# Chapter 9: Heat Transfer by Radiation¶

## Ex9.1: Page 547¶

In [8]:
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 1"
#Temperature of the tungsten filament in Kelvin
T=1400;

print "a)Wavelength at which the monochromatic emissive power of the given tungsten filament is maximum in meters"
#Wavelength in m
lamda_max=2.898e-3/T
print  "{:.2e}".format(lamda_max)

print "b)Monochromatic emissive power at calculated maximum wavelength in W/m**3"
#Emissive power in W/m3
Eb_max=12.87e-6*(T**5)
print  "{:.2e}".format(Eb_max)
#Given wavelength in meters
lamda=5e-6;
#Product of wavelength and temperature in m-K
lamda_T=lamda*T;

print "c)Monochromatic emissive power at given wavelength in W/m**3"
#Emissive power in W/m3
Eb_lamda=Eb_max*(2.898e-3/(lamda_T))**5*(((math.e**4.965)-1)/((math.e**(0.014388/lamda_T)-1)))
print "{:.2e}".format(Eb_lamda)
print "Thus ,Monochromatic emissive power at 5e-6 m wavelength is 25.4% of the Monochromatic emissive power at maximum wavelength"

Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 1
a)Wavelength at which the monochromatic emissive power of the given tungsten filament is maximum in meters
2.07e-06
b)Monochromatic emissive power at calculated maximum wavelength in W/m**3
6.92e+10
c)Monochromatic emissive power at given wavelength in W/m**3
1.76e+10
Thus ,Monochromatic emissive power at 5e-6 m wavelength is 25.4% of the Monochromatic emissive power at maximum wavelength


## Ex9.2: Page 549¶

In [10]:
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 2"
#Temperature at which sun is radiating as a blackbody in K
T=5800;

#Lower limit of wavelength for which glass is transparent in microns
lamda_l=0.35;
#lower limit of product of wavelength and temperature in micron-K
lamda_l_T=lamda_l*T;
#Lower limit of wavelength for which glass is transparent in microns
lamda_u=2.7;
#lower limit of product of wavelength and temperature in micron-K
lamda_u_T=lamda_u*T;

# For lamda_T= 2030, ratio of blackbody emission between zero and lamda_l to the total emission in terms of percentage
r_l=6.7;
# For lamda_T= 15660, ratio of blackbody emission between zero and lamda_u to the total emission in terms of percentage
r_u=97;

#Total radiant energy incident upon the glass from the sun in the wavelength range between lamda_l and lamda_u
print "Percentage of solar radiation transmitted through the glass in terms of percentage"

Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 2
Percentage of solar radiation transmitted through the glass in terms of percentage
83.1


## Ex9.3: Page 552¶

In [23]:
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 3"
#Area of the flat black surface in m**2
A_1=10e-4;
#Radiation emitted by the flat black surface in W/m** sr
I_1=1000;
# Another surface having same area as A1 is placed relative to A1 such that length of radiation ray connecting dA_1 and dA_2 in meters
r=0.5;
#Area in m**2
A_2=10e-4;
# Since both areas are quite small, they can be approximated as differential surface areas and the solid angle can be calculated as
#d_omega21=dA_n2/r**2 where dA_n2 is the projection of A2 in the direction normal to the incident radiation for dA_1,thus

#Angle between the normal n_2 ant the radiation ray connecting dA_1 and dA_2
theta_2=30;

#Therefore solid angle in sr
d_omega21=(A_2*math.cos(theta_2*math.pi/180)/(r**2));

print "Irradiation of A_2 by A_1 in watt"
q_r12= I_1*A_1*math.cos(90*math.pi/180-theta_2*math.pi/180)*d_omega21
print round(q_r12,5)

Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 3
Irradiation of A_2 by A_1 in watt
0.00173


