In [1]:

```
import math
# Variables
I = 1350; #Solar Irradiation in W/m**2
L = (1.5*10**8); #Approximate dismath.tance in km
D = (1.39*10**6); #Approximate diameter in km
# Calculations
E = (I*(L*1000)**2*3.14)/((3.14/4)*(D*1000)**2); #Emissive power of Earth
Ts = (E/(5.67*10**-8))**0.25; #Surface temperature of sun in K
# Results
print 'Surface temperature of sun is %d K'%(Ts)
```

In [3]:

```
# Variables
S = 1; #Side of a square in m
L = 0.4; #Distance between the plates in m
T1 = 900; #Temperature of one plate in degree C
T2 = 400; #Temperature of the other plate in degree C
# Calculations
R = (S/L); #Ratio of the side of the square to the distance between plates
F12 = 0.415; #From Fig.10.4 on page no.409
Q = (5.67*10**-8*S*S*F12*((T1+273)**4-(T2+273)**4))/1000; #The net heat transfer in kW
# Results
print 'The net exchange of energy due to radiation between the plates is %3.1f kW'%(Q)
```

In [4]:

```
# Variables
A51 = 2; #Ratio of areas A5 and A1
A21 = 1; #Ratio of areas A2 and A1
F56 = 0.15; #Shape factor
F53 = 0.11; #Shape factor
F26 = 0.24; #Shape Factor
F23 = 0.2; #Shape Factor
# Calculations
F14 = (A51*(F56-F53))-(A21*(F26-F23)); #Shape factor
# Results
print 'Shape factor F14 is %3.2f'%(F14)
```

In [4]:

```
# Variables
Th = 40.; #Radiating heating panel in degree C
Tb = 5.; #Temperature of black plane in degree C
Tc = 31.; #Temperature of ceiling in degree C
A = (10.*12); #Area in m**2
# Calculations
F56 = 0.075; #Using Fig.10.2 on page no. 408
F63 = 0.04; #Using Fig.10.2 on page no. 408
F12 = 0.052; #Shape factor
F1w = (1-F12); #Shape factor between the floor and all the walls but the window
Q12 = (A*F12*5.67*10**-8*((Th+273)**4-(Tb+273)**4)); #Heat exchange between the floor and window in W
#Q1 = (5.67*10**-8*A*((Th+273.15)**4-((F12*(Th+273.15)**4)-(F1w*(Tb+273.15)**4))))/1000; #Net heat given up by the floor in kW
Q1 = (5.67*10**-8*A*((Th+273)**4-((F12*(Th+273)**4)-(F1w*(294)**4))))/1000; #Net heat given up by the floor in kW
# Results
print 'Heat exchange between the floor and window is %3.0f W \n \
Net heat given up by the floor is %3.1f kW'%(Q12,Q1)
# Note : rounding off error.
```

In [5]:

```
# Variables
A2 = (6.*2); #Area of windows in m**2
A1 = (10.*12); #Area of floor in m**2
Th = 40.; #Radiating heating panel in degree C
Tb = 5.; #Temperature of black plane in degree C
F12 = 0.052; #Shape factor
# Calculations
F12a = ((A2-(A1*F12**2))/(A1+A2-(2*A1*F12))); #Shape factor
Q12 = (A1*F12a*5.67*10**-8*((Th+273)**4-(Tb+273)**4)); #Net heat exchange in W
X = (((A2/A1)-F12)/(1-F12)); #X value for equilibrium temperature
T = (((Th+273)**4+(X*(Tb+273)**4))/(X+1))**0.25; #Equilibrium temperature in K
# Results
print 'Net heat exchange is %3.0f W Equilibrium temperature is %3.2f K'%(Q12,T)
```

In [8]:

```
# Variables
D = 0.2; #Diameter of each disc in m
L = 2; #Distance between the plates in m
T = 800+273,300+273; #Temperatures of the plates in K
e = [0.3,0.5] ; #Emissivities of plates
# Calculations
e1 = (e[0]*e[1]); #Equivalent emissivity
R = (D/L); #Ratio between diameter and distance between the plates
F = 0.014; #F value from Fig.10.4 from page no. 409
Q = (e1*(3.14/4)*D**2*F*5.67*10**-8*((T[0]**4-(T[1]**4)))); #Radiant heat exchange for the plates in W
# Results
print 'Radiant heat exchange for the plates is %3.2f W'%(Q)
```

