Chapter 6:Introduction to Heat Transfer¶

Example 6.1 Page no:221¶

In [2]:
#Input data
t1=270		#Temperature inside surface of the furnace wall in degree centigrade
t3=20		#Temperature outside surface is dissipating heat by convection into air in degree centigrade
L=0.04		#Thickness of the wall in m
K=1.2		#Thermal conductivity of wall in W/m-K
t2=70		#Temperature of outside surface should not exceed in degree centigrade
A=1		 #Assuming area in m**2

#Calculations
Q1=(K*A*(t1-t2))/(L)	#Heat transfer through the furnace wall in W
hc=(Q1)/(A*(t2-t3))		#Heat transfer coefficient in W/m**2K

#Output
print"The minimum value of heat transfer coefficient at the outer surface is",hc,"W/m**2K"

The minimum value of heat transfer coefficient at the outer surface is 120.0 W/m**2K


Example 6.2 page no:222¶

In [82]:
#Input data
t1=30		 #Normal temperature of black body in degree centigrade
t2=100		#Heated temperature of black body in degree centigrade
s=20.52*10**-8	#Stefan Boltzmann constant in kJ/hrK**4
A=1		   #Assume area in m**2

#Calculations
T1=273+t1	#Black body temperatures in kelvin K
T2=273+t2	#Heated temperature of black body in kelvin K
E=s*(T2**4-T1**4)	#Increase of emissive power in kJ/hr

#Output
print'The change in its emissive power',round(E,4),"kJ/hr"

The change in its emissive power 2242.4228 kJ/hr


Example 6.3 page no:222¶

In [83]:
#Input data
L=0.012   	#Wall thickness of a mild steel tank in m
t1=100.0		#Temperature of water in tank in degree centigrade
t4=20.0	 	#Atmospheric temperature of air in degree centigrade
K=50.0		  #Thermal conductivity of mild steel in W/m-K
hi=2850.0		#Convection heat transfer coefficient on water side in W/m**2-K
ho=10.0		#Convection heat transfer coefficient on air side in W/m**2-K
Q1=60.0    	#Heat trasfer from the incandicent lamp in W
s=5.67*10**-8	#Stefan boltzmann constant in W/m**2/K**4
T1=2500.0	   #Lamp surface temperature in K
T2=300.0		#Room temperature in K
A=1.0		   #Assuming area in m**2

#Calculations
T=t1-t4		#Temperature difference in degree centigrade
Q=(T)/((1/hi)+(L/K)+(1/ho))	#Rate of heat loss per m**2 area of surface of tank in W
t3=(Q/(ho*A))+(t4)		#Temperature of the outside surface in degree centigrade
U=(Q)/(A*T)		#Overall Heat transfer coefficient in W/m**2/K
a=(Q1)/(s*(T1**4-T2**4))	#surface area of the coil in m**2
a1=a*10**6#Surface area of the coil in mm**2

#Output
print'(a) The rate of heat loss per sq m area of the tank is',round(Q,2)," W "
print '(b) Overall heat transfer coefficient is ',round(U,2)," W/m**2/K"
print '(c) Temperature of the outside surface of tank is ',round(t3,2),"C"
print '(d)The surface area of the coil is ',round(a1,3),"mm**2"

(a) The rate of heat loss per sq m area of the tank is 795.3  W
(b) Overall heat transfer coefficient is  9.94  W/m**2/K
(c) Temperature of the outside surface of tank is  99.53 C
(d)The surface area of the coil is  27.096 mm**2


Example 6.4 page no:225¶

In [84]:
#Input data
A1=3.5		#Area of the boiler plate in m**2
X2=0.02		#Thickness of the plate in m
K2=50.0		#Thermal conductivity of plate in W/m-K
X1=0.002	  #Thickness of layer inside boiler in m
K1=1.0		  #Thermal conductivity of layer in W/m-K
t1=250.0		#The hot gas temperature of the plate in degree centigrade
t3=200.0		#Temperature of cold air in degree centigrade