## Ex9.4: Page 559¶

In [24]:
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 4"
#Hemispherical emissivity of an aluminum paint at wavelengths below 3 microns
epsilon_lamda_1=0.4;
#Hemispherical emissivity of an aluminum paint at longer wavelengths
epsilon_lamda_2=0.8;
#At room temperature 27 degree celcius, product of lamda and T in micron-K
lamda_T_1=3*(27+273);
#At elevated temperature 527 degree celcius, product of lamda and T  in micron-K
lamda_T_2=3*(527+273);
#From Table 9.1
# For lamda_T_1, ratio of blackbody emission between zero and lamda_l to the total emission
r_1=0.00016;
# For lamda_T_2, ratio of blackbody emission between zero and lamda_u to the total emission
r_2=0.14;
print "Thus, the emissivity at 27°C"
#Emissivity
epsilon=0.8
print epsilon
print "emissivity at 527°C"
#Emissivity at higher temp.
epsilon=(r_2*epsilon_lamda_1)+(epsilon_lamda_2*0.86)
print epsilon
print "The reason for the difference in the total emissivity is that at the higher temperature,the percentage of the total emissive power in the low-emittance region of the paint is appreciable, while at the lower temperature practically all the radiation is emittedat wavelengths above 3 microns"

Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 4
Thus, the emissivity at 27°C
0.8
emissivity at 527°C
0.744
The reason for the difference in the total emissivity is that at the higher temperature,the percentage of the total emissive power in the low-emittance region of the paint is appreciable, while at the lower temperature practically all the radiation is emittedat wavelengths above 3 microns


## Ex9.5: Page 560¶

In [26]:
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 5"
#Temperature of the sun in K
T=5800;
#For the case of Solar irradiation, value of the product of lamda and T in micron-K
lamda_T_1=3*T;# value of lamda is taken from Example 9.4
#From table 9.1
# For lamda_T_1, ratio of blackbody emission between zero and lamda_l to the total emission
r_1=0.98;
#This means that 98% of the solar radiation falls below 3 microns
#Hemispherical emissivity of an aluminum paint at wavelengths below 3 microns
epsilon_lamda_1=0.4;
#Hemispherical emissivity of an aluminum paint at longer wavelengths
epsilon_lamda_2=0.8;
print "Effective absorptivity for first case"
#Effective absorptivity
alpha_1=(r_1*epsilon_lamda_1)+(epsilon_lamda_2*0.02)
print round(alpha_1,3)
#For the case second with source at 800 K, value of the product of lamda and T in micron-K
lamda_T_2=3*800;
# For lamda_T_2, ratio of blackbody emission between zero and lamda_l to the total emission
r_2=0.14;
print "Effective absorptivity for second case"
#Effective absorptivity
alpha_2=(r_2*epsilon_lamda_1)+(epsilon_lamda_2*0.86)
print round(alpha_2,3)

Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 5
Effective absorptivity for first case
0.408
Effective absorptivity for second case
0.744


## Ex9.6: Page 562¶

In [29]:
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 6"
#Stefan–Boltzmann constant in W/m**2 K**4
sigma=5.67e-8;
#Temperature of the painted surface in K
T=1000;
#Temperature of the sun in K
T_s=5800;
#Given, below 2 microns the emissivity  of the surface is 0.3,so
lamda_1=2; #wavelength in microns
epsilon_1=0.3; #emissivity

#Given, between 2 and 4 microns emmisivity is 0.9,so
lamda_2=4;#wavelength in microns
epsilon_2=0.9;#emissivity

#Given, above 4 microns emmisivity is 0.5, so
epsilon_3=0.5;#emissivity

#value of the product of lamda_1 and T in micron-K
lamda_1_T=2e-3*T;

#From table 9.1
# For lamda_1_T, ratio of blackbody emission between zero and lamda_l to the total emission
r_1=0.0667; #1st ratio

#value of the product of lamda_2 and T in micron-K
lamda_2_T=2e-3*T;
#From table 9.1
# For lamda_2_T, ratio of blackbody emission between zero and lamda_l to the total emission
r_2=0.4809; #2nd ratio

print "a)Effective emissivity over the entire spectrum"
#Effective emissivity
epsilon_bar=epsilon_1*r_1+epsilon_2*(r_2-r_1)+epsilon_3*(1-r_2)
print round(epsilon_bar,4)

print "b)Emissive power in W/m**2"
#Emissive power in W/m**2
E=epsilon_bar*sigma*T**4
print "{:.1e}".format(E)

