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)"
```

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."
```

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
Hr=4*sigma*math.pow(Tm,3)*e; #radiative heat transfer coefficient,W/(m**2*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."
```

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"
```

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"
```

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."
```

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
```