#Calculations
T=t1-t3		#Temperature difference in degree centigrade
Q=(T*A1)/((X1/K1)+(X2/K2))	#Rate of heat loss in W
Q1=Q/1000		#Rate of heat loss in kJ/s
Q2=Q1*3600		#Rate of heat loss in kJ/hr

#Output
print'(a)Rate of heat loss is',round(Q1,2)," kJ/s "
print'(b)Rate of heat loss per hour is',round(Q2,2),"kJ/hr"

(a)Rate of heat loss is 72.92  kJ/s
(b)Rate of heat loss per hour is 262500.0 kJ/hr


Example 6.5 page no:226¶

In [85]:
#Input data
L1=0.225		#Thickness of the brick in m
K1=4.984		#Thermal conductivity of brick in kJ/hr m C/m
L2=0.125		#Thickness of insulating brick in m
K2=0.623		#Thermal conductivity of insulating brick in kJ/hr m C /m
Ti=1650.0		#Temperature inside the furnace in degree centigrade
hl=245.28		#Conductance at inside wall in kJ/hr m**2 C
ho=40.88		#Conductance at outside wall in kJ/hr m**2 C
To=27.0		#Temperature of surrounding atmosphere in degree centigrade

#Calculations
R=((1.0/hl)+(L1/K1)+(L2/K2)+(1.0/ho))	#Total resistance of the wall in C hr/kJ
q=(Ti-To)/R			#Rate of heat loss per m**2 of the wall in kJ/hr m**2
T1=Ti-(q*(1.0/hl))			#Inner surface temperature in degree centigrade
T3=Ti-(q*((1.0/hl)+(L1/K1)+(L2/K2)))	#Outer surface temperature in degree centigrade

#Output
print'(a)The rate of heat loss per sq m of the wall is',round(q,1), "kJ/hr m**2"
print'(b)The temperature at the inner surface is',round(T1,2),"C"
print'(c)The temperature at the outer surface is',round(T3,2),"C"

(a)The rate of heat loss per sq m of the wall is 5916.3 kJ/hr m**2
(b)The temperature at the inner surface is 1625.88 C
(c)The temperature at the outer surface is 171.72 C


Example 6.6 page no:227¶

In [86]:
#Input data
x=0.3		#Thickness of the wall in degree centigrade
t1=24.0		#Inside surface temperature of the wall in degree centigrade
t2=-6		#Outside temperature of wall in degree centigrade
h=2.75		#Height of the wall in m
L=6.1		#Length of the wall in m
K=2.6		#Coefficient of conductivity of brick in kJ/hr m C

#Calculations
A=h*L		#Area of the wall in m**2
A=round(A,1)
T=t2-t1		#Temperature difference in degree centigrade
q=(K*A*(-T))/(x)	#Heat transfer by conduction in kJ/hr
R=(t1-t2)/q	#Resistance of the wall in C hr/kJ
C=1.0/R		#Conductance of the wall in kJ/m C

#Output
print'(a)The heat transfer by conduction through the wall is',q, "kJ/hr"
print'(b)Resistance of the wall is',round(R,5),"C hr/kJ"
print'(C)Conductance of the wall is',round(C,1),"kJ/m C"

(a)The heat transfer by conduction through the wall is 4368.0 kJ/hr
(b)Resistance of the wall is 0.00687 C hr/kJ
(C)Conductance of the wall is 145.6 kJ/m C


Example 6.8 page no:230¶

In [87]:
T=300		#Temperature of the earth as a black body in K
s=20.52*10**-8	#Stefan Boltzmann constant in kJ/hr m**2 T**4

#Calculations
Q=s*T**4	#Heat received by unit area on the earths surface perpendicular to solar rays in kJ/hr

#Output
print'Heat received by the unit area of earths surface',round(Q,2),"kJ/hr"