#value of the product of lamda_1 and T_s in micron-K
lamda_1_T_s=2e-3*T_s;
#From table 9.1
# For lamda_1_T_s, ratio of blackbody emission between zero and lamda_l to the total emission
r_1_s=0.941;
#value of the product of lamda_2 and T_s in micron-K
lamda_2_T_s=2e-3*T_s;
#From table 9.1
# For lamda_2_T_s, ratio of blackbody emission between zero and lamda_l to the total emission
r_2_s=0.99;
print "c) Average solar absorptivity"
#Average solar absorptivity
alpha_s=epsilon_1*r_1_s+epsilon_2*(r_2_s-r_1_s)+epsilon_3*(1-r_2_s)
print round(alpha_s,3)

# the answer in textbook is slightly different due to approximation

Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 6
a)Effective emissivity over the entire spectrum
0.6523
b)Emissive power in W/m**2
3.7e+04
c) Average solar absorptivity
0.331


## Ex9.7: Page 569¶

In [31]:
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 7"
#Temperature of the oxidised surface in Kelvin
T=1800;
#Area of the oxidised surface in m**2
A=5e-3;
#Stefan–Boltzmann constant in W/m**2 K**4
sigma=5.67e-8;
print "a)Emissivity perpendicular to the surface"
#Emissivity
epsilon_zero=0.70*math.cos(0)
print round(epsilon_zero,3)
print "b)Hemispherical emissivity"
#Hemispherical emissivity
epsilon_bar=((-1.4)/3)*((math.cos(90*math.pi/180))**3-(math.cos(0))**3)
print round(epsilon_bar,3)
print "c)Emissive Power in Watt"
#Emissive Power in W
E=epsilon_bar*A*sigma*T**4
print round(E,3)

# the answer in textbook is slightly different due to approximation

Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 7
a)Emissivity perpendicular to the surface
0.7
b)Hemispherical emissivity
0.467
c)Emissive Power in Watt
1388.832


## Ex9.8: Page 574¶

In [34]:
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 8"

# Theoretical Proof
print "The given example is theoretical and does not involve any numerical computation"

Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 8
The given example is theoretical and does not involve any numerical computation


## Ex9.9: Page 578¶

In [33]:
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 9"
#Window arrangement consists of a long opening with dimensions
#Height in meters
h=1;
#Length in meters
l=5;
#width of table in meters
w=2;
#Assuming that window and table are sufficiently long and applying crossed string method, we get
#Distance ab in m
ab=0;
#Distance cb in m
cb=w;
#Distance cd in m
cd=math.sqrt(l);

print "Shape factor between the window and the table"
#Shape factor between the window and the table
print round(F_12,4)

Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 9
Shape factor between the window and the table
0.382


## Ex9.10: Page 580¶

In [37]:
print "Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 10"
#Window area in ft**2
A1=6*20;
#Second area in ft**2
A2=4*20;
#Assuming A5=A1+A2
#Area in ft**2
A5=A1+A2;

#From Fig. 9.27
#Shape Factors required
F56=0.19;
F26=0.32;
F53=0.08;
F23=0.19;

print "Shape factor"
#Shape factor
F14=(A5*F56-A2*F26-A5*F53+A2*F23)/A1
print round(F14,3)

print "Thus,only about 10% of the light pasmath.sing through the window will impinge on the floor area A4"

Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 10
Shape factor
0.097
Thus,only about 10% of the light pasmath.sing through the window will impinge on the floor area A4


## Ex9.11: Page 590¶

In [40]:
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.11 "

#Absolute boiling temperature of liquid oxygen in R
T1 = 460-297.0;
#Absolute temperature of sphere in R
T2 = 460+30;
#Diameter of inner sphere in ft
D1 = 1;
#Area of inner sphere in ft2
A1 = (math.pi*D1)*D1;
#Diameter of outer sphere in ft
D2 = 1.5;
#Area of outer sphere in ft2
A2 = (math.pi*D2)*D2;
#Stefans constant
sigma = 0.1714;
#Emissivity of Aluminium
epsilon1 = 0.03;#Sphere1
epsilon2 = 0.03;#Sphere2

#Umath.sing Eq. 9.74
print "Rate of heat flow by radiation to the oxygen in Btu/h is"
#Rate of heat flow by radiation to the oxygen in Btu/h
q = ((A1*sigma)*((T1/100.0)**4-(T2/100.0)**4))/(1/epsilon1+(A1/A2)*((1-epsilon2)/epsilon2))
print round(q,1)

Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.11
Rate of heat flow by radiation to the oxygen in Btu/h is
-6.4


## Ex9.12: Page 594¶

In [28]:

print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.12 "

# As the example involves no calculations and the code for matlab is already given in the textbook thus following the guidelines this exapmle is to be skipped

Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.12


## Ex9.13: Page 597¶

In [29]:

print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.13 "

# As the example involves no calculations and the code for matlab is already given in the textbook thus following the guidelines this exapmle is to be skipped

Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.13


## Ex9.14: Page 598¶

In [16]:

print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.14 "

#Absolute temperature of first plate in degree R
Ta = 2040+460.0;
#Absolute temperature of second plate in degree R
Tb = 540+460.0;
#Stefans constant
sigma = 0.1718;

#For first radiation band, heat transfer is calculated
#Emissivity of A
epsilonA = 0.1;
#Emissivity of B
epsilonB = 0.9;
#Shape factor
Fab = 1/(1/epsilonA+1/epsilonB-1);
#The percentage of the total radiation within a given band is obtained from Table 9.1.
#Coefficients of T**4
A = 0.375;
#Coefficients of T**4
B = 0.004;

#Rate of heat transfer in first band in Btu/h ft2
q1 = (Fab*sigma)*(A*((Ta/100.0)**4)-B*((Tb/100.0)**4));

#Similarly for other two bands, heat transfer in Btu/h ft2
q2 = 23000;
#heat transfer in Btu/h ft2
q3 = 1240;

print "Total rate of radiation heat transfer in Btu/h ft2"
#heat transfer in Btu/h ft2
q = q1+q2+q3
print round(q)
# The answer is slightly different in textbook due to approximation

Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.14
Total rate of radiation heat transfer in Btu/h ft2
26728.0


## Ex9.15: Page 608¶

In [13]:
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.15 "

#Temperature in degree K
T = 800;
#Diameter of sphere in m
D = 0.4;
#Partial pressure of nitrogen in atm
PN2 = 1;
#Partial pressure of H2O in atm
PH2O = 0.4;
#Partial pressure of CO2 in atm
PCO2 = 0.6;

#The mean beam length for a spherical mass of gas is obtained from Table 9.7
#Beam length in m
L = (2/3)*D;

#The emissivities are given in Figs. 9.46 and 9.47
#Emissivity of H2O
epsilonH2O = 0.15;
#Emissivity of CO2
epsilonCO2 = 0.125;

#N2 does not radiate appreciably at 800 K, but math.since the total gas pressure
#is 2 atm, we must correct the 1-atm values for epsilon.
#From Figs. 9.48 and 9.49 the pressure correction factors are
#Pressure correction factor for H2O
CH2O = 1.62;
#Pressure correction factor for CO2
CCO2 = 1.12;

#From fig. 9.50
#Chnage in emissivity
deltaEpsilon = 0.014;

#Finally, the emissivity of the mixture can be obtained from Eq. (9.114):
print "Emissivity of the mixture is"
#Emissivity of the mixture
epsilonMix = CH2O*epsilonH2O+CCO2*epsilonCO2-deltaEpsilon
print round(epsilonMix,3)

Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.15
Emissivity of the mixture is
0.369


## Ex9.16: Page 609¶

In [11]:
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.16 "

#Total pressure in atm
Pt = 2;
#Temperature in degree K
TH2O = 500.0;
#Mean beam length in m
L = 0.75;
#Partial pressure of water vapor in atm
PH2O = 0.4;
#Source temperature in degree K
Ts = 1000.0;

#Since nitrogen is transparent, the absorption in the mixture is due to the water vapor alone.

