Chapter 7: Humidification Operations

Ex7.1: Page 222

In [1]:
# Illustration 7.1
# Page: 222

print'Illustration 7.1 - Page: 222\n\n'

# Solution

import math
from scipy.optimize import fsolve
# ****Data****#
Temp1 = 273+26.1;# [K]
P1 = 100;# [mm Hg]
Temp2 = 273+60.6;# [K]
P2 = 400;# [mm Hg]
P = 200;# [mm Hg]
#*****#

def f12(T):
    return ((1/Temp1)-(1/T))/((1/Temp1)-(1/Temp2))-((math.log(P1)-math.log(P))/(math.log(P1)-math.log(P2)))
T = fsolve(f12,37);# [K]
print"At",round(T-273,1)," degree C, the vapour pressure of benzene is 200 mm Hg\n"
Illustration 7.1 - Page: 222


At 42.4  degree C, the vapour pressure of benzene is 200 mm Hg

Ex7.2: Page 223

In [5]:
# Illustration 7.2:
# Page: 223

print'Illustration 7.2 - Page: 223\n\n'
#part(a) and part(b) are table based and doesn't require an calculation

print'Illustration 7.2 (c)\n\n'

# Solution (c)

# Reference: H20
# At 25 OC
m = 0.775;
Mr = 18.02;# [kg/kmol]
lambdar = 2443000;# [N/m.kg]
M = 78.05;# [kg/kmol]
# From Eqn. 7.6:
Lambda = m*lambdar*Mr/M;# [N/m.kg]
print"Latent Heat of Vaporization at 25 degree C is",round(Lambda/1000,2)," kN/m.kg\n"
# the answer is slightly different in textbook due to approximation 
Illustration 7.2 - Page: 223


Illustration 7.2 (c)


Latent Heat of Vaporization at 25 degree C is 437.13  kN/m.kg

Ex7.3: Page 226

In [8]:
# Illustration 7.3
# Page: 226

print'Illustration 7.3 - Page: 226\n\n'

# solution

# ****Data****#
m = 10;# [kg]
Cvap = 1.256;# [kJ/kg.K]
Cliq = 1.507;# [kJ/kg.K]
Temp1 = 100;# [OC]
Temp4 = 10;# [OC]
#******#

# Using Fig 7.2 (Pg 224):
Temp2 = 25;# [OC]
# Using the notation of Fig. 7.3:
H1_diff_H2 = Cvap*(Temp1-Temp2);# [kJ/kg]
# From Illustration 7.2:
H2_diff_H3 = 434;# [Latent Heat of Vaporisation, kJ/kg]
H3_diff_H4 = Cliq*(Temp2-Temp4);# [kJ/kg]
H1_diff_H4 = H1_diff_H2+H2_diff_H3+H3_diff_H4;# [kJ/kg]
H = m*H1_diff_H4;# [kJ]
print"Heat evolved for 10 kg Benzene is ",int(H)," kJ\n"
Illustration 7.3 - Page: 226


Heat evolved for 10 kg Benzene is  5508  kJ

Ex7.4: Page 227

In [11]:
# Illustration 7.4
# Page: 227

print'Illustration 7.4 - Page: 227\n\n'

# solution

#****Data****#
# A = benzene vapour; B = Nitrogen Gas
P = 800.0;# [mm Hg]
Temp = 273.0+60;# [K]
pA = 100.0;# [mm Hg]
#******#

pB = P-pA;# [mm Hg]
MA = 78.05;# [kg/kmol]
MB = 28.08;# [kg/kmol]

# Mole Fraction
print"On the Basis of Mole Fraction\n"
yAm = pA/P;
yBm = pB/P;
print"Mole Fraction of Benzene is ",yAm
print"\nMole Fraction of Nitrogen is ",yBm
print"\n"

# Volume Fraction
print"On the Basis of Volume Fraction\n"
# Volume fraction is same as mole Fraction
yAv = yAm;
yBv = yBm;
print"Volume Fraction of Benzene is ",yAv
print"\n Volume Fraction of Nitrogen is ",yBv
print"\n"

