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

Question 10 of 10: Peach Puree Evaporator and Condenser — Flow Rates and Duties

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 10: Peach Puree Evaporator and Condenser — Flow Rates and Duties (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.

Peach puree evaporator data
QuantitySymbolValue
Concentrated product rate\(\dot m_2\)70 kg/h
Feed solids, temperature\(x_F,\,T_F\)10.9%, 15°C
Product solids, temperature\(x_P,\,T_P\)40.1%, 40°C
Steam temperature\(T_{S}\)120°C
Condensate (of vapour) leaving temperature—37°C
Cooling water in / out—20°C / 30°C
Specific heat, solids / water\(C_{p,s},\,C_{p,w}\)2.09, 4.19 kJ/(kg·K)

Find. (a) feed and vapour/condensate flow rates; (b) steam consumption; (c) cooling-water flow rate.

Evaporator(40 C)CondenserPeach puree m1=257.5 kg/h10.9% solids, 15 CSteam m3=216.5 kg/h, 120 CConcentrate m2=70 kg/h40.1% solids, 40 CVapour m4=187.5 kg/h, 40 CCooling water m5=10827 kg/h20 C inCondensate m4, 37 C /Water out, 30 C
Fig. F — single-effect evaporator feeding a surface condenser; the vapour \(\dot m_4\) leaving the evaporator is the condenser's only heat input.

Approach. The concentrated-product rate is given directly, so a solids balance yields the feed rate, and an overall mass balance then yields the vapour (=condensate) rate. An energy balance on the evaporator (sensible heat to bring the feed to boiling, plus the latent heat of the vapour generated) sized against the steam's latent heat gives the steam rate; an energy balance on the condenser (condensing AND subcooling the vapour) sized against the cooling water's allowed temperature rise gives the water rate.

  1. Feed rate (solids balance). \(\dot m_1 x_F = \dot m_2 x_P \Rightarrow \dot m_1 = \dot m_2\,x_P/x_F = 70(0.401/0.109) = \boxed{257.5\ \text{kg/h}}\).
  2. Vapour / condensate rate (overall mass balance). \(\dot m_4 = \dot m_1-\dot m_2 = 257.5-70 = \boxed{187.5\ \text{kg/h}}\).
  3. Evaporator duty and steam rate. Feed specific heat \(C_{pF}=0.109(2.09)+0.891(4.19)=3.961\ \text{kJ/(kg}\cdot\text{K)}\); duty \(Q=\dot m_1 C_{pF}(40-15)+\dot m_4\,h_{fg,40}=257.5(3.961)(25)+187.5(2406.7)=\boxed{4.77\times10^5\ \text{kJ/h}}\). With \(h_{fg,120}=2202.6\ \text{kJ/kg}\): \(\dot m_3 = Q/h_{fg,120}=\boxed{216.5\ \text{kg/h}}\).
  4. Condenser duty and cooling-water rate. The vapour condenses at 40°C then subcools to the 37°C condensate exit: \(Q_{cond}=\dot m_4\big[h_{fg,40}+C_{p,w}(40-37)\big]=187.5[2406.7+4.19(3)]=\boxed{4.537\times10^5\ \text{kJ/h}}\). \(\dot m_5 = \dfrac{Q_{cond}}{C_{p,w}(30-20)}=\boxed{10{,}827\ \text{kg/h}}\).
Final results
QuantityValue
(a) Feed rate \(\dot m_1\)257.5 kg/h
(a) Vapour/condensate rate \(\dot m_4\)187.5 kg/h
(b) Steam consumption \(\dot m_3\)216.5 kg/h
(c) Cooling-water rate \(\dot m_5\)10,827 kg/h
Check: the evaporator energy balance neglects boiling-point elevation and any heat of dilution/concentration specific to peach solids (standard simplifying assumptions at this level, consistent with the specific-heat-only data the question supplies).
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