# 35 Phase Change and Heat Capacity

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

## Energy Balance Equipment

- Almost all energy exchange equipment is either a heat exchanger or jacket
- Heat exchangers and jackets often work by transferring heat with a stream of fluid
- Electrical heating elements are also available, but their high cost per unit energy may limit their use in industry

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 [latex]T (^{\circ}C)[/latex] |
Outlet [latex]T (^{\circ}C)[/latex] |
P ([latex]psia[/latex]) |
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

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 ([latex]\Delta\hat{H}_{vap}[/latex]) |

Solid → Liquid | Heat of Fusion ([latex]\Delta\hat{H}_{fus}[/latex]) |

Gas → Liquid | (-) Heat of Vapourization ([latex]-\Delta\hat{H}_{vap}[/latex]) |

Liquid → Solid | (-) Heat of Fusion ([latex]-\Delta\hat{H}_{fus}[/latex]) |

*Note that heat of vapourization is also referred to as latent heat*

## Heat Capacities

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:

[latex]Q = \Delta\hat{U}[/latex]

The heat capacity at constant volume can be described as follows, where [latex]C_{V}[/latex] is a function of temperature [latex]T[/latex]:

[latex]C_{V}(T) = \bigg(\frac{\delta\hat{U}}{\delta T}\bigg)_{V}[/latex]

Notation: [latex]\big(\;\;\big)_{V}[/latex] means that the volume is kept constant during the process

This expression can be rearranged and integrated to obtain the following:

[latex]d\hat{U} = C_{V}(T)dT[/latex]
[latex]\Delta\hat{U} = \int^{T_{2}}_{T_{1}}C_{V}dT[/latex] |

If the [latex]C_{V}[/latex] is constant:

[latex]\Delta\hat{U} = C_{V}*(T_{2}-T_{1})[/latex]

**For an open system**, the heat capacity is defined under constant pressure conditions:

[latex]C_{P}(T) = \bigg(\frac{\delta\hat{H}}{\delta T}\bigg)_{P}[/latex] [latex]\Delta\hat{H} = \int^{T_{2}}_{T_{1}}C_{P}dT[/latex] |

If the [latex]C_{P}[/latex] is constant:

[latex]\Delta\hat{H} = C_{P}*(T_{2}-T_{1})[/latex]

Example: Phase Changes and Heat Capacity in Energy Balances

Compound |
Formula |
Boiling Point at 1 atm ([latex]^{\circ}C[/latex]) |
Heat of Vapourization at 1 atm ([latex]kJ/mol[/latex]) |

EDC | [latex]C_{2}H_{4}Cl_{2}[/latex] | 84 | 35 |

HCl | [latex]HCl[/latex] | -85 | 16 |

VC | [latex]C_{2}H_{3}Cl[/latex] | -13 | 21 |

The mixture exits a reactor at [latex]200^{\circ}C[/latex] and has the following properties:

Compound |
Formula |
Flow to Separate (tonne/hr) |
Flow to Separate (tonne-mol/hr) |
[latex]y_{i}[/latex] (gas mole fraction) |

EDC | [latex]C_{2}H_{4}Cl_{2}[/latex] | 47.5008 | 0.48 | 0.25 |

HCl | [latex]HCl[/latex] | 26.2512 | 0.72 | 0.375 |

VC | [latex]C_{2}H_{3}Cl[/latex] | 45 | 0.72 | 0.375 |

Recall

Raoult’s Law:[latex]x_{i} = \frac{p_{i}*}{y_{i}×P }[/latex]

CompoundFormulaFlow to Separate (tonne/hr)Flow to Separate (tonne-mol/hr)[latex]y_{i}[/latex] (gas mole fraction)[latex]x_{i}[/latex] (liquid mole fraction)EDC [latex]C_{2}H_{4}Cl_{2}[/latex] 47.5008 0.48 0.25 0.94 HCl [latex]HCl[/latex] 26.2512 0.72 0.375 0.005 VC [latex]C_{2}H_{3}Cl[/latex] 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 [latex]46.4^{\circ}C[/latex], therefore we must use a heat exchanger to bring the mixture from [latex]200^{\circ}C[/latex] to [latex]46.4^{\circ}C[/latex]

*How much energy is removed to cool this stream from from [latex]200^{\circ}C[/latex] to [latex]46.4^{\circ}C[/latex]?*

*ideal gas*heat capacities for the mixture described above are listed below:

Compound |
Formula |
Flow to Separate (tonne/hr) |
Flow to Separate (tonne-mol/hr) |
Cp (J/mol-K) |

EDC | [latex]C_{2}H_{4}Cl_{2}[/latex] | 47.5008 | 0.48 | 29 |

HCl | [latex]HCl[/latex] | 26.2512 | 0.72 | 29 |

VC | [latex]C_{2}H_{3}Cl[/latex] | 45 | 0.72 | 29 |

- EDC:

[latex]\Delta\dot{H} = \dot{n}\int^{T_{2}}_{T_{1}} C_{P}(T)dT[/latex]

[latex]\Delta\dot{H} = 0.48\frac{tonne-mol}{h}*1,000,000\frac{mol}{tonne-mol}*29\frac{J}{mol-K}*(200^{\circ}C-46.4^{\circ})*\frac{1K}{1^{\circ}}[/latex]