In [9]:

```
# Variables
e = 0.8; #Emissivity of brick wall
D1 = [6,4]; #Width and Height in m
L = 0.04; #Distance from the wall in m
D2 = [0.2,0.2]; #Dimensions of the furnace wall in m
D3 = [1,1]; #Dimensions at lower and left of the centre of the wall in m
T = [1523+273,37+273]; #Furnace temperature and wall temperature in degree C
# Calculations
F12 = 0.033; #Shape factor from Fig.10.3 on page no. 409
F13 = 0.05; #Shape factor from Fig.10.3 on page no. 409
F14 = 0.12; #Shape factor from Fig.10.3 on page no. 409
F15 = 0.08; #Shape factor from Fig.10.3 on page no. 409
Fow = (F12+F13+F14+F15); #Shape factor between opening and wall
Q = (e*L*Fow*5.67*10**-8*(T[0]**4-T[1]**4))/1000; #Net radiation exchange in kW
# Results
print 'Net radiation exchange between the opening and the wall is %3.1f kW'%(Q)
```

In [10]:

```
# Variables
D = [2,1,1]; #Dimensions of the math.tank in m
A = 8; #Area of the tank in m**2
e = 0.9; #Surface emissivity
Ts = 25+273; #Surface temperature in K
Ta = 2+273; #Ambient temperature in K
e1 = 0.5; #Emissivity of aluminium
# Calculations
Q = (e*A*5.67*10**-8*(Ts**4-Ta**4))/1000; #Heat lost by radiation in kW
r = ((e-e1)/e)*Q; #Reduction in heat loss if the tank is coated with an aluminium paint in kW
# Results
print 'Heat lost by radiation is %3.2f kW \n \
Reduction in heat loss if the tank is coated with an aluminium paint is %3.3f kW'%(Q,r)
```

In [5]:

```
# Variables
D = 0.2; #Outer diameter of the pipe in m
Ta = 30+273; #Temperature of the air in K
Ts = 400+273; #Surface temperature in K
e = 0.8; #Emissivity of the pipe surface
D1 = 0.4; #Diamter of brick in m
e1 = 0.91; #Emissivity of brick
# Calculations
Q = (e*3.14*D*5.67*10**-8*(Ts**4-Ta**4))/1000; #Loss of heat by thermal radiation in kW/m
e2 = (1./((1./e)+((D/D1)*((1./e1)-1)))); #Equivalent emissivity
Q1 = (e2*3.14*D*5.67*10**-8*(Ts**4-Ta**4))/1000; #Heat loss when brick is used in kW/m
r = (round(Q,2)-round(Q1,2))*1000; #Reduction in heat loss in W/m
# Results
print 'Loss of heat by thermal radiation is %3.1f*10**3 W/m \n \
Reduction in heat loss is %3.0f W/m'%(Q,r)
```

In [16]:

```
# Variables
e = 0.03; #Emissivity of silver
T2 = -153.+273; #Temperature of the outer surface of the inner wall in K
T1 = 27.+273; #Temperature of the inner surface of the outer wall in K
D1 = 0.42; #Diamter of first sphere in m
D2 = 0.6; #Diamter of the second sphere in m
V = 220.; #Rate of vapourization in kJ/kg
# Calculations
e1 = (1./((1./e)+((D1/D2)**2*((1./e)-1)))); #Equivalent emissivity
A = (4*3.14*(D1/2)**2); #Area in m**2
Q = (e1*A*5.67*10**-8*(T1**4-T2**4))/(1000./3600); #Radiation heat transfer through walls into the vessel in kJ/h
R = (Q/V); #Rate of evaporation in kg/h
# Results
print 'Radiation heat transfer through walls into the vessel is %3.3f kJ/h \n \
Rate of evaporation of liqiud oxygen is %3.4f kg/h'%(Q,R)
```

