In [1]:

```
# Variables
l = 5; #Length of the wall in m
h = 4; #Height of the wall in m
L = 0.25; #Thickness of the wall in m
T = [110,40]; #Temperature on the inner and outer surface in degree C
k = 0.7; #Thermal conductivity in W/m.K
x = 0.20; #Distance from the inner wall in m
# Calculations
A = l*h; #Arear of the wall in m**2
Q = (k*A*(T[0]-T[1]))/L; #Heat transfer rate in W
T = (((T[1]-T[0])*x)/L)+T[0]; #Temperature at interior point of the wall, 20 cm distant from the inner wall in degree C
# Results
print 'a)Heat transfer rate is %i W \n \
b)Temperature at interior point of the wall, 20 cm distant from the inner wall is %i degree C'%(Q, T)
```

In [3]:

```
import math
# Variables
Di = 0.05; #Inner diameter of hollow cylinder in m
Do = 0.1; #Outer diameter of hollow cylinder in m
T = [200,100]; #Inner and outer surface temperature in degree C
k = 70; #Thermal conductivity in W/m.K
# Calculations
ro = (Do/2); #Outer radius of hollow cylinder in m
ri = (Di/2); #Inner radius of hollow cylinder in m
Q = ((2*3.14*k*(T[0]-T[1]))/(math.log(ro/ri))); #Heat transfer rate in W
r1 = (ro+ri)/2; #Radius at halfway between ro and ri in m
T1 = T[0]-((T[0]-T[1])*(math.log(r1/ri)/(math.log(ro/ri)))); #Temperature of the point halfway between the inner and outer surface in degree C
# Results
print 'Heat transfer rate is %3.1f W /m \n \
Temperature of the point halfway between the inner and outer surface is %3.1f degree C'%(Q,T1)
```

In [4]:

```
# Variables
Di = 0.1; #Inner diameter of hollow sphere in m
Do = 0.3; #Outer diameter of hollow sphere in m
k = 50. #Thermal conductivity in W/m.K
T = [300,100]; #Inner and outer surface temperature in degree C
# Calculations
ro = (Do/2); #Outer radius of hollow sphere in m
ri = (Di/2); #Inner radius of hollow sphere in m
Q = ((4*3.14*ro*ri*k*(T[0]-T[1]))/(ro-ri))/1000; #Heat transfer rate in W
r = ri+(0.25*(ro-ri)); #The value at one-fourth way of te inner and outer surfaces in m
T = ((ro*(r-ri)*(T[1]-T[0]))/(r*(ro-ri)))+T[0]; #Temperature at a point a quarter of the way between the inner and outer surfaces in degree C
# Results
print 'Heat flow rate through the sphere is %3.2f kW \n \
Temperature at a point a quarter of the way between the inner and outer surfaces is %i degree C'%(Q,T)
```

In [1]:

```
# Variables
L = 0.4; #Thickness of the furnace in m
T = [300,50]; #Surface temperatures in degree C
#k = 0.005T-5*10**-6T**2
# Calculations
q = ((1./L)*(((0.005/2)*(T[0]**2-T[1]**2))-((5*10**-6*(T[0]**3-T[1]**3))/3))); #Heat loss per square meter surface area in W/m**2
# Results
print 'Heat loss per square meter surface area is %3.0f W/m**2'%(q)
```

In [6]:

```
# Variables
L = 0.2; #Thickness of the wall in m
T = [1000,200]; #Surface temperatures in degree C
ko = 0.813; #Value of thermal conductivity at T = 0 in W/m.K
b = 0.0007158; #Temperature coefficient of thermal conductivity in 1./K
# Calculations
km = ko*(1+((b*(T[0]+T[1]))/2)); #Constant thermal conductivity in W/m.K
q = ((km*(T[0]-T[1]))/L); #Rate of heat flow in W/m**2
# Results
print 'Rate of heat flow is %3.0f W/m**2'%(q)
```