Heat received by the unit area of earths surface 1662.12 kJ/hr


Example 6.9 page no:230¶

In [88]:
#Input data
D=0.07		  # Diameter of the steel tube in m
L=3.0		   #Length of the steel tube
t1=227.0		#Temperature of the steel tube in m
t2=27.0	 	#Temperature of the room in degree centigrade
s=20.52*10**-8	#Stefan Boltzmann constant in kJ/hr m**2 T**4
pi=3.1428		#Constant value of pi

#Calculations
A=2*pi*D*L	  #Surface area of the tube in m**2
Q=(A)*(s)*((t1+273)**4-(t2+273)**4)	#Loss of heat by radiation in kJ/hr
Q1=Q/3600.0	 #Loss of heat by radiation in kW

#Output
print'The loss of heat by radiation from steel tube is',round(Q1,3),"KW"

The loss of heat by radiation from steel tube is 4.093 KW


Example 6.10 page no:231¶

In [89]:
#Input data
T1=7.0		 #Inside temperature of refrigerator in degree centigrade
T0=28.0		#Temperature in the kitchen in degree centigrade
K1=40.0		#Thermal conductivity of mild steel in W/mC
x1=0.03		#Thickness of mild sheets in m
K3=40.0		#Thermal conductivity of the mild steel in W/mC
x3=0.03		#Thickness of another side mild sheet in m
x2=0.05		#Thickness of glass wool insulated in m
hi=10.0		#Heat transfer coefficient in the inner surface of refrigerator in W/m**2 C
ho=12.5		#Heat transfer coefficient in the outer surface of refrigerator in W/m**2 C
K2=0.04		#Thermal conductivity of glass in W/mC

#Calculations
Q=(T1-T0)/((1/hi)+(x1/K1)+(x2/K2)+(x3/K3)+(1/ho))	#Heat transfer per unit area in W/m**2

#Output
print'The rate of heat removed from the refrigirator ',round(Q,3),"W/m**2"

The rate of heat removed from the refrigirator  -14.67 W/m**2


Example 6.11 page no:232¶

In [90]:
#Input data
x1=0.2	 	#Thickness of the fire brick
x2=0.2	 	#Thickness of the common brick
Ti=1400.0  	#Temperature of hot gases in the inner surface of the brick in degree centigrade
To=50.0		#Temperature of gases in the outer surface of the brick in degree centigrade
h1=16.5		#Convection heat transfer coefficient on gas side in W/mC
h2=17.5		#radiation heat transfer coefficient on gas side in W/mC
h3=12.5		#Convection heat transfer coefficient on outer side in W/mC
h4=6.5 		#Radiation heat transfer coeeficient on outer side in W/mC
K1=4.0 		#Thermal conductivity of fire brick in W/mC
K2=0.65		#Thermal conductivity of common brick in W/mC

#Calculations
hi=h1+h2		#Total heat transfer coefficient in inner
ho=h3+h4		#Total heat transfer coefficient in outer
Q=(Ti-To)/((1/hi)+(x1/K1)+(x2/K2)+(1/ho))	#Heat flow through the furnace composite wall per unit area in W/m**2
Q1=Q/1000
T1=Ti-(Q/hi)	#Temperature at the inside of the fire brick
T2=T1-(Q*(x1/K1))#Maximum temperature to which common brick is subjected in degree centigrade

#Output
print'(a)Heat loss per m**2 area of the furnace wall is',round(Q)/1000,"kW/m**2"
print'(b)Maximum temperature to which common brick is subjec',round(T1,3),"C"
print'(c)similarly on other side is',round(T2,3),"C"

(a)Heat loss per m**2 area of the furnace wall is 3.07 kW/m**2
(b)Maximum temperature to which common brick is subjec 1309.705 C
(c)similarly on other side is 1156.204 C


Example 6.12 page no:234¶

In [91]:
#Input data
K1=0.93		#Thermal conductivity of fire clay in W/mC
K2=0.13		#Thermal conductivity of diatomite brick in W/mC
K3=0.7		#Thermal conductivity of red brick in W/mC
x1=0.12		#Thickness of fire clay in m
x2=0.05		#Thickness of diatomite in m
x3=0.25		#Thickness of brick in m
T=1		#Assume the difference between temperature in degree centigrade

