Chapter 18: HEAT TRANSFER BY RADIATION

Example 18.01, page: 472

In [1]:
from __future__ import division
import math

# Initialization  of  Variable
E = 525 #W/m2
a = 0.85
p = 0.15
Glambda= 600 #W/m2.um

#calculations:
lambda1 = 0.5 #um
lambda2 = 1.0 #um
lambda3 = 2.5 #um
#total irradiation
G = 0.5*Glambda*(lambda1 - 0) + Glambda*(lambda2 - lambda1) + 0.5*Glambda*(lambda3 - lambda2) + 0
#Absorbed rad
Gabs = a*G
#total radiosity
J =E + p*G
#net radiative flux leaving the surface
qradnet2 = E - Gabs

#Results
print  "a) The total irradiation and the absorbed total irradiation are", round(G,0),"W/m2 and", round(Gabs,0),"W/m2"
print  "b)total radiosity is",round(J,0),"W/m2"
print  "c)net radiative flux leaving the surface is", round(qradnet2,0),"W/m2"

a) The total irradiation and the absorbed total irradiation are 900.0 W/m2 and 765.0 W/m2
b)total radiosity is 660.0 W/m2
c)net radiative flux leaving the surface is -240.0 W/m2

Example 18.02, page: 477

In [2]:
from __future__ import division
import math

# Initialization  of  Variable
T = 2000 #K
x = 0.1
sigma = 5.67E-8 #W/m2.K4
lambda1T = 2200 #um.K
lambda2T = 9382 #um.K
C1 = 3.742E8 #W.um4/m2
lambda_max = 1.45 #um
C2 = 1.439E4 #um.K

#calculations:
#blackbody spectral emissive power
E = sigma*T**4
#wavelength 1 and 2
lambda1 = lambda1T/T
lambda2 = lambda2T/T
#spectral emissive power corresponding to the peak of the blackbody curve
Elambda_max_T = C1/lambda_max**5/(math.e**(C2/lambda_max/T) - 1)
#Irradiation
G = E

#Results
print  "a)Emissive power of a small aperture on the enclosure is", round(E,0),"W/m2"
print  "b)Wavelengths below which and above which 10% of the radiation is concentrated is", round(lambda1,1),"um and",round(lambda2,1),"um"
print  "c)Maximum spectral emissive power and wavelength at which it occurs is", round(Elambda_max_T,0),"W/m2.um"
print  "d)Irradiation on a small object inside the enclosure is", round(G,0),"W/m2"

a)Emissive power of a small aperture on the enclosure is 907200.0 W/m2
b)Wavelengths below which and above which 10% of the radiation is concentrated is 1.1 um and 4.7 um
c)Maximum spectral emissive power and wavelength at which it occurs is 411448.0 W/m2.um
d)Irradiation on a small object inside the enclosure is 907200.0 W/m2

Example 18.03, page: 478

In [3]:
from __future__ import division
import math

# Initialization  of  Variable
l1 = 2 #um
l2 = 4 #um
sigma = 5.67E-8 #W/m2.K4
T = 1500 #K

#calculations:
#from Table 18.2
F02 = 0.273
F04 = 0.738
#Stefan-Boltzmann law
El1l2 = (F04 - F02)*sigma*T**4

#Results
print "Rate of emission per unit area is", round(El1l2,0),"W/m2"

Rate of emission per unit area is 133475.0 W/m2

Example 18.04, page: 480

In [4]:
from __future__ import division
import math

# Initialization  of  Variable
T = 1600 #k
sigma = 5.67E-8 #W/m2.K4
e1 = 0.4
e2 = 0.8

#calculations:
#from Table 18.2
F02 = 0.318
F05 = 0.856
#total emissivity for this spectrally selective material at 1600 K
e = e1*F02 + e2*(F05 - F02)
#total emissive power
E = e*sigma*T**4

