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22-Agric-B8 Food Process Engineering (Part 1) · May 2014

Question 8 of 10: Double-Effect Evaporator, Reverse Feed — Solids Leaving the Second Effect

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

Notes on this paper

Paper format. 04-Agric-B8 Food Process Engineering (Part 1), National Exams May 2014 — a three-hour open-book exam (any non-communicating calculator permitted). Ten questions are set in four sections (I–IV), each with a "do one/any N of M" instruction; a candidate following the choice rules answers six questions for a 100-mark paper. All ten are worked here so the set is a complete study resource.

Reference texts. R.T. Toledo, Fundamentals of Food Process Engineering, 3rd ed. (thermal-process lethality, D and z values, Ball/Stumbo process calculation, aseptic holding-tube residence time, evaporator design — this is the exam's own appendix source); C.J. Geankoplis, Transport Processes and Separation Process Principles, 4th ed. (evaporator heat and mass balances, multiple-effect steam economy); R.P. Singh and D.R. Heldman, Introduction to Food Engineering, 5th ed. (freezing-time estimation, modified Plank and Cleland-Earle equations, unsteady-state heat transfer in canned foods); A.C. Cleland, Food Refrigeration Processes: Analysis, Design and Simulation (Plank/Cleland-Earle freezing-time correlations); F.P. Incropera and D.P. DeWitt, Fundamentals of Heat and Mass Transfer (transient conduction, Heisler charts, composite-wall resistance).

Check: this paper's four roman-numeral section headers ("I. Heat transfer", "II. Food freezing and freeze concentration", "III. Thermal processing", "IV. Several assumptions (retort come-up correction factor, reference temperature for spore D-values, evaporator steam temperature reused for Question 9) are flagged inline where the source leaves a value implicit.

Question 8: Double-Effect Evaporator, Reverse Feed — Solids Leaving the Second Effect (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.

Reverse-feed double-effect evaporator data
QuantitySymbolValue
Feed rate\(F\)100 kg/min
Feed solids, temperature\(x_F,\,T_F\)10%, 55°C
Final product solids, temperature\(x_P,\,T_P\)30%, 75°C
Steam temperature (Effect 1)\(T_{S1}\)100°C
Effect 1 / Effect 2 temperature—75°C / 70°C
Specific heat, solids / water\(C_{p,s},\,C_{p,w}\)2.095, 4.186 kJ/(kg·K)

Find. The solids content of the liquid stream leaving the second effect (the intermediate stream flowing on to Effect 1).

Effect 2(70 C)Effect 1(75 C)Feed F=100 kg/min10% solids, 55 CB=68.04 kg/min liquid14.70% solids, 70->75 CSteam S1=36.28 kg/min, 100 CV1=34.70 kg/min vapour, 75 C -> to Effect-2 coilProduct P=33.33 kg/min30% solids, 75 C<- V1 vapour from Effect-1 coilV2=31.96 kg/minvapour, 70 C
Fig. D — reverse-feed double-effect evaporator: liquid flows Effect 2→Effect 1 while the heating vapour flows Effect 1→Effect 2 (opposite directions, hence "reverse feed"); vapour-line stubs shown at each effect's coil.

Approach. An overall solids balance fixes the final product rate directly. The unknown vapour split between the two effects is then found from an energy balance on Effect 2 alone (heat supplied by V1 condensing in its coil equals the sensible heat to raise the fresh feed to 70°C plus the latent heat of the vapour V2 it generates); a mass balance on Effect 2 then gives the intermediate liquid stream B and its solids content.

  1. Overall solids balance. Solids are conserved from feed to final product: \(F\,x_F = 100(0.10)=10\ \text{kg solids/min}\), so \(P = 10/0.30 = \boxed{33.33\ \text{kg/min}}\) and \(C_{pF}=x_F C_{p,s}+(1-x_F)C_{p,w}=0.10(2.095)+0.90(4.186)=\boxed{3.977\ \text{kJ/(kg}\cdot\text{K)}}\).
  2. Mass balances around each effect. Effect 2: \(F=B+V_2\) (feed splits into the liquid B leaving for Effect 1, and vapour \(V_2\)). Effect 1: \(B=P+V_1\), so \(V_2=F-P-V_1=66.67-V_1\).
  3. Energy balance on Effect 2. Vapour V1 condenses in Effect 2's coil at 75°C, supplying \(V_1\,h_{fg,75}\); this raises the fresh feed from 55 to 70°C and evaporates \(V_2\): \(V_1 h_{fg,75} = F\,C_{pF}(70-55) + V_2\,h_{fg,70}\), with \(h_{fg,75}=2321.4\), \(h_{fg,70}=2333.8\ \text{kJ/kg}\) (steam tables). Substituting \(V_2=66.67-V_1\) and solving the single linear equation for \(V_1\): \(V_1=\boxed{34.70\ \text{kg/min}}\).
  4. Intermediate stream and its solids content. \(B=P+V_1=33.33+34.70=\boxed{68.04\ \text{kg/min}}\); \(x_B = \dfrac{10}{68.04}=\boxed{14.70\%}\) — between the feed's 10% and the final product's 30%, as physically expected for an intermediate concentration step. \(V_2=66.67-34.70=\boxed{31.96\ \text{kg/min}}\).
  5. Steam consumption (energy balance, Effect 1). With \(C_{pB}=0.1470(2.095)+0.8530(4.186)=3.879\ \text{kJ/(kg}\cdot\text{K)}\) and \(h_{fg,100}=2257.0\ \text{kJ/kg}\): \(S_1=\dfrac{B\,C_{pB}(75-70)+V_1 h_{fg,75}}{h_{fg,100}}=\boxed{36.28\ \text{kg/min}}\).
Final results
QuantityValue
Final product \(P\)33.33 kg/min
\(V_1\) (Effect 1 vapour)34.70 kg/min
\(B\) (liquid leaving Effect 2)68.04 kg/min
Solids content leaving Effect 2, \(x_B\)14.70%
\(V_2\) (Effect 2 vapour)31.96 kg/min
Steam consumption \(S_1\)36.28 kg/min
Check: the intermediate stream is assumed to enter Effect 1 already at Effect 2's 70°C boiling temperature and to be raised to Effect 1's 75°C before flash/evaporation there, consistent with the figure's own labelled internal temperatures; the condensate return lines are omitted from the diagram for clarity.