In [7]:

```
from __future__ import division
import math
#Variables
T1=20; #Entering Temperature of Water, K
T2=40; #Exit Temperature of water, K
m=25/60 #Condensation rate of steam, kg/s
T3=60; #Condensation Temperature,K
A=12; #area of exchanger, m**2
h=2358.7*math.pow(10,3); #latent heat, J/kg
Cp=4174; #Specific heat of water, J/kg K
#Calculations
U=(m*h)/(A*((T2-T1)/math.log((T3-T1)/(T3-T2))));#Overall heat transfer coefficient, W/(m^2*K)
Mh=(m*h)/(Cp*(T2-T1)); #Required flow of water, kg/s
#Results
print "Overall heat transfer coefficient is :",round(U,1),"W/(m^2*K)\n"
print "Required flow of water is :",round(Mh,2),"kg/s\n"
```

In [8]:

```
from __future__ import division
import math
#Variables
m=5.795; #flow rate of oil, kg/s
T1=454; #Entering Temperature of oil, K
T2=311; #Exit Temperature of oil, K
T3=305; # Entering Temperature of water, K
T4=322; #Exit Temperature of water, K
c=2282; #heat capacity, J/(kg*K)
U=416; #overall heat transfer coefficient , J/(m**2*K*s)
F=0.92; #Correction factor for 2 shell and 4 tube-pass exchanger,
#since R=(T1-T2)/(T4-T3)=8.412 >1, P=(T4-T3)/(T1-T2)=0.114,we can get this value of F by using value of P =R*0.114
#Calculations
A=(m*c*(T1-T2))/(U*F*((T1-T4-T2+T3)/math.log((T1-T4)/(T2-T3))));#Area for heat exchanger, m^2.
#Results
print "Area for heat exchanger is :",round(A,3),"m^2\n"
```

In [2]:

```
from __future__ import division
import math
#Variables
T1=313; #entering temperature of cold water, K
T2=423; #Entering temperature of hot water, K
Cc=20000; #heat capacity of cold water, W/K
Ch=10000; #heat capacity of hot water, W/K
A=30; #area, m**2
U=500; #overall heat transfer coefficient, w/(m**2*K)
e=0.596; #no. of transfer units(NTU)=(U*A)/Ch=1.5, the effectiveness of heat exchanger e can be found by using this value of NTU
#Calculations
Q=e*Ch*(T2-T1); #Heat transfer, W
Q1=Q/1000 #Heat transfer, KW
Texh=T2-Q/Ch; #exit hot water temperature, K
Tn1=Texh-273; #exit hot water temperature, C
Texc=T1+Q/Cc #exit cold water temperature, K
Tn2=Texc-273; #exit cold water temperature, C
#Results
print "Heat transfer is :",Q1,"KW\n"
print "The exit hot water temperature is:",Tn1,"C\n"
print "The exit cold water temperature is :",Tn2,"C\n"
```

In [19]:

```
from __future__ import division
import math
#Variables
T1=313; #entering temperature of cold water, K
T2=423; #Entering temperature of hot water, K
T3=363; #Exit temperature of hot water, K
Cc=20000; #heat capacity of cold water, W/K
Ch=10000; #heat capacity of hot water, W/K
U=500; #overall heat transfer coefficient, w/(m**2*K)
#Calculations
T4=T1+(Ch/Cc)*(T2-T3); #Exit cold fluid temp. K
e=(T2-T3)/(T2-T1); #Effectiveness method
NTU=1.15; #No. of transfer unit
A1=Ch*(NTU)/U; # since NTU=1.15=U*A/Ch, Area can be found by using this formula
#another way to calculate the area is by using log mean diameter method
LMTD=(T2-T1-T3+T4)/math.log((T2-T1)/(T3-T4)); #Logarithmic mean temp. difference
A2=Ch*(T2-T3)/(U*LMTD); #Aera by method 2, in meters^2.
#Results
print "Area is :",A1,"m^2\n"
print "Area is :",round(A2,3),"m^2\n"
print "There is difference of 1 percent in answers which reflects graph reading inaccuracy."
# we can see that area calulated is same in above 2 methods.
```