Example 3:
a. What is the rate of heat transfer to concentrate 8000 kg/h, 80 oC, and dilute liquid feed with 0.2% solute content to 2.5% liquid product? Saturated steam is supplied at 200 kPa abs., and the single evaporator operates at atmospheric pressure.
b. What is the steam economy of the evaporator?
Solution:
F = 8000 kg/hr
Tf = 80 oC = 353.15 K
xf = 0.2/100 = 0.002
xl = 2.5/100 = 0.025
Let Reference state, P and T be that of the Liquid Product.
Operating Pressure = 1 atm = 0.101325 MPa
Steam Pressure = 200 kPa abs. = 0.2MPa
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.
Ts = 393.08 K
hc = hl = 9.0712 
= 503.45 
hs = hv = 48.742 = 2705.3
The solution is dilute, so it will behave as pure water. At 1 atm, the solution will boil. Also, its specific heat capacity, cp, is that of liquid water – 4.186 kJ/ kg.K .
Tb of Solution at 0.101325 MPa:
| Thermodynamic Properties of Water: 8th Perry’s Handbook, T 2 – 30, p. 2-413 | |||
| p(MPa) | T(K) | hl(kJ/mol) | hv(kJ/mol) |
|---|---|---|---|
| 0.090535 | 370 | 7.3121 | 48.111 |
| 0.12885 | 380 | 8.0725 | 48.393 |
| 0.101325 | – | – | – |
Evaluate the Temperature at p=0.101325 MPa by interpolating. Refer to this example of interpolation.
To = To = 372.82 K
hl = 7.5262 = 417.70
hv = 48.190 = 2674.54
F = L + V
V = 8000 – L
Solute Balance:
xfF = xlL
0.1/100 * (8000) = 2.5/100 * L
L = 320 kg/h
Therefore, V = 8000 – 320 = 7680 kg/h
Energy Balance:
Fhf + Shs = Vhc + Shc + Lhl
Fcp(Tf – Tref) + S(hs – hc) = V
+ Lcp(Tl – Tref)
(8000)(4.186)( 353.15– 372.82) + S(2705.3 – 503.45) = 7680(2674.5 – 417.70) + 320(0)
S = 8170.8 kg/hr
a:
q = S (hs – hc) = 8170.8(2705.3 – 503.45)
q = 17, 991 MJ/hr = 4997.5 kW
b:
Steam Economy = 
Steam Economy = = 0.9514