EVAPORATION

Multi-Effect Evaporator

Evaporation is carried out in cascade of evaporators. The set-up was invented by Norbert Rillieux which aims to lower the cost of production by utilizing the steam produced per effect and use it to heat the succeeding vessel.
 

Rillieux Principle:
One pound of steam will evaporate about as many pounds of water as there are effects in series. Hence, a pound of steam will evaporate about 4 pounds of water in a quadruple-effect evaporator.

Over-all Heat Transfer in Series:
q = UoA T T = T1 + T2 + …+ Tn = Ts – Tsn – BPR	q – heat transfer rate T – Over-all Temperature Gradient Uo – Over-all Heat Transfer Coefficient BPR – total Boiling point elevation across the series Tsn – temperature of the steam ejected at the last effect

Example 4:
A three-effect evaporator in series with forward feed is to concentrate solution with 7% solute content to 35%.
a. If feed rate is 1000 kg/hr, determine the total amount of vapor produced.
b. What must be the minimum operating pressure in the vapor space of the last effect to enable the operation? Saturated steam passes through the first effect at a temperature of 120 ^oC. The total boiling point rise along the cascade of evaporators is 50 ^oC.
 

Solution:
OMB:
F = 1000 kg/hr
x_f = 0.07
x_l = 0.35
T_s = 120 ^oC = 393.15K
BPR = 50 ^oC = 50 K
 

OMB:
F = L + V_T
1000 = V_T + L          eq. 1

Solute Balance:
x_fF = x_lL
0.07(1000) = 0.35L
L = 200 kg/h
Therefore, V = 8000 – 320 = 7680 kg/h

a: Substituting derived values to equation 1:
1000 = V_T + 200
V_T = 800 kg/hr

b: In order for the heat transfer operation to proceed, the driving force should be greater than zero.
T > 0
T_s – T_s3 – BPR > 0
393.15 – T_s3 – 50 > 0
T_s3 < 343.15K  

Evaluate the Pressure at T = 343.15 by interpolating. Refer to this example of interpolation.
p = 0.0318 MPa = 0.3138 atm = 238.52 mmHg