Chapter 3 : OneDimensional Steady State Heat Conduction

Example 3.1 Page No : 45

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

Example 3.5 Page No : 55

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

Example 3.6 Page No : 58

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)
Heat loss per unit length is 66.45 kW/m

Example 3.8 Page No : 63

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)
Thickness of the insulation required is 0.161 m

Example 3.9 Page No : 64

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)
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 [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)
Heat transfer through the composite wall is 40.8 kW

Example 3.11 Page No : 66

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

Example 3.12 Page No : 67

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

Example 3.13 Page No : 70

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)
Heat loss from an insulated pipe is 114.43 W/m

Example 3.14 Page No : 71

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

Example 3.16 Page No : 73

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)
Mass of steam condensed in 12 hours is 284.43 kg

Example 3.17 Page No : 74

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)
Rate of heat flow through the sphere is 11.24 kW

Example 3.18 Page No : 77

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.
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 [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)
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 [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)
Surface temperature is 60 degree C 
Maximum temperature in the wall is 63.333 degree C

Example 3.21 Page No : 85

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)
Temperature at the outer surface is 57.44 degree C

Example 3.22 Page No : 88

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

Example 3.24 Page No : 93

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

Example 3.25 Page No : 95

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)
Temperature at the centre of the orange is 50 degree C

Example 3.26 Page No : 102

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)
Total heat dissipated by the rod is 6.841 W

Example 3.27 Page No : 103

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

Example 3.28 Page No : 103

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)
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 [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 )
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 [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)
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 [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)
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 [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)
Heat loss per fin is 60.94 W

Example 3.34 Page No : 117

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)
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 [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)
Measurement error is  16 degree C