# Chapter 11: Radiation¶

### ILLUSTRATIVE EXAMPLE 11.3, Page number: 181¶

In :
from sympy import symbols, integrate,oo,exp,pi

#Variable declaration:
l = symbols('l')                        #Wavelength (mu.m)
I = 40*exp(-l**2)                       #Intensity of radiation (Btu/h.ft^2.mu.m)

#Calculation:
E = integrate(I, (l,0,oo)).evalf()      #Total emissive power (Btu/h.ft^2)

#Result:
print "The total emissive power is :",round(E,1)," Btu/h.ft^2 ."

The total emissive power is : 35.4  Btu/h.ft^2 .


### ILLUSTRATIVE EXAMPLE 11.4, Page number: 182¶

In :
#Variable declaration:
l = 0.25                            #Wavelength (mu.m)
#From equation 11.4:
lT = 2884                           #Product of wavelength and absolute temperature (mu.m.°R)

#Calculation:
T = lT/l                            #Sun's temperature (°R)

#Result:
print "The Sun's temperature is :",round(T,-2)," °R ."
print "The Sun's temperature in fahrenheit scale is :",round(T-460,-3)," °F ."

The Sun's temperature is : 11500.0  °R .
The Sun's temperature in fahrenheit scale is : 11000.0  °F .


### ILLUSTRATIVE EXAMPLE 11.5, Page number: 188¶

In :
#Variable declaration:
T1 = 1500.0+460.0                   #Absolute temperature 1 (°R)
T2 = 1000.0+460.0                   #Absolute temperature 2 (°R)

#Calculation:
X = T1**4/T2**4                     #Ratio of quantity of heat transferred
x = 100*(T1**4-T2**4)/T2**4         #Percentage increase in heat transfer (%)

#Result:
print "The ratio of the quantity/rate of heat transferred is :",round(X,2)," ."
print "The percentage increase in heat transfer is :",round(x),"%"

The ratio of the quantity/rate of heat transferred is : 3.25  .
The percentage increase in heat transfer is : 225.0 %


### ILLUSTRATIVE EXAMPLE 11.6, Page number: 189¶

In :
#Variable declaration:
T1 = 1200.0+460.0                       #Absolute temperature of wall 1 (°R)
T2 = 800.0+460.0                        #Absolute temperature of wall 2 (°R)

#Calculation:
#From equation 11.23:
X = 0.173*((T1/100.0)**4-(T2/100.0)**4) #Heat removed from colder wall (Btu/h.ft^2)

#Result:
print "The heat removed from the colder wall to maintain a steady-state is :",round(X)," Btu/h.ft^2 ."

The heat removed from the colder wall to maintain a steady-state is : 8776.0  Btu/h.ft^2 .


### ILLUSTRATIVE EXAMPLE 11.7, Page number: 190¶

In :
#Variable declaration:
s = 0.173                       #Stefan-Boltzmann constant (Btu/h.ft^2.°R)
EH = 0.5                        #Energy transferred from hotter body (Btu/h.ft^2)
EC = 0.75                       #Energy transferred to colder body (Btu/h.ft^2)
TH = 1660.0                     #Absolute temperature of hotter body (°R)
TC = 1260.0                     #Absolute temperature of colder body (°R)

#Calculation:
E = s*((TH/100.0)**4-(TC/100.0)**4)/((1.0/EH)+(1.0/EC)-1.0) #Net energy exchange per unit area (Btu/h.ft^2)

#Result:
print "The net energy exchange per unit area is :",round(E,-1)," Btu/h.ft^2 ."

The net energy exchange per unit area is : 3760.0  Btu/h.ft^2 .


### ILLUSTRATIVE EXAMPLE 11.8, Page number: 191¶

In :
#Variable declaration:
#From example 11.6-11.7:
E1 = 8776.0                     #Energy exchange between black bodies (Btu/h.ft^2)
E2 = 3760.0                     #Energy exchange between non-black bodies (Btu/h.ft^2)

#Calculation:
D = (E1-E2)/E1*100              #Percent difference in energy (%)

#Result:
print "The percent difference relative to the black body is:",round(D,1)," % ."

The percent difference relative to the black body is: 57.2  % .


### ILLUSTRATIVE EXAMPLE 11.9, Page number: 192¶

In :
#Variable declaration:
s = 0.173*10**-8                        #Stefan-Boltzmann constant (Btu/h.ft^2.°R)
TH = 300.0+460.0                        #Absolute temperature of external surface (°R)
TC = 75.0+460.0                         #Absolute temperature of duct (°R)
#From Table 6.2:
AH = 0.622                              #External surface area of pipe (ft^2)
#From Table 11.2:
EH = 0.44                               #Emissivity of oxidized steel
AC = 4.0*1.0*1.0                        #External surface area of duct (ft^2)
EC = 0.23                               #Emissivity of galvanized zinc

#Calculation:
FE = 1.0/(1.0/EH+((AH/AC)*(1.0/EC-1.0))) #Emissivity correction factor
Q = FE*AH*s*(TH**4-TC**4)                #Net radiation heat transfer (Btu/h.ft)

#Result:
print "The net radiation heat transfer is :",round(Q,2)," Btu/h.ft^2 ."
print "There is a calculation error in book."