In [7]:

```
import math
# Variables
r = [0.01,0.02]; #Inner and outer radius of a copper cylinder in m
T = [310,290]; #Inner and Outer surface temperature in degree C
ko = 371.9; #Value of thermal conductivity at T = 0 in W/m.K
b = (9.25*10**-5); #Temperature coefficient of thermal conductivity in 1./K
# Calculations
Tm = ((T[0]-150)+(T[1]-150))/2; #Mean temperature in degree C
km = ko*(1-(b*Tm)); #Constant thermal conductivity in W/m.K
q = ((2*3.14*km*(T[0]-T[1]))/math.log(r[1]/r[0]))/1000; #Heat loss per unit length in kW/m
# Results
print 'Heat loss per unit length is %3.2f kW/m'%(q)
```

In [1]:

```
# Variables
L1 = 0.5; #Thickness of the wall in m
k1 = 1.4; #Thermal conductivity in W/m.K
k2 = 0.35; #Thermal conductivity of insulating material in W/m.K
q = 1450.; #Heat loss per square metre in W
T = [1200,15]; #Inner and outer surface temperatures in degree C
# Calculations
L2 = (((T[0]-T[1])/q)-(L1/k1))*k2; #Thickness of the insulation required in m
# Results
print 'Thickness of the insulation required is %3.3f m'%(L2)
```

In [13]:

```
# Variables
L1 = 0.006; #Thickness of each glass sheet in m
L2 = 0.002; #Thickness of air gap in m
Tb = -20; #Temperature of the air inside the room in degree C
Ta = 30; #Ambient temperature of air in degree C
ha = 23.26; #Heat transfer coefficient between glass and air in W/m**2.K
kglass = 0.75; #Thermal conductivity of glass in W/m.K
kair = 0.02; #Thermal conductivity of air in W/m.K
# Calculations
q = ((Ta-Tb)/((1./ha)+(L1/kglass)+(L2/kair)+(L1/kglass)+(1./ha))); #Rate of heat leaking into the room per unit area of the door in W/m**2
# Results
print 'Rate of heat leaking into the room per unit area of the door is %3.1f W/m**2'%(q)
```

In [14]:

```
# Variables
LA = 0.05; #Length of section A in m
LB = 0.1; #Length of section A in m
LC = 0.1; #Length of section A in m
LD = 0.05; #Length of section A in m
LE = 0.05; #Length of section A in m
kA = 50; #Thermal conductivity of section A in W/m.K
kB = 10; #Thermal conductivity of section B in W/m.K
kC = 6.67; #Thermal conductivity of section C in W/m.K
kD = 20; #Thermal conductivity of section D in W/m.K
kE = 30; #Thermal conductivity of section E in W/m.K
Aa = 1; #Area of section A in m**2
Ab = 0.5; #Area of section B in m**2
Ac = 0.5; #Area of section C in m**2
Ad = 1; #Area of section D in m**2
Ae = 1; #Area of section E in m**2
T = [800,100]; #Temperature at inlet and outlet temperatures in degree C
# Calculations
Ra = (LA/(kA*Aa)); #Thermal Resistance of section A in K/W
Rb = (LB/(kB*Ab)); #Thermal Resistance of section B in K/W
Rc = (LC/(kC*Ac)); #Thermal Resistance of section C in K/W
Rd = (LD/(kD*Ad)); #Thermal Resistance of section D in K/W
Re = (LE/(kE*Ae)); #Thermal Resistance of section E in K/W
Rf = ((Rb*Rc)/(Rb+Rc)); #Equivalent resistance of section B and section C in K/W
R = Ra+Rf+Rd+Re; #Equivalent resistance of all sections in K/W
Q = ((T[0]-T[1])/R)/1000; #Heat transfer through the composite wall in kW
# Results
print 'Heat transfer through the composite wall is %3.1f kW'%(Q)
```