#Results
print  "a)Total emissivity is", round(e,3)
print  "b)Total Emissive power is", round(E/1000,0),"kW/m2"

a)Total emissivity is 0.558
b)Total Emissive power is 207.0 kW/m2

Example 18.05, page: 485

In [5]:
from __future__ import division
import math

# Initialization  of  Variable
al1 = 0.8
al2 = 0.1
Tsi = 300 #K
Tf = 1200 #K
sigma = 5.67E-8 #W/m2.K4

#calculations:
#from Table 18.2
F0l1 = 0.738
F0l2 = 1
#total absorptivity
ai = al1*F0l1 + al2*(F0l2 - F0l1)
#from Table 18.2
F0l1 = 0.014
F0l2 = 1
#total emissivity
ei = al1*F0l1 + al2*(F0l2 - F0l1)
#net radiative heat flux leaving the sphere
qradnet2 = ei*sigma*Tsi**4 - ai*sigma*Tf**4
#at final condition
af = ai
ef = af

#Results
print  "a)For the Initial Condition, Total absorptivity is", round(ai,2),", emissivity of the coating is", round(ei,2)," and net radiative heat flux is", round(qradnet2/1000,0),"kW/m2"
print  "b)For the Final Condition, Total absorptivity is", round(af,2)," and emissivity of the coating is", round(ef,2)

a)For the Initial Condition, Total absorptivity is 0.62 , emissivity of the coating is 0.11  and net radiative heat flux is -72.0 kW/m2
b)For the Final Condition, Total absorptivity is 0.62  and emissivity of the coating is 0.62

Example 18.06, page: 487

In [6]:
from __future__ import division
import math

# Initialization  of  Variable
Tp = 300 #K
Gs = 1353 #W/m2
Tds = 0 #K
l1 = 0.3 #um
l2 = 1.5 #um
Ts = 5800 #K
sigma = 5.67E-8 #W/m2.K4

#calculations:
#from Fig E18.6a
al1 = 1 - 0
al2 = 1 - 0.95
al3 = 1 - 0.05
F0l1 = 0.03345
F0l2 = 0.8808
#solar absorptivity
aS = al1*F0l1 + al2*(F0l2 - F0l1) + al3*(1 - F0l2)
pl3 = 0.05
#total emissivity
al3 = 1 - pl3
e = al3
#net radiative heat flux leaving the panel surface is
qradnet2 = e*sigma*Tp**4 - aS*Gs

#Results
print "net  radiative  heat  flux  leaving  is", round(qradnet2,0),"W/m2"
#answer wrong in book

net  radiative  heat  flux  leaving  is 181.0 W/m2

Example 18.07, page: 491

In [7]:
from __future__ import division
import math

# Initialization  of  Variable
La = 1 #m
Ra = 1 #m
Db = 1 #m
r3c = 0.5 #m
Lc = 1 #m
r1c = 0.5 #m
Dc = 2*r3c #m

#calculations:
#a)
A1a = 2*Ra*La
A2a = 2*math.pi*Ra*La/4
#F12a = F13a, from symmetry
F11a = 0
F12a = 0.5
#reciprocity relation between surfaces A1 and A2
F21a = A1a/A2a*F12a
#b)
A1b = math.pi*(Db/2)**2/4
A2b = math.pi*Db**2/2
F12b = 1.0
#the reciprocity relation between surfaces A1 and A2
F21b = A1b/A2b*F12b
#based upon a symmetry argument that A2 sees as much of itself as it does of A1 and A3 combined
F22b = 0.50
#Applying the summation rule to surface A2
F23b = 1 - F21b - F22b
#c)
A1c = (math.pi*Dc**2)/4
A2c = math.pi*Dc*Lc
#By inspection,
F11c = 0
#from the chart for the coaxial, parallel disks,
F13c = 0.17
#Substituting numerical values into the summation rule
F12c = 1 - F11c - F13c
#From the reciprocity relation between surfaces A1 and A2,
F21c = A1c/A2c*F12c

#Results
print  "a) for long open channel, F12 = ", round(F12a,1)," and F21 = ", round(F21a,3)
print  "b) for Hemispherical-disk arrangement, F12 = ", round(F12b,0),",F21 = ", round(F21b,3),", F22 = ", round(F22b,2)," and F23 = ", round(F23b,3)
print  "c) for End and side of a circular tube of equal length and diameter, F12 = ", round(F12c,2)," and F21 = ", round(F21c,2)

a) for long open channel, F12 =  0.5  and F21 =  0.637
b) for Hemispherical-disk arrangement, F12 =  1.0 ,F21 =  0.125 , F22 =  0.5  and F23 =  0.375
c) for End and side of a circular tube of equal length and diameter, F12 =  0.83  and F21 =  0.21

