22-Agric-B8 Food Process Engineering (Part 1) · May 2017
Question 10 of 10: Single-effect evaporator — rating at two different feed rates
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-Agric-B8, Food Process Engineering (Part 1) — National Exams, May 2017. 3 hours duration, closed book (one aid sheet permitted). Ten questions are printed, grouped into four sections (I–IV); a complete exam paper requires six. All ten are solved below as a complete study set.
Reference texts: Toledo, R.T., Fundamentals of Food Process Engineering (this exam's own cited source for its Stumbo g-table and steam-table appendix); Geankoplis, C.J., Transport Processes and Separation Process Principles (evaporator design and steam economy); Incropera, F.P. & DeWitt, D.P., Fundamentals of Heat and Mass Transfer (unsteady-state Heisler-chart conduction, composite cylindrical walls); Singh, R.P. & Heldman, D.R., Introduction to Food Engineering, and Cleland, A.C., Food Refrigeration Processes (freezing-time prediction, modified Plank equation).
Question 10: Single-effect evaporator — rating at two different feed rates (20 marks)
Approach. Part (a): solids and overall mass balances give V and L directly (independent of U); the accompanying energy balance then backs out U from the known area and temperature driving force. Part (b): since A and U carry over unchanged, Q=UAΔT is now FIXED, and the energy balance is solved the other way — for V given the higher feed rate's larger sensible-heat demand.
Part (a) — mass and solids balance. F=4535/3600=1.260 kg/s. Solids=1.260(0.02)=0.02519 kg/s (constant). $$L=\frac{\text{solids}}{x_P}=\frac{0.02519}{0.03}=0.8398\ \text{kg/s} = \boxed{3023\ \text{kg/h}}$$
$$V=F-L=1.260-0.8398=0.4199\ \text{kg/s}=\boxed{1512\ \text{kg/h}}$$
Part (a) — energy balance and U. $$Q=FC_p(T_b-T_F)+V\,h_{fg,100^\circ C}=(1.260)(4100)(100-38)+(0.4199)(2\,257\,000)$$
$$=320\,300+947\,600=1\,268\,000\ \text{W}=1268\ \text{kW}$$
$$U=\frac{Q}{A(T_s-T_b)}=\frac{1\,268\,000}{(69.7)(110-100)}=\boxed{1819\ \text{W/(m}^2\cdot\text{K)}}$$
Part (b) — the duty is now fixed by the SAME U, A, ΔT. Since U, A, and the temperature driving force (110°C steam, 100°C boiling, unchanged) are all identical to part (a), the evaporator can only deliver the same Q=1268 kW — it does not automatically scale up with the larger feed. Solve the energy balance for V at the new feed rate F₂=6804/3600=1.890 kg/s:
$$Q=F_2C_p(100-38)+Vh_{fg,100^\circ C} \;\Rightarrow\; 1\,268\,000=(1.890)(4100)(62)+V(2\,257\,000)$$
$$1\,268\,000=480\,400+2\,257\,000\,V \;\Rightarrow\; V=\frac{787\,600}{2\,257\,000}=\boxed{0.349\ \text{kg/s}=1256\ \text{kg/h}}$$
Part (b) — liquid product and its concentration. $$L=F_2-V=1.890-0.349=1.541\ \text{kg/s}=\boxed{5548\ \text{kg/h}}$$
$$x_P=\frac{F_2\,x_F}{L}=\frac{(1.890)(0.02)}{1.541}=\boxed{2.45\%}$$
Doubling-and-a-half the feed rate, without any change in area or heat source, only manages to evaporate about the same absolute amount of water as before (0.349 vs 0.420 kg/s) — so the outlet concentration actually falls (2.45% vs the 3% target of part a). This illustrates why an evaporator sized for one throughput cannot simply be "fed harder" to get proportionally more concentrated product.
