in each stream as indicated in the question.
Step 2: Express for the individual component flowrate and total flowrate in each stream. For feed:
Step 3: Reactor Feed:
Note: If we reference an empty cell (like the flowrates in the recycle stream), the value will be calculated as 0. This will be updated after we input the formulas for the flow rate in the recycle stream.
Step 4: Reactor effluent:
Note: The 1/2 in the flowrate calculations for water and ethanol comes from the mass balance for the conversion of grain. 1/2 is the ratio of and
Step 5: Slurry:
Step 6: Grain extract:
Step 7: Purge:
Step 8: Recycle:
Because the recycle stream is the last step in the loop, we can see the other values being updated as we input the formulas for flow rates in the recycle stream.
Now the flow rates of all streams and all components in each stream are fully solved.
Finding this data can be very challenging. Companies typically want to keep pricing information secret (for competition reasons). The best way I have found to search for chemical pricing data is as follows:
This process is tedious and often takes some time, even for people experienced in this process. So don’t get discouraged when looking for this data and feel free to ask questions (through the discussion board, in tutorials or office hours, etc.).
A. Select ‘Water’ from the ‘Please select the species of interest:’ drop-down menu
B. Choose the steam table units you’d like to work with, in step 2.
C. In Step 3, choose what kind of data you’re looking to obtain. For an isothermal system, select ‘Isothermal properties’. For a constant pressure system, select ‘Isobaric properties’.
D. Select the desired standard state convention. This course will most likely only use the ‘Default for fluid’ convention.
EPI Suite – Estimation Program Interface Suite is a program we can use to get parameters to assess the environmental fate of substances. It is only available to be run on windows and can be downloaded from https://www.epa.gov/tsca-screening-tools/epi-suitetm-estimation-program-interface.
Steps to find Chemical Properties Using EPI Suite
1. Input the CAS number of the molecule of interest in the “Input CAS #” field and click “Search CAS”. EPI Suite will find the structure based on the CAS Number. To find the CAS number, we can search for the compound on NIST, and the CAS number can be found on the general session on top of the page for the substance.
2. Click “calculate” to calculate the physical properties.
3. The calculated properties will be shown in the pop-up window or on the bottom of the EPI suite window. We can scroll between tabs to see subsets of results.
Here is a thorough table of control symbology:
First Letter (X) | Second or Third Letter (Y) | |
---|---|---|
A | Analysis | Alarm |
B | Burner Flame | – |
C | Conductivity | Control |
D | Density or Specific Gravity | – |
E | Voltage | Element |
F | Flowrate | – |
H | Hand (Manually Initiated) | High |
I | Current | Indicate |
J | Power | – |
K | Time or Time Schedule | Control Station |
L | Level | Light or Low |
M | Moisture or Humidity | Middle or Intermediate |
O | – | Orifice |
P | Pressure or Vacuum | Point |
Q | Quantity or Event | – |
R | Radioactivity or Ratio | Record or Print |
S | Speed or Frequency | Switch |
T | Temperature | Transmit |
V | Viscosity | Valve, Damper, or Louver |
W | Weight | Well |
Y | – | Relay or Compute |
Z | Position | Drive |
Below are a variety of scholarly societies relevant for CHBE and CHML students. All of these societies provide relevant opportunities for undergraduate students and I try to provide a short list of key opportunities below.
Note that I typically call all students in our department CHBE students, as this reflects the name of our department. I know this can be confusing with the two programs (CHBE and CHML). Rest assured that everything in this course is relevant to both programs.
Joining AIChE is highly recommended:
Joining CSChE is recommended:
IChemE might be relevant to you depending on where you are in the future:
The notebooks are initially created using Jupyter notebook through https://ubc.syzygy.ca. The cells in Jupiter notebook are written in markdown. The advantage of this platform is the ability to create books using multiple cells and run the cells separately, which can be good when revisiting the notebooks for editing, but may cause formatting inconsistency when converting on pressbook.
If the book is written in Jupyter notebook (HTML/Python) and will upload to Pressbooks as an HTML file, here are some useful guides to coding things in HTML so that it will translate to Pressbooks nicely: Writing Markdown Cells in Jupypter Notebooks.
The notebooks are uploaded to GitHub, then reviewed and edited by all members of the group before being uploaded to pressbook.
Lecture notebooks are uploaded to pressbook in HTML format. Usually some formatting needs to be revisited after uploading, and text boxes are added to distinguish sections of learning objectives, examples, and exercises. The notebooks can also be originally created on pressbook to avoid conversion issues.
H5P is a set of programs that can be added to the chapters in pressbook. In our book, the flashcards in the unit review sessions are created using H5P. H5P is also capable of creating various format of practice questions and forms of data display.
Here are some resources that can be helpful for the process of creating a pressbook:
Some useful resources from BC campus for general questions about pressbook: