Design problem
A 5% aqueous solution of a high molecular weight solute has to be concentrated to 40% in a forward-feed double effect evaporator at the rate of 8000 kg.h-1 . The feed temperature is 40°C. Saturated steam at 3.5 kg.cm2 is available for heating. A vacuum of 600 mm Hg is maintained in the second effect. Calculate the area requirements, if calandria of equal area are used. The overall heat transfer coefficients are 550 and 370 kcal.h-1 m-2 °C-1 in the first and the last effect respectively. The specific heat of the concentrated liquor is 0.87 kcal.kg-1 °C-1 .
SOLUTION :
Part 1. Thermal design:
Pressure in effect I to be decided.
Pressure in effect II = 760 - 600 = 160 mm Hg
Boiling point at this pressure = 60°C (from steam table)
(high molecular wt. solute, BPE is neglected)
Latent heat vapor generated in effect II at 160 mm Hg (0.2133bar) = 563 kcal.kg-1 (λs2)
Heating steam is at 3.5 kg.cm-2 gauge; temp ( Ts)=148 °C; Latent heat (λs)= 506 kcal.kg-1
Feed rate = 8000 kg.h-1, Solute content = 5%
Final concentration = 40%
Solid in = 8000 × 0. 05 = 400 kg.h-1, water in = 8000-400= 7600 kg.h-1
Product out (40% solid) = 400/0.40 kg.h-1
= 1000 kg.h-1
Water out with the product = 1000 (1-0.40) kg.h-1
= 600 kg.h-1
Total evaporation rate in two effects ( 
) = 7600- 600 = 7000 kg.h-1 (3.A)
Allow equal areas to two effects, i.e., 
= 148-60 = 88°C
UD1 = 550, UD2 = 370 kcal.h-1 m-2 °C -1 ⇒
= 35.4 °C & 
= 52.6°C (3.11)
Temperature of the vapor leaving effect I (Tb1 ) =148 – 35.4 = 112.6°C
Latent heat vapor generated in effect I at 112.6 °C = 531 kcal.kg-1 ( λs1)
Energy balance for effect I :
(3.2)
Enthalpy values: reference temperature = 112.6°C (temp of solution leaving effect I)
Hf = (40 - 112.6)(1 kcal/ kg °C) = -72.6 kcal/ kg
H1= 0 kcal/ kg (w.r.t. the reference temperature of 112.6 °C)
λs = 506 kcal/ kg; λs1 = 531 kcal/ kg
∴ (8000)(-72.6) + ms(506) = (8000- ms1 )(0) + ms1 (531)
⇒ms = 1.05 ms1 + 1148 (3.B)