In [5]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.1 "
#Temperature Inside in F
Ti = 55;
#Temperature outside in F
To = 45;
#Thickness of the wall in ft
t = 1;
#Heat loss through the wall in Btu/h-ft2
q = 3.4;
#Converting Btu/h-ft2 to W/m2
print "Heat loss through the wall in W/m2 is"
#Heat loss through the wall in W/m2
print "qdash = ",(q*0.2931)/0.0929
#Heat loss for a 100ft2 surface over a 24-h period
print "Heat loss for a 100ft2 surface over a 24-h period in Btu is"
#Heat loss for a 100ft2 surface over a 24-h period in Btu
print "Q = (q*100)*24
#Q in SI units i.e. kWh
print "Q = ",(Q*0.2931)/1000;
#At price of 10c/kWh, the total price shall be
print "So, the total price in c are"
#Total price in c
print "Price = ",round(10*Q)
```

In [12]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.2 "
#Thermal conductivity of window glass in W/m-K
k = 0.81;
#Height of the glass in m
h = 1;
#Width of the glass in m
w = 0.5;
#Thickness of the glass in m
t = 0.005;
#Outside temperature in C
T2 = 24;
#Inside temperature in C
T1 = 24.5;
#Assume that steady state exists and that the temperature is uniform over the inner and outer surfaces
#Cross sectional area in m2
A = h*w;
print "Thermal resistance to conduction in K/W is"
#Thermal resistance to conduction in K/W
R=t/(k*A)
print "R = ",round(R,4)
#The rate of heat loss from the interior to the exterior surface is
#obtained by dividing temperature difference from the thermal resistence
print "Heat loss in W from the window glass is"
#Heat loss in W
print "q = ",int((T1-T2)/R)
```

In [14]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.3 "
#Area of room in m2 is given as
A = 20*20;
#Air temperature in C
Tair = -3;
#Roof temperature in C
Troof = 27;
#Heat transfer coefficient in W/m2-K
h = 10;
#Assume that steady state exists and the direction of heat flow is from the
#roof to the air i.e higher to lower temperature (as it should be).
print " The rate of heat transfer by convection from the roof to the air in W"
#The rate of heat transfer by convection from the roof to the air in W
print "q = ",(-h*A)*(Troof-Tair)
```

In [19]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.4 "
#Diameter of rod in m
d = 0.02;
# Emissivity and temperautre of rod in K
epsilon = 0.9;
T1 = 1000;
#Temperature of walls of furnace
T2 = 800;
#Assuming steady state has been reached.
#Since the walls of the furnace completely enclose the heating rod, all the radiant energy emitted by the surface of the rod is intercepted by the furnace walls
#From eq. 1.17, net heat loss can be given
print "Net heat loss per unit length considering 1m length in W"
#Area in m2
A =(math.pi*d)*1;
#Constant sigma in W/m2-K4
sigma = 0.0000000567;
#Net heat loss per unit length considering 1m length in W
q=((A*sigma)*epsilon)*(T1**4-T2**4)
print" q = ",round(q)
#From eq. 1.21 radiation heat transfer coefficient in W/m2-K is
print "Radiation heat transfer coefficient in W/m2-K is"
#Radiation heat transfer coefficient in W/m2-K
print "hr = ",round(((epsilon*sigma)*(T1**4-T2**4))/(T1-T2))
```

