23-Chem-A5 Chemical Plant Design and Economics · December 2013
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 2013 — 04-Chem-A5 Chemical Plant Design and Economics. Three-hour, open-book exam; any non-communicating calculator permitted. Six equally weighted questions are posed and the candidate answers any five; only the first five are marked. All six are answered below for completeness. Questions 1, 3 and 6 are conceptual design / management questions answered as organised prose; questions 2, 4 and 5 contain the numerical work (production capacity and pricing, simple- and compound-interest loan accounting, and sinking-fund depreciation) and every boxed figure.
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, interest and investment, depreciation, profitability, process synthesis, and plant safety); R. Turton et al., Analysis, Synthesis, and Design of Chemical Processes (4th ed., Prentice Hall) — flowsheet synthesis, separation selection, and safety; W.D. Seider et al., Product and Process Design Principles (3rd ed., Wiley) — separation-train synthesis; supporting Canadian tax practice from the Canada Revenue Agency Capital Cost Allowance classes and the half-year rule.
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.
Working capital is the additional money, over and above the fixed-capital investment in physical plant, that a company must have on hand to operate the plant. It funds the raw-materials and product inventories, the goods in process, the accounts receivable (product shipped but not yet paid for), the cash for wages and supplies, and the accounts payable and taxes. Unlike fixed capital, working capital is not consumed or depreciated: it is tied up while the plant runs and is recovered in full at the end of the project life. The total capital investment is the sum of the fixed-capital investment and the working capital, so here the fixed-capital investment is $\$5{,}000{,}000-\$750{,}000=\$4{,}250{,}000$.
The turnover ratio is a rapid economic yardstick defined as the ratio of gross annual sales to the fixed-capital investment, $\text{turnover ratio}=\dfrac{\text{gross annual sales}}{\text{fixed-capital investment}}$. Its reciprocal is the capital ratio. A high turnover ratio means each dollar of plant generates many dollars of sales per year (typical of high-throughput, low-value bulk chemicals); a low ratio is characteristic of capital-intensive, high-value products. It is used for order-of-magnitude estimates of either the sales a given investment should support or the investment a target sales figure requires.
Given.
| Quantity | Value |
|---|---|
| Total capital investment, $C_T$ | $5,000,000 |
| Working capital, $C_{WC}$ | $750,000 |
| Production capacity | 32 metric tonnes/day |
| Down-time | two shutdowns × 10 days = 20 days/yr |
| Turnover ratio | 0.8 |
Find. (ii) the annual production capacity in tonnes/year, and (iii) the unit selling price of product ($/tonne) consistent with a turnover ratio of 0.8.
Approach. Convert the daily rate to an annual tonnage using the actual on-stream days, obtain the fixed-capital investment by removing working capital from the total, use the turnover-ratio definition to get gross annual sales, and divide sales by annual production to get the unit price.
| Quantity | Value |
|---|---|
| On-stream time | 345 days/yr |
| Annual production capacity | 11,040 t/yr |
| Fixed-capital investment | $4,250,000 |
| Gross annual sales (at turnover 0.8) | $3,400,000/yr |
| Unit selling price | $307.97/t ≈ $308/t ($0.308/kg) |