Steam Consumption and Area of a Single-Effect Evaporator: 23-Chem-A2 Unit Operations and Separation Processes, December 2015, Q5

Question B2: Steam Consumption and Area of a Single-Effect Evaporator (25 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. A single-effect evaporator concentrating a solution, with a cold feed, sub-cooled condensate leaving the steam chest, and stream-specific heat capacities.

Quantity	Value
Feed rate $F$ / composition	7 kg/s, 10% → 50% solids
Steam: $T_s$ / $H_s$	394 K / 2530 kJ/kg
Boiling point at 13.5 kN/m²	325 K
Vapour enthalpy at 325 K	2594 kJ/kg
Feed / condensate temperatures	294 K / 352.7 K
$c_{p,10}$ / $c_{p,50}$	3.76 / 3.14 kJ/kg·K
Overall coefficient $U$	3 kW/m²K

Find. (a) the steam rate $D$; (b) the required heating surface area $A$.

Figure B2 — Single-effect evaporator flowsheet: cold feed enters, vapour leaves overhead, thick product leaves the base; condensing steam supplies the duty and leaves sub-cooled.

Approach. A solids balance fixes the vapour and product rates; an overall enthalpy balance (single liquid datum) solves for the steam rate; the heat load and steam-to-liquor $\Delta T$ then give the area.

Material balance. $$P=\frac{F\,x_F}{x_P}=\frac{7(0.10)}{0.50}=1.4\ \text{kg/s},\qquad V=F-P=5.6\ \text{kg/s}.$$
Stream enthalpies (273 K datum). $$h_F=Fc_{p,10}(294-273)=7(3.76)(21)=552.7\ \text{kW},$$ $$h_P=Pc_{p,50}(325-273)=1.4(3.14)(52)=228.6\ \text{kW},\quad H_V=V(2594)=14526\ \text{kW},$$ $$h_{\text{cond}}=c_{p,w}(352.7-273)=4.18(79.7)=333.1\ \text{kJ/kg}.$$
Part (a): steam rate from the energy balance. $h_F+D(H_s-h_{\text{cond}})=h_P+H_V$ gives $$D=\frac{H_V+h_P-h_F}{H_s-h_{\text{cond}}}=\frac{14526+228.6-552.7}{2530-333.1}=\frac{14202}{2196.9}.$$ $D = 6.46\ \text{kg/s}$ — economy $V/D=5.6/6.46=0.87$ (below unity because the cold feed must first be heated to its boiling point).
Part (b): heat load and area. The duty crossing the surface is what the steam gives up, and the driving force is steam-saturation minus boiling-liquor temperature: $$Q=D(H_s-h_{\text{cond}})=6.46(2196.9)=1.420\times10^{4}\ \text{kW},\quad \Delta T=394-325=69\ \text{K},$$ $$A=\frac{Q}{U\,\Delta T}=\frac{1.420\times10^{4}}{3(69)}.$$ $A = 68.6\ \text{m}^2$

Quantity	Result
Product / vapour rates	1.4 / 5.6 kg/s
(a) Steam rate $D$	6.46 kg/s
Steam economy $V/D$	0.87
Heat load $Q$	$1.42\times10^{4}$ kW
(b) Heating area $A$	68.6 m²

Concept & Theory

Topic. Energy and material balance on a single-effect evaporator and the area from the rate equation.

Key relations. Solids balance $P=Fx_F/x_P$, $V=F-P$; enthalpy balance $h_F+D(H_s-h_{\text{cond}})=h_P+H_V$; rate equation $Q=UA\,\Delta T$ with $\Delta T=T_s-T_{bp}$; economy $=V/D$.

Why this works. Every stream is written on one liquid datum as $\dot m c_p\Delta T$ or $\dot m H$, so the balance closes on the single unknown steam rate. The economy falls below one whenever the feed enters cold, because part of the steam only warms the liquor rather than boiling it — the motivation for feed pre-heating and multiple-effect operation. For area, the relevant heat is the full duty given up by the steam and the relevant $\Delta T$ is the isothermal condensing side minus the boiling liquor.

Common pitfall. Using the latent heat of steam instead of $H_s-h_{\text{cond}}$ (which includes condensate sub-cooling), or taking $\Delta T$ as steam-to-feed rather than steam-to-boiling-liquor. Any boiling-point elevation would raise the vapour enthalpy datum and shrink the effective $\Delta T$.

Source. Coulson & Richardson, Chemical Engineering Vol. 2, ch. 14 (evaporation; worked Example 14.1); McCabe, Smith & Harriott ch. 16.

Question 5 of 6: Steam Consumption and Area of a Single-Effect Evaporator

Question B2: Steam Consumption and Area of a Single-Effect Evaporator (25 marks)