In [22]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.5 "
#Thickness of inside steel in m and thermal conductivity in W/m-k
t1 = 0.005;
k1 = 40;
#Thickness of outside brick in m and thermal conductivity in W/m-k
t2 = 0.1;
k2 = 2.5;
#Inside temperature in C
T1 = 900;
#Outside temperature in C
To = 460;
#Assuming the condition of steady state and umath.sing Eq. 1.24
print "The rate of heat loss per unit area in W/m2 is"
#The rate of heat loss per unit area in W/m2
qk = (T1-To)/(t1/k1+t2/k2)
print int(qk)
print "Temperature at the interface in K is given as"
#Temperature at the interface in K
T2 = T1-(qk*t1)/k1
print round(T2,1)
```

In [29]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.6 "
#Thermal conductivity of aluminium in W/m-K
k = 240.0;
#Thickness of each plate in m
t = 0.01;
#Temperature at the surfaces of plates in C is given as
Ts1 = 395.0;
Ts3 = 405.0;
#From Table 1.6 the contact resistance at the interface in K/W is
R2 = 0.000275;
#Thermal resistance of the plates in K/W is
R1 = t/k;
R3 = t/k;
print "Heat flux in W/m2-K is"
#Heat flux in W/m2-K
q = (Ts3-Ts1)/(R1+R2+R3)
print "{:.2e}".format(q)
#Since the temperature drop in each section of this one-dimensional system is propor-tional to the resistance.
print "Temperature drop due to contact resistance in degree C is"
#Temperature drop due to contact resistance in degree C
deltaT = (R2/(R1+R2+R3))*(Ts3-Ts1)
print round(deltaT,2)
```

In [32]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.7 "
#Because of symmetry, we need to calculate for only one half of the system
#Thickness of firebrick in inches
L1 = 1.0;
#Thermal conductivity of firebrick in Btu/h-ft-F
kb = 1.0;
#Thickness of steel plate in inches
L3 = 1/4.0;
#Thermal conductivity of steel in Btu/h-ft-F
ks = 30;
#Average height of asperities in inches is given as
L2 = 1/32.0;
#Temperature difference between the steel plates in F is
deltaT = 600.0;
#The thermal resistance of the steel plate is, on the basis of a unit wall area, equal to
R3 = L3/(12*ks);#12 is added to convert ft to in
#The thermal resistance of the brick asperities is, on the basis of a unit wall area, equal to
R4 = L2/((0.3*12)*kb);#Considering the 30 percent area
#At temperature of 300F, thermal conductivity of air in Btu/h-ft-F is
ka = 0.02;
# Thermal resistance of the air trapped between the asperities, is, on the basis of a unit area, equal to
R5 = L2/((0.7*12)*ka);#Considering the other 70 percent area
#Since R4 and R5 are in parallel, so there combined resistance is
R2 = (R4*R5)/(R4+R5);
#The thermal resistance of half of the solid brick is
R1 = L1/(12*kb);
#The overall unit conductance for half the composite wall in Btu/h-ft2-F is then
kk = 0.5/(R1+R2+R3);
print "The rate of heat flow per unit area in Btu/h-ft2 is"
#The rate of heat flow per unit area in Btu/h-ft2
q = kk*deltaT
print round(q,2)
# the answer is slightly different in textbook due to approximation
```

In [34]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.8 "
#Length for heat transfer for stainless steel in m
Lss = 0.1;
#Area for heat transfer for stainless steel in m2
A = 0.01;
#Thermal conductivity for stainless steel in W/m-K
kss = 144;
#Length for heat transfer for Duralumin in m
La1 = 0.02;
#Area for heat transfer for Duralumin in m2
A = 0.01;
#Thermal conductivity for Duralumin in W/m-K
ka1 = 164;
#Resistance in case of steel in K/W
Rk1 = Lss/(A*kss);
#Resistance in case of Duralumin in K/W
Rk2 = La1/(A*ka1);
#From Fig. 1.20, contact resistance in K/W
Ri = 0.05;
#Total resistance to heat transfer in K/W
Rtotal = Rk1+Rk2+Ri;
#Temperature diff. is given in K
deltaT = 40;
print "Maximum allowable rate of heat dissipation in W is"
#Maximum allowable rate of heat dissipation in W
q = deltaT/Rtotal
print int(q)
```