# Absolute Humidity
print"On the basis of Absolute humidity\n"
Y = pA/pB;# [mol benzene/mol nitrogen]
Y_prime = Y*(MA/MB);# [kg benzene/kg nitrogen]
print"The concentration of benzene is ",round(Y_prime,3)," kg benzene/kg nitrogen\n"
Illustration 7.4 - Page: 227


On the Basis of Mole Fraction

Mole Fraction of Benzene is  0.125

Mole Fraction of Nitrogen is  0.875


On the Basis of Volume Fraction

Volume Fraction of Benzene is  0.125

 Volume Fraction of Nitrogen is  0.875


On the basis of Absolute humidity

The concentration of benzene is  0.397  kg benzene/kg nitrogen

Ex7.5: Page 228

In [14]:
# Illustration 7.5
# Page: 228

print'Illustration 7.5 - Page: 228\n\n'

print'Illustration 7.5 (a)\n\n'
# solution(a)

#****Data****#
# A = benzene vapour; B = Nitrogen Gas
P = 1.0;# [atm]
#*****#

MA = 78.05;# [kg/kmol]
MB = 28.02;# [kg/kmol]
# Since gas is saturated, from Fig. 7.2 (Pg 224):
pA = 275.0/760;# [atm]
Y = pA/(P-pA);# [kmol benzene/kmol nitrogen]
Y_prime = Y*(MA/MB);# [kg benzene/kg nitrogen]
print"The concentration of benzene is ",round(Y_prime,3)," kg benzene/kg nitrogen\n\n"

print'Illustration 7.5 (b)\n\n'
# solution(b)

# A = benzene vapour; B = CO2
MA = 78.05;# [kg/kmol]
MB = 44.01;# [kg/kmol]
# Since gas is saturated, from Fig. 7.2:
pA = 275.0/760;# [atm]
Y = pA/(P-pA);# [kmol benzene/kmol CO2]
Y_prime = Y*(MA/MB);# [kg benzene/kg CO2]
print"The concentration of benzene is",round(Y_prime,3)," kg benzene/kg CO2\n"
Illustration 7.5 - Page: 228


Illustration 7.5 (a)


The concentration of benzene is  1.579  kg benzene/kg nitrogen


Illustration 7.5 (b)


The concentration of benzene is 1.006  kg benzene/kg CO2

Ex7.6: Page 234

In [23]:
# Illustration 7.6
# Page: 234

print'Illustration 7.6 - Page: 234\n\n'

# solution

#****Data****#
# A = water vapour; B = air
TempG = 55;# [OC]
P = 1.0133*10**(5);# [N/square m]
Y_prime = 0.030;# [kg water/kg dry air]
#******#

MA = 18.02;# [kg/kmol]
MB = 28.97;# [kg/kmol]

# Percent Humidity
# From psychrometric chart, at 55 OC
Ys_prime = 0.115;# [kg water/kg dry air]
percent_Humidity = (Y_prime/Ys_prime)*100;
print"The sample has percent Humidity =",round(percent_Humidity,1),"%"

# Molal Absolute Humidity
Y = Y_prime*(MB/MA);# [kmol water/kmol dry air]
print"\n Molal Absolute Humidity of the sample is",round(Y,4)," kmol water/kmol dry air\n"

# Partial Pressure
pA = Y*P/(1+Y);# [N/square m]
print"The Partial Pressure Of Water is",int(pA)," N/square m\n"

# Relative Humidity
pa = 118*133.3;# [vapour pressure of water at 55 OC,N/square m]
relative_Humidity = (pA/pa)*100;
print"The sample has relative Humidity = ",round(relative_Humidity,1)," %\n"

# Dew Point
# From psychrometric chart,
dew_point = 31.5;# [OC]
print"Dew point Of the Sample is",dew_point," degree C\n"

# Humid Volume
# At 55 OC
vB = 0.93;# [specific volume of dry air,cubic m/kg]
vsB = 1.10;# [specific volume of saturated air,cubic m/kg]
vH = vB+((vsB-vB)*(percent_Humidity/100));# [cubic m/kg]
print"The Humid Volume of the Sample is ",round(vH,3)," cubic m/kg\n"

