Cost Estimation: 23-Chem-A5 Chemical Plant Design and Economics, May 2014, Q2

Question 2: Cost Estimation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Target exchanger area $A_2 = 800~\text{ft}^2$. A geometrically similar exchanger of one-quarter that area, $A_1 = 800/4 = 200~\text{ft}^2$, costs $\$4000$ (both in 2014, so no cost-index escalation is needed). Cost–capacity exponent $n = 0.60$ over 100–400 ft² and $n = 0.81$ over 400–2000 ft².

Find. The 2014 purchase cost of the 800 ft² exchanger.

Approach. Apply the power-law ("six-tenths") cost–capacity rule, but because the reference (200 ft²) and the target (800 ft²) sit on opposite sides of the 400 ft² exponent break, scale in two steps that meet at 400 ft², using the correct exponent in each range.

State the cost–capacity rule. For similar equipment the purchase cost scales with size raised to a capacity exponent: $$C_2 = C_1\left(\frac{A_2}{A_1}\right)^{n}$$ where $C_1$ is the known cost at area $A_1$ and $n$ is the exponent valid over that size range. Here the reference size is $A_1 = 200~\text{ft}^2$ at $C_1 = \$4000$.
Scale 200 → 400 ft² at $n = 0.60$. This step stays entirely inside the 100–400 ft² band, so use $n = 0.60$: $$C_{400} = \$4000\left(\frac{400}{200}\right)^{0.60} = \$4000\,(2)^{0.60} = \$4000(1.5157) = \boxed{\$6063}$$
Scale 400 → 800 ft² at $n = 0.81$. This step lies inside the 400–2000 ft² band, so switch to $n = 0.81$ and start from the just-computed $C_{400}$: $$C_{800} = \$6063\left(\frac{800}{400}\right)^{0.81} = \$6063\,(2)^{0.81} = \$6063(1.7532) = \boxed{\$10{,}630}$$ The purchase cost of the 800 ft² exchanger in 2014 is therefore about $\$10{,}630$.

Quantity	Value
Reference area / cost	200 ft² / $\$4000$
Cost at 400 ft² (via $n=0.60$)	$\$6063$
Cost at 800 ft² (via $n=0.81$)	$\$10{,}630$

Check: the two-step bridge is essential. Applying the single exponent $n=0.81$ blindly over the full four-fold jump, $\$4000\,(800/200)^{0.81}=\$4000(4)^{0.81}=\$12{,}295$, over-states the cost by about 16 % because it wrongly charges the steeper large-area exponent to the 200–400 ft² portion, where the shallower $n=0.60$ actually applies.

Concept & Theory

Topic. Power-law (six-tenths) cost–capacity scaling with a size-dependent exponent.

Key relations. $C_2 = C_1 (A_2/A_1)^{n}$; the "six-tenths rule" is the special case $n\approx0.6$. When the equipment's cost data are known to change slope, the exponent is applied piecewise across its valid size range. Cross-year comparisons would add a cost-index ratio $C_2 = C_1 (A_2/A_1)^{n}\,(I_2/I_1)$ using CEPCI, but here both costs are 2014 so $I_2/I_1 = 1$.

Why this works. Economies of scale mean cost rises less than linearly with capacity; the exponent captures the degree of that scale economy for a given equipment type. Because the exponent itself differs between 100–400 and 400–2000 ft², integrating "in one shot" would mis-price the first decade of area — hence the bridge at 400 ft².

Common pitfall. Using the reference area as 800 ft² instead of recognising that "one quarter the heating surface" is the known-cost unit (200 ft²); and applying one exponent across the break point instead of scaling piecewise.

Source. Peters, Timmerhaus & West, Ch. 6 "Analysis of Cost Estimation" (cost–capacity exponents and the six-tenths rule).

Question 2 of 6: Cost Estimation

Question 2: Cost Estimation (20 marks)