# Chapter 3 : OneDimensional Steady State Heat Conduction¶

## Example 3.1 Page No : 45¶

In :
# 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-T))/L;	    		#Heat transfer rate in W
T = (((T-T)*x)/L)+T;			#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)

a)Heat transfer rate is 3920 W
b)Temperature at interior point of the wall, 20 cm distant from the inner wall is 54 degree C


## Example 3.2 Page No : 48¶

In :
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-T))/(math.log(ro/ri)));			#Heat transfer rate in W
r1 = (ro+ri)/2;			#Radius at halfway between ro and ri in m
T1 = T-((T-T)*(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)

Heat transfer rate is 63420.9 W /m
Temperature of the point halfway between the inner and outer surface is 141.5 degree C


## Example 3.3 Page No : 51¶

In :
# 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-T))/(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-T))/(r*(ro-ri)))+T;			#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)

Heat flow rate through the sphere is 9.42 kW
Temperature at a point a quarter of the way between the inner and outer surfaces is 200 degree C


## Example 3.4 Page No : 55¶

In :
# 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**2-T**2))-((5*10**-6*(T**3-T**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)

Heat loss per square meter surface area is 435 W/m**2


## Example 3.5 Page No : 55¶

In :
# 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+T))/2));			#Constant thermal conductivity in W/m.K
q = ((km*(T-T))/L);			        #Rate of heat flow in W/m**2

# Results
print 'Rate of heat flow is %3.0f W/m**2'%(q)

Rate of heat flow is 4649 W/m**2


## Example 3.6 Page No : 58¶

In :
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-150)+(T-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-T))/math.log(r/r))/1000;			#Heat loss per unit length in kW/m

# Results
print 'Heat loss per unit length is %3.2f kW/m'%(q)

Heat loss per unit length is 66.45 kW/m


## Example 3.8 Page No : 63¶

In :
# 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-T)/q)-(L1/k1))*k2;			#Thickness of the insulation required in m

# Results
print 'Thickness of the insulation required is %3.3f m'%(L2)

Thickness of the insulation required is 0.161 m


## Example 3.9 Page No : 64¶

In :
# 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)

Rate of heat leaking into the room per unit area of the door is 247.5 W/m**2


## Example 3.10 Page No : 65¶

In :
# 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-T)/R)/1000;			#Heat transfer through the composite wall in kW

# Results
print 'Heat transfer through the composite wall is %3.1f kW'%(Q)

Heat transfer through the composite wall is 40.8 kW


## Example 3.11 Page No : 66¶

In :
# 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)

The surface temperature is 398 degree C
The convective conductance is 72.4 W/m**2.K


## Example 3.12 Page No : 67¶

In :
# 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-T)/R;			#Heat transfer in W/m**2
T2 =  T-(q*R1);			#Temperature in degree C
T3 = T+(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 )

Amount of heat loss is 1106.7 W/m**2
Thickness of diatomite is 0.0932 m


## Example 3.13 Page No : 70¶

In :
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)

Heat loss from an insulated pipe is 114.43 W/m


## Example 3.14 Page No : 71¶

In :
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-T))/(((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;			#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)

Loss of heat per metre length of pipe is 89.6 W/m
Interface temperature is 96.3 degree C


## Example 3.15 Page No : 72¶

In :
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-T))/(((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)

Loss of heat per metre length of pipe is 105.71 W/m


## Example 3.16 Page No : 73¶

In :
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)

Mass of steam condensed in 12 hours is 284.43 kg


## Example 3.17 Page No : 74¶

In :
# 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-T))/(((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)

Rate of heat flow through the sphere is 11.24 kW


## Example 3.18 Page No : 77¶

In :
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-T)/((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;			#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)


Heat loss per metre of pipe at critical radius is 589 W/m
Outer surface temperature is 113.93 degree C


## Example 3.19 Page No : 78¶

In :
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)

The maximum safe current limit is 1.074 A
As the critical radius of insulation is much greater,the current carrying capacity of the conductor can be raised upto 6.2 cm
considerably in increasing the radius of plastic covering


## Example 3.20 Page No : 83¶

In :
# 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)

Surface temperature is 60 degree C
Maximum temperature in the wall is 63.333 degree C


## Example 3.21 Page No : 85¶

In :
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)

Temperature at the outer surface is 57.44 degree C


## Example 3.22 Page No : 88¶

In :
# 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+((q*r*r)/(4*k));			#Maximum temperature in the wire in degree C
h = ((r*q)/(2*(T-T)));		#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)

Maximum temperature in the wire is 200.08 degree C
Heat transfer coefficient is 10.73 W/m**2.K


## Example 3.23 Page No : 89¶

In :
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)

Diameter of nichrome wire is 2.3078 mm
Rate of current flow is 204 A


## Example 3.24 Page No : 93¶

In :
# 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)

Temperature drop from the centre line to the surface is -1171.875 degree C


## Example 3.25 Page No : 95¶

In :
# 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)

Temperature at the centre of the orange is 50 degree C


## Example 3.26 Page No : 102¶

In :
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)

Total heat dissipated by the rod is 6.841 W


## Example 3.27 Page No : 103¶

In :
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-Ti)/(T-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)

Thermal conductivity of the rod material is 406.8 W/m.K


## Example 3.28 Page No : 103¶

In :
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)

Temperature distribution is given by :
T-Ti  =  -370 cosh[69.970.06-x)] cosh[69.970.06)]
Heat transfer rate is -280.4 W


## Example 3.29 Page No : 104¶

In :
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 )

Rate of heat transfer is 723.99 W
Temperature at the end of the fin is 146.74 degree C


## Example 3.31 Page No : 109¶

In :
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)

Temperature distribution is T = 40+257.1(6.0132√x)
Rate of heat flow per unit  mass of the fin is 164.10 W/kg


## Example 3.32 Page No : 116¶

In :
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)

Efficiency of fin when aluminium is used is 0.8983
Efficiency of fin when steel is used is 0.524


## Example 3.33 Page No : 117¶

In :
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)

Heat loss per fin is 60.94 W


## Example 3.34 Page No : 117¶

In :
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)

This large fin only produces an increase of 13 percent in heat dissipation,  so naturally this configuration is undesirable


## Example 3.35 Page No : 119¶

In :
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)

Measurement error is  16 degree C