#Parameters required
#A Parameter in atm-m
A = PH2O*L;
#B Parameter in atm
B = (Pt+PH2O)/2.0;

#From Figs. 9.46 and 9.48 we find
#For water, C factor in SI units
CH2O = 1.4;
#Emissivity of water
epsilonH2O = 0.29;

#From Eq. (9.115) the absorptivity of H2O is
print "Absorptivity of H2O is"
alphaH2O = (CH2O*epsilonH2O)*((TH2O/Ts)**0.45)
print round(alphaH2O,2)

Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.16
Absorptivity of H2O is
0.3


## Ex9.17: Page 609¶

In [6]:
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.17 "

#Temperature of flue gas in degree F
Tgas = 2000.0;
#Inner-wall surface temperature in degree F
Tsurface = 1850.0;
#Partial pressure of water in atm
p = 0.05;
#Convection heat transfer coefficient in Btu/h ft2 F
h = 1.0;
#Length of square duct in ft
L = 2.0;
#Volume in ft3
V = L*L;
#Surface area in ft2
A = 4*L;

#The rate of heat flow from the gas to the wall by convection per unit
#length in Btu/h ft is
qc = (h*A)*(Tgas-Tsurface);

#Effective beam length in m
L = ((0.3058*3.4)*V)/A;

#Product of partial pressure and L
k = p*L;

#From Fig. 9.46, for pL=0.026 and T=2000F, we find

#Emissivity
epsilon = 0.035;
#Absorptivity
alpha = 0.039;
#stefans constant
sigma = 0.171;

#Assuming that the brick surface is black, the net rate of heat flow from the gas to the wall by radiation is, according to Eq. (9.117)
qr = (sigma*A)*(epsilon*(((Tgas+460)/100.0)**4)-alpha*(((Tsurface+460.0)/100)**4));#Btu/h

print "Total heat flow from the gas to the duct in Btu/h"
#Total heat flow from the gas to the duct in Btu/h
q = qc+qr
print round(q,2)
# The answer is slightly different in textbook due to approximation

Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.17
Total heat flow from the gas to the duct in Btu/h
3543.12


## Ex9.18: Page 611¶

In [4]:
print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.18 "

#Emissivity
epsilon = 0.8;
#Stefan's constant
sigma = 0.1714;
#Temperature of walls in degree F
Twall = 440;
#Temperature indicated ny thermocouple in degree F
Tt = 940;
#Heat transfer coefficient in Btu/h ft2 F
h = 25;

#The temperature of the thermocouple is below the gas temperature because the couple loses heat by radiation to the wall.

#Under steady-state conditions the rate of heat flow by radiation from the thermocouple junction to the wall equals the rate of heat flow by convection from the gas to the couple.

#Umath.sing this heat balance, q/A in Btu/h ft2
q = (epsilon*sigma)*(((Tt+460)/100)**4-((Twall+460)/100)**4);

print "True gas temperature in degree F"
#True gas temperature in degree F
Tg = Tt+q/h

print round(Tg)
# The answer is slightly different in textbook due to approximation

Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.18
True gas temperature in degree F
1115.0


## Ex9.19: Page 612¶

In [2]:

print "Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.19 "

#Emissivity of thermocouple
epsilonT = 0.8;
#Emissivity of shield
epsilonS = 0.3;
#Stefan''s constant
sigma = 0.1714;
#Temperature of walls in degree F
Tw = 440;
#Temperature indicated ny thermocouple in degree F
Tt = 940.0;
#Heat transfer coefficient of thermocouple in Btu/h ft2 F
hrt = 25.0;
#Heat transfer coefficient of shield in Btu/h ft2 F
hrs = 20.0;

#Area for thermocouple be unity ft2
At = 1.0;
#Corresponding area of shield in ft2
As = 4.0;#Inside dia=4*dia of thermocouple

#From Eq. (9.76)
#View factors Fts and Fsw
Fts = 1/((1-epsilonT)/(At*epsilonT)+1/At+(1-epsilonS)/(As*epsilonS));
Fsw = As*epsilonS;

#Assuming a shield temperature of 900°F, we have, according to Eq. (9.118)
#Temperature in degree F
Ts = 923;

#Coeffcients for heat balance are as following
#A parameter Btu/h-F
A = 9.85;#A=hrt*At
#B parameter Btu/h-F
B = 13.7;#B=hrs*As

#Umath.sing heat balance
print "Correct temperature of gas in degree F"
#Correct temperature of gas in degree F
Tg = Ts+(B*(Ts-Tw)-A*(Tt-Ts))/((hrs*2)*As)
print round(Tg,2)
# The answer is slightly different in textbook due to approximation

Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 9 Example # 9.19
Correct temperature of gas in degree F
963.31