In [2]:

```
# Variables
T1 = 2000; #Temperature of hot gas in degree C
Ta = 45; #Room air temperature in degree C
Qr1 = 23.260; #Heat flow by radiation from gases to inside surface of the wall in kW/m**2
h = 11.63; #Convective heat transfer coefficient in W/m**2.
C = 58; #Thermal conductance of the wall in W/m**2.K
Q = 9.3; #Heat flow by radiation from external surface to the surrounding in kW.m**2
T2 = 1000; #Interior wall temperature in degree C
# Calculations
qr1 = Qr1; #Haet by radiation in kW/m**2
qc1 = h*((T1-T2)/1000); #Heat by conduction in kW/m**2
q = qc1+qr1; #Total heat entering the wall in kW/m**2
R = (1./C); #Thermal resistance in m**2.K/W
T3 = T2-(q*R*1000); #External wall temperature in degree C
Ql = q-Q; #Heat loss due to convection kW/m**2
h4 = (Ql*1000)/(T3-Ta); #Convective conductance in W/m**2.K
# Results
print 'The surface temperature is %i degree C \n \
The convective conductance is %3.1f W/m**2.K'%(T3,h4)
```

In [17]:

```
# Variables
L1 = 0.125; #Thickness of fireclay layer in m
L2 = 0.5; #Thickness of red brick layer in m
T = [1100,50]; #Temperatures at inside and outside the furnaces in degree C
k1 = 0.533; #Thermal conductivity of fireclay in W/m.K
k2 = 0.7; #Thermal conductivity of red brick in W/m.K
# Calculations
R1 = (L1/k1); #Resismath.tance of fireclay per unit area in K/W
R2 = (L2/k2); #Resistance of red brick per unit area in K/W
R = R1+R2; #Total resistance in K/W
q = (T[0]-T[1])/R; #Heat transfer in W/m**2
T2 = T[0]-(q*R1); #Temperature in degree C
T3 = T[1]+(q*R2*0.5); #Temperature at the interface between the two layers in degree C
km = 0.113+(0.00023*((T2+T3)/2)); #Mean thermal conductivity in W/m.K
x = ((T2-T3)/q)*km; #Thickness of diatomite in m
# Results
print 'Amount of heat loss is %3.1f W/m**2 \n \
Thickness of diatomite is %3.4f m'%(q,x )
```

In [3]:

```
import math
# Variables
Di = 0.1; #I.D of the pipe in m
L = 0.01; #Thickness of the wall in m
L1 = 0.03; #Thickness of insulation in m
Ta = 85; #Temperature of hot liquid in degree C
Tb = 25; #Temperature of surroundings in degree C
k1 = 58; #Thermal conductivity of steel in W/m.K
k2 = 0.2; #Thermal conductivity of insulating material in W/m.K
ha = 720; #Inside heat transfer coefficient in W/m**2.K
hb = 9; #Outside heat transfer coefficient in W/m**2.K
D2 = 0.12; #Inner diameter in m
r3 = 0.09; #Radius in m
# Calculations
q = ((2*3.14*(Ta-Tb))/((1./(ha*(Di/2)))+(1./(hb*r3))+(math.log(D2/Di)/k1)+(math.log(r3/(D2/2))/k2))); #Heat loss fro an insulated pipe in W/m
# Results
print 'Heat loss from an insulated pipe is %3.2f W/m'%(q)
```

In [21]:

```
import math
# Variables
Di = 0.1; #I.D of the pipe in m
Do = 0.11; #O.D of the pipe in m
L = 0.005; #Thickness of the wall in m
k1 = 50; #Thermal conductivity of steel pipe line in W/m.K
k2 = 0.06; #Thermal conductivity of first insulating material in W/m.K
k3 = 0.12; #Thermal conductivity of second insulating material in W/m.K
T = [250,50]; #Temperature at inside tube surface and outside surface of insulation in degree C
r3 = 0.105; #Radius of r3 in m as shown in fig.3.14 on page no.71
r4 = 0.155; #Radius of r4 in m as shown in fig.3.14 on page no.71
# Calculations
r1 = (Di/2); #Radius of the pipe in m
r2 = (Do/2); #Radius of the pipe in m
q = ((2*3.14*(T[0]-T[1]))/(((math.log(r2/r1))/k1)+((math.log(r3/r2))/k2)+((math.log(r4/r3))/k3))); #Loss of heat per metre length of pipe in W/m
T3 = ((q*math.log(r4/r3))/(2*3.14*k3))+T[1]; #Interface temperature in degree C
# Results
print 'Loss of heat per metre length of pipe is %3.1f W/m \n \
Interface temperature is %3.1f degree C'%(q,T3)
```