# Humid Heat
CB = 1005;# [J/kg.K]
CA = 1884;# [J/kg.K]
Cs = CB+(Y_prime*CA);# [J/kg]
print"The Humid Heat is ",round(Cs,1)," J/kg dry air.K\n"

# Enthalpy
HA = 56000;# [J/kg dry air]
HsA = 352000;# [J/kg dry air]
H_prime = HA+((HsA-HA)*(percent_Humidity/100));# [J/kg dry air]
print"The Enthalphy of the sample is ",round(H_prime/1000,1),"KJ/kg dry air\n"
# the answer is slightly different in textbook due to approximation 
Illustration 7.6 - Page: 234


The sample has percent Humidity = 26.1 %

 Molal Absolute Humidity of the sample is 0.0482  kmol water/kmol dry air

The Partial Pressure Of Water is 4662  N/square m

The sample has relative Humidity =  29.6  %

Dew point Of the Sample is 31.5  degree C

The Humid Volume of the Sample is  0.974  cubic m/kg

The Humid Heat is  1061.5  J/kg dry air.K

The Enthalphy of the sample is  133.2 KJ/kg dry air

Ex7.7: Page 236

In [24]:
# Illustration 7.7
# Page: 236

print'Illustration 7.7 - Page: 236\n\n'

# solution

#****Data****#
# A = water vapour; B = air
V = 100;# [m**3]
Tempi = 55;# [OC]
Tempf = 110;# [OC]
#*****#

# From Illustration 7.6
vH = 0.974;# [m**3/kg]
Cs = 1061.5;# [J/kg]
WB = V/vH;# [kg]
Q = WB*Cs*(Tempf-Tempi);# [J]
print"Heat required is ",round(Q,3)," J\n"
# the answer is slightly different in textbook due to approximation in book
Illustration 7.7 - Page: 236


Heat recquired is  5994096.509  J

Ex7.9: Page 240

In [31]:
# Illustration 7.9
# Page:240
from scipy.optimize import fsolve 
print'Illustration 7.9 - Page:240\n\n'

# solution

#****Data****#
Tempw = 35;# [OC]
Tempg = 65;# [OC]
#******#

# From psychrometric chart
lambda_w = 2419300;# [J/kg]
Y_prime_w = 0.0365;# [kg H2O/kg dry air]
# From fig 7.5(a)
hG_by_kY = 950;# [J/kg]
# From Eqn. 7.26
def f13(Y_prime):
    return (Tempg-Tempw)-((lambda_w*(Y_prime_w-Y_prime))/hG_by_kY)
Y_prime = fsolve(f13,2);# [kg H2O/kg dry air]
print"Humidity of air is",round(Y_prime[0],4),"kg H2O/kg dry air\n"
Illustration 7.9 - Page:240


Humidity of air is 0.0247 kg H2O/kg dry air

Ex7.10: Page 241

In [32]:
# Illustration 7.10
# Page:241

print'Illustration 7.10 - Page:241\n\n'

# solution

#****Data****#
Tg = 60;# [OC]
Y_prime = 0.050;# [kg toulene/kg air]
#*****#

# Wet Bulb temparature
Dab = 0.92*10**(-5);# [square m/s]
density_air = 1.060;# [kg/cubic cm];
viscocity_G = 1.95*10**(-5);# [kg/m.s]
Sc = viscocity_G/(density_air*Dab);
# From Eqn. 7.28
hG_by_kY = 1223*(Sc**0.567);# [J/kg.K]
# Soln. of Eqn. 7.26 by trial & error method:
# (Tg-Tw) = (Yas_prime-Y_prime)*(lambda_w/hG_by_kY)
Tw = 31.8;# [OC]
print"Wet Bulb Temparature:",Tw," degree C\n"

