# Chapter 3 - Heat exchanger design¶

### Example 3.3, Page number: 113¶

In :
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"

Overall heat transfer coefficient is : 2838.4 W/(m^2*K)

Required flow of water is : 11.77 kg/s



### Example 3.4, Page number: 117¶

In :
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"

Area for heat exchanger is : 121.216 m^2



### Example 3.5, Page number: 112¶

In :
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"

Heat transfer is : 655.6 KW

The exit hot water temperature is: 84.44 C

The exit cold water temperature is : 72.78 C



### Example 3.6, Page number: 123¶

In :
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.

Area is : 23.0 m^2

Area is : 22.73 m^2

There is difference of 1 percent in answers which reflects graph reading inaccuracy.