Chapter 7 - Radiation Heat Transfer

Example No : 7.1 - Page : 212

In [20]:
from __future__ import division
from math import exp
# given data
lamda=2*10**-6   # in m
C1=0.374*10**-15#
T=2000+273   # in K'
C2=1.4388*10**-2#

#(a)
# Formula Eb_lamda= (C1*lamda**-5)/(exp(C2/(lamda*T))-1)
Eb_lamda= (C1*lamda**-5)/(exp(C2/(lamda*T))-1)#
print "Monochromatic emissive power at 2 micro wavelength = %0.2e W/m**2" %Eb_lamda

#(b)
# Formula lamda_max * T =2898 # in micro m K
lamda_max= 2898/T   # in micro m
print "Wave-length at which the emission is maximum = %0.3f micro m" %lamda_max

#(c)
Elamdab_max=1.285*10**-5*T**5   # in W/m**2-m
print "Maximum emissive power = %0.4e W/m**2-m" %Elamdab_max

#(d)
sigma=5.67*10**-8#
E=sigma*T**4#
print "Total emissive power = %0.3e W/m**2" %E

#Note: Answer of  part (a) in the book is wrong

Monochromatic emissive power at 2 micro wavelength = 5.15e+11 W/m**2
Wave-length at which the emission is maximum = 1.275 micro m
Maximum emissive power = 7.7965e+11 W/m**2-m
Total emissive power = 1.513e+06 W/m**2

Example No : 7.2 - Page : 213

In [21]:
from numpy import pi
# given data
lamda=2*10**-6   # in m
C1=0.374*10**-15#
T=2000+273   # in K'
C2=1.4388*10**-2#

epsilon=0.3#
sigma=5.67*10**-8#
T1=300   # in K
T2=200   # in K
del_T=T1-T2#
h=12   # in W/m**2 degree C
d=4*10**-2   # diameter in m
l=1   # in m
A=pi*d*l#
# Heat transfer rate by radiation,
q_r= epsilon*sigma*A*(T1**4-T2**4)   # in W
# Heat transfer rate by convection,
q_c=h*A*del_T   # in W
# Total heat transfer,
q=q_r+q_c#
# Formula q=U*A*del_T
U=q/(A*del_T)   # Overall heat tranfer coefficient
print "Overall heat tranfer coefficient = %0.3f W/m**2 degree C" %U
#Note: Value of q_c is wrong in the book, so the answer in the book is wrong

Overall heat tranfer coefficient = 13.106 W/m**2 degree C

Example No : 7.3 - Page : 214

In [22]:
import math
# given data
epsilon=0.5#
T1=1200   # in K
T2=300   # in K
#(a) Heat transfer rate between the two plates is 
# Formula Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))
epsilon1=epsilon#
epsilon2=epsilon#
A1byA2=1#
Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1byA2)#
# Formula q12= sigma*A*Fg12*(T1**4-T2**4)  
sigma=5.67*10**-8#
q12byA=sigma*Fg12*(T1**4-T2**4)   # in W/m**2
print "Heat transfer rate between the two plates = %0.4e W/m**2" %q12byA

#(b)
epsilon3=.05#
Fg13=1/(1/epsilon1+(1/epsilon3-1)*A1byA2)#
Fg32=1/(1/epsilon3+(1/epsilon2-1)*A1byA2)#
# q13=q32
# sigma*A*Fg13*(T1**4-T3**4) = sigma*A*Fg32*(T3**4-T2**4)  
T3= ((T1**4+T2**4)/2)**(1/4)#
T3=math.floor(T3)#
q13byA=sigma*Fg13*(T1**4-T3**4)   # in W/m**2
print "Heat transfer rate if a radiation shield with an emissivity of 0.05 on both sides is placed "
print "between the two plates = %0.3e W/m**2 " %q13byA