In [22]:

```
import math
# Variables
Di = 0.1; #I.D of the pipe in m
Do = 0.11; #O.D of the pipe in m
L = 0.005; #Thickness of the wall in m
k1 = 50; #Thermal conductivity of steel pipe line in W/m.K
k3 = 0.06; #Thermal conductivity of first insulating material in W/m.K
k2 = 0.12; #Thermal conductivity of second insulating material in W/m.K
T = [250,50]; #Temperature at inside tube surface and outside surface of insulation in degree C
r3 = 0.105; #Radius of r3 in m as shown in fig.3.14 on page no.71
r4 = 0.155; #Radius of r4 in m as shown in fig.3.14 on page no.71
# Calculations
r1 = (Di/2); #Radius of the pipe in m
r2 = (Do/2); #Radius of the pipe in m
q = ((2*3.14*(T[0]-T[1]))/(((math.log(r2/r1))/k1)+((math.log(r3/r2))/k2)+((math.log(r4/r3))/k3))); #Loss of heat per metre length of pipe in W/m
# Results
print 'Loss of heat per metre length of pipe is %3.2f W/m'%(q)
```

In [24]:

```
import math
# Variables
D1 = 0.1; #O.D of the pipe in m
P = 1373; #Pressure of saturated steam in kPa
D2 = 0.2; #Diameter of magnesia in m
k1 = 0.07; #Thermal conductivity of magnesia in W/m.K
k2 = 0.08; #Thermal conductivity of asbestos in W/m.K
D3 = 0.25; #Diameter of asbestos in m
T3 = 20; #Temerature under the canvas in degree C
t = 12; #Time for condensation in hours
l = 150; #Lemgth of pipe in m
T1 = 194.14; #Saturation temperature of steam in degree C from Table A.6 (Appendix A) at 1373 kPa on page no. 643
hfg = 1963.15; #Latent heat of steam in kJ/kg from Table A.6 (Appendix A) at 1373 kPa on page no. 643
# Calculations
r1 = (D1/2); #Radius of the pipe in m
r2 = (D2/2); #Radius of magnesia in m
r3 = (D3/2); #Radius of asbestos in m
Q = (((2*3.14*l*(T1-T3))/((math.log(r2/r1)/k1)+(math.log(r3/r2)/k2)))*(3600./1000)); #Heat transfer rate in kJ/h
m = (Q/hfg); #Mass of steam condensed per hour
m1 = (m*t); #Mass of steam condensed in 12 hours
# Results
print 'Mass of steam condensed in 12 hours is %3.2f kg'%(m1)
```

In [26]:

```
# Variables
D1 = 0.1; #I.D of the first pipe in m
D2 = 0.3; #O.D of the first pipe in m
k1 = 70; #Thermal conductivity of first material in W/m.K
D3 = 0.4; #O.D of the second pipe in m
k2 = 15; #Thermal conductivity of second material in W/m.K
T = [300,30]; #Inside and outside temperatures in degree C
# Calculations
r1 = (D1/2); #Inner Radius of first pipe in m
r2 = (D2/2); #Outer Radius of first pipe in m
r3 = (D3/2); #Radius of second pipe in m
Q = ((4*3.14*(T[0]-T[1]))/(((r2-r1)/(k1*r1*r2))+((r3-r2)/(k2*r2*r3))))/1000; #Rate of heat flow through the sphere in kW
# Results
print 'Rate of heat flow through the sphere is %3.2f kW'%(Q)
```