In [17]:

```
# Variables
T = [800+273,300+273]; #Temperatures of the plates in K
e = [0.3,0.5]; #Emissivities of the plates
# Calculations
Q = ((5.67*10**-8*(T[0]**4-T[1]**4))/((1./e[0])+((1./e[1]))-1))/1000; #Net radiant heat exchange in kW/m**2
# Results
print 'Net radiant heat exchange is %3.2f kW/m**2'%(Q)
```

In [18]:

```
# Variables
T1 = 127+273; #Temperature of the outer side of the brick setting in K
T2 = 50+273; #Temperature of the inside of the steel plate in K
e1 = 0.6; #Emissivity of steel
e2 = 0.8; #Emissivity of fireclay
# Calculations
Q = ((5.67*10**-8*(T1**4-T2**4))/((1./e1)+((1./e2))-1)); #Net radiant heat exchange in W/m**2
# Results
print 'Net radiant heat exchange is %3.0f W/m**2'%(Q)
```

In [19]:

```
# Variables
D = 1; #Dimension of the plate in m
L = 0.5; #Distance between the plates in m
Ts = 27+273; #Surface temperature of the walls in K
T = [900+273,400+273]; #Temperature of the plates in K
e = [0.2,0.5]; #Emissivities of the plates
# Calculations
F12 = 0.415; #From Fig.10.4 on page no.409
F13 = (1-F12); #Shape factor
F23 = (1-F12); #Shape factor
R1 = (1-e[0])/(e[0]*D*D); #Resistance for 1
R2 = (1-e[1])/(e[1]*D*D); #Resistance for 2
R3 = 0; #Resistance for 3
A1F12I = (1./(D*D*F12)); #Inverse of the product of area and Shape factor
A1F13I = (1./(D*D*F13)); #Inverse of the product of area and Shape factor
A2F23I = (1./(D*D*F23)); #Inverse of the product of area and Shape factor
Eb1 = (5.67*10**-8*T[0]**4)/1000; #Emissive power of 1 in kW/m**2
Eb2 = (5.67*10**-8*T[1]**4)/1000; #Emissive power of 2 in kW/m**2
Eb3 = (5.67*10**-8*Ts**4); #Emissive power of 3 in W/m**2
J1 = 25; #Radiosity at node 1 in kW/m**2
J2 = 11.53; #Radiosity at node 2 in kW/m**2
J3 = 0.46; #Radiosity at node 3 in kW/m**2
Q1 = ((Eb1-J1)/R1); #Total heat loss by plate 1 in kW
Q2 = ((Eb2-J2)/R2); #Total heat loss by plate 2 in kW
Q3 = ((J1-J3)/(A1F13I))+((J2-J3)/(A2F23I)); #Total heat received by the room in kW
# Results
print 'Total heat loss by plate 1 is %3.1f kW Total heat loss by plate 2 is %3.1f kW \n \
Total heat received by the room is %3.2f kW'%(Q1,Q2,Q3)
```

In [20]:

```
# Variables
T = [800+273,300+273]; #Temperatures of the plates in K
e = [0.3,0.5]; #Emissivities of the plates
e3 = 0.05; #Emissivity of aluminium
# Calculations
q = ((5.67*10**-8*(T[0]**4-T[1]**4))/((1./e[0])+(1./e[1])-1))/1000; #Heat transfer without the shield in kW/m**2
R1 = (1-e[0])/e[0]; #Resistance in 1
R2 = (1-e[1])/e[1]; #Resistance in 2
R3 = (1-e3)/e3; #Resistance in 3
R = (R1+(2*R2)+(2*R3)); #Total resismath.tance
q1 = ((5.67*10**-8*(T[0]**4-T[1]**4))/R)/1000; #Heat transfer with shield in kW/m**2
r = ((q-q1)*100)/q; #Reduction in heat transfer
X1 = ((1./e3)+(1./e[1])-1); #X1 for tempearture T3
X2 = ((1./e[0])+(1./e3)-1); #X1 for tempearture T3
T3 = (((X1*T[0]**4)+(X2*T[1]**4))/(X2+X1))**0.25; #Temperature of the sheild in K
T3c = T3-273; #Temperature of the sheild in degree C
# Results
print 'Percentage reduction in heat transfer is %3.0f percent \n \
Temperature of the sheild is %3.2f degree C'%(r,T3c)
```