# Adiabatic Saturation Temparature
C_air = 1005;# [J/kg.K]
C_toulene = 1256;# [J/kg.K]
Cs = C_air+(C_toulene*Y_prime);# [J/kg.K]
# Soln. of Eqn. 7.21 by trial & error method:
# (Tg-Tas) = (Yas_prime-Y_prime)*(lambda_as/Cs)
Tas = 25.7;# [OC]
print"Adiabatic Saturation Temparature: ",round(Tas,1)," degree C\n"
Illustration 7.10 - Page:241


Wet Bulb Temparature: 31.8  degree C

Adiabatic Saturation Temparature:  25.7  degree C

Ex7.11: Page 249

In [81]:
# Illustration 7.11
# Page: 249

print'Illustration 7.11 - Page: 249\n\n'

# solution


#****Data****#
L_min = 2.27;# [kg/square m.s]
G_min = 2;# [kg/square m.s]
L2_prime = 15;# [kg/s]
Q = 270.0;# [W]
Templ2 = 45.0;# [OC]
Tempg1 = 30.0;# [OC]
Tempw1 = 24.0;# [OC]
Kya = 0.90;# [kg/cubic m.s]
#*******#

H1_prime = 72;# [kJ/kg dry air]
Y1_prime = 0.0160;# [kg water/kg dry air]
Templ1 = 29;# [OC]
Cal = 4.187;# [kJ/kg]

# Tower cross section Area:
Al = L2_prime/L_min;# [square m]
Ag = Gs/G_min;# [square m]
A = min(Al,Ag);# [square m]
Area = 3.25;
# From Eqn. 7.54
def f16(Z):
    return  Area-(Kya*Z/G_min)
Z = fsolve(f16,2);
print"The height of tower is",round(Z,2)," m\n"
NtoG = 3.25;
HtoG = G_min/Kya;# [m]

# Make up water
# Assuming the outlet air is essentially saturated:
Y2_prime = 0.0475;# [kg water/kg dry air]
E = G_min*(A)*(Y2_prime-Y1_prime);# [kg/s]
# Windage loss estimated as 0.2 percent
W = 0.002*L2_prime;# [kg/s]
ppm_blowdown = 2000;# [ppm]
ppm_makeup = 500;# [ppm]
# Since the weight fraction are proportional to the corresponding ppm values:
B = (E*ppm_makeup/(ppm_blowdown-ppm_makeup))-W;# [kg/s]
M = B+E+W;# [kg/s]
print"The makeup water requirement is estimated to be",round(M,2)," kg/s\n"
Illustration 7.11 - Page: 249


The height of tower is 7.22  m

The makeup water requirement is estimated to be 0.46  kg/s

Ex7.13: Page 254

In [137]:
# Illustration 7.13
# Page: 254


print'Illustration 7.13\n\n'

# solution

import math
from scipy.optimize import fsolve 
# Given
Tempg1=65;# [OC]
Y1_prime=0.0170;# [kg water/kg dry air]
# Using adiabatic satursion line on Fig. 7.5 (Pg 232)
Tempas=32;# [OC]
Yas_prime=0.0309;# [kg water/kg dry air]
Tempg2=45;# [OC]
Z=2;# [m]
#*******#

Y2_prime=0.0265;# [kg water/kg dry air]
def f19(Kya_by_Gs):
    return math.log((Yas_prime-Y1_prime)/(Yas_prime-Y2_prime))-(Kya_by_Gs*Z)
Kya_by_Gs=fsolve(f19,1);# [1/m]

# For the extended chamber:
Z=4;# [m]
def f20(Y2_prime):
    return math.log((Yas_prime-Y1_prime)/(Yas_prime-Y2_prime))-(Kya_by_Gs*Z)
Y2_prime=fsolve(f20,0.029);#[kg water/kg dry air] 
# With the same adiabatic curve:
Tempg2=34;# [OC] from the curve
print"The Outlet Conditions are:\n"
print"Absolute Humidity is",round(Y2_prime,4)," kg water/kg dry air\n"
print"Dry Bulb Temperature is",round(Tempg2), "degree C\n"
Illustration 7.13


The Outlet Conditions are:

