In [3]:

```
#Heat Transfer Area in a Single-Effect Evaporator
#Variable declaration
F = 9072. #Feed rate to the evaporator, kg/hr
xF = 1. #weight percent of salt, kg/kg
TF = 311.0 #Temperature of feed, K
xL = 1.5 #Weight percentage of concentrated product, kg/kg
P = 101.325 #Pressure of vapor space, kPa
Ps = 143.3 #Pressure of saturated steam used for heating, kPa
U = 1704. #Overall heat transfer coefficient, W/m2K
cpF = 4.140 #Specific heat of Feed to the evaporator, kJ/(kg.K)
Hv373 = 2257. #Latent heat of evaporation of water at 373.2 K
Hv143 = 2230. #Latent heat of evaporation of saturated water at 143.3 kPa
Ts = 383.2 #Temperature of Saturated steam at 143.3 kPa
TL = 373.2 #Temperature of concentrated Liquid enthalpy calculation, K
Tref = 373.2 #Reference temperature for enthalpy calculation, K
#Calculation
L = F*xF/xL
V = F-L
#Energy Balance
hF = F*cpF*(TF-Tref)
hL = L*cpF*(TL-Tref)
hV = Hv373*V
#hF + q = hL + hV
q = (hL+hV-hF)
S = q/(Hv143)
qsi = q*1000/3600
A = qsi/(U*(Ts-TL))
#Result
print 'Concetrated liquid Rate: %5.1f kg/h'%L
print 'Vapor rate from evaporator: %5.1f kg/h'%V
print 'Heat Transfer area required: %4.1f m2'%(A)
```

In [4]:

```
#Use of Duhring Chart for Boiling point rise
#Variable declaration
P = 25.6 #Pressure in the evaporator, kPa
C = 30 #Weight % of NaOH in solution, K
Tb256 = 65.6 #Boiling point of water at 25.6 kPa, °C or 150 °C
#Calculation
#From Fig. 8.4-2 for 30% wt of NaoOH and boiling point of water 150 °F
Ts = 175 #Boiling point of solution °F
TsC = 79.5 #Boiling point of solution °C
BPR = TsC-Tb256
#Result
print "Boiling point rise: ",BPR,"°C"
```

In [1]:

```
#Evaporation of NaOH Solution
# Variable declaration
F = 4536. #Feed rate to the evaporator, kg/hr
xF = 0.2 #weight percent of NaOH, kg/kg
TF = 60.0 #Temperature of feed, °C
xL = 0.5 #Weight percentage of concentrated product, kg/kg
P = 11.7 #Pressure of vapor space, kPa
Ps = 172.4 #Pressure of saturated steam used for heating, kPa
U = 1560. #Overall heat transfer coefficient, W/m2K
cpS = 1.884 #Specific heat of steam 11.7 kPa , kJ/(kg.K)
Hv895 = 2667. #Enthalpy of steam at 11.7 kPa and 89.5°C
Hv489 = 2590. #Enthalp of steam at 11.7 kPa and 49.5°C
LambdaV1724 = 2214 #Latent heat of evaporation of water at 172.4 kPa and at 115.6°C
Ts = 115.6 #Saturation temperature of 172.4kPa in °C
# Calculation
L = F*xF/xL
V = F - L
BPw117 = 48.9 #BP of Water From steam table, °C
BPs117 = 89.5 #BP of Soln From Duhring Chart Fig 8.4-2, °C
BPR = BPs117-BPw117
hF = 214. #Enthalpy of 20% feed at 60°C, KJ/Kg from fig 8.4-3
hL = 505. #Enthalpy of 50% feed at 89.5°C, KJ/Kg
HV = Hv489 + cpS*BPR
S = (L*hL+V*HV-F*hF)/LambdaV1724
Q = S*LambdaV1724/3600 #Heat addition rate in kW
#Newtons law of cooling
A = Q*1000/(U*(Ts-BPs117))
SteamEco = V/S
#Result
print "The steam used",round(S,1),"kg steam/h"
print "The calculated steam economy is ",round(SteamEco,3)
print "The heating surface area is ", round(A,1),"m2"
print 'Difference in answers is due to machin precision'
```

