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23-Chem-A5 Chemical Plant Design and Economics · December 2015

Question 3 of 6: Rate of Return

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

Notes on this paper

National Exams — December 2015 — 04-Chem-A5 Chemical Plant Design and Economics. Three-hour, closed-book exam; one two-sided aid sheet and an approved calculator permitted. Six equally weighted (20-mark) questions are posed and the candidate answers any five; only the first five are marked. All six are answered below for completeness. Question 1 is a conceptual flowsheet-synthesis question (hydrodealkylation of toluene to benzene) answered with a process flow sheet and organised prose; questions 2, 3 and 6 are numerical (capacity-scaled and index-escalated plant cost, yield-improvement rate of return, and evaporator heat-transfer area); questions 4 and 5 are qualitative essays on materials selection against the common corrosion mechanisms and on process-hazard classification.

Reference texts: M.S. Peters, K.D. Timmerhaus & R.E. West, Plant Design and Economics for Chemical Engineers (5th ed., McGraw-Hill) — the exam's named primary text (cost estimation Ch. 6, interest and profitability Ch. 7–10, materials of construction Ch. 12, plant safety and loss prevention Ch. 3); J.M. Douglas, Conceptual Design of Chemical Processes (McGraw-Hill) — the hydrodealkylation (HDA) flowsheet-synthesis case study used in Question 1; R. Turton et al., Analysis, Synthesis, and Design of Chemical Processes (4th ed., Prentice Hall) — flowsheet synthesis and equipment cost correlations; AIChE, Dow’s Fire & Explosion Index Hazard Classification Guide (7th ed.) — the process-hazard checklist behind Question 5; supporting Canadian practice from CCOHS/WHMIS 2015, the Canadian Environmental Protection Act (CEPA), and CSA/ASME materials and pressure-vessel codes.

Question 3: Rate of Return (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. Fixed product output $P=10{,}000$ t/yr; yield rises from $0.70$ to $0.75$ kg product/kg raw material; raw-material price $\$500$/t; product price $\$900$/t (unchanged); extra investment $\Delta I=\$1{,}250{,}000$; extra operating cost $\approx0$.

Find. Whether the incremental return on the extra investment justifies the modification.

Approach. Because the output is fixed and the selling price is unchanged, revenue does not move — the benefit is entirely the raw material saved by the higher yield. Value that saving as the annual incremental profit, then form the rate of return and payback on the extra $\$1.25$M investment. The product price is a red herring.

  1. Raw material required, before and after. To make the fixed 10,000 t of product, $$m_{\text{old}} = \frac{10{,}000}{0.70} = 14{,}286\ \text{t/yr},\qquad m_{\text{new}} = \frac{10{,}000}{0.75} = 13{,}333\ \text{t/yr}$$
  2. Raw material saved. $$\Delta m = 14{,}286 - 13{,}333 = 952.4\ \text{t/yr}$$
  3. Annual saving (= incremental profit). At $\$500$/t and negligible extra operating cost, $$\Delta\Pi = 952.4\ \tfrac{\text{t}}{\text{yr}}\times\$500\ \tfrac{\$}{\text{t}} = \boxed{\$476{,}190\text{/yr}}$$
  4. Rate of return and payback on the extra investment. $$\text{ROI} = \frac{\Delta\Pi}{\Delta I} = \frac{476{,}190}{1{,}250{,}000} = \boxed{38.1\%},\qquad \text{payback} = \frac{\Delta I}{\Delta\Pi} = \boxed{2.6\text{ yr}}$$ A 38 % return, far above any normal hurdle rate of 10–15 %, with a 2.6-year payback — the modification is clearly worth making.
QuantityValue
Raw material saved$952.4$ t/yr
Annual saving (incremental profit)$\$476{,}190$/yr
Incremental rate of return$38.1\%$
Payback period$\approx2.6$ yr
DecisionMake the modification
Check: the selling price ($\$900$/t) never enters the arithmetic — with output and price fixed, revenue is identical in both cases, so the only cash effect is the raw-material saving. The one genuine ambiguity is whether the plant holds output or feed fixed. Holding output fixed (the reading taken here, since the question states the plant “is producing 10,000 metric tons per year”) the benefit is the raw material saved. If instead the feed were held at $14{,}286$ t/yr, the higher yield would lift output to $14{,}286\times0.75 = 10{,}714$ t/yr and the benefit would be $714\times\$900 = \$642{,}900$/yr of extra sales, i.e. $\text{ROI}=51.4\%$ and a $1.9$-year payback. Both readings clear any normal hurdle rate by a wide margin, so the accept decision is unchanged; state whichever assumption is used, as NOTE 1 of the paper requires.