In [8]:

```
import math
# Variables
Di = 0.1; #I.D of a steam pipe in m
Do = 0.25; #I.D of a steam pipe in m
k = 1.; #Thermal conductivity of insulating material in W/m.K
T = [200.,20]; #Steam temperature and ambient temperatures in degree C
h = 8.; #Convective heat transfer coefficient between the insulation surface and air in W/m**2.K
# Calculations
ri = (Di/2); #Inner Radius of steam pipe in m
ro = (Do/2); #Outer Radius of steam pipe in m
rc = (k/h)*100; #Critical radius of insulation in cm
q = ((T[0]-T[1])/((math.log(ro/ri)/(2*3.14*k)+(1./(2*3.14*ro*h))))); #Heat loss per metre of pipe at critical radius in W/m
Ro = (q/(2*3.14*ro*h))+T[1]; #Outer surface temperature in degree C
# Results
print 'Heat loss per metre of pipe at critical radius is %i W/m \n \
Outer surface temperature is %3.2f degree C'%(q,Ro)
# Note : Answer in book is wrong. Please check manually.
```

In [6]:

```
import math
# Variables
Di = 0.001; #Diameter of copper wire in m
t = 0.001; #Thickness of insulation in m;
To = 20; #Temperature of surrondings in degree C
Ti = 80; #Maximum temperature of the plastic in degree C
kcopper = 400; #Thermal conductivity of copper in W/m.K
kplastic = 0.5; #Thermal conductivity of plastic in W/m.K
h = 8; #Heat transfer coefficient in W/m**2.K
p = (3*10**-8); #Specific electric resistance of copper in Ohm.m
# Calculations
r = (Di/2); #Radius of copper tube in m
ro = (r+t); #Radius in m
R = (p/(3.14*r*r*0.01)); #Electrical resistance per meter length in ohm/m
Rth = ((1./(2*3.14*ro*h))+(math.log(ro/r)/(2*3.14*kplastic))); #Thermal resistance of convection film insulation per metre length
Q = ((Ti-To)/Rth); #Heat transfer in W
I = math.sqrt(Q/R); #Maximum safe current limit in A
rc = ((kplastic*100)/h); #Critical radius in cm
# Results
print 'The maximum safe current limit is %3.3f A \n \
As the critical radius of insulation is much greater,the current carrying capacity of the conductor can be raised upto %3.1f cm \n \
considerably in increasing the radius of plastic covering'%(I,rc)
```

In [7]:

```
# Variables
L = 0.1; #Thickness of the wall in m
Q = (4*10**4); #Heat transfer rate in W/m**3
h = 50; #Convective heat transfer coefficient in W/m**2.K
T = 20; #Ambient air temperature in degree C
k = 15; #Thermal conductivity of the material in W/m.K
# Calculations
Tw = (T+((Q*L)/(2*h))); #Surface temperature in degree C
Tmax = (Tw+((Q*L*L)/(8*k))); #Maximum temperature in the wall in degree C
# Results
print 'Surface temperature is %i degree C \nMaximum temperature in the wall is %3.3f degree C'%(Tw,Tmax)
```

In [31]:

```
import math
# Variables
Do = 0.006; #Outer diameter of hallow cylinder in m
Di = 0.004; #Inner diameter of hallow cylinder in m
I = 1000; #Current in A
T = 30; #Temperature of water in degree C
h = 35000; #Heat transfer coefficient in W/m**2.K
k = 18; #Thermal conductivity of the material in W/m.K
R = 0.1; #Electrical reisitivity of the material in ohm.mm**2/m
# Calculations
ro = (Do/2); #Outer radius of hallow cylinder in m
ri = (Di/2); #Inner radius of hallow cylinder in m
V = ((3.14*(ro**2-ri**2))); #Vol. of wire in m**2
Rth = (R/(3.14*(ro**2-ri**2)*10**6)); #Resistivity in ohm/mm**2
q = ((I*I*Rth)/V); #Heat transfer rate in W/m**3
To = T+(((q*ri*ri)/(4*k))*((((2*k)/(h*ri))-1)*((ro/ri)**2-1)+(2*(ro/ri)**2*math.log(ro/ri)))); #Temperature at the outer surface in degree C
# Results
print 'Temperature at the outer surface is %3.2f degree C'%(To)
```

