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22-Agric-B8 Food Process Engineering (Part 1) · May 2017

Question 9 of 10: Steam economy of a double-effect evaporator with two feed preheaters

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 9: Steam economy of a double-effect evaporator with two feed preheaters (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. The source figure (reproduced below, from the page immediately following the question in the source paper) shows a forward-feed double effect with two feed preheaters recovering heat from the vapour and condensate streams.

Heater 1Heater 2Effect I125CEffect II100CFeed 0.25 kg/s16C75C110CSteam ms140C sat.V1=0.059 kg/s125C (heats coil)liquid 0.191 kg/sV2=0.066 kg/s100Ccondensate 125CProduct 0.125 kg/s100C
Double-effect evaporator with two feed preheaters: Heater 1 uses Effect II's vapour V2 (100°C); Heater 2 uses the condensate from Effect I's vapour V1 after it gives up its latent heat in Effect II's coil (125°C).
Stream data read from the figure
StreamValue
Feed0.25 kg/s at 16°C
Steam to Effect I140°C, saturated
Effect I boiling temperature125°C (vapour V1 leaves at 125°C)
Effect II boiling temperature100°C (vapour V2 leaves at 100°C)
Feed path16°C → Heater 1 → 75°C → Heater 2 → 110°C → Effect I
Product0.125 kg/s at 100°C

Find. The steam economy V₁+V₂)/mṊ.

Approach. Mass balance fixes total vapour (V1+V2) directly from feed and product flow — independent of the V1/V2 split. To get the steam rate, however, effect I's own energy balance still needs V1 individually, which follows from effect II's own energy balance (coil duty from V1 condensing, plus flash heat from the liquid entering at 125°C and boiling at 100°C, supplies V2's evaporation).

Check: the liquid stream leaving Effect I is taken as the mass-balance quantity (0.25−V1) rather than literally as printed, and Heater 2 is taken as using the condensate of V1 (still at 125°C, cooling by sensible heat as it gives up the last of its usefulness) to raise the feed from 75°C to 110°C, consistent with the figure's overall heat-integration layout.
  1. Overall mass balance (independent of the V1/V2 split). $$V_1+V_2 = F-P = 0.25-0.125 = \boxed{0.125\ \text{kg/s}}$$
  2. Effect II energy balance fixes the split. V1 (at 125°C) condenses in Effect II's coil; the liquid arriving from Effect I (at 125°C, quantity F−V1) also flashes down to Effect II's 100°C boiling point, releasing further sensible heat: $$V_1 h_{fg,125}+(F-V_1)C_{p}(125-100)=V_2 h_{fg,100}=(0.125-V_1)h_{fg,100}$$ Using steam-table values h᷇ᵤ(125°C)=2188.5 kJ/kg, h᷇ᵤ(100°C)=2257.0 kJ/kg and Cᶲ=4.186 kJ/(kg·K) for the dilute feed, and solving for V1: $$V_1 = \boxed{0.0590\ \text{kg/s}}, \qquad V_2 = 0.125-0.0590 = \boxed{0.0660\ \text{kg/s}}$$
  3. Effect I energy balance gives the steam rate. The feed enters Effect I already preheated to 110°C, so steam only needs to supply the last 15°C of sensible heating plus V1's latent heat: $$\dot m_s h_{fg,140} = F C_p(125-110)+V_1 h_{fg,125}$$ $$\dot m_s(2\,144\,700) = (0.25)(4186)(15)+(0.0590)(2\,188\,500) = 15\,700+129\,100=144\,800$$ $$\dot m_s = \boxed{0.0675\ \text{kg/s}}$$
  4. Steam economy. $$\text{Economy}=\frac{V_1+V_2}{\dot m_s}=\frac{0.125}{0.0675}=\boxed{1.85\ \text{kg vapour/kg steam}}$$ An economy above 1.0 (rather than the ≈1.0 of a single effect) is exactly the payoff expected from a double effect with heat-integrated preheating — each kg of live steam ultimately drives nearly two kg of evaporation once the vapour and condensate are re-used through the two preheaters.
Final results
QuantityValue
V₁, V₂0.0590, 0.0660 kg/s
Steam rate, mṊᵣ0.0675 kg/s
Steam economy1.85 kg vapour/kg steam