Absolute Humidity is 0.0295  kg water/kg dry air

Dry Bulb Temperature is 34.0 degree C

Ex7.14: Page 256

In [1]:
# Illustration 7.14
# Page: 256

print'Illustration 7.14 - Page: 256\n\n'

# solution

import math
from scipy.optimize import fsolve
#****Data****#
# a = N2 b = CO
# Entering gas
Y1_prime = 0.0;# [kg water/kg dry air]
Pt = 1.0;# [atm]
Tempg1 = 315.0;# [OC]
G_prime = 5.0;# [square m/s]

# Temp of the tower:
Templ2 = 18.0;# [OC]
Density_L2 = 1000.0; #[kg/square m]
viscocity_L2 = 1.056*10**(-3);# [kg/m.s]
Tempg2 = 27.0;# [OC]

Mb = 28.0;# [kg/kmol]
Ma = 18.02;# [kg/kmol]
Density_G1 = (Mb/22.41)*(273/(Tempg1+273));# [kg/square m]
G1 = G_prime*(Density_G1);# [kg/s]

# Since the outlet gas is nearly saturated:
Y_prime = 0.024;# [kg water/kg dry air]
Y2_prime = 0.022;# [kg water/kg dry air, assumed]
G2 = G1*(1+Y2_prime);# [kg/s]
Mav = (1+Y2_prime)/((1/Mb)+(Y2_prime/Ma));# [kg/kmol]
Density_G2 = (Mav/22.4)*(273.0/(Templ2+273));# [kg/square m]
L2_by_G2 = 2.0;
abcissa = L2_by_G2*(Density_G2/(Density_L2-Density_G2))**(1/2);
# From Fig. 6.34:
# For a gas pressure drop of 400 N/square m/m
ordinate = 0.073;
# From Table 6.3:
Cf = 65.0;
J = 1.0;
def f21(G2_prime):
    return ((G2_prime**2)*Cf*(viscocity_L2**0.1)*J/(Density_G2*(Density_L2-Density_G2)))-ordinate
# Tentative data:
G2_prime = fsolve(f21,2);# [kg/square m.s]
Area = G1/G2_prime;# [square m]
dia = math.sqrt(4*Area/math.pi);# [m]

# Final data:
dia = 1.50;# [m]
Area = math.pi*dia**2.0/4;# [square m]
Gs_prime = G1/Area;# [kg/square m.s]
G2_prime = G2/Area;# [kg/square m.s]
L2_prime = L2_by_G2*G2_prime;# [kg/square m.s]
# From Eqn. 7.29:
def f22(L1_prime):
    return (L2_prime-L1_prime)-(Gs_prime*(Y2_prime-Y1_prime))
L1_prime = fsolve(f22,2);
Cb = 1089;# [J/kg.K]
Ca = 1884;# [J/kg.K]
Cs1 = Cb+(Y1_prime*Ca);# [J/(kg dry air).K]
Cs2 = Cb+(Y2_prime*Ca);# [J/(kg dry air).K]
Tempo = Templ2;# [base temp.,K]
Lambda = 2.46*10**6;# [J/kg]
CaL = 4187;# [J/kg K]
# From Eqn. 7.31:
def f23(Templ1):
    return ((L2_prime*CaL*(Templ2-Tempo))+(Gs_prime*Cs1*(Tempg1-Tempo)))-((L1_prime*CaL*(Templ1-Tempo))+(Gs_prime*(Cs2*(Tempg2-Tempo))+(Y2_prime*Lambda)))
Templ1 = fsolve(f23,2);
# At Templ1 = 49.2 OC
viscocity_L = 0.557*10**(-3);# [kg/m.s]
Density_L = 989.0;# [kg/square m]
K = 0.64;# [w/m.K]
Prl = CaL*viscocity_L/K;

# For Entering Gas:
viscocity_G1 = 0.0288*10**(-3);# [kg*/m.s]
Dab = 0.8089*10**(-4);# [square m/s]
ScG = viscocity_G1/(Density_G1*Dab);
PrG = 0.74;

