23-Chem-A2 Unit Operations and Separation Processes · May 2015
Question 5 of 6: Maximum Feed Rate to a Double-Effect Evaporator
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 04-CHEM-A2 Mechanical and Thermal Operations, May 2015. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt two from each section. All six are worked below for completeness.
Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 1 (fluid flow, heat transfer) and Vol. 2 (particle technology, evaporation); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Perry's Chemical Engineers' Handbook (9th ed.).
Note on the charts and tables. Page 3 (Fanning friction-factor chart), page 5 (LMTD correction-factor plot) and page 6 (Table B.2, saturated water by pressure) were read from the printed figure. The smooth-pipe friction factor is computed from the Colebrook equation, which the chart plots; the B3 condenser is isothermal on the steam side, so the correction factor is exactly $F=1$; and the B2 saturation temperatures are read from the supplied Table B.2.
Question B2: Maximum Feed Rate to a Double-Effect Evaporator (25 marks)
Given. Forward-feed, two equal-area effects with fixed $U$; the total available temperature driving force is fixed by the steam and condenser saturation temperatures less the boiling-point rise, and the feed enters cold (310 K).
Find. The maximum permissible feed rate $F$ (kg/s).
Figure B2 — Double-effect forward-feed evaporator: live steam drives effect 1; its vapour $V_1$ drives effect 2; feed and liquor flow co-currently with the vapour.
Approach. The two effects are thermally coupled — the vapour raised in effect 1 is the heating medium for effect 2 — so solve the coupled mass and enthalpy balances for the temperature split ($\Delta T_1,\Delta T_2$) and feed rate simultaneously (by iteration), rather than assuming equal duties.
Available temperature driving force. $\Delta T_{avail}=T_s-T_c-\mathrm{BPR}=141.3-51.5-3=86.8\ \mathrm{K}$, to be split between the two effects, $\Delta T_1+\Delta T_2=86.8$.
Heat-transfer rate per effect. Each effect transfers $q_i=U_iA\,\Delta T_i$: $q_1=2.8(10)\Delta T_1$, $q_2=1.7(10)\Delta T_2$ (kW, with $\Delta T$ in K).
Effect-1 enthalpy balance (cold feed). Steam condensing in effect 1 first heats the feed from 36.85 °C to the effect-1 boiling temperature $T_1=T_s-\Delta T_1$, then evaporates $V_1$: $$q_1=U_1A\Delta T_1 = F\,c_p(T_1-T_f)+V_1\lambda,\qquad V_1=\frac{U_2A\,\Delta T_2}{\lambda}.$$
Effect-2 enthalpy balance (couple & close). The vapour $V_1$ condenses in effect 2, heating the effect-1 liquor from $T_1$ to $T_{2}=T_c+\mathrm{BPR}$ and evaporating $V_2$: $$V_1\lambda=(F-V_1)c_p(T_{2}-T_1)+V_2\lambda,\qquad V_2=0.8F-V_1,$$ where the solids balance $0.10F=0.50\,P$ gives total evaporation $V_1+V_2=F(1-0.10/0.50)=0.8F$.
Solve the coupled set (iterate on the split). Bisection on $\Delta T_2$ satisfying the effect-2 balance yields $$\Delta T_1=38.5\ \mathrm{K},\quad \Delta T_2=48.2\ \mathrm{K},\quad T_1=102.7\ \mathrm{°C},$$ $$V_1=0.352,\quad V_2=0.403\ \mathrm{kg/s},\quad F=\frac{U_1A\Delta T_1-V_1\lambda}{c_p(T_1-T_f)}=0.943\ \mathrm{kg/s}.$$ $F_{max}=0.943\ \mathrm{kg/s}\approx 3.40\times10^{3}\ \mathrm{kg/h}$ $\Delta T_2>\Delta T_1$ because the cold feed absorbs part of effect-1's duty as sensible heat, leaving less for evaporation there.
Consistency checks. Steam demand $S=U_1A\Delta T_1/\lambda=0.463\ \mathrm{kg/s}$; economy $=(V_1+V_2)/S=0.755/0.463=1.63$ (typical of a double effect); and $V_1+V_2=0.755=0.8F$ closes the solids balance.
