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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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 "
```

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"
```

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"
```

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 "
```

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"
```

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 "
```