1. SINGLE EFFECT EVAPORATOR
Group 6
Buensuceso, Mary Mae
Camacho, Imiefer
Ogilvie, Jeremy James
Quijano, Crystel
Simbulan, Rose

2. A single effect evaporator is used to concentrate KCl solution from 10% to 40% solids. The feed enters at 25deg Celsius and steam at 180 deg C. The vapor space is maintained at 54 deg C. The over-all heat transfer to coefficient is 3000 W/m2-K. Cp is constant at kJ/kg-K. The evaporation rate is 1000 kg/hr.
Determine:
Heat transfer Area
economy

3. Evaporate (E)
= 1000 kg/hr
T1 = 54 deg C Vo
To = 180 deg C Vo
Feed(F)
xF = 0.10
TF = 25 deg C
CpF= kJ/kg-K
Liquor (L)
xL = 0.40
U = 3000 W/m2-K

4. Sol’n:
SB: F(xF) = L(xL) 0.10F = 0.40L F= 4L OMB: F = E + L 4L = L L= F= Economy = E/Vo VoLo = F(CpF)(TI – TF) + E(L1 – 2.303BPR)

5. Sol’n: At To = 180 deg C Lo = 2013. 1 J/kg At T1 = 54 deg C
Economy = E/Vo
VoLo = F(CpF)(TI – TF) + E(L1 – 2.303BPR)
At To = 180 deg C
Lo = J/kg
At T1 = 54 deg C
L1 = J/kg
T1 = 54 deg C = deg F
Find TI
TI = T1 + BPR
Assume TI= 150 deg F
BPR Chart = 13.5 deg F
BPR Calc = 150 – = 20.8 deg F
TI= deg F= deg F
BPR chart = 13.5 deg F = 7.5 deg C
TI = = deg C

6. Sol’n: A= 7.2 m2 Vo= 1271. 5187 Economy = 0.786 UA(To- TI)(3.6)= VoLo
At To = 180 deg C
Lo = J/kg
At T1 = 54 deg C
L1 = J/kg
BPR chart = 7.5 deg C
TI = deg C
VoLo = F(CpF)(TI – TF) + E(L1 – 2.303BPR)
Vo(2013.1)= (4.187)( ) ( – 2.303(7.5))
Vo=
Economy = E/ Vo
Economy=1000/
Economy = 0.786
UA(To- TI)(3.6)= VoLo
(3000)(A)(180 – 61.5)(3.6) = (2013.1)
A= 7.2 m2

7. end