#Calculations
Q=(T)/((x1/K1)+(x2/K2)+(x3/K3))	#The heat flow per unit area in W/m**2
X3=K3*((T/Q)-(x1/K1))		#Thickness of the red brick layer in m
X=X3*100			#Thickness of the red brick layer in cm

#Output
print'The thickness of the red brick layer is',round(X,3),"cm"

The thickness of the red brick layer is 51.923 cm


Example 6.13 page no:235¶

In [92]:
#Input data
R1=0.06		#Thickness of material layer in m
R2=0.12		#Thickness of the two insulating materials in m
R3=0.16		#Thickness of material layers with pipe in m
K1=0.24		#Thermal conductivity of one layer in W/mC
K2=0.4		 #Thermal conductivity of another layer in W/mC
L=60.0		 #Length of the pipe in m
hi=60.0		#Heat transfer coefficient inside in W/m**2C
ho=12.0		#Heat transfer coefficient outside in W/m**2C
ti=65.0		#Temperature of hot air flowing in pipe in degree centigrade
to=20.0		#Atmospheric temperature in degree centigrade
pi=3.1428	  #Constant value of pi

#Calculations
Q=(ti-to)*(2*pi*L)/((1/(hi*R1))+(math.log(R2/R1)/(K1))+(math.log(R3/R2)/(K2))+(1/(ho*R3)))	#Rate of heat loss in W
Q1=Q/1000	#Rate of heat loss in kW

#Output
print'The rate of heat loss is',round(Q1,4),"kW"

The rate of heat loss is 3.8519 kW


Example 6.14 page no:237¶

In [93]:
#Input data
R1=8.0		  #Inner radius of the pipe in cm
R2=8.5		  #Outter radius of the pipe in cm
x1=3.0		  #Thickness of first layer in cm
x2=5.0		  #Thickness of second layer in cm
T1=300.0		#Inner surface temperature of the steam pipe in degree centigrade
pi=3.1428	   #Constant value of pi
T4=50.0		 #Temperature at outer surface of insulation in degree centigrade
L=1.0		   #Length of the pipe in m
K1=50.0	 	#Thermal conductivity of pipe in W/mC
K2=0.15	 	#Thermal conductivity of first layer in W/mC
K3=0.08	 	#Thermal conductivity of second layer in W/mC
h=2751.0		#Enthalpy of dry and saturated steam at 300 degree centigrade in kJ/kg
q=40.0	  	#Quantity of steam flow in gm/hr
hf=1345.0		#Enthalpy of fluid at 300 degree centigrade in kJ/kg
hfg=1406.0		#enthalpy at 300 degree centigrade in kJ/kg

#Calculations
R3=R2+x1	#Radius of pipe with first layer
R4=R3+x2	#Radius of pipe with two layers
Q=(2*pi*L*(T1-T4))/((math.log(R2/R1)/(K1))+(math.log(R3/R2)/(K2))+(math.log(R4/R3)/(K3)))
Q1=Q/1000	#Quantity of heat loss per meter length of pipe in kW
Q2=Q1*3600	#Quantity of heat loss per meter length of pipe in kJ/hr
hg=((h)-(Q2/q))	#Enthalpy of steam in kJ/kg
x=(hg-hf)/(hfg)	#Dryness fraction of steam

#Output
print'(a)The quantity of heat lost per meter length of steam pipe is',round(Q,2),"W/m or",round(Q*3600/1000),"kJ/hr"
print'(b)The quantity of steam coming out of one meter length pipe is',round(x,4),"gm/h"

(a)The quantity of heat lost per meter length of steam pipe is 240.68 W/m or 866.0 kJ/hr
(b)The quantity of steam coming out of one meter length pipe is 0.9846 gm/h


