Forward-Feed Double-Effect Evaporator — Steam Economy: 22-Agric-B8 Food Process Engineering (Part 1), December 2013, Q6

Question 6: Forward-Feed Double-Effect Evaporator — Steam Economy (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.

Double-effect evaporator data
Quantity	Symbol	Value
Feed rate	\(F\)	5 kg/s
Feed specific heat	\(c_p\)	4.18 kJ/(kg·K)
Total evaporation	\(V_1+V_2\)	0.5\(F\) = 2.5 kg/s
Steam / effect-2 boiling temperature	\(T_s,\,T_2\)	122°C / 100°C
\(U_2/U_1\), \(A_1=A_2\), \(Q_1=Q_2\)	—	0.75, given
Latent heat (all streams)	\(\lambda\)	2230 kJ/kg

Find. Effect-1 boiling temperature \(T_1\), steam flow rate, vapour flow from each effect, and the steam economy.

Forward-feed double effect: effect 1's vapour heats effect 2; the liquid leaving effect 1 also flashes as it drops from \(T_1\) to \(T_2\).

Approach. Use \(Q_1=Q_2\) with \(A_1=A_2\) and \(U_2=0.75U_1\) to find \(T_1\); since the feed enters effect 1 pre-heated to \(T_1\), effect 1's steam duty goes entirely into evaporating \(V_1\); effect 2 is driven by \(V_1\)'s condensing heat plus the flash heat released as the effect-1 liquid \(L_1=F-V_1\) drops from \(T_1\) to \(T_2\).

Effect-1 boiling temperature. \(U_1A_1(T_s-T_1)=U_2A_2(T_1-T_2)\) with \(A_1=A_2\), \(U_2=0.75U_1\) gives \((122-T_1)=0.75(T_1-100)\), so \(T_1 = \dfrac{122+75}{1.75} = \boxed{112.57^\circ\text{C}}\).
Effect 1: steam duty evaporates \(V_1\) directly. Feed enters already at \(T_1\) (external preheat), so \(Q_1 = \dot m_s\lambda = V_1\lambda\), giving \(\boxed{\dot m_s = V_1}\) (equal latent heats on both sides of the coil).
Effect 2: condensing \(V_1\) plus flash from \(L_1\). \(L_1=F-V_1\) enters effect 2 at \(T_1\) but flashes down to \(T_2=100^\circ\text{C}\), releasing sensible heat that also evaporates liquid: \(V_2\lambda = V_1\lambda + L_1 c_p (T_1-T_2)\). With \(V_1+V_2=2.5\) kg/s, substituting \(L_1=5-V_1\) and \(T_1-T_2=12.57^\circ\text{C}\): \(2230V_1+(5-V_1)(4.18)(12.57)=(2.5-V_1)(2230)\) \(\Rightarrow V_1 = \boxed{1.205\ \text{kg/s}}\), \(V_2=2.5-1.205=\boxed{1.295\ \text{kg/s}}\).
Steam flow and economy. \(\dot m_s=V_1=\boxed{1.205\ \text{kg/s}}\) (checked: \(Q_1=\dot m_s\lambda=2687.8\ \text{kW}=Q_2=V_1\lambda\), confirming \(Q_1=Q_2\)). Steam economy \(=\dfrac{V_1+V_2}{\dot m_s}=\dfrac{2.5}{1.205}=\boxed{2.074\ \text{kg vapour/kg steam}}\).

Final results
Quantity	Value
Effect-1 boiling temperature, \(T_1\)	112.57°C
Steam flow rate, \(\dot m_s\)	1.205 kg/s
Vapour from effect 1 / effect 2	1.205 kg/s / 1.295 kg/s
Steam economy	2.074 kg vapour/kg steam

Concept & Theory

Topic: Multiple-effect evaporation with flash contribution (forward feed).

Key relations: equal-duty design \(U_1A_1\Delta T_1=U_2A_2\Delta T_2\); effect energy balance \(V_i\lambda = (\text{heating-medium duty}) + L_{i-1}c_p(T_{i-1}-T_i)\).

Why this works: because the liquid leaving effect 1 is still at effect 1's higher boiling point when it enters the lower-pressure effect 2, it is momentarily superheated relative to effect 2's boiling point and must flash to re-establish equilibrium; that flash vapour is "free" additional evaporation beyond what \(V_1\)'s condensing heat alone would produce, which is why the computed steam economy (2.07) comfortably exceeds the naive single-effect value of 1.0 and even the idealized loss-free 2-effect value without flash.

Common pitfall: forgetting the flash term entirely and setting \(V_2\lambda=V_1\lambda\) (i.e. \(V_2=V_1\)), which both violates the stated total-evaporation constraint and understates the true economy.

Source: Geankoplis, Transport Processes and Separation Process Principles, Ch. 8.5 (multiple-effect evaporators, forward feed).

Question 6 of 10: Forward-Feed Double-Effect Evaporator — Steam Economy

Question 6: Forward-Feed Double-Effect Evaporator — Steam Economy (20 marks)