# From Illustration 6.7:
a = 53.1;# [square m/square m]
Fga = 0.0736;# [kmol/square m]
Hga = 4440.0;# [W/square m.K]
Hla = 350500.0;# [W/square m.K]
# At the bottom, by several trial:
Tempi = 50.3;# [OC]
pai = 93.9/760;# [atm]
paG = 0;# [atm]
# By Eqn. 7.64:
dY_prime_by_dZ = -(Ma*Fga/Gs_prime)*math.log((1-(pai/Pt))/(1-(paG/Pt)));# [(kg H2O/kg dry gas)/m]
Hg_primea = -(Gs_prime*Ca*dY_prime_by_dZ)/(1-math.exp((Gs_prime*Ca*dY_prime_by_dZ)/(Hga)));# [W/square m.K]
dTempg_by_dZ = -(Hg_primea*(Tempg1-Tempi)/(Gs_prime*Cs1));# [OC/m]
Tempi = (Templ1)+((Gs_prime*(Cs1*dTempg_by_dZ)+((Ca*(Tempg1))-(CaL*(Templ1))+(((CaL-Ca)*(Tempo))+Lambda))*dY_prime_by_dZ)/((Gs_prime*CaL*dY_prime_by_dZ)-Hla));#[OC]
# Assume:
delta_Tempg = -30;# [OC]
delta_Z = delta_Tempg/(dTempg_by_dZ);# [m]
Tempg = Tempg1+delta_Tempg;# [OC]
Y_prime = Y1_prime+(dY_prime_by_dZ)*delta_Z;# [kg H2O/kg dry gas]
paG = Y_prime/(Y_prime+(Ma/Mb));# [atm]
Cs = Cb+Ca*(Y_prime);# [J/(kg dry air).K]
# Water balance, From Eqn. 7.29:
def f24(L_prime):
    return (L2_prime-L_prime)-(Gs_prime*(Y_prime-Y1_prime))
L_prime = fsolve(f24,2);# [kg/square m.s]

def f25(Templ):
    return ((L_prime*CaL*(Templ-Tempo))+(Gs_prime*Cs1*(Tempg1-Tempo)))-((L1_prime*CaL*(Templ1-Tempo))+(Gs_prime*(Cs*(Tempg-Tempo))+(Y_prime*Lambda)))
Templ = fsolve(f25,2);

# This process is repeated several times until gas temp falls to Tempg2
Z = 1.54;# [m] Z = sum of all delta_Z
# The value of Y2_prime was calculated to be 0.0222 which is sufficiently close to the assumed value.
print"The diameter of tower is ",dia," m\n"
print"The packed height is",Z, "m\n"
Illustration 7.14 - Page: 256


The diameter of tower is  1.5  m

The packed height is 1.54 m

Ex7.15: Page 267

In [1]:
# Illustration 7.15
# Page: 267

print'Illustration 7.15 - Page: 267\n\n'

from scipy.optimize import fsolve
import math
import numpy
# solution

#***Data***#
w = 0.75;# [m]
OD = 19.05/1000;# [m]
l = 3.75;# [m]
n = 20;
t = 1.65/1000;# [m]
Ws = 2.3;# [kg/s]
Wal = 10.0;# [kg/s]
Wt = 4.0;# [kg/s]
Density = 800;# [kg/cubic m]
viscocity = 0.005;# [kg/m.s]
K = 0.1436;# [W/m.K]
Ct = 2010.0;# [J/kg.K]
Cal = 4187.0;# [J/kg.K]
Y1_prime = 0.01;# [kg H2O/kg dry air]
Y2_prime = 0.06;# [kg H2O/kg dry air]
TempT = 95.0;# [OC]
#*****#