In [21]:

```
# Variables
Q = 79; #Reduction in net radiation from the surfaces
e1 = 0.05; #Emissivity of the screen
e2 = 0.8; #Emissivity of the surface
# Calculations
n = (((Q*((2/e2)-1))-((2/e2)+1))/((2/e1)-1)); #Number of screens to be placed between the two surfaces to achieve the reduction in heat exchange
# Results
print 'Number of screens to be placed between the two surfaces to achieve the reduction in heat exchange is%3.0f'%(n)
```

In [22]:

```
# Variables
e = 0.8; #Emissivity of the pipe
D = 0.275; #Diameter of the pipe in m
Ts = 500+273; #Surface temperature in K
Te = 30+273; #Temperature of enclosure in K
D1 = 0.325; #Diamter of the steel screen in m
e1 = 0.7; #Emissivity of steel screen
Tsc = 240+273; #Temperature of screen in K
#CALCUATIONS
Q = (e*5.67*10**-8*3.14*D*(Ts**4-Te**4))/1000; #Loss of heat per unit length by radiation in kW/m
e2 = (1./((1./e)+((D/D1)*((1./e1)-1)))); #Equivalent emissivity
Q1 = (e2*5.67*10**-8*3.14*D*(Ts**4-Tsc**4))/1000; #Radiant heat exchange per unit length of header with screen in kW/m
R = (Q-Q1); #Reduction in heat by radiation due to the provision of the screen in kW/m
# Results
print 'Loss of heat per unit length by radiation is %3.1f kW/m \n \
Reduction in heat by radiation due to the provision of the screen is %3.2f kW/m'%(Q,R)
```

In [23]:

```
# Variables
e = 0.6; #Emissivity of thermocouple
Ta = 20+273; #Ambient temperature in K
Tt = 500+273; #Temperature from the thermocouple in K
e = 0.3; #Emissivity of radiation shield
h = 200; #Convective heat transfer coefficient in W/m**2.K
Ts = 833; #Temperature in K
# Calculations
T = ((5.67*10**-8*e*(Tt**4-Ta**4))/(h*1000))+Tt; #Temperature of the shield in K
T1 = (Ts-T); #Error between the thermocouple temperature and gas temperature in K
Ts = 825.; #Surface temperature with radiation shield in K
Tc = 829.; #Thermocouple temperature with radiation shield in K
e = (Tc-Ts); #Error between the thermocouple temperature and gas temperature with the shielded thermocouple arrangement in K
# Results
print 'Error between the thermocouple temperature and gas temperature is%3.0f K \n \
Error between the thermocouple temperature and gas temperature with the shielded thermocouple arrangement is%3.0f K'%(T1,e)
```

In [24]:

```
# Variables
D = 0.2; #Diameter of pipe in m
Ta = 30+273; #Temperature of air in K
Ts = 200+273; #Temperature of surface in K
e = 0.8; #Emissivity of the pipe
# Calculations
Q = (e*5.67*10**-8*3.14*D*(Ts**4-Ta**4)); #Heat lost by thermal radiation in W/m
T = (Ta+Ts)/2; #Film temperature in degree C
k = 0.03306; #Thermal conductivity in W/m.K
v1 = (24.93*10**-6); #Kinematic viscosity in m**2/s
b = (1./388); #Coefficient of thermal expansion in 1./K
Pr = 0.687; #Prantl number
Ra = ((9.81*b*D**3*(Ts-Ta)*Pr)/(v1**2)); #Rayleigh number
Nu = (0.53*(Ra)**0.25); #Nussults number
h = (k*Nu)/D; #Heat transfer coefficient in W/m**2.K
Q1 = (h*3.14*D*(Ts-Ta)); #Heat lost by convection in W/m
Q2 = (Q+Q1); #Total heat lost per meter length in W/m
# Results
print 'Heat lost by thermal radiation is %3.0f W/m \n \
Heat lost by convection is %3.1f W/m'%(Q,Q1)
```