The net radiation heat transfer is : 96.96  Btu/h.ft^2 .
There is a calculation error in book.


### ILLUSTRATIVE EXAMPLE 11.10, Page number: 193¶

In :
#Variable declaration:
TH = 140.0+460.0                    #Absolute outside temperature of pipe (ft^2)
TC = 60.0+460.0                     #Absolute temperature of surrounding atmosphere (ft^2)
A = 10.0                            #Area of pipe (ft^2)
E = 0.9                             #Emissivity of pipe

#Calculation:
Q = E*A*0.173*((TH/100.0)**4-(TC/100.0)**4) #Heat loss due to radiation (Btu/h)

#Result:
print "The heat loss due to radiation is :",round(Q,-1)," Btu/h ."

The heat loss due to radiation is : 880.0  Btu/h .


### ILLUSTRATIVE EXAMPLE 11.11, Page number: 193¶

In :
#Variable declaration:
#Froma example 11.10:
Q = 880.0                   #Heat loss due to radiation (Btu/h)
A = 10.0                    #Area of pipe (ft^2)
TH = 140.0                  #Absolute outside temperature of pipe (°F)
TC = 60.0                   #Absolute temperature of surrounding atmosphere (°F)

#Calculation:
hr = Q/(A*(TH-TC))          #Radiation heat transfer coefficient (Btu/h.ft^2.°F)

#Result:
print "The radiation heat transfer coefficient is :",round(hr,1)," Btu/h.ft^2.°F ."

The radiation heat transfer coefficient is : 1.1  Btu/h.ft^2.°F .


### ILLUSTRATIVE EXAMPLE 11.12, Page number: 194¶

In :
from math import pi

#Variable declaration:
D = 0.0833                          #Diameter of tube (ft)
L = 2.0                             #Length of tube (ft)
h = 2.8                             #Heat transfer coefficient (Btu/h.ft^2.°F)
Ta1 = 1500.0+460.0                  #Temperature of hot air in furnace (°R)
Ta2 = 1350.0+460.0                  #Temperature of hot air in the furnace brick walls (°R)
Tt = 600.0+460.0                    #Surface temperature of tube (°R)
E = 0.6                             #Surface emissivity of tube
s = 0.1713*10**-8                   #Stefan-Boltzmann constant

#Calculation:
#Case 1:
A = pi*D*L                          #Area of tube (ft^2)
Qc = round(h*A*(Ta1-Tt),-1)         #Convection heat transfer from air to tube (Btu/h)
Qr = round(E*s*A*(Ta2**4-Tt**4),-2) #Radiation feat transfer from wall to tube (Btu/h)
Q = Qr+Qc                           #Total heat transfer (Btu/h)
#Case 2:
Qp = Qr/Q*100                       #Radiation percent
#Case 3:
hr = Qr/(A*(Ta2-Tt))                #Radiation heat transfer coefficient (Btu/h.ft^2.°F)
#Case 4:
T = Ta2-Tt                          #Temperature difference (°F)

#Result:
print "1. The convective heat transferred to the metal tube is :",Qc," Btu/h ."
print "   The radiative heat transferred to the metal tube is :",Qr," Btu/h ."
print "   The total heat transferred to the metal tube is :",Q," Btu/h ."
print "2. The percent of total heat transferred by radiation is :",round(Qp,1)," % ."
print "3. The radiation heat transfer coefficient is :",round(hr,1)," Btu/h.ft^2.°F ."
if (T > 200):
print "4. The use of the approximation Equation (11.30), hr = 4EsTav^3, is not appropriate."
elif (T < 200):
print "4. The use of the approximation Equation (11.30), hr = 4EsTav^3, is appropriate."

1. The convective heat transferred to the metal tube is : 1320.0  Btu/h .
The radiative heat transferred to the metal tube is : 5100.0  Btu/h .
The total heat transferred to the metal tube is : 6420.0  Btu/h .
2. The percent of total heat transferred by radiation is : 79.4  % .
3. The radiation heat transfer coefficient is : 13.0  Btu/h.ft^2.°F .
4. The use of the approximation Equation (11.30), hr = 4EsTav^3, is not appropriate.