Example 18.08, page: 494

In [8]:
from __future__ import division
import math

# Initialization  of  Variable
L = 0.15 #m
D = 0.075 #m
T1 = 1350 #degC
T2 = 1650 #degC
Tsur = 300 #K
T3 = Tsur
sigma = 5.67E-8 #W/m2.K4

#calculations:
#from fig 18.22b
F23 = 0.06
#From the summation rule for surface A2,
F22 = 0
F21 = 1 - F22 - F23
#using the reciprocity relation
A2 = math.pi*D**2/4
A1 = math.pi*D*L
F12 = A2/A1*F21
#From symmetry considerations,
F13 = F12
#Power, Pe
Pe = A1*F13*sigma*((T1 + 273)**4 - T3**4) + A2*F23*sigma*((T2 + 273)**4 - T3**4)

#Results
print  "Electrical power required to maintain prescribed temperatures is", round(Pe,0),"W"
#answer wrong in book

Electrical power required to maintain prescribed temperatures is 1837.0 W

Example 18.09, page: 499

In [9]:
from __future__ import division
import math

# Initialization  of  Variable
sigma = 5.67E-8 #W/m2.K4
T2 = 300 #K
D2 = 0.05 #m
e2 = 0.05
T1 = 77 #K
D1 = 0.02 #m
e1 = 0.02
D3 = 0.035 #m
e3 = 0.02
L = 1 #m

#calculations:
F13 = 1
F32 = 1
#heat rate from the cryogenic fluid per unit tube length is
qwo1 = sigma*math.pi*D1*(T1**4 - T2**4)/(1/e1 + (1-e2)/e2*D1/D2)
#b) Total Resistance
Rtot = (1 - e1)/e1/math.pi/D1/L + 1/math.pi/D1/L/F13 + 2*(1-e3)/e3/math.pi/D3/L + 1/math.pi/D3/L/F32 + (1-e2)/e2/math.pi/D2/L
#the desired heat rate
qw1 = sigma*(T1**4 - T2**4)/Rtot/L
#percentage change in the heat transfer to the cryogenic fluid is
dqw1 = (qw1 - qwo1)/qwo1*100

#Results
print  "a)Heat transfer by the cryogenic fluid passing through the inner tube without the radiation shield is", round(qwo1,2),"W/m"
print  "b)Percentage change in heat transfer with the radiation shield inserted midway between inner and outer tubes is",round(dqw1,0),"%"

a)Heat transfer by the cryogenic fluid passing through the inner tube without the radiation shield is -0.5 W/m
b)Percentage change in heat transfer with the radiation shield inserted midway between inner and outer tubes is -50.0 %

Example 18.10, page: 501

In [10]:
from __future__ import division
import math

# Initialization  of  Variable
e3 = 0.8
W = 1 #m
T1 = 1200 #K
e1 = 0.8
T2 = 500 #K
e2 = 0.4
sigma = 5.67E-8 #W/m2.K4
L = 1 #m

#calculations:
#From symmetry considerations
F12 = 0.5
F1R = 0.5
F2R = 0.5
A1 = W*L
A2 = W*L
#The rate at which energy must be supplied to the heated surface
q1_dash = sigma*(T1**4 - T2**4)/((1-e1)/e1/A1 + 1/(A1*F12 + (1/A1/F1R + 1/A2/F2R)**-1) + (1-e2)/e2/A2)/L
#the surface energy balance
J1 = sigma*T1**4 - (1-e1)/e1/W*q1_dash
J2 = sigma*T2**4 - (1-e2)/e2/W*-1*q1_dash
#From the energy balance for the reradiating surface
JR = (J1 + J2)/2
#Temp
TR = (JR/sigma)**0.25

#Results
print  "a)Rate at which energy must be supplied per unit length of duct is", round(q1_dash/1000,0),"kW/m"
print  "b)Temperature of the insulated surface is", round(TR,0),"K"

a)Rate at which energy must be supplied per unit length of duct is 37.0 kW/m
b)Temperature of the insulated surface is 1102.0 K