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

Question 1 of 6: Distillation Column Heat Integration

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 1: Distillation Column Heat Integration (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. Two simple columns (reboiler duty = condenser duty at every pressure) whose temperatures and duties rise with pressure, per the tables above. Utilities: MP steam at $T_s = 200$ °C, cooling water returned at $T_{cw} = 30$ °C, and a minimum approach temperature $\Delta T_{min} = 10$ °C for any heat exchange.

Find. The minimum external heating (steam) and cooling (cooling-water) duties for (a) the un-integrated base case at 1 bar, (b) forward integration (condenser of column #1 drives reboiler of column #2), and (c) backward integration (condenser of column #2 drives reboiler of column #1).

Approach

Heat integration between two columns is feasible only when a condenser is at least $\Delta T_{min}$ hotter than the reboiler it is asked to drive. For each case we raise the pressure of the heat-source column just enough to clear that 10 °C hurdle, then the recovered duty is the smaller of the two matched duties; whatever the reboiler still needs comes from steam and whatever the condensers cannot give up goes to cooling water.

  1. Confirm the utilities can serve both columns. Steam at 200 °C must exceed every reboiler temperature by 10 °C and cooling water (30 °C) must sit 10 °C below every condenser temperature. At 1 bar the reboilers are 120 and 130 °C (steam 200 °C is ample) and the condensers are 90 and 110 °C (both > 40 °C), so all utility matches are valid.
  2. (a) Base case — both columns at 1 bar, no integration. Check whether integration is even possible at 1 bar: condenser #1 = 90 °C cannot drive reboiler #2 = 130 °C, and condenser #2 = 110 °C cannot drive reboiler #1 = 120 °C (neither clears $+10$ °C). So both reboilers take steam and both condensers reject to cooling water: $$Q_{heat} = Q_{reb,1} + Q_{reb,2} = 3000 + 5000, \qquad Q_{cool} = Q_{cond,1} + Q_{cond,2} = 3000 + 5000$$ ==**Base case: 8000 kW heating and 8000 kW cooling.**==
  3. (b) Forward integration — condenser #1 heats reboiler #2. Column #2 stays at 1 bar ($T_{reb,2} = 130$ °C). We raise column #1 until its condenser is $\ge 130 + 10 = 140$ °C. From the table the first pressure that satisfies $T_{cond,1} \ge 140$ °C is 3 bar ($T_{cond,1} = 140$ °C, $Q_1 = 4000$ kW).
  4. Match the duties and close the balance. The heat that condenser #1 can hand over is $\min(Q_{cond,1}, Q_{reb,2}) = \min(4000, 5000) = 4000$ kW. Reboiler #2 still needs $5000 - 4000 = 1000$ kW of steam; reboiler #1 (now 170 °C) takes its full 4000 kW from steam. Condenser #1 is fully absorbed by reboiler #2, so only condenser #2 (5000 kW) rejects to cooling water: $$Q_{heat} = \underbrace{4000}_{reb\,1} + \underbrace{1000}_{reb\,2\ \text{top-up}} = 5000, \qquad Q_{cool} = \underbrace{5000}_{cond\,2} = 5000$$ ==**Forward integration: 5000 kW heating and 5000 kW cooling (col. #1 at 3 bar, 4000 kW recovered).**==
    5. Column #1(3 bar)Column #2(1 bar)MP steam ->reb#1 4000 kWcond#1 -> reb#24000 kW recoveredtop-up steam1000 kWcond#2 -> CW5000 kW
    Figure 1.1 — Forward heat integration. Column #1 is pressurised to 3 bar so its 140 °C condenser can drive the 130 °C reboiler of column #2 (1 bar), recovering 4000 kW; reboiler #1 takes 4000 kW steam, reboiler #2 a 1000 kW top-up, and condenser #2 rejects 5000 kW to cooling water. Net: 5000 kW heating / 5000 kW cooling.
  5. (c) Backward integration — condenser #2 heats reboiler #1. Column #1 stays at 1 bar ($T_{reb,1} = 120$ °C). Raise column #2 until its condenser is $\ge 120 + 10 = 130$ °C; the first qualifying pressure is 2 bar ($T_{cond,2} = 130$ °C, $Q_2 = 6000$ kW). The recoverable heat is capped by the small reboiler #1: $\min(6000, 3000) = 3000$ kW.
  6. Close the backward balance. Reboiler #1 is fully met by recovered heat (0 kW steam); reboiler #2 (now 153 °C) needs its full 6000 kW from steam. Condenser #2 gives 3000 kW to reboiler #1 and dumps the remaining 3000 kW to cooling water; condenser #1 (3000 kW) also goes to cooling water: $$Q_{heat} = \underbrace{6000}_{reb\,2} = 6000, \qquad Q_{cool} = \underbrace{3000}_{cond\,2\ \text{excess}} + \underbrace{3000}_{cond\,1} = 6000$$ ==**Backward integration: 6000 kW heating and 6000 kW cooling (col. #2 at 2 bar, 3000 kW recovered).**==
    7. Column #2(2 bar)Column #1(1 bar)MP steam ->reb#2 6000 kWcond#2 -> reb#13000 kW recoveredcond#2 excess-> CW 3000 kWcond#1 -> CW3000 kW
    Figure 1.2 — Backward heat integration. Column #2 is pressurised to 2 bar so its 130 °C condenser can drive the 120 °C reboiler of column #1 (1 bar); recovery is capped at reboiler #1's 3000 kW. Reboiler #2 takes 6000 kW steam; condenser #2's 3000 kW excess and condenser #1's 3000 kW go to cooling water. Net: 6000 kW heating / 6000 kW cooling.
  7. Compare the schemes. Relative to the 8000 kW base case, forward integration saves 3000 kW of both steam and cooling water while backward integration saves 2000 kW. Forward integration is preferred because the hotter, higher-duty condenser of column #1 can drive the larger reboiler #2, recovering 4000 kW versus only 3000 kW for the reverse match.
CasePressuresRecovered heatHeating (steam)Cooling (CW)
(a) Base, no integration#1 @1 bar, #2 @1 bar08000 kW8000 kW
(b) Forward (cond#1→reb#2)#1 @3 bar, #2 @1 bar4000 kW5000 kW5000 kW
(c) Backward (cond#2→reb#1)#1 @1 bar, #2 @2 bar3000 kW6000 kW6000 kW
Check: the analysis assumes the recovered heat exactly offsets duty kW-for-kW (a condenser giving up 4000 kW supplies 4000 kW to the reboiler it drives), which holds because both are latent-heat exchangers operating at essentially constant temperature; the only feasibility constraint is the 10 °C approach. Raising a column's pressure also raises its reboiler duty (more reflux, higher relative-volatility penalty), so the modest duty growth from 1→3 bar on column #1 is already embedded in the table values used.
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