In [1]:

```
#Evaporation of Sugar Solution in a Triple-Effect Evaporator
import numpy as np
from scipy.interpolate import interp1d
# Variable declaration
xF = 0.1 #wt fraction of sugar in Feed
xP = 0.5 #wt fraction of sugar in Product
Ts1 = 121.1 #Saturation temperature of the steam at 205.5kPa in °C
F = 22680. #Feed Rate kg/hr
TF = 26.7 #Temperature of the feed, °C
U1 = 3123. #Overall heat transfer coefficient for evaporator 1, W/(m2K)
U2 = 1987. #Overall heat transfer coefficient for evaporator 2, W/(m2K)
U3 = 1136. #Overall heat transfer coefficient for evaporator 3, W/(m2K)
T = np.array([50.,60.,70.,80.,90.,100.,110.,120,126.])
H = np.array([2592.2,2609.7,2626.9,2643.8,2660.1,2676.0,2691.3,2706.0,2714.4])
h = np.array([209.3,251.1,293.0,334.9,376.9,419.1,461.3,503.7,529.2])
lamb = np.array([2382.9,2358.6,2334.0,2308.8,2283.2,2256.9,2230.0,2202.2,2185.2])
cp = np.array([])
# Calculation
def BPRDegC(xw):
return 1.78*xw+6.22*xw*xw
def SpecHeat(xw):
return 4.19-2.35*xw
fH = interp1d(T,H)
fh = interp1d(T,h)
flamb = interp1d(T,lamb)
#Step 1
BPR3 = BPRDegC(xP)
Tvs3 = 51.67 #Saturation temperature of steam at 13.4 kPa for 3rd evaporator
T3 = Tvs3 + BPR3
#Step 2
L3 = F*xF/xP #Concentrated product rate, kg/hr
V = F - L3 #Total water vaporised from three evaporators (V = V1 + v2 +v3 ), kg/hr
V1 = V/3 #Assuming V1 = V2 = V3
V2 = V1
V3 = V1
#Making liquid balance on each evaporator
L1 = F - V1 #Concentrated liquid rate from evaporator 1, kg/hr
L2 = L1 - V2 #Concentrated liquid rate from evaporator 2, kg/hr
L3 = L2 - V3
#Making Solid balance on each evaporator
x1 = F*xF/L1
x2 = L1*x1/L2
x3 = xP
#Step 3
BPR1 = BPRDegC(x1)
BPR2 = BPRDegC(x2)
BPR3 = BPRDegC(x3)
SDelT = Ts1 - Tvs3 - (BPR1+BPR2+BPR3)
ISU13 = 1/U1+1/U2+1/U3
DelT1 = SDelT*(1/U1)/ISU13
DelT2 = SDelT*(1/U2)/ISU13
DelT3 = SDelT*(1/U3)/ISU13
T1 = Ts1 - DelT1
T2 = T1 - BPR1 - DelT2
Ts2 = T1 - BPR1
T3 = T2 -BPR2 - DelT3
Ts3 = T2 - BPR2
Ts4 = T3 - BPR3
#Step4:
CpF = SpecHeat(xF) #Calculate Heat Capacities
Cp1 = SpecHeat(x1)
Cp2 = SpecHeat(x2)
Cp3 = SpecHeat(x3)
H1 = fH(Ts2) + 1.884*BPR1 #for effect 1
lambdas1 = fH(Ts1)-fh(Ts1)
H2 = fH(Ts3) + 1.884*BPR2 #for effect 2
lambdas2 = fH(Ts2)-fh(Ts2)
H3 = fH(Ts4) + 1.884*BPR3 #for effect 3
lambdas3 = fH(Ts3)-fh(Ts3)
a11 = Cp1*T1 - lambdas2 - H2
a12 = -(Cp2*T2 - H2)
b1 = -F*lambdas2
a21 = lambdas3
a22 = Cp2*T2 - lambdas3 - H3
b2 = L3*Cp3*T2 - L3*H3
a = np.array([[a11,a12], [a21,a22]])
b = np.array([b1,b2])
L1,L2 = np.linalg.solve(a, b)
V1 = F - L1
V2 = L1 - L2
V3 = L2 - L3
S = (L1*Cp1*T1 + V1*H1 - F*CpF*TF)/lambdas1
#Step5:
q1 = S*lambdas1*1000/3600
q2 = V1*lambdas2*1000/3600
q3 = V2*lambdas3*1000/3600
A1 = q1/(U1*DelT1)
A2 = q2/(U2*DelT2)
A3 = q3/(U3*DelT3)
Am = Am1 = (A1+A2+A3)/3
#Step6:
x1 = F*xF/L1
x2 = L1*x1/L2
x3 = L2*x2/L3
#Step7:
BPR1 = BPRDegC(x1)
BPR2 = BPRDegC(x2)
BPR3 = BPRDegC(x3)
SDelT = Ts1 - Tvs3 - (BPR1+BPR2+BPR3)
DelT1 = DelT1*A1/Am
DelT2 = DelT2*A2/Am
DelT3 = DelT3*A3/Am
T1 = Ts1 - DelT1
T2 = T1 - BPR1 - DelT2
Ts2 = T1 - BPR1
T3 = T2 -BPR2 - DelT3
Ts3 = T2 - BPR2
Ts4 = T3 - BPR3
#Step8:
CpF = SpecHeat(xF) #Calculate Heat Capacities
Cp1 = SpecHeat(x1)
Cp2 = SpecHeat(x2)
Cp3 = SpecHeat(x3)
H1 = fH(Ts2) + 1.884*BPR1 #for effect 1
lambdas1 = fH(Ts1)-fh(Ts1)
H2 = fH(Ts3) + 1.884*BPR2 #for effect 2
lambdas2 = fH(Ts2)-fh(Ts2)
H3 = fH(Ts4) + 1.884*BPR3 #for effect 3
lambdas3 = fH(Ts3)-fh(Ts3)
a11 = Cp1*T1 - lambdas2 - H2
a12 = -(Cp2*T2 - H2)
b1 = -F*lambdas2
a21 = lambdas3
a22 = Cp2*T2 - lambdas3 - H3
b2 = L3*Cp3*T2 - L3*H3
a = np.array([[a11,a12], [a21,a22]])
b = np.array([b1,b2])
L1,L2 = np.linalg.solve(a, b)
V1 = F - L1
V2 = L1 - L2
V3 = L2 - L3
S = (L1*Cp1*T1 + V1*H1 - F*CpF*TF)/lambdas1
q1 = S*lambdas1*1000/3600
q2 = V1*lambdas2*1000/3600
q3 = V2*lambdas3*1000/3600
A1 = q1/(U1*DelT1)
A2 = q2/(U2*DelT2)
A3 = q3/(U3*DelT3)
Am = (A1+A2+A3)/3
SEconomy = (V1+V2+V3)/S
#Results
print 'Area A1 A2 and A3 for 1st 2nd and 3rd effect are %4.3f, %4.3f,and %4.3f'%(A1,A2,A3)
print "The Average area of each effect",round(Am,2),"as compared to average area",round(Am1,2),"in first trial"
print "Steam Economy for tripple effect evaporator:", round(SEconomy,3)
print 'Difference in answers is due to machin precision'
```