Heat transfer rate between the two plates = 3.9038e+04 W/m**2
Heat transfer rate if a radiation shield with an emissivity of 0.05 on both sides is placed 
between the two plates = 2.789e+03 W/m**2

Example No : 7.4 - Page : 215

In [23]:
# given data
T1=800+273   # in K
A= 5*6   # in square meter
epsilon=0.45#
sigma=5.67*10**-8#
q=epsilon*sigma*A*T1**4   #in watt
print "Energy emitted by a grey surface = %0.5e watt" %q

Energy emitted by a grey surface = 1.01465e+06 watt

Example No : 7.5 - Page : 215

In [24]:
from __future__ import division
# given data
A=5   # in m**2
intensity=660   # in W/m**2
alpha=6/11#
rho=6/22#
toh=6/33#
energy_absorbed= intensity*alpha*A   # in watt
print "Energy absorbed = %0.f watt" %energy_absorbed
energy_transmitted=intensity*rho*A   #in watt
print "Energy transmitted = %0.f watt" %energy_transmitted
energy_emitted= intensity*toh*A   # in watt
print "Energy emitted = %0.f watt" %energy_emitted

Energy absorbed = 1800 watt
Energy transmitted = 900 watt
Energy emitted = 600 watt

Example No : 7.5 - Page : 216

In [25]:
# given data
T1=200+273   # in K
T2=100+273   # in K
A= 1*2   # in square meter
sigma=5.67*10**-8#
x_D= 1/4#
y_D= 1/2#
Fg12= 0.033#
q12= Fg12*sigma*A*(T1**4-T2**4)   # in watt
print "The net heat exchange between two surfaces = %0.2f watt" %q12

The net heat exchange between two surfaces = 114.88 watt

Example No : 7.6 - Page : 216

In [26]:
from numpy import pi
# given data
d=20*10**-2   #diameter of pipe in m
l=1   # length of pipe in m
s=30*10**-2   # side of duct in m
A1=pi*d*l   # area of pipe in m**2
A2=4*s*s   # area of duct in m**2
epsilon1=0.8#
epsilon2=0.9#
T1=200+273   # in K
T2=20+273   # in K
# Formula Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))
Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))#
# Heat transfer rate between pipe and duct
sigma=5.67*10**-8#
q12=sigma*Fg12*A1*(T1**4-T2**4)   # in W
print "Heat transfer rate between pipe and duct = %0.3f W" %q12
#Note: Answer in the book is wrong

Heat transfer rate between pipe and duct = 1053.148 W

Example No : 7.10 - Page : 220

In [27]:
from numpy import pi
# given data
D=150*10**-3   # in m
H=400*10**-3   # in m
T1=500   # in K
epsilon=0.7#
# Formula F11=(4*H)/(4*H+D)
F11=(4*H)/(4*H+D)#
sigma=5.67*10**-8#
A1=pi*D*H#
q=sigma*A1*epsilon*T1**4*((1-F11)/(1-F11*(1-epsilon)))#
print "Heat Heat loss for cavity = %0.3f W" %q
#Note: Answer in the book is not accutate

Heat Heat loss for cavity = 55.227 W

Example No : 7.11 - Page : 221

In [28]:
from numpy import pi
# given data
epsilon1=.04#
epsilon2=epsilon1#
T1=-153+273   # in K
T2=27+273   # in K
h_fg=209   # in kJ/kg
h_fg=h_fg*10**3   # in J/kg
d1=20*10**-2   # in m
d2=30*10**-2   # in m
A1=d1**2   # in square meter
A2=d2**2   # in square meter
A=4*pi*(d2-d1)**2#
Fg12=1/((1/epsilon1+(1/epsilon2-1)*A1/A2))#
sigma=5.67*10**-8#    
q12=sigma*A*Fg12*(T1**4-T2**4)   # in W
print "Net radiant heat transfer rate = %0.3f watt" %q12
print "Negative sign indicates that heat flows into the sphere"
q12=-q12 #
m=q12*60/h_fg #
print "Rate of evaporation per minutes = %0.3e kg/min" %m