Example 6.15 page no:238¶

In [94]:
#Input data
x=0.3		#Thickness of brick wall in m
ti=24.0		#Inside surface temperature of wall in degree centigrade
to=-6.0		#Outside surface temperature of wall in degree centigrade
h=2.75		#Height of the wall in m
L=6.1		#Length of the wall in m
K=2.6		#Thermal conductivity of brick material in kJ/m hr C

#Calculations
T=ti-to		#Temperature difference across the wall in degree centigrade
A=h*L		#Area of the wall in m**2
Q=(K*A*T)/(x)	#Heat transfer through conduction by the wall per hour in kJ/hr
R=T/Q		#Resistance of the wall in hr C/kJ
C=1.0/R		#Conductance of the wall in kJ/hr C

#Output
print'(a)The heat transfer by conduction through the wall is',round(Q),"kJ/hr "
print'(b)The resistance of the wall is ',round(R,5),"Chr/Kj "
print'  The conductance of the wall is',round(C,1), "kJ/hr C "

(a)The heat transfer by conduction through the wall is 4362.0 kJ/hr
(b)The resistance of the wall is  0.00688 Chr/Kj
The conductance of the wall is 145.4 kJ/hr C


Example 6.16 page no:240¶

In [95]:
#Input data
x1=0.3     	#Thickness of refractory bricks in m
K1=5.66		#Thermal conductivity of refractory bricks in kJ/hr mC
t1=1650.0		#Inner surface temperature of the wall in degree centigrade
t2=320.0	 	#Outside surface temperature of the wall in degree centigrade
x2=0.3	 	#Thickness of insulating brick in m
K2=1.26		#Thermal conductivity of insulating brick in kJ/hr mC
A=1.0	  	#unit surface area in m**2
t3=27.0		#Outside surface temperature of the brick in degree centigrade

#Calculations
T1=t1-t2		#Temperature difference in degree centigrade
Q1=(K1*A*T1)/(x1)	#Heat loss without insulation in kJ/hr/m**2
R1=(K1*A)/(x1)	#Heat loss for the change in temperature for refractory brick wall material in kJ/hrC
R2=(K2*A)/(x2)	#Heat loss for the change in temperature for insulated brick wall material kJ/hrC
Q2=(t1-t3)/((1.0/R1)+(1.0/R2))	#Heat loss with insulation in kJ/hr/m**2
Q3=Q1-Q2		#Reduction in heat loss through the wall in kJ/hr/m**2

#Output
print'The reduction in heat loss through the wall is ',round(Q3,1),"kJ/hr/m**2"
print"\nNOTE:Answer wrongly written in book as 1951.4"

The reduction in heat loss through the wall is  19517.2 kJ/hr/m**2

NOTE:Answer wrongly written in book as 1951.4


Example 6.17 page no:241¶

In [96]:
#Input data
L=4.6	 	  #Length of the wall in m
b=2.3	   	#Breadth of the wall in m
x1=0.025		#Thickness of the wood in m
x2=0.075		#Thickness of the cork slabbing in m
x3=0.115		#Thickness of the brick in m
t1=18.0		 #Exterior temperature of the wall in degree centigrade
t4=-20.0		#Interior temperature of the wall in degree centigrade
K1=7.5 		 #Thermal conductivity of the wood in kJ/hr mC
K2=1.9 		 #Thermal conductivity of the wood in kJ/hr.mC
K3=41.0		 #Thermal conductivity of the brick in kJ/hr mC

#Calculations
A=L*b		#Area of the wall in m**2
R1=(K1*A)/(x1)	#Heat loss for the change in temperature for insulated wood material in kJ/hrC
R2=(K2*A)/(x2)	#Heat loss for the change in temperature for cork material in kJ/hrC
R3=(K3*A)/(x3)	#Heat loss for the change in temperature for brick in kJ/hrC
Q=(t1-t4)/((1.0/R1)+(1.0/R2)+(1.0/R3))	#Heat loss with insulation in kJ/hr
Q1=Q*24.0		#Heat loss with insulation in kJ/24hr
t2=t1-(Q/R1)	#Interface temperature t2 in degree centigrade
t3=t2-(Q/R2)	#Interface temperature t3 in degree centigrade

