EVAPORATION

Example 4:
a. What is the rate of heat transfer to concentrate 8000 kg/h, 80 ^oC, and 20% NaOH solution to 30% liquid product? Saturated steam is supplied at 200 kPa abs., and the single evaporator operates at atmospheric pressure.
b. What is heating surface area if the over-all heat transfer coefficient is assumed to 1200 W/m².K?
 

Solution:
For concentrated solutions, enthalpies are usually provided by empirical data.
F = 8000 kg/hr
T_f = 80 ^oC = 353.15 K
x_f = 20/100 = 0.20
x_l = 30/100 = 0.30
Operating Pressure = 1 atm = 0.101325 MPa
Steam Pressure = 200 kPa abs. = 0.2MPa
 

ENTHALPY OF STEAM AND CONDENSTATE:
Ts of Water at 0.2MPa:

Thermodynamic Properties of Water: 8th Perry’s Handbook, T 2 – 30, p. 2-413
p(MPa)	T(K)	hl(kJ/mol)	hv(kJ/mol)
0.17964	390	8.8354	48.665
0.24577	400	9.6013	48.924
0.2	–	–	–

Evaluate the Temperature at p=0.2 MPa by interpolating. Refer to this example of interpolation.
T_s = 393.08 K = 119.93^oC
h_c = h_l = 9.0712 = 503.45
h_s = h_v = 48.742 = 2705.3
 

ENTHALPY of FEED:

From the plot: hf 290 kJ/kg 125 Btu/lbm

 

ENTHALPY OF WATER VAPOR AND LIQUID PRODUCT:
Tb of 0.15 NaOH Solution at 0.101325 MPa:

Thermodynamic Properties of Water: 8th Perry’s Handbook, T 2 – 30, p. 2-413
p(MPa)	T(K)
0.0905435	370
0.12885	380
0.101325	–

Evaluate the Temperature at p=0.101325 MPa by interpolating. Refer to this example of interpolation.
T = 372.82 K = 99.67 ^oC
 

Boiling Point of Solution with BPR: Tb 108o

 

ENTHALPY OF LIQUID PRODUCT:

hL 190 Btu/lbm 442 kJ/kg

 

ENTHALPY OF SUPERHEATED VAPOR:
Enthalpy of Steam at T= 108^oC = 381.15K and p = 0.101325 Pa
First Interpolation:

Thermodynamic Properties of Water(Superheated Properties): 8th Perry’s Handbook, T 2 – 30, p. 2-413
p(MPa)	T(K)	h(kJ/mol)
0.1	372.76	48.190
	400	49.189
(interpolation)	381.15	48.498
1	453.03	50.030
	500	52.086
(exrapolation)	381.15	46.884

 
2nd Interpolation:

Thermodynamic Properties of Water: 8th Perry’s Handbook, T 2 – 30, p. 2-413
p(MPa)	T(K)
0.1	48.498
1	46.884
0.101325	–

Refer to this example of interpolation.
h_v = 48.496 = 2601.5
 

OMB:
F = L + V
V = 8000 – L

Solute Balance:
x_fF = x_lL
0.2(8000) = 0.30L
L = 5333 kg/h
Therefore, V = 8000 – 5333 = 2667 kg/h

Energy Balance:
Fh_f + Sh_s = Vh_V + Sh_c + Lh_l
Fh_f + S(h_s – h_c )= Vh_V + Lh_l
8000*290 + S(2705.3 – 503.45) = 2667*2601.5 + 5333*442
S = 3168 kg/hr
 

a: q = S(h_s – h_c) = 3168(2705.3 – 503.45)
q = 6975.5 MJ/hr = 1937.6 kW

b: q = UA(T_s -T_b)
1937.6*10³ = 1200A(119.93 – 108)
A = 135.35 m²