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23-Chem-A5 Chemical Plant Design and Economics · December 2019

Question 4 of 6: Most Economical Particulate-Control Device by EUAR Analysis

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams / EGBC — December 2019 — 16-Chem-A5 Chemical Plant Design and Economics. Three-hour closed-book examination; one aid sheet (both sides) and an approved Sharp/Casio calculator are permitted. Six questions are printed and any five constitute a complete paper (each worth 20 marks); all six are solved below for completeness. The two calculation questions (Q3, Q4) are worked with explicit engineering-economy factors; the four discussion questions (Q1, Q2, Q5, Q6) are answered as structured lists with supporting description, as the paper directs.

Reference texts: M. S. Peters, K. D. Timmerhaus & R. E. West, Plant Design and Economics for Chemical Engineers (5th ed., McGraw-Hill) — profitability measures (rate of return, incremental analysis), straight-line depreciation, after-tax cash flow, and the anatomy of a process/economic study; R. Turton, R. C. Bailie, W. B. Whiting & J. A. Shaeiwitz, Analysis, Synthesis, and Design of Chemical Processes (4th ed., Prentice Hall) — the process flow diagram and its information content, equipment/economics; G. Towler & R. Sinnott, Chemical Engineering Design (Coulson & Richardson Vol. 6, 2nd ed.) — utilities, offsites and storage; O. Levenspiel, Chemical Reaction Engineering (3rd ed.) and H. S. Fogler, Elements of Chemical Reaction Engineering — reactor scale-up. Engineering-economy factors follow the standard notation $(A/P,i,n)$ and $(P/A,i,n)$; as the question specifies straight-line depreciation, that method is used throughout (rather than the Canadian CCA declining-balance system).

Question 4: Most Economical Particulate-Control Device by EUAR Analysis (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. Three particulate-control options with unequal lives, zero salvage, straight-line depreciation and $t=0.52$:

ParameterWet ScrubberESPFabric Filter
Life (years)102015
Revenue $R$$\$0$$\$290{,}000$$\$290{,}000$
Total capital investment $P$$\$5{,}300{,}000$$\$9{,}750{,}000$$\$7{,}870{,}000$
Total annual cost $C$$\$2{,}770{,}000$$\$1{,}840{,}000$$\$2{,}345{,}000$
Depreciation $D=P/n$$\$530{,}000$$\$487{,}500$$\$524{,}667$

Find. The most economical device (highest EUAR, i.e. least negative net annual worth) at an after-tax hurdle rate of (a) 6% and (b) 18%.

02468101214161820-$9.75M capital-$0.49M/yr net op.-$0.49M/yr net op.ESP after-tax cash flow (net operating cost + capital, 20-yr life)period (year)
Figure 3 — After-tax cash flow of the ESP option (20-year life): a $\$9{,}750{,}000$ capital outlay at year 0 and a net after-tax operating outflow of about $\$490{,}500$ each year. The EUAR converts this whole stream to a single equivalent annual figure at the chosen hurdle rate.

Approach. Because the lives are unequal, compare the options on an annual basis: form each device’s after-tax operating cash flow, annualise its capital cost with the capital-recovery factor over its own life, and combine to get the EUAR. The largest (least-negative) EUAR is the most economical, evaluated separately at each hurdle rate.

  1. After-tax operating cash flow. With the depreciation tax shield, the annual operating figure (excluding capital) is $(R-C)(1-t)+tD$. It is the same in both scenarios: Wet Scrubber $=(0-2{,}770{,}000)(0.48)+0.52(530{,}000)=-1{,}054{,}000$; ESP $=(290{,}000-1{,}840{,}000)(0.48)+0.52(487{,}500)=-490{,}500$; Fabric Filter $=(290{,}000-2{,}345{,}000)(0.48)+0.52(524{,}667)=-713{,}573$ per year.
  2. Capital-recovery factors over each life. $(A/P,i,n)=i(1+i)^n/[(1+i)^n-1]$. At 6%: $(A/P,6\%,10)=0.13587$, $(A/P,6\%,20)=0.08719$, $(A/P,6\%,15)=0.10296$. At 18%: $(A/P,18\%,10)=0.22251$, $(A/P,18\%,20)=0.18682$, $(A/P,18\%,15)=0.19640$.
  3. EUAR at 6% (scenario a). $\text{EUAR}=(R-C)(1-t)+tD-P\,(A/P,i,n)$. Wet Scrubber $=-1{,}054{,}000-5{,}300{,}000(0.13587)=-1{,}774{,}100$; ESP $=-490{,}500-9{,}750{,}000(0.08719)=\boxed{-\$1{,}340{,}500}$; Fabric Filter $=-713{,}573-7{,}870{,}000(0.10296)=-1{,}523{,}900$. The ESP has the highest (least-negative) EUAR.
  4. EUAR at 18% (scenario b). Repeating with the 18% factors, Wet Scrubber $=-1{,}054{,}000-5{,}300{,}000(0.22251)=\boxed{-\$2{,}233{,}300}$; ESP $=-490{,}500-9{,}750{,}000(0.18682)=-2{,}312{,}000$; Fabric Filter $=-713{,}573-7{,}870{,}000(0.19640)=-2{,}259{,}300$. Now the Wet Scrubber has the highest EUAR.
  5. Interpret the reversal. At the low 6% rate the capital-heavy ESP wins because its high first cost is cheap to carry and its dust revenue and low operating cost dominate. At the high 18% rate the cost of capital dominates, so the low-first-cost wet scrubber becomes the most economical despite earning no dust revenue.
DeviceEUAR @ 6%EUAR @ 18%
Wet Scrubber (10 yr)$-\$1{,}774{,}100$$-\$2{,}233{,}300$
ESP (20 yr)$-\$1{,}340{,}500$$-\$2{,}312{,}000$
Fabric Filter (15 yr)$-\$1{,}523{,}900$$-\$2{,}259{,}300$
Most economicalESPWet Scrubber

Recommendation: at a 6% after-tax hurdle rate choose the electrostatic precipitator; at an 18% after-tax hurdle rate choose the high-energy wet scrubber.