### ILLUSTRATIVE EXAMPLE 11.13, Page number: 194¶

In :
#Variable declaration:
Q = 5.0                         #Radiation heat transfer (W)
E = 1.0                         #Emissivity of filament
s = 5.669*10**-8                #Stefan-Boltzmann constant
T1 = 900.0+273.0                #Light bulb temperature (K)
T2 = 150.0+273.0                #Glass bulb temperature (K)

#Calculation:
A = Q/(E*s*(T1**4-T2**4))       #Surface area of the filament (m^2)

#Result:
print "The surface area of the filament is :",round(A*10**4,2), "cm^2"

The surface area of the filament is : 0.47 cm^2


### ILLUSTRATIVE EXAMPLE 11.14, Page number: 195¶

In :
from math import pi

#Variable declaration:
T1 = 127.0+273.0                    #Surface temperature (K)
T2 = 20.0+273.0                     #Wall temperature (K)
T3 = 22.0+273.0                     #Air temperature (K)
s = 5.669*10**-8                    #Stefan-Boltzmann constant
e = 0.76                            #Surface emissivity of anodized aluminium
D = 0.06                            #Diameter of pipe (m)
L = 100.0                           #Length of pipe (m)
h = 15.0                            #Pipe convective heat transfer coefficient (W/m^2.K)

#Calculation:
Eb = s*T1**4                        #Emissive energy of pipe (W/m^2)
E = e*Eb                            #Emissive power from surface of pipe (W/m^2)
A = pi*D*L                          #Surface area of pipe (m^2)
Qc = h*A*(T1-T3)                    #Convection heat transfer to air (W)
Qr = e*s*A*(T1**4-T2**4)            #Radiation heat transfer rate (W)
Q = Qc+Qr                           #Total heat transfer rate (Btu/h)
Tav = (T1+T2)/2.0                   #Average temperature (K)
hr = 4*e*s*Tav**3                   #Radiation heat transfer coefficient (W/m^2.K)

#Result:
print "The emissive power from surface of pipe is :",round(E)," W/m^2 ."
print "The convection heat transfer to air is :",round(Qc/10**3,1)," kW ."
print "The radiation heat transfer rate is :",round(Qr/10**3,1)," kW "
print "The radiation heat transfer coefficient is :",round(hr,1)," W/m^2.K ."

The emissive power from surface of pipe is : 1103.0  W/m^2 .
The convection heat transfer to air is : 29.7  kW .
The radiation heat transfer rate is : 14.8  kW
The radiation heat transfer coefficient is : 7.2  W/m^2.K .


### ILLUSTRATIVE EXAMPLE 11.15, Page number: 196¶

In :
#Variable declaration:
#From example 11.14:
Qc = 15.0                       #Convection heat transfer coefficient (W/m^2.K)
hr = 7.2                        #Radiation heat transfer coefficient (W/m^2.K)

#Calculation:
X = hr/(Qc+hr)*100.0            #Percent heat transfer by radiation (%)

#Result:
print "The percent heat transfer by radiation is :",round(X,1)," % ."

The percent heat transfer by radiation is : 32.4  % .


### ILLUSTRATIVE EXAMPLE 11.16, Page number: 200¶

In :
#Variable declaration:
FV = 1.0                            #Correction factor
#From example 11.9:
FE = 0.358                          #Emissivity correction factor
TH = 300.0+460.0                    #Absolute temperature of external surface (°R)
TC = 75.0+460.0                     #Absolute temperature of duct (°R)
AH = 0.622                          #Area of pipe (ft^2)
s = 0.173*10**-8                    #Stefan-Boltzmann constant

#Calculation:
Q = FV*FE*AH*s*(TH**4-TC**4)        #Heat transfer rate (Btu/h.ft)

#Result:
print "The heat transfer rate is :",round(Q,2)," Btu/h.ft"
print "Since, 'Q' obtained in (11.9) is 96.96 Btu/h.ft, the solution does not match with book."

The heat transfer rate is : 96.96  Btu/h.ft
Since, 'Q' obtained in (11.9) is 96.96 Btu/h.ft, the solution does not match with book.


### ILLUSTRATIVE EXAMPLE 11.17, Page number: 200¶

In :
#Variable declaration:
#From figure 11.2:
L = 1.0                                 #Space between plates (m)
X = 0.5                                 #Length of plate (m)
Y = 2.0                                 #Width of plate (m)
s = 5.669*10**-8                        #Stefan-Boltzmann constant
TH = 2000.0+273.0                       #Temperature of hotter plate (K)
TC = 1000.0+273.0                       #Temperature of colder plate (K)
Btu = 0.2934*10**-3                     #Btu/h in a KW

#Calculation:
A = X*Y                                 #Area of plate (m^2)
Z1 = Y/L                                #Ratio of width with space
Z2 = X/L                                #Ratio of length with space
#From figure 11.2:
FV = 0.18                               #Correction factor
FE = 1.0                                #Emissivity correction factor
Q1 = FV*FE*s*A*(TH**4-TC**4)            #Net radiant heat exchange between plates (kW)
Q2 = Q1/Btu                             #Net radiant heat exchange between plates in Btu/h (Btu/h)

#Result:
print "The net radiant heat exchange between plates is :",round(Q1,-2)," kW ."
print "The net radiant heat exchange between plates in Btu/h is :",round(Q2/10**8,2)," x 10^8 Btu/h ."

The net radiant heat exchange between plates is : 245600.0  kW .
The net radiant heat exchange between plates in Btu/h is : 8.37  x 10^8 Btu/h .