# 46 Purchased Equipment (PE) Cost

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**.

## Effect of Capacity

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:

[latex]\frac{C_{a}}{C_{b}}=\Big(\frac{A_{a}}{A_{b}}\Big)^n[/latex]

Where

[latex]C_{a}[/latex] – desired equipment cost

[latex]C_{b}[/latex] – base cost for known equipment

[latex]A_{a}[/latex] – desired capacity

[latex]A_{b}[/latex] – base capacity

[latex]n[/latex] – 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):

[latex]ln(C_{a})=ln(K)+n ln(A_{a})[/latex]

Where

[latex]K=\frac{C_{b}}{A_{b}^n}[/latex]

If you are curious, we show how this equation is derived below:

Since [latex]C_{b}[/latex] and [latex]A_{b}[/latex] are known values for equipment, we can treat them as constants and take use [latex]K[/latex] to represent [latex]\frac{C_{b}}{A_{b}^n}[/latex], 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 | [latex]kW[/latex] | 0.85 |

Shell and tube heat exchanger stainless steel | 1.9 – 1860 | [latex]m^2[/latex] | 0.60 |

Horizontal tank carbon steel | 0.5-74 | [latex]m^3[/latex] | 0.30 |

centrifugal pump stainless steel | 1-70 | [latex]hp[/latex] | 0.67 |

Crystalizer | 0.2-3.8 | [latex]m^3[/latex] | 0.47 |

Note: heat exchanger capacities are measured by the area of heat exchange, thus the unit is [latex]m^2[/latex].

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.[latex]^{[1]}[/latex]

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 [latex]m^3[/latex] crystalizer costs $35,000, how much does a piece of similar equipment with a capacity of 3.0 [latex]m^3[/latex]? Compare the results using the given cost exponent and the six-tenth rule.

### Solution

**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).

## Effect of Time

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.

[latex]C_{2}=C_{1}\Big(\frac{I_{2}}{I_{1}}\Big)[/latex]

where

[latex]C_{1}[/latex] – known cost of equipment/plant at a known (past) time

[latex]C_{2}[/latex] – cost of equipment/plant at a time of interest

[latex]I_{1}[/latex] – the cost index at a known time

[latex]I_{2}[/latex] – 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.[latex]^{[1]}[/latex]

**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?

### Solution

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?

### Solution

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?

### Solution

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*}

## References

[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].

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