Question 3 of 6: Single-Effect Evaporator — Steam Economy and Area at Two Feed Temperatures
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (Wiley) — variable-conductivity conduction, resistance networks, convection correlations; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — double-pipe exchanger sizing, evaporator heat load, diffusion through stagnant films; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular-diffusion flux relations and two-film interphase transfer; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — gas absorption. Property data from Perry's Chemical Engineers' Handbook (9th ed.).
Exam format: six problems worth 25 points each — three in Part A (Heat Transfer) and three in Part B (Mass Transfer, printed as Problems 1–3 of Part B and numbered Questions 4–6 here). At least two problems from each part must be attempted and only the first two from each section are marked. Open-book, three-hour paper. All six problems are solved here for completeness.
Paper note
This is the 16-CHEM-A3 Heat and Mass Transfer paper of May 2019. The printed paper is five pages (cover notes, Part A Problems 1–3, Part B Problems 1–3), followed by two sheets of graph paper and a periodic table.
Part A — Heat Transfer
Question 3: Single-Effect Evaporator — Steam Economy and Area at Two Feed Temperatures (25 points)
Find. The steam economy (kg vapour per kg steam) and the heat-transfer area $A$ for feed at (a) 20 °C and (b) 35 °C.
Figure 3 — Single-effect evaporator. A solids balance fixes the product and vapour rates; the steam condenses on the heating surface (driving force $120-50=70$ K) and supplies the feed pre-heat plus the latent heat of the 24,000 kg/hr of vapour boiled off.
Approach. A solids balance gives the vapour rate; an enthalpy balance (sensible pre-heat of the feed to the boiling point plus latent heat of vaporization) gives the heat load, hence the steam rate and economy; finally $A=Q/(U\,\Delta T)$. Only the sensible pre-heat term differs between the two feed temperatures.
Mass balance (solids tie). Solids in $=0.10(30{,}000)=3000$ kg/hr are conserved, so the product is $L=3000/0.50=6000$ kg/hr and the vapour boiled off is$$V=F-L=30{,}000-6000=\boxed{24{,}000\ \text{kg/hr}}.$$
Heat load. The feed is first heated to the boiling point (sensible), then $V$ is vaporized (latent), so $Q=Fc_p(T_{bp}-T_f)+V\lambda_{\text{vap}}$ with $V\lambda_{\text{vap}}=24{,}000(2383)=5.719\times10^{7}$ kJ/hr:$$\text{(a) }T_f=20^\circ\text{C: }Q=30{,}000(3.98)(30)+5.719\times10^{7}=3.58\times10^{6}+5.719\times10^{7}=6.077\times10^{7}\ \text{kJ/hr}$$$$\text{(b) }T_f=35^\circ\text{C: }Q=30{,}000(3.98)(15)+5.719\times10^{7}=1.79\times10^{6}+5.719\times10^{7}=5.898\times10^{7}\ \text{kJ/hr}$$
Steam rate and economy. The condensing steam supplies $Q$ through its latent heat, $S=Q/\lambda_s$:$$\text{(a) }S=\frac{6.077\times10^{7}}{2202}=27{,}600\ \text{kg/hr},\quad \text{economy}=\frac{24{,}000}{27{,}600}=\boxed{0.870}$$$$\text{(b) }S=\frac{5.898\times10^{7}}{2202}=26{,}787\ \text{kg/hr},\quad \text{economy}=\frac{24{,}000}{26{,}787}=\boxed{0.896}$$A single effect gives an economy below 1, because part of the steam heats the cold feed instead of boiling water, and because $\lambda_s<\lambda_{\text{vap}}$ here. The warmer 35 °C feed needs 813 kg/hr less steam.
Heat-transfer area. The driving force is $\Delta T=T_{\text{sat}}-T_{bp}=120-50=70$ K in both cases, so $U\Delta T=2.9(70)=203$ kW/m²:$$\text{(a) }Q=\frac{6.077\times10^{7}}{3600}=16{,}882\ \text{kW},\quad A=\frac{16{,}882}{203}=\boxed{83.2\ \text{m}^2}$$$$\text{(b) }Q=\frac{5.898\times10^{7}}{3600}=16{,}384\ \text{kW},\quad A=\frac{16{,}384}{203}=\boxed{80.7\ \text{m}^2}$$