In [8]:

```
# Variables
D = 0.025; #Diameter of annealed copper wire in m
I = 200; #Current in A
R = (0.4*10**-4); #Resistance in ohm/cm
T = [200,10]; #Surface temperature and ambient temperature in degree C
k = 160; #Thermal conductivity in W/m.K
# Calculations
r = (D/2); #Radius of annealed copper wire in m
Q = (I*I*R*100); #Heat transfer rate in W/m
V = (3.14*r*r); #Vol. of wire in m**2
q = (Q/V); #Heat loss in conductor in W/m**2
Tc = T[0]+((q*r*r)/(4*k)); #Maximum temperature in the wire in degree C
h = ((r*q)/(2*(T[0]-T[1]))); #Heat transfer coefficient in W/m**2.K
# Results
print 'Maximum temperature in the wire is %3.2f degree C \nHeat transfer coefficient is %3.2f W/m**2.K'%(Tc,h)
```

In [33]:

```
import math
# Variables
p = 100.; #Resistivity of nichrome in µ ohm-cm
Q = 10000.; #Heat input of a heater in W
T = 1220.; #Surface temperature of nichrome in degree C
Ta = 20.; #Temperature of surrounding air in degree C
h = 1150.; #Outside surface coeffient in W/m**2.K
k = 17.; #Thermal conductivity of nichrome in W/m.K
L = 1.; #Length of heater in m
# Calculations
d = (Q/((T-Ta)*3.14*h))*1000; #Diameter of nichrome wire in mm
A = (3.14*d*d)/4; #Area of the wire in m**2
R = ((p*10**-8*L)/A); #Resistance of the wire in ohm
I = math.sqrt(Q/R)/1000; #Rate of current flow in A
# Results
print 'Diameter of nichrome wire is %3.4f mm \n \
Rate of current flow is %i A'%(d,I)
```

In [34]:

```
# Variables
Do = 0.025; #O.D of the rod in m
k = 20; #Thermal conductivity in W/m.K
Q = (2.5*10**6); #Rate of heat removal in W/m**2
# Calculations
ro = (Do/2); #Outer radius of the rod in m
q = ((4*Q)/(ro)); #Heat transfer rate in W/m**3
T = ((-3*q*ro**2)/(16*k)); #Temperature drop from the centre line to the surface in degree C
# Results
print 'Temperature drop from the centre line to the surface is %3.3f degree C'%(T)
```

In [35]:

```
# Variables
Q = 300; #Heat produced by the oranges in W/m**2
s = 0.08; #Size of the orange in m
k = 0.15; #Thermal conductivity of the sphere in W/m.K
# Calculations
q = (3*Q)/(s/2); #Heat flux in W/m**2
Tc = 10+((q*(s/2)**2)/(6*k)); #Temperature at the centre of the sphere in degree C
# Results
print 'Temperature at the centre of the orange is %i degree C'%(Tc)
```

In [36]:

```
import math
# Variables
To = 140; #Temperature at the junction in degree C
Ti = 15; #Temperature of air in the room in degree C
D = 0.003; #Diameter of the rod in m
h = 300; #Heat transfer coefficient in W/m**2.K
k = 150; #Thermal conductivity in W/m.K
# Calculations
P = (3.14*D); #Perimeter of the rod in m
A = (3.14*D**2)/4; #Area of the rod in m**2
Q = math.sqrt(h*P*k*A)*(To-Ti); #Total heat dissipated by the rod in W
# Results
print 'Total heat dissipated by the rod is %3.3f W'%(Q)
```