#Output
print'(a)The leakage through the wall per 24 hours is',round(Q,2),"kJ/hr=",round(Q,2)*24
print'(b)Temperature at the interface is',round(t2,3),"C"
print'(c)Temperature at interface ',round(t3,3),"C"

(a)The leakage through the wall per 24 hours is 8814.37 kJ/hr= 211544.88
(b)Temperature at the interface is 15.223 C
(c)Temperature at interface  -17.663 C


Example 6.18 page no:243¶

In [97]:
#Input data
L=0.3		#Thickness of the wall in m
ti=320		#Inner surface temperature in degree centigrade
to=38		#Outer surface temperature in degree centigrade
A=1		#Assume unit area in m**2

#Calculations
Q=(A/L)*((0.01256/2)*(ti**2-to**2)-(4.2/3)*10**-6*(ti**3-to**3))  #Heat loss per sq metre of surface area

#Output
print'The heat loss per sq metre of surface area for a furnace wall is',round(Q,2),"kJ/hr/m**2 "

The heat loss per sq metre of surface area for a furnace wall is 1960.68 kJ/hr/m**2


Example 6.19 page no:245¶

In [98]:
#Input data
d=11.5		#Outer diameter of steam pipe line in cm
t1=5.0		#Thickness of first layer in cm
K1=0.222		#Thermal conductivity of first layer in kJ/hr mC
t2=3.0		#Thickness of second layer in cm
pi=3.1428		#Constant value of pi
K2=3.14		#Thermal conductivity of second layer in kJ/hr mC
T1=235		#Outside surface temperature of steam pipe in degree centigrade
T3=38		#Outer surface of lagging in degree centigrade
L=1.0		#Length of the pipe in m

#Calculations
I=math.log((d+(2*t1))/d)			#For inner layer calculation
O=math.log((d+(2*t1)+(2*t2))/(d+(2*t1)))		#For outer layer calculations
R1=(2.0*pi*L*K1)/I		#Heat loss for change in temperature for first insulated material in kJ/hC
R2=(2.0*pi*L*K2)/O		#Heat loss for the change in temperature for second insulated material in kJ/hC
Q=(T1-T3)/(1.0/R1+1.0/R2)	#Heat loss per metre length of pipe per hr in kJ/hr
T2=T1-(Q/R1)#Temperature between the two layers of insulation in degree centigrade

#Output
print'(a)The heat loss per metre length of pipe per hr is',round(Q,2)," kJ/hr"
print'(b)Temperature between the two layers of insulation is',round(T2,2),"C"

(a)The heat loss per metre length of pipe per hr is 427.45  kJ/hr
(b)Temperature between the two layers of insulation is 43.33 C


Example 6.20 page no:247¶

In [21]:
#Input data
t1=24.0    	#Temperature at the outside surface in degree centigrade
t4=-15.0   	#Temperature at the inner surface in degree centigrade
A=1.0  		#Assuming unit area in m**2
K1=23.2		#Thermal conductivity of steel in W/mC
K2=0.014		#Thermal conductivity of glasswood in W/mC
K3=0.052		#Thermal conductivity of plywood in W/mC
x1=0.0015		#Thickness of steel sheet at outer surface in m
x2=0.02		#Thickness of glasswood in between in m
x3=0.01		#Thickness of plywood at a inner surface in m

#Calculations
R1=(K1*A)/x1	#Heat loss for the change in temperature for first insulated material
R2=(K2*A)/x2	#Heat loss for the change in temperature for second insulated material
R3=(K3*A)/x3	#Heat loss for the change in temperature for third insulated material
Q=(t1-t4)/(1/R1+1/R2+1/R3)	#The rate of heat flow in W/m**2

#Output
print'The rate of heat flow is',round(Q,2),"W/m**2 "

The rate of heat flow is 24.06 W/m**2