# Chapter 8: Evaporation¶

## Example 8.4-1, Page number 498¶

In :
#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)

Concetrated liquid Rate: 6048.0 kg/h
Vapor rate from evaporator: 3024.0 kg/h
Heat Transfer area required: 149.3 m2


## Example 8.4-2, Page Number 499¶

In :
#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"

Boiling point rise:  13.9 °C


## Example 8.4-3, Page number 501¶

In :
#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'

The steam used 3253.2 kg steam/h
The calculated steam economy is  0.837
The heating surface area is  49.1 m2


## Example 8.5-1, Page number 505¶

In :
#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'

Area A1 A2 and A3 for 1st 2nd and 3rd effect are 111.421, 113.714,and 111.640
The Average area of each effect 112.26 as compared to average area 112.08 in first trial
Steam Economy for tripple effect evaporator: 1.999