# Chapter 2: Heat conduction concepts, thermal resistance, and the overall heat transfer coefficient¶

### Example 2.3, Page number: 64¶

In [1]:
from __future__ import division

#Variables
l=1;                                          #  tube  length,  m
m=0.01;                                       #  mass  fraction
D12=2.84*10**-5;                              #  diffusivity,  m**2/s
a=1.18;                                       #  density,  kg/m**3

#Calculations
J=a*D12*m/l;								  #steady  state  flux  of  water  from  one  side  to  the  other,kg/(m**2*s)

#Results
print "Steady flux of water is",J,"kg/(m^2*s)"

Steady flux of water is 3.3512e-07 kg/(m^2*s)


### Example 2.7, Page number: 72¶

In [3]:
from __future__ import division
import math

#Variables
h=20;                                  #convective  heat  transfer  coefficient,  W/(m**2*K)
k=0.074;                               #thermal  conductivity,  J/(m*K)

#Calculations
Ro=k/h;                                #  formula  for  critical  thickness  of  insulation, m.

#Results
print "Critical thickness of insulation is :",Ro,"m\n"
print "Insulation will not even start to do any good until ratio of outer radius and inner radius is 2.32 or outer radius is  0.0058 m."

Critical thickness of insulation is : 0.0037 m

Insulation will not even start to do any good until ratio of outer radius and inner radius is 2.32 or outer radius is  0.0058 m.


### Example 2.8, Page number: 76¶

In [2]:
from __future__ import division
import math

#Variables
P=0.1;                                               #dissipating  power,W
D=0.0036;                                            #outer  diameter  of  cylinder,  m
l=0.01;                                              #length  of  cylinder,  m
T=308;                                               #temperature  of  air  in  the  cabinet,K
Test=323;                                            #Estimated temp of resistor, K
h=13;                                                #convection  coefficient,  W/(m**2*K)
e=0.9;						                         #emmisivity
A=1.33*math.pow(10,-4);                              #area  of  ressistor's  surface,  m**2
sigma=5.67*math.pow(10,-8);                          #Stefan-Boltzmann constant, Wm**-2K**-4

#Calculations
Tm=(T+Test)/2;                                       #ressistor's  temperature  at  50  K
Rteq=1/(A*(Hr+h));                                   #Equivalent thermal resistance K/W
Tres=T+P*Rteq;                                       #Resistor's temp. C
#we  guessed  a  ressistor's  temperature  of  323K  in  finding  Hr,recomputing  with  this  higher  temperature,
#we  have  Tm=327K  and  Hr=7.17W/(m**2*K).  if  we  repeat  the  rest  of  calculations,  we  get  a  new  value  Tres=345.3K,
#since  the  use  of  hr  is  an  approximation,  we  should  check  its  applicability:  1/4*((345.3-308)/327)**2=0.00325<<1,
#in  this  case,  the  approximation  is  a  very  good  one.
Tr=Tres-273.06;                                      #Resistor's temp. , K

#Results
print "Temperature of ressistor is :",round(Tr,3),"K\n"
print "Since 1/4*(temperature diffference/mean temperature)= 1/4*((72.3-35)/327)^2=0.00325<<1, in this case, the approximation is a very good one."

Temperature of ressistor is : 73.676 K

Since 1/4*(temperature diffference/mean temperature)= 1/4*((72.3-35)/327)^2=0.00325<<1, in this case, the approximation is a very good one.


### Example 2.9, Page number: 77¶

In [4]:
from __future__ import division
import math

#Variables
k=10;                                         #thermal  conductivity  of  ressistor,  W/(m*K)
a=2000;                                       #density  of  ressistor,  kg/m**3
l=0.01;                                       #length  of  cylinder,  m
A=1.33*math.pow(10,-4);                       #area  of  ressistor's  surface,  m**2
T1=308;                                       #temperature  of  air  in  the  cabinet,K
Cp=700;                                       #heat  capacity  of  ressistor,  J/kg/K
Heff=18.44;                                   #the  effective  heat    transfer  coefficient  of  parallel  convection  and  radiation  process,    W/(m**2*K)
D=0.0036;                                     #outer  diameter  of  cylinder,  m

#Calculations
Bi=Heff*(D/2)/k;                              #Biot no.
T=a*Cp*math.pi*l*math.pow(D,2)/(4*Heff*A);    #since  from  previous  example,To=72.3C,  we  have  Tres=T1+(To-T)*exp(-t/T),Tres=308+(37.3)*.exp(-t/T).  95%  of  the  temperature  drop  has  occured  when  t=T*3=174s.
t=3*T;                                        #Time for 95 percent cooling of ressistor, seconds.

#Results
print "Time for 95 percent cooling of ressistor is :",t,"s\n"

Time for 95 percent cooling of ressistor is : 174.313737829 s



### Example 2.10, Page number: 79¶

In [6]:
from __future__ import division
import math

#Variables
h1=200;                              #convective  heat  transfer  coefficient,  W/(m**2*K)
a=1/160000;                          #1/a=l/Kal,  l=0.001m,  Kal=160  W/(m*K)
h2=5000;                             #convective  heat  transfer  coefficient  during  boiling,W/(m**2*K)

#Calculations
U=1/(1/h1+a+1/h2);   				 #Overall heat transfer coefficient,W/(m^2*K)

#Results
print "Overall heat transfer coefficient is :",round(U,3),"W/(m^2*K)\n"

Overall heat transfer coefficient is : 192.077 W/(m^2*K)



### Example 2.12, Page number: 85¶

In [7]:
from __future__ import division
import math

#Variables
Rf=0.0005;                            #fouling  ressistance,m**2*K/W
U=5;                                  #heat  transfer  coefficient,W/(m**2*K)

#Calculations
Ucor=(U*Rf+1)/(U);					  #Corrected heat transfer coefficient, W/m^2.K

#Results
print "Corrected heat transfer coefficient is :", Ucor,"W/(m^2*K)\n Therefore the fouling is entirely irrelevant to domestic heat holds."

Corrected heat transfer coefficient is : 0.2005 W/(m^2*K)
Therefore the fouling is entirely irrelevant to domestic heat holds.


### Example 2.13, Page number: 85¶

In [8]:
from __future__ import division

#Variables
U1=4000;                                       # overall  heat  transfer  coefficient  of  water  cooled  steam  condenser,  W/(m**2*K)
Rf1=0.0006;                                    # lower  limit  of  fouling  ressistance  of  water  side,  m**2*K/W
Rf2=0.0020;                                    # upper  limit  of  fouling  ressistance  of  water  side,  m**2*K/W

#Calculations
U2=U1/(U1*Rf1+1);							   #Upper  limit of the corrected overall heat transfer coefficient
U3=U1/(U1*Rf2+1);							   #Lower limit of corrected overall heat transfer coefficient

#Results
print "Upper  limit of the corrected overall heat transfer coefficient is :",round(U2,3),"W/(m^2*K)\n"
print "Lower limit of corrected overall heat transfer coefficient is :",round(U3,3),"W/m^2/K, U is reduced from 4000 to between 444 and 1176 W/(m^2*K),fouling is crucial in this case and engineering was in serious error.\n"
#end

Upper  limit of the corrected overall heat transfer coefficient is : 1176.471 W/(m^2*K)

Lower limit of corrected overall heat transfer coefficient is : 444.444 W/m^2/K, U is reduced from 4000 to between 444 and 1176 W/(m^2*K),fouling is crucial in this case and engineering was in serious error.