In [38]:

```
import math
# Variables
D = 0.025 #Diameter of the rod in m
Ti = 22.; #Temperature of air in the room in degree C
x = 0.1; #Dismtance between the points in m
T = [110.,85.]; #Temperature sat two points in degree C
h = 28.4; #Heat transfer coefficient in W/m**2.K
# Calculations
m = -math.log((T[1]-Ti)/(T[0]-Ti))/x; #Calculation of m for obtaining k
P = (3.14*D); #Perimeter of the rod in m
A = (3.14*D**2)/4; #Area of the rod in m**2
k = ((h*P)/((m)**2*A)); #Thermal conductivity of the rod material in W/m.K
# Results
print 'Thermal conductivity of the rod material is %3.1f W/m.K'%(k)
```

In [15]:

```
import math
# Variables
L = 0.06; #Length of the turbine blade in m
A = (4.65*10**-4); #Cross sectional area in m**2
P = 0.12; #Perimeter in m
k = 23.3; #Thermal conductivity of stainless steel in W/m.K
To = 500; #Temperature at the root in degree C
Ti = 870; #Temperature of the hot gas in degree C
h = 442; #Heat transfer coefficient in W/m**2.K
# Calculations
m = math.sqrt((h*P)/(k*A)); #Calculation of m for calculating heat transfer rate
X = (To-Ti)/math.cosh(m*L); #X for calculating tempetarure distribution
Q = math.sqrt(h*P*k*A)*(To-Ti)*math.tanh(m*L); #Heat transfer rate in W
# Results
print 'Temperature distribution is given by :\n T-Ti = %i cosh[%3.2f%3.2f-x)] cosh[%3.2f%3.2f)] \n \
Heat transfer rate is %3.1f W'%(To-Ti,m,L,m,L,Q)
```

In [12]:

```
import math
# Variables
W = 1; #Length of the cylinder in m
D = 0.05; #Diameter of the cylinder in m
Ta = 45; #Ambient temperature in degree C
n = 10; #Number of fins
k = 120; #Thermal conductivity of the fin material in W/m.K
t = 0.00076; #Thickness of fin in m
L = 0.0127; #Height of fin in m
h = 17; #Heat transfer coefficient in W/m**2.K
Ts = 150; #Surface temperature of cylinder in m
# Calculations
P = (2*W); #Perimeter of cylinder in m
A = (W*t); #Surface area of cyinder in m**2
m = round(math.sqrt((h*P)/(k*A)),2); #Calculation of m for determining heat transfer rate
Qfin = (math.sqrt(h*P*k*A)*(Ts-Ta)*((math.tanh(m*L)+(h/(m*k)))/(1+((h/(m*k))*math.tanh(m*L))))); #Heat transfer through the fin in kW
Qb = h*((3.14*D)-(n*t))*W*(Ts-Ta); #Heat from unfinned (base) surface in W
Q = ((Qfin*10)+Qb); #Total heat transfer in W
Ti = ((Ts-Ta)/(math.cosh(m*L)+((h*math.sinh(m*L))/(m*k)))); #Ti to calculate temperature at the end of the fin in degree C
T = (Ti+Ta); #Temperature at the end of the fin in degree C
# Results
print 'Rate of heat transfer is %3.2f W \nTemperature at the end of the fin is %3.2f degree C'%(Q,T )
```

In [10]:

```
import math
# Variables
t = 0.025; #Thickness of fin in m
L = 0.1; #Length of fin in m
k = 17.7; #Thermal conductivity of the fin material in W/m.K
p = 7850; #Density in kg/m**3
Tw = 600; #Temperature of the wall in degree C
Ta = 40; #Temperature of the air in degree C
h = 20; #Heat transfer coefficient in W/m**2.K
I0 = 2.1782; #Io value taken from table 3.2 on page no.108
I1 = 1.48871; #I1 value taken from table 3.2 on page no. 108
# Calculations
B = math.sqrt((2*L*h)/(k*t)); #Calculation of B for determining temperature distribution
X = ((Tw-Ta)/2.1782); #Calculation of X for determining temperature distribution
Y = (2*B); #Calculation of Y for determining temperature distribution
Q = (math.sqrt(2*h*k*t)*(Tw-Ta)*((1.48871))/(2.1782));
m = ((p*t*L)/2); #Mass of the fin per meter of width in kg/m
q = (Q/m); #Rate of heat flow per unit mass in W/kg
# Results
print 'Temperature distribution is T = %i+%3.1f(%3.4f√x) \n \
Rate of heat flow per unit \
mass of the fin is %3.2f W/kg'%(Ta,X,Y,q)
```