Net radiant heat transfer rate = -1.577 watt
Negative sign indicates that heat flows into the sphere
Rate of evaporation per minutes = 4.526e-04 kg/min

Example No : 7.12 - Page : 221

In [29]:
# given data
T1=500   # in K
T2=300   # in K
sigma=5.67*10**-8#
A=2   # surface area of each plate in m**2
#(a) If the plates are perfectly black
F12=1#
q12=sigma*A*F12*(T1**4-T2**4)#
print "Radiation heat transfer between two black parellel plates = %0.4e watt" %q12

#(b) If the plates are gray surface
#in this case
F12=1#
#A1=A2, so
A1byA2=1
epsilon1=.4#
epsilon2=epsilon1#
#Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1byA2)#
Fg12=1/((1-epsilon1)/epsilon1 + 1/F12 + ((1-epsilon2)/epsilon2)*A1byA2)#
q12=sigma*A*Fg12*(T1**4-T2**4)   # in W
print "Heat transfer rate = %0.3e watt" %q12

Radiation heat transfer between two black parellel plates = 6.1690e+03 watt
Heat transfer rate = 1.542e+03 watt

Example No : 7.13 Page No - 222

In [30]:
from numpy import pi
# given data
T1=800   # in K
T3=200   # in K
sigma=5.67*10**-8#
d1=20*10**-2   # in m
d2=30*10**-2   # in m
d3=40*10**-2   # in m
A1=4*pi*(d1/2)**2   # in m**2
A2=4*pi*(d2/2)**2   # in m**2
A3=4*pi*(d3/2)**2   # in m**2
epsilon1=0.2#
epsilon2=epsilon1
epsilon3=epsilon1
Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1/A2)#
Fg23=1/(1/epsilon2+(1/epsilon3-1)*A2/A3)#
# Under steady state condition 
# q12 = q23
# A1*Fg12*sigma*(T1**4-T2**4) = A2*Fg23*sigma*(T2**4-T3**4)
T2 = ((A2*Fg23*T3**4/(A1*Fg12)+T1**4)/(A2*Fg23/(A1*Fg12) + 1))**(1/4)
print "Steady state temperature of the intermediate sphere = %0.f K" %T2

Steady state temperature of the intermediate sphere = 604 K

Example No : 7.14 Page No : 223

In [31]:
# given data
T1=400   # in K
T2=500   # in K
T3=1200   # in K
alpha1=0.70#
alpha2=0.6#
alpha3=0.4#
# First part
sigma=5.67*10**-8#
qa=alpha3*sigma*T3**4#
print "The rate of energy absorption = %0.2e W/m**2" %qa
# Second part
qa=alpha1*sigma*T1**4#
print "The rate of emission of radiation energy = %0.2f W/m**2" %qa
# Note : Answer of the first part in the book is wrong

The rate of energy absorption = 4.70e+04 W/m**2
The rate of emission of radiation energy = 1016.06 W/m**2

Example No : 7.15 Page No : 223

In [32]:
from numpy import pi
# given data
d1=100   # in mm
d1=d1*10**-3   # in m
d2=100+10*2   # in mm
d2=d2*10**-3   # in m
l=1   # in m
A1byA2=d1**2/d2**2#
A1=pi*d1*l   # in m**2
sigma=5.67*10**-8#
T1=120+273   # in K
T2=35+273   # in K
epsilon1=.8#
epsilon2=.1#
Fg12=1/(1/epsilon1+(1/epsilon2-1)*A1byA2)#
# Radiant heat transfer from the tube
q=A1*Fg12*sigma*(T1**4-T2**4)
print " Radiant heat transfer from the tube = %0.3f W/m" %q
#Note: Answer in the book is wrong

 Radiant heat transfer from the tube = 35.282 W/m