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).
Given. Forward-feed double-effect evaporator; last effect at 12.35 kPa (Tᵣᵏᵤ₂=50°C); constant λ=2260 kJ/kg and Cᶲ=3.75 kJ/(kg·K) everywhere.
Given data
Quantity
Value
Feed rate, F
2.5 kg/s at 15°C, 10% solids
Product concentration
50% solids
Last-effect vapour-space pressure
12.35 kPa ⇒ Tᵣᵏᵤ₂=50°C
Boiling-point rise (last effect)
5°C ⇒ T₂(boiling)=55°C
Area (each effect)
50 m²
U₁, U₂
2.8, 1.7 kW/(m²·K)
λ, Cᶲ
2260 kJ/kg (constant), 3.75 kJ/(kg·K) (constant)
Find. The saturated-steam pressure (temperature) required.
Forward-feed double effect: fresh feed enters Effect I, vapour V1 heats Effect II's coil, product leaves Effect II at 50% solids.
Approach. Overall solids balance fixes total evaporation. Effect II's coil duty must equal U₂A(T₁−T₂) AND supply V₂'s latent heat plus the sensible "flash" heat released as effect I's liquid cools into effect II — solving that single equation for T₁ also fixes V₁. Effect I's own energy balance (feed sensible heat + V₁'s latent heat = U₁A(Tₛ−T₁)) then gives the steam temperature Tₛ.
Effect II: relate T₁ and V₁ through the area-limited coil duty. All of V₁ condenses in Effect II's coil, so its duty is both U₂A(T₁−T₂) and V₁λ:
$$V_1 = \frac{U_2 A (T_1-T_2)}{\lambda}$$
Effect II's own energy balance (coil duty + flash heat from the incoming liquid cooling from T₁ to T₂ = vapour V₂'s latent heat) is:
$$V_1\lambda + (F-V_1)C_p(T_1-T_2) = V_2\lambda = (2.0-V_1)\lambda$$
Substituting the first relation for V₁ and solving numerically for T₁ (T₂=55°C fixed):
$$T_1 = \boxed{80.7\ ^\circ\text{C}}, \qquad V_1=0.967\ \text{kg/s}, \qquad V_2 = 2.0-0.967 = 1.033\ \text{kg/s}$$
Effect I: energy balance sizes the steam temperature. $$Q_1 = FC_p(T_1-T_{feed})+V_1\lambda = (2.5)(3750)(80.7-15)+(0.967)(2\,260\,000)$$
$$=615\,900+2\,185\,400=2\,801\,300\ \text{W} = 2801\ \text{kW}$$
This same duty is delivered through the fixed area at U₁A(Tₛ−T₁):
$$T_s = T_1+\frac{Q_1}{U_1A}=80.7+\frac{2\,801\,300}{(2800)(50)}=80.7+20.0=\boxed{100.7\ ^\circ\text{C}}$$
Convert to saturation pressure. Interpolating the steam table between 100°C (101.3 kPa) and 105°C (120.8 kPa) at 100.7°C: $$P_s \approx 101.3+\frac{120.8-101.3}{5}(0.7)\approx\boxed{104\ \text{kPa (absolute)}}$$ — only slightly above atmospheric pressure, a realistic and gentle steam supply for a food-grade double-effect evaporator train.