In [11]:

```
import math
# Variables
t = 0.002; #Thickness of fin in m
L = 0.015; #Length of fin in m
k1 = 210.; #Thermal conductivity of aluminium in W/m.K
h1 = 285.; #Heat transfer coefficient of aluminium in W/m**2.K
k2 = 40.; #Thermal conductivity of steel in W/m.K
h2 = 510.; #Heat transfer coefficient of steel in W/m**2.K
# Calculations
Lc = (L+(t/2)); #Corrected length of fin in m
mLc1 = Lc*math.sqrt((2*h1)/(k1*t)); #Calculation of mLc for efficiency
n1 = math.tanh(mLc1)/mLc1; #Efficiency of fin when aluminium is used
mLc2 = Lc*math.sqrt((2*h2)/(k2*t)); #Calculation of mLc for efficiency
n2 = math.tanh(mLc2)/mLc2; #Efficiency of fin when steel is used
# Results
print 'Efficiency of fin when aluminium is used is %3.4f \n \
Efficiency of fin when steel is used is %3.3f'%(n1,n2)
```

In [51]:

```
import math
# Variables
k = 200; #Thermal conductivity of aluminium in W/m.K
t = 0.001; #Thickness of fin in m
L = 0.015; #Width of fin in m
D = 0.025; #Diameter of the tube in m
Tb = 170; #Fin base temperature in degree C
Ta = 25; #Ambient fluid temperature in degree C
h = 130; #Heat transfer coefficient in W/m**2.K
# Calculations
Lc = (L+(t/2)); #Corrected length of fin in m
r1 = (D/2); #Radius of tube in m
r2c = (r1+Lc); #Corrected radius in m
Am = t*(r2c-r1); #Corrected area in m**2
x = Lc**(3/2)*math.sqrt(h/(k*Am)); #x for calculating efficiency
n = 0.82; #From fig. 3.18 on page no. 112 efficiency is 0.82
qmax = (2*3.14*(r2c**2-r1**2)*h*(Tb-Ta)); #Maximum heat transfer in W
qactual = (n*qmax); #Actual heat transfer in W
# Results
print 'Heat loss per fin is %3.2f W'%(qactual)
```

In [52]:

```
import math
# Variables
k = 16; #Thermal conductivity of fin in W/m.K
L = 0.1; #Length of fin in m
D = 0.01; #Diameter of fin in m
h = 5000; #Heat transfer coefficient in W/m**2.K
# Calculations
P = (3.14*D); #Perimeter of fin in m
A = (3.14*D**2)/4; #Area of fin in m**2
m = math.sqrt((h*P)/(k*A)); #Calculation of m for determining heat transfer rate
n = math.tanh(m*L)/math.sqrt((h*A)/(k*P)); #Calculation of n for checking whether installation of fin is desirable or not
x = (n-1)*100; #Conversion into percentage
# Results
print 'This large fin only produces an increase of %i percent in heat dissipation, \
so naturally this configuration is undesirable'%(x)
```

In [16]:

```
import math
# Variables
k = 55.8; #Thermal conductivity of steel in W/m.K
t = 0.0015; #Thickness of steel tube in m
L = 0.12; #Length of steel tube in m
h = 23.3; #Heat transfer coefficient in W/m**2.K
Tl = 84; #Temperature recorded by the thermometer in degree C
Tb = 40; #Temperature at the base of the well in degree C
# Calculations
m = math.sqrt(h/(k*t)); #Calculation of m for determining the temperature distribution
x = 1./math.cosh(m*L); #Calculation of x for determining the temperature distribution
Ti = ((Tl-(x*Tb))/(1-x)); #Temperature distribution in degree C
T = (Ti-Tl); #Measurement error in degree C
# Results
print 'Measurement error is %3.0f degree C'%(T)
```