In [42]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.9 "
#Cross sectional area in m2
A = 1.0;
#Heat transfer coefficient on hot side in W/m2-K
hchot = 10.0;
#Heat transfer coefficient on cold side in W/m2-K
hccold = 40.0;
#Length for heat transfer in m
L = 0.1;
#Thermal conductivity in W/m-K
k = 0.7;
#Resistances in K/w
R1 = 1/(hchot*A);
R2 = L/(k*A);
R3 = 1/(hccold*A);
#Total resistance
Rtotal = R1+R2+R3;
#Temperature on hot side in K
T1 = 330.0;
#Temperature on cold side in K
T2 = 270.0;
print "Rate of heat transfer per unit area in W is"
#Rate of heat transfer per unit area in W
q = (T1-T2)/(R1+R2+R3)
print round(q)
```

In [46]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.10 "
#diameter of pipe in m
d = 0.5;
#Epsilon is given as
epsilon = 0.9;
#sigma(constant) in SI units is
sigma = 0.0000000567;
#Temperatures in K are given as
T1 = 500;
T2 = 300;
#Radiation heat transfer coefficient in W/m2K
hr = ((sigma*epsilon)*(T1*T1+T2*T2))*(T1+T2);
#Convection heat transfer coefficient in W/m2K
hc = 20;
#total heat transfer coefficient in W/m2K
h = hc+hr;
print "Rate of heat loss per meter in W/m is"
#Rate of heat loss per meter in W/m
q = ((math.pi*d)*h)*(T1-T2)
print round(q,2)
# the answer is slightly different due to approximation
```

In [54]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.11 "
#Hot-gas temperature in K
Tgh = 1300.0;
#Heat transfer coefficient on hot side in W/m2K
h1 = 200.0;
#Heat transfer coefficient on cold side in W/m2K
h3 = 400.0;
#Coolant temperature in K
Tgc = 300.0;
#Max temp. in C
Tsg = 800.0;
#Maximum permissible unit thermal resistance per square meter of the metal wall in K/W
R2 = ((Tgh-Tgc)*(1/h1)/(Tgh-Tsg))-1/h1-1/h3;
print "Maximum permissible unit thermal resistance per square meter of the metal wall in m2.K/W is"
print R2
# The answer is wrong in the textbook
```

In [57]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.12 "
# total length of metal sheet in m
L = 0.625/39.4;
# we estimate the thermal conductivity of the metal sheets to be approximately 43 W/m K
k = 43;
# therefore the resistance in K/W offered by metal sheey
R = L/k;
#heat loss in W/m2 is given as
q = 1200;
# overall heat transfer coefficient between the gas and the door is given
# in W/m2K
U = 20;
#The temperature drop between the gas and the interior surface of the door at the specified heat flux is
deltaT1 = q/U;
#Hence, the temperature of the Inconel will be in degree C
T = 1200-deltaT1;
#The heat transfer coefficient between the outer surface of the door and
#the surroundings at 20°C in W/m2K
h = 5;
#The temperature drop at the outer surface in degree C is
deltaT2 = q/h;
#Selecting milled alumina-silica chips as insulator (Fig 1.31 on page 48)
# Hence, temperature difference across the insulation is
deltaT3 = T-deltaT1-deltaT2;
#thermal conductivity for milled alumina-silica chips in W/mK is
k = 0.27;
print "The insulation thickness in m is"
#The insulation thickness in m
L = (k*deltaT3)/q
print round(L,1)
```

In [59]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.13 "
#Temperature of air in degree K
Tair = 300;
#Heat transfer coefficient in W/m2K
h = 10.0;
print "Part a"
#Radiation solar flux in W/m2
q = 500.0;
#Ambient temperature in K
Tsurr = 50.0;
print "Solving energy balance equaiton by trial and error for the roof temperature, we get temp. in degree K"
#Room temperature in degree K
Troof = 303
print Troof
print "Part b"
#No heat flux, energy balance equaiton is modified
print "Room temperature in degree K"
#Room temperature in degree K
Troof = 270
print Troof
```

In [60]:

```
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 1 Example # 1.14 "
print "The given example is theoretical and does not involve any numerical computation"
```