# Chapter 11: Introduction to Radiation Heat Transfer¶

### Example 11.1 Page NO.554¶

In [13]:
import math
dA2=(1*1)/144.0
Beta1=40*math.pi/180.0
r=4
dw2_1=dA2*math.cos(Beta1)/r**2
dA3=dA2
Beta2=0
dw3_1=dA3*cos(Beta2)/r**2

theta2=math.pi*50/180.0
theta3=math.pi*60/180.0
I_theta2=2000*(1-0.4*(sin(theta2))**2)
I_theta3=2000*(1-0.4*(sin(theta3))**2)

dA1=1/144.0
dq1_2=I_theta2*dA1*math.cos(theta2)*dw2_1          #In book calculation mistake
dq1_3=I_theta3*dA1*math.cos(theta2)*dw3_1

print"(a)The solid angle subtended by area dA2 with respect to dA1 is ",round(dw2_1,4),"sr"
print" The solid angle subtended by area dA3 with respect to dA1 is  ",round(dw3_1,4),"sr"
print"(b) The intensity of radiation emitted from dA1 in the direction of dA2 is ",round(I_theta2,0),"BTU/(hr.sq.ft.sr)"
print" The intensity of radiation emitted from dA1 in the direction of dA3 is",round(I_theta3,0),"BTU/(hr.sq.ft.sr)"
print"(c)The rate at which radiation emitted by dA1 is intercepted by dA2 is ",round(dq1_2,4),"BTU/hr"
print" The rate at which radiation emitted by dA1 is intercepted by dA3 is ",round(dq1_3,4),"BTU/hr"

(a)The solid angle subtended by area dA2 with respect to dA1 is  0.0003 sr
The solid angle subtended by area dA3 with respect to dA1 is   0.0004 sr
(b) The intensity of radiation emitted from dA1 in the direction of dA2 is  1531.0 BTU/(hr.sq.ft.sr)
The intensity of radiation emitted from dA1 in the direction of dA3 is 1400.0 BTU/(hr.sq.ft.sr)
(c)The rate at which radiation emitted by dA1 is intercepted by dA2 is  0.0023 BTU/hr
The rate at which radiation emitted by dA1 is intercepted by dA3 is  0.0027 BTU/hr


### Example 11.2 Page NO.557¶

In [16]:
dA2=0.02*0.02
Beta=0
r=1

import math
dw2_1=dA2*math.cos(Beta)/r**2
dA1=dA2
theta=math.pi*30/180.0
I_theta=1000# The intensity of radiation leaving dA1 in any direction is 1 000 W/(m**2.sr
dq1_2=I_theta*dA1*cos(theta)*dw2_1
dQ1_2=dq1_2/dA2

print"(a)The solid angle subtended by area dA2 with respect to dA1 is ",round(dw2_1,4),"sr"
print"(b)The rate at which radiation emitted by dA1 is intercepted by dA2 is",round(dq1_2,5),"W"
print"(c)The irradiation associated with dA2 is ",round(dQ1_2,3),"W/sq.m"

(a)The solid angle subtended by area dA2 with respect to dA1 is  0.0004 sr
(b)The rate at which radiation emitted by dA1 is intercepted by dA2 is 0.00014 W
(c)The irradiation associated with dA2 is  0.346 W/sq.m


### Example 11.3 Page NO. 563¶

In [23]:
D=2.5/12.0          # diameter in ft
L=4.5/12.0          # length in ft

import math
A=(2*math.pi*D**2/4)+(math.pi*D*L)
A_hole=math.pi*(1/(8.0*12.0))**2/4.0
f=A_hole/A          # fraction of area removed
emissivity=0.039
emissivity_hole=emissivity/(emissivity+(1-emissivity)*f)

sigma=0.1714e-8      # stefan Boltzmann constant in BTU/(hr~ft**2 degree R)
T=150+460            # temperature in degree R
qe=emissivity_hole*sigma*T**4
Qe=A_hole*qe

print"(a)The emissivity of the hole is %.4f",round(emissivity_hole,4)
print"(b)The heat lost by the hole is ",round(Qe,4),"BTU/hr"

(a)The emissivity of the hole is %.4f 0.9933
(b)The heat lost by the hole is  0.0201 BTU/hr


### Example 11.4 Page NO.568¶

In [27]:
T=2800              #Temprature
lambda1=4e-7        #Wavelength
lambda2=7e-7
hT=lambda1*T

lambdaT=lambda2*T
I1=0.0051           #Fraction of Total Radiation Emitted for lower Wavelength-Temperature Product from Table 11.1
I2=0.065            #Fraction of Total Radiation Emitted for upper Wavelength-Temperature Product from Table 11.1
dI=I2-I1

print"The percentage of total emitted energy that lies in the visible range is",round(dI*100,0),"percant"

The percentage of total emitted energy that lies in the visible range is 6.0 percant


### Example 11.5 Page NO. 570¶

In [29]:
lambda_max=0.5e-6   # maximum wavelength in m
T=2.898e-3/lambda_max
sigma=5.675e-8      # value of Stefan-Boltzmann constant in W/(m**2.K**4)
q=sigma*T**4

print"The Surface Temperature of the Sun is ",round(T,2),"K"
print"The heat flux emitted is ",round(q,0),"W/sq.m"

The Surface Temperature of the Sun is  5796.0 K
The heat flux emitted is  64044136.0 W/sq.m


### Example 11.6 Page NO.575¶

In [32]:
lambda1=300e-9         # lower limit of wavelength
lambda2=380e-9         # upper limit of wavelength
T=5800
lambda1_T=lambda1*T
lambda2_T=lambda2*T
I1=0.101               #Fraction of Total Radiation Emitted for lower Wavelength-Temperature Product from Table 11.1
I2=0.0334              #Fraction of Total Radiation Emitted for upper Wavelength-Temperature Product from Table 11.1
dI=abs(I2-I1)
t=dI*0.68               # transmissivity
q_in=t*q                # energy transmitted from the sun through the glass

Tb=311                 # temperature of black body source in K
lambda1_Tb=lambda1*Tb
lambda2_Tb=lambda2*Tb
dI_b=0                # Table 11.1 gives negligibly small values of the corresponding integrals.
t_b=dI_b*0.68         # transmissivity
q_out=t_b*q

print"(a)The energy transmitted from the sun through the glass is ",round(q_in,1),"W/sq.m"
print"(b)the rate at which radiant energy from the car interior is transmitted through the glass windshield is",q_out,"W/sq.m"

(a)The energy transmitted from the sun through the glass is  50.6 W/sq.m
(b)the rate at which radiant energy from the car interior is transmitted through the glass windshield is 0.0 W/sq.m

In [ ]: