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23-Chem-A5 Chemical Plant Design and Economics · May 2016

Question 3 of 6: Coagulant Dosage — Minimum-Cost Optimisation

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

Notes on this paper

National Exams — May 2016 — 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. Questions 1–3 are numerical (distillation heat integration, after-tax net present worth, and a coagulant-dosage cost optimisation); questions 4–6 are qualitative essays on materials of construction, plant startup/shutdown safety, and equipment-selection factors.

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 (interest and profitability Ch. 7–10, materials of construction Ch. 12, plant safety and loss prevention Ch. 3, equipment selection throughout Ch. 14–22); R. Smith, Chemical Process Design and Integration (2nd ed., Wiley) and B. Linnhoff et al., A User Guide on Process Integration (IChemE) — pinch analysis and column heat integration behind Question 1; R.K. Sinnott & G. Towler, Chemical Engineering Design (Coulson & Richardson vol. 6) — materials selection and equipment sizing; supporting Canadian practice from CCOHS, provincial OH&S process-safety-management regulations and the CSA Z767 (PSM) framework.

Question 3: Coagulant Dosage — Minimum-Cost Optimisation (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. Turbidity model $T(F) = 37.0893 - 7.739F + 0.7263F^2 - 0.0233F^3$; backwash model $B(F) = -0.549F + 1.697T$; unit costs of water $c_w = \$0.0608$/m3 and coagulant $c_c = \$0.183$/kg; per 1000 m3 of product the chemical charge is $c_c F$ (dosage in mg/L $\equiv$ kg per 1000 m3) and the backwash-water charge is $c_w B$.

Find. (a) the dosage $F^{*}$ minimising total cost per 1000 m3 and the minimum cost $C_{min}$; (b) the percentage cost saving relative to the current 12 mg/L dosage.

Approach

Assemble total cost as one cubic in $F$ (chemical cost plus water cost of backwash, the backwash itself being linear in $F$ and in the cubic turbidity), then set $dC/dF = 0$ and solve the resulting quadratic; the physically meaningful root with positive curvature is the minimum. Part (b) is a direct ratio of costs at 12 mg/L versus the optimum.

246810122.02.22.42.62.8min: F=6.15 mg/L, $2.11Coagulant dosage F (mg/L)Total cost ($ / 1000 m3)
Figure 3.1 — Total cost (chemical + backwash water) per 1000 m3 versus coagulant dosage. The convex curve has an interior minimum at $F^{*}\approx6.15$ mg/L, $C_{min}\approx\$2.11$; the current 12 mg/L dosage sits well up the right-hand branch at $\$2.68$.
  1. Build the total-cost function. Per 1000 m3 of product, $$C(F) = c_c F + c_w B(F) = c_c F + c_w\big(-0.549F + 1.697\,T(F)\big)$$ Substituting $T(F)$ and collecting powers of $F$ (with $c_c = 0.183$, $c_w = 0.0608$) gives the cubic $$C(F) = a_3 F^3 + a_2 F^2 + a_1 F + a_0,\quad a_3 = -0.002404,\; a_2 = 0.07494,\; a_1 = -0.9195,\; a_0 = 3.8272$$
  2. Stationary condition. Differentiate and set to zero: $$\frac{dC}{dF} = 3a_3 F^2 + 2a_2 F + a_1 = -0.007212\,F^2 + 0.14988\,F - 0.9195 = 0$$
  3. Solve the quadratic. The roots are $F = 6.15$ and $F = 20.6$ mg/L. The second derivative $C''(F) = 6a_3 F + 2a_2$ is positive only at the smaller root ($C''(6.15) > 0$), so $$F^{*} = 6.15 \text{ mg/L}$$ ==**Optimum coagulant dosage $F^{*} \approx 6.15$ mg/L.**==
  4. Minimum cost. At $F^{*} = 6.15$ mg/L: $T = 11.55$, $B = -0.549(6.15) + 1.697(11.55) = 16.22$ m3/1000 m3, so $$C_{min} = 0.183(6.15) + 0.0608(16.22) = \$2.11 \text{ per } 1000\text{ m}^3$$ ==**Minimum total cost $\approx \$2.11$ per 1000 m3 of product water.**==
  5. (b) Cost at the current 12 mg/L dosage. $T(12) = 8.55$, $B(12) = -0.549(12) + 1.697(8.55) = 7.91$, so $$C_{12} = 0.183(12) + 0.0608(7.91) = \$2.677 \text{ per } 1000\text{ m}^3$$
  6. Percentage saving. $$\text{saving} = \frac{C_{12} - C_{min}}{C_{12}}\times 100 = \frac{2.677 - 2.111}{2.677}\times 100 = 21.1\%$$ At the plant flow of 189,250 m3/day this is $(2.677-2.111)\times189.25 \approx \$107$/day, about $\$39{,}000$/yr. ==**Switching from 12 mg/L to the optimum saves about 21 % of total cost (≈$107/day).**==
QuantityValue
Optimum dosage $F^{*}$6.15 mg/L
Minimum total cost $C_{min}$$2.11 / 1000 m3
Cost at current 12 mg/L$2.68 / 1000 m3
Percentage saving21.1 % (≈$107/day)
Check: a numerical sweep of $C(F)$ over 2–12 mg/L (the range where the fitted turbidity model is valid) confirms the calculus optimum at $F \approx 6.15$ mg/L, $C \approx \$2.11$. The larger quadratic root (20.6 mg/L) is a mathematical maximum of the cubic and lies outside the fitted data range, so it is discarded on physical grounds.