[latex]\Delta\dot{H} = 2.14 x 10^{9} \frac{J}{h}[/latex]

- HCl and VC:

[latex]\Delta\dot{H} = \dot{n}\int^{T_{2}}_{T_{1}} C_{P}(T)dT[/latex]

[latex]\Delta\dot{H} = 0.72\frac{tonne-mol}{h}*1,000,000\frac{mol}{tonne-mol}*29\frac{J}{mol-K}*(200^{\circ}C-46.4^{\circ})*\frac{1K}{1^{\circ}}[/latex]

[latex]\Delta\dot{H} = 3.21 x 10^{9} \frac{J}{h}[/latex]

There is no exchange of energy in the form of work in a heat exchanger. Therefore, the total heat removed [latex]\dot{Q}=\Sigma\Delta\dot{H}[/latex] is the sum of the change in enthalpy for each species:

[latex]\dot{Q}=\Sigma\Delta\dot{H}=8.56 x 10^{9} \frac{J}{h}[/latex]

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 [latex]150 ^{\circ}C[/latex] and needs to be cooled to [latex]85^{\circ}C[/latex] 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:

[latex]C_{P} = A + BT + CT^{2} + DT^{3}[/latex]

where [latex]C_{P}[/latex] 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?

### Solution

**Step 1: **Calculate the change in enthalpy for each compound using the heat capacities taking 1 kmol/hr as the molar flow.

[latex]\Delta\dot{H} = \dot{n}\int^{T_{2}}_{T_{1}} C_{P}(T)dT[/latex]

where [latex]T_{1} = 150 ^{\circ}C=423K[/latex] and [latex]T_{2} = 85 ^{\circ}C=358 K[/latex]

[latex]\Delta\dot{H} = \dot{n}\int^{T_{2}}_{T_{1}} (A + BT + CT^{2} + DT^{3})dT[/latex]

[latex]\Delta\dot{H} = \dot{n}*(AT + \frac{B}{2}T^{2} + \frac{C}{3}T^{3} + \frac{D}{4}T^{4}[/latex]

For n-hexane:

[latex]\Delta\dot{H} = 0.5\frac{kmol}{h}*(-4.413*T + \frac{0.528}{2}T^{2} + \frac{-3.119E-04}{3}T^{3} + \frac{6.494E-8}{4}T^{4})\bigg|^{423 K}_{358 K}[/latex]

[latex]\Delta\dot{H} = 0.5\frac{kmol}{h}*1000 \frac{mol}{kmol}*1859\frac{J}{mol}*\frac{kJ}{1000 J}[/latex]

[latex]\Delta\dot{H} = 929.5 \frac{kJ}{h}[/latex]

Similarly for n-heptane:

[latex]\Delta\dot{H} = 0.5\frac{kmol}{h}*1000 \frac{mol}{kmol}*2424\frac{J}{mol}*\frac{kJ}{1000 J}[/latex]

[latex]\Delta\dot{H} = 1212 \frac{kJ}{h}[/latex]

**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*}

### Choosing Utilities for System

Option |
Utility |
Inlet T ([latex]^{\circ}C[/latex]) |
Outlet T ([latex]^{\circ}C[/latex]) |
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*}

## Process Paths

Recall that [latex]\hat{U}[/latex] and [latex]\hat{H}[/latex] 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:

[latex]\Delta\hat{H} = \Sigma_{i}\Delta\hat{H}_{i}[/latex] |

Example: EDC Process Path

Consider the enthalpy change of [latex]EDC[/latex] transforming from vapour at [latex]200^{\circ}C[/latex] and 3 atm to liquid at [latex]25^{\circ}C[/latex] and 1 atm:

In this process, there are 3 variables changing:

- pressure
- temperature
- phase (vapour to liquid)

The following steps are taken in the process path and added together to get the overall change:

[latex]\Delta\hat{H}_{1}[/latex] is the enthalpy change from going from 3 atm to 1 atm at a constant temperature of 200°C in the vapour phase ([latex]v[/latex])

[latex]\Delta\hat{H}_{2}[/latex] is the enthalpy change from going from 200°C to 84°C at a constant pressure of 1 atm in the vapour phase ([latex]v[/latex])

[latex]\Delta\hat{H}_{3}[/latex] is the enthalpy change for a phase change, going from the vapour phase ([latex]v[/latex]) to the liquid phase ([latex]l[/latex]) at a constant pressure and temperature of 1 atm and 84°C

[latex]\Delta\hat{H}_{4}[/latex] is the enthalpy change from going from 84°C to 25°C at a constant pressure of 1 atm in the liquid phase ([latex]l[/latex])

When we add these changes together, we can get the overall change in enthalpy:

[latex]\Delta\hat{H} = \Delta\hat{H}_{1} + \Delta\hat{H}_{2} + \Delta\hat{H}_{3} + \Delta\hat{H}_{4}[/latex]

Exercise: Process Path

What process paths can be taken to calculate the change in enthalpy for acetone going from [latex]25^{\circ}C[/latex] in the liquid phase to [latex]60^{\circ}C[/latex] in the vapour phase?

### Solution

**Step 1:** Bring the acetone to the boiling point temperature from [latex]25^{\circ}C[/latex] without changing the phase using the [latex]C_{P}[/latex] 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 [latex]60^{\circ}C[/latex] from the boiling point temperature using the [latex]C_{P}[/latex] in the vapour phase.

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