Free_area = (w-(n*OD))*l;# [square m]
Gs_min = 2.3/Free_area;# [kg/square m.s]
M1 = 1.461;Yav_prime = (Y1_prime+Y2_prime)/2;# [kg H2O/kg dry air]
# From Eqn. 7.86:
ky = 0.0493*(Gs_min*(1+Yav_prime))**0.905;# [kg/square m.s.delta_Y_prime]
# From Fig. 7.5:
H1_prime = 56000.0;# [J/kg]
Ao = 400*math.pi*OD*l;# [square m]
# Cooling water is distributed over 40 tubes & since tubes are staggered
geta = Wal/(40.0*2*l);# [kg/m.s]
geta_by_OD = geta/OD;# [kg/square m.s]
# Assume:
TempL = 28.0;# [OC]
# From Eqn. 7.84:
hL_prime = (982+(15.58*TempL))*(geta_by_OD**(1/3));# [W/square m.K]
# From Eqn. 7.85:
hL_dprime = 11360;# [W/square m.K]
# From Fig. 7.5 (Pg 232)
m = 5000.0;# [J/kg.K]
Ky = 1.0/((1/ky)+(m/hL_dprime));
ID = (OD-(2.0*t));# [m]
Ai = math.pi*(ID**2)/4;# [square m]
Gt_prime = Wt/(n*Ai);# [kg/square m.s]
M2 = -0.7204;Re = ID*Gt_prime/viscocity;
Pr = Ct*viscocity/K;
# From a standard correlation:
hT = 364.0;# [W/square m.K]
Dav = (ID+OD)/2.0;# [m]
Zm = (OD-ID)/2;# [m]
Km = 112.5;# [W/m.K]
# From Eqn. 7.67:
Uo = 1/((OD/(ID*hT))+((OD/Dav)*(Zm/Km))+(1/hL_prime));# [W/square m.K]
# From Eqn. 7.75:
alpha1 = -(((Uo*Ao)/(Wt*Ct))+((Uo*Ao)/(Wal*Cal)));
alpha2 = m*Uo*Ao/(Wt*Ct);
# From Eqn. 7.76:
beeta1 = Ky*Ao/(Wal*Cal);
beeta2 = -((m*Ky*Ao/(Wal*Cal))-(Ky*Ao/Ws));
def f26(r):
    return (r**2)+((alpha1+beeta2)*r)+((alpha1*beeta2)-(alpha2*beeta1))
r1 = fsolve(f26,10);
r2 = fsolve(f26,0);
beeta2 = 1.402;
# From Eqn. 7.83:
# N1-(M1*(r1+alpha1)/beeta1) = 0............................................(1)
# N2-(M2*(r2+alpha2)/beeta2) = 0............................................(2)
# From Eqn. 7.77:
# At the top:
x1 = 1.0;
# TempL2+(M1*exp(r1*x1))+(M2*exp(-(r2*x1))) = TempL.........................(3)
# From Eqn. 7.78:
# At the bottom:
x2 = 0.0;
# H1_star-N1-N2 = H1_prime..................................................(4)
# From Eqn. 7.80:
# ((M1/r1)*(exp(r1)-1))+((M2*r2)*(exp(r2)-1)) = (Tempt-TempL)...............(5)
# From Eqn. 7.81:
# ((N1/r1)*(exp(r1)-1))+((N2*r2)*(exp(r2)-1)) = (H1_star-H1_prime)..........(6)
# From Eqn. 7.91 & Eqn. 7.92:
# Uo*Ao*(TempT-TempL)=Ky*Ao*(H1_star-H1_prime)..............................(7)

# Elimination of M's & N's by solving Eqn. (1) to (4) and (7) simultaneously:
# and from Fig. 7.5 (Pg 232):
TempL1=28.0;# [OC]
H1_star=(Uo*Ao*(TempT-TempL)/(Ky*Ao))+H1_prime;# [J/kmol]


N1 = 3594.0*M1
N2 =-43288.0*M2;

# By Eqn. 5
delta_Temp = ((M1/r1)*(math.exp(r1)-1))+((M2*r2)*(math.exp(r2)-1));# [OC]
Q = Uo*delta_Temp*Ao;
TempT1 = TempT-(Q/(Wt*Ct));# [OC]
H2_prime = Q/(Ws)+H1_prime;# [J/kg]
print"Temperature to which oil was cooled:",int(TempT1)," degree C\n"
# The solution in the textbook is wrong 
Illustration 7.15 - Page: 267


Temperature to which oil was cooled: 57  degree C