In [25]:

```
# Variables
Ts = 200+273; #Temperature of stream main in K
Ta = 30+273; #Rooom temperature in K
h = 17.98; #Heat transfer coefficient in W/m**2.K
e = 0.8; #Emissivity of the pipe surface
# Calculations
q = (5.67*10**-8*e*(Ts**4-Ta**4)); #Heat transfer by radiation in W/m**2
hr = (q/(Ts-Ta)); #Heat transfer coefficient due to radiation in W/m**2.K
hc = (h-hr); #Heat transfer coefficient due to convection in W/m**2.K
# Results
print 'Heat transfer coefficient due to radiation is %3.1f W/m**2.K \n \
Heat transfer coefficient due to convection is %3.2f W/m**2.K'%(hr,hc)
```

In [26]:

```
import math
# Variables
t = 0.05; #Thickness of the gas layer in m
r = 0.1; #Remaining radiation intensity
# Calculations
a = (-1./t)*2.3*(math.log(r)/math.log(10)); #Extinction coefficient per m
# Results
print 'Extinction coefficient is %3.2f/m'%(a)
```

In [27]:

```
# Variables
A = 30.; #Total surface area in m**2
V = 10.; #Volume in m**3
Ts = 1000.; #Temperature of the furnace in degree C
p = 2.; #Total pressure in atm
ph2o = 0.1; #Partial pressure of water vapour in atm
pco2 = 0.3; #Partial pressure of CO2
# Calculations
lms = (3.6*V)/A; #Mean beam length in m
pco2lms = (pco2*lms); #pco2lms in m.atm
eco2 = 0.16; #From Fig.10.23 on page no. 458
cco2 = 1.11; #From Fig.10.23 on page no. 458
cco2eco2 = (cco2*eco2); #cco2eco2 value
ph2olms = (ph2o*lms); #ph2olms in m.atm
eh2o = 0.12; #From Fig.10.24 on page no. 459
P = (p+ph2o)/2; #P value in atm
ch2o = 1.43; #From Fig.10.26 on page no. 460
ch2oeh2o = (ch2o*eh2o); #ch2oeh2o value
P1 = (ph2o/(ph2o+pco2)); #Ratio of pressures
X = (pco2lms+ph2olms); #X value in m.atm
e = 0.035; #Error value from Fig. 10.27 on page no.461
et = (cco2eco2+ch2oeh2o-e); #Total emissivity of the gaseous mixture
# Results
print 'Emissivity of the gaseous mixture is %3.4f'%(et)
```

In [6]:

```
# Variables
Tg = 950+273; #Flue gas temperature in K
p = 1; #Total pressure in atm
pco2 = 0.1; #Percent of co2
ph2o = 0.04; #Percent of h2o
D = 0.044; #Diameter of the tube in m
e = 0.8; #Emissivity of grey surface
Tw = 500+273.; #Uniform temperature in K
# Calculations
lms = (3*0.044); #lms value from Table 10.2 on page no. 457
pco2lms = (pco2*lms); #pco2lms in m.atm
ph2olms = (ph2o*lms); #ph2olms in m.atm
eco2 = 0.05; #From Fig.10.23 on page no. 458
eh2o = 0.005; #From Fig.10.24 on page no. 459
b = 1.05; #Correction factor from Fig. 10.28 on page no. 461
eg = 0.061; #Total emissivity of gaseous mixture
ag = ((0.056*(Tg/Tw)**0.65)+(b*0.021)); #Absorbtivity of the gases
q = (0.5*(e+1)*5.67*10**-8*((eg*Tg**4)-(ag*Tw**4))); #Heat transfer rate by radiation in W/m**2
hr = (q/(Tg-Tw)); #Radiation heat transfer coefficient in W/m**2.degree C
# Results
print 'Net radiation exchange between the gas and the tube walls is %3.0f W/m**2 \n \
Radiation heat transfer coefficient is %3.2f W/m**2.degree C'%(q,hr)
```