Plate Evaporator — Milk Concentration and Fouling: 22-Agric-B8 Food Process Engineering (Part 1), December 2013, Q4

Question 4: Plate Evaporator — Milk Concentration and Fouling (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.

Plate evaporator data
Quantity	Symbol	Value
Feed rate	\(F\)	1500 kg/h
Feed / product solids	\(x_F,\,x_P\)	10% / 30%
Feed / evaporating temperature	\(T_F,\,T_{evap}\)	20°C / 75°C
Steam pressure (abs.) / sat. temp.	\(P_s,\,T_s\)	200 kPa / 120.2°C
Feed specific heat	\(c_{p,F}\)	3.86 kJ/(kg·K)
Overall \(U\), plate area	\(U,\,A_{plate}\)	650 W/(m²·K), 0.44 m²
Fouling film	\(x_f,\,k_f\)	0.1 mm, 0.1 W/(m·K)

Find. Number of plates required, and the % reduction in evaporator capacity once the fouling film forms.

Mass and energy streams around the single-effect plate evaporator.

Approach. Mass balance for the vapour load, a lumped sensible+latent heat balance for the steam-side duty (ignoring boiling-point rise), size \(A=Q/(U\Delta T)\), then add the fouling resistance in series to get the new (lower) \(U\) and express the capacity drop as a percentage.

Mass balance. Overall solids balance: \(P = F\,x_F/x_P = 1500(0.10/0.30) =\boxed{500\ \text{kg/h}}\) concentrate, so \(V=F-P=\boxed{1000\ \text{kg/h}}\) water evaporated.
Heat duty. Sensible heat to bring the whole feed to boiling, plus latent heat to evaporate \(V\) at 75°C (\(h_{fg,75^\circ C}=2321.4\ \text{kJ/kg}\)): \(Q = F c_{p,F}(T_{evap}-T_F) + V h_{fg} = 1500(3.86)(55) + 1000(2321.4) = 318{,}450+2{,}321{,}400 = 2{,}639{,}850\ \text{kJ/h} =\boxed{733.3\ \text{kW}}\).
Heat transfer area and plate count. \(\Delta T = T_s-T_{evap}=120.2-75=45.2^\circ\text{C}\). \(A = \dfrac{Q}{U\Delta T} = \dfrac{733{,}290}{650\times45.2}=24.94\ \text{m}^2\). Number of plates \(=A/0.44 = 56.7 \to \boxed{57\ \text{plates}}\) (round up, a partial plate cannot be built).
Effect of fouling on \(U\). The fouling film adds a series resistance \(R_f=x_f/k_f = 1\times10^{-4}/0.1=1.0\times10^{-3}\ \text{m}^2\text{K/W}\) to the clean resistance \(R_{clean}=1/U=1.538\times10^{-3}\ \text{m}^2\text{K/W}\): \(U_{foul}=1/(R_{clean}+R_f)=1/(2.538\times10^{-3})=\boxed{393.9\ \text{W/(m}^2\text{K)}}\). With the number of plates (hence \(A\)) and \(\Delta T\) fixed, capacity is directly proportional to \(U\), so the reduction is \((U-U_{foul})/U = (650-393.9)/650 = \boxed{39.4\%}\).

Final results
Quantity	Value
Concentrate / vapour rates	500 kg/h / 1000 kg/h
Heat duty, \(Q\)	733.3 kW
Heating area / plate count	24.9 m² / 57 plates
Fouled \(U\)	393.9 W/(m²·K)
Capacity reduction from fouling	39.4%

Concept & Theory

Topic: Single-effect evaporator sizing and fouling-resistance degradation.

Key relations: solids mass balance \(Fx_F=Px_P\); heat duty \(Q=Fc_p\Delta T+V\lambda\); \(A=Q/(U\Delta T)\); series thermal resistance \(1/U_{foul}=1/U_{clean}+x_f/k_f\).

Why this works: a plate evaporator's heat duty splits cleanly into raising the whole feed stream to its boiling point and then vaporizing only the fraction that leaves as vapour; because plates are added in whole units, the area calculation must be rounded up, and because fouling acts as an extra conduction layer in series with the existing film/wall/film resistances, its effect on \(U\) (and hence on capacity, at fixed area and \(\Delta T\)) is additive in resistance, not in \(U\) itself.

Common pitfall: subtracting the fouling resistance from \(U\) directly (\(U-R_f\)) instead of adding it to \(1/U\) before re-inverting — because \(U\) and \(R\) are reciprocals, this understates the true capacity loss.

Source: Geankoplis, Transport Processes and Separation Process Principles, Ch. 8 (evaporation); Earle, Unit Operations in Food Processing, Ch. 8.

Question 4 of 10: Plate Evaporator — Milk Concentration and Fouling

Question 4: Plate Evaporator — Milk Concentration and Fouling (20 marks)