# Chapter 7 : radiation heat transfer¶

## Example 7.3 Page No : 215¶

In :
import math
# Variables
Ts = 5780.              			#K, surface temp.

# Calculations and Results
#(a)
lamda1 = 0.4         			#micrometer, starting visible spectrum range
lamda2 = 0.7         			#micrometer,ending visible spectrum range
E1 = lamda1*Ts       			#micrometer K,
E2 = lamda2*Ts       			#micrometer K,
#from table 7.2
#fraction of radiation lying between 0 and lamda1
F1 = 0.1229
#fraction of radiation lying between 0 and lamda2
F2 = 0.4889
#the fraction of radiation falls betweem lamda1 & lamda 2
F3 = F2-F1
print "the fraction of radiation falls in visible range is %.3f "%(F3)
#(b)
F4 = F1
print "the fraction of radiation on the left of visible range is %.4f "%(F4)
#(c)
F5 = 1-F2
print "the fraction in right of visible range is %.4f "%(F5)
#(d)
#from wein's print lacement law
lmax = 2898/Ts
print "The maximum wavelength is %.4f micrometer is"%(lmax)
c = 2.998*10**8      			#m/s, speed of light
mu = c/lmax
print "The frequency is %1.2e s**-1"%(mu)
#(e)
#from eq. 7.4
h = 6.6256*10**-34         			#Js planck's consmath.tant
k = 1.3805*10**-23         			#J/K, boltzman consmath.tant
Eblmax = (2*math.pi*h*c**2*(lmax*10**-6)**-5)/((math.exp(h*c/(lmax*10**-6*k*Ts)))-1)
print "the maximum spectral emissive power is %1.3e W/m**2"%(Eblmax)
#(f)
s = 5.668*10**-8           			#stephen math.cosmath.tant
Eb = s*Ts**4
print "the hemispherical total emissive power is %1.3e W/m**2"%(Eb)

# note : rounding off error.

the fraction of radiation falls in visible range is 0.366
the fraction of radiation on the left of visible range is 0.1229
the fraction in right of visible range is 0.5111
The maximum wavelength is 0.5014 micrometer is
The frequency is 5.98e+08 s**-1
the maximum spectral emissive power is 8.298e+13 W/m**2
the hemispherical total emissive power is 6.326e+07 W/m**2


## Example 7.4 Page No : 216¶

In :
#Variables
Eb = 4000.           			#W/m sq, Total emmisive power
s = 5.669*10**-8  			#Stephen boltzman consmath.tant

#Calculation
T = (Eb/s)**0.25  			#k, surface temp. of black body
ym = 2898./T      			#micro meter,
#By weins law : Max. wavelength of emmision is inversaly proportional
#to temprature. and consmath.tant is 2898 micrometer.

#Result
print "Surface temp. is %.0f C"%(T)
print "wavength is  %.2f micrometer "%(ym)
print " from fig 7.1 it  falls in the infrared region of spectrum."

Surface temp. is 515 C
wavength is  5.62 micrometer
from fig 7.1 it  falls in the infrared region of spectrum.


## Example 7.5 Page No : 219¶

In :
# Variables
T = 1500.       			#K, surface temprature
#from fig 7.7
e1 = 0.2     			#emissivity ,when wavelength(l1) is 0<l1<2 micrometer
e2 = 0.6     			#emissivity ,when wavelength(l2) is 2<l2<6 micrometer
e3 = 0.1     			#emissivity ,when wavelength(l3) is 6<l3<10 micrometer
e4 = 0       			#emissivity ,when wavelength(l4) is l4>10 micrometer
#from table 7.2
F1 = 0.2733        			#fraction of energy in  wavelength (l1)
F2 = 0.89-F1       			#fraction of energy in  wavelength (l2)
F3 = 0.9689-0.89   			#fraction of energy in  wavelength (l3)

#Calculation and Result
s = 5.669*10**-8    			#stephen's consmath.tant
Eb = s*T**4         			#emissive power
E = (e1*F1+e2*F2+e3*F3)*Eb
print "total hemispherical) emissive power is %1.3e W/m**2"%(E)
#(b)
e = E/(s*T**4)
print "total hemispherical) emissivity of the surface is %.4f"%(e)

total hemispherical) emissive power is 1.241e+05 W/m**2
total hemispherical) emissivity of the surface is 0.4326


## Example 7.6 Page No : 226¶

In :
from scipy.integrate import quad

# Variables
ri = 5.         			#cm ,inside radius  of ring
w = 3.          			#cm, width
ro = ri+w      			#cm, outside radius
L = 20.         			#cm, surface dismath.tance

# Calculations
def f4(r):
return 20.**2*r/(20.**2+r**2)**2

#view factor along surface dA1-A2"

def f5(r):
return 20**2*r/(20**2+r**2)**2

# Results
print "fraction of radiation passes through hole  %.4f "%(F1)
print "fraction of radiation intercepted by the ring %.4f "%(F2)

fraction of radiation passes through hole  0.0588
fraction of radiation intercepted by the ring 0.0791


## Example 7.8 Page No : 232¶

In :
import math
#Variables
F11 = 0        			#view factor
d = 1.           			#let it be
print "view factor F11  =  %.0f" %(F11)

#Calculation and Result
F12 = 1-F11   			#view factor
print  "view factor F22  = %.0f"%(F12)

A1 = ((math.pi)*d**2)/4   			#sq m, area
A2 = ((math.pi)*d**2)/2   			#sq m, area
F21 = A1/A2            			#from eq . 7.26
print "view factor F21  = %.1f"%( F21)
F22 = 1-F21
#Results
print "view factor = %.1f"%(F22)

view factor F11  =  0
view factor F22  = 1
view factor F21  = 0.5
view factor = 0.5


## Example 7.9 Page No : 233¶

In :
import math
#Variable
s = 3.         			#no. of surface
tvf = s**2    			#total view factor

#using the result of example 7.8
F11 = 0
F33 =  0.5
print "view factor F11  = %.0f"%(F11)
print "view factor F33  = %.1f"%(F33)

#Calculation & Results
R1 = 0.25     			#R = d/2*h &h = 2d
R2 = 0.25
X = 1+((1+R2**2)/(R1**2))
F14 = (0.5)*(X-math.sqrt((X**2)-4*(R2/R1)**2))
print "view factor F14  = %.3f"%(F14)
F13 = F14
print "view factor F13  = %.3f"%(F13)
F12 = 1-F11-F13  			# from eq. 7.31 for surface 1
print "view factor F12  = %.3f"%(F12)

d = 1.            			#say
A1 = (math.pi*(d**2))/4.
A3 = (math.pi*(d**2))/2.
F31 = A1*F13/(A3)
print "view factor F31  = %.3f"%(F31)

# from eq. 7.31 for surface 3
F33 = 0.5
F32 = 1-F31-F33
print "view factor F32  = %.3f"%(F32)

#for surface 2
A2 = 2*math.pi*d**2
F21 = A1*F12/A2
print "view factor F21  = %.3f"%(F21)
F23 = A3*F32/A2
print "view factor F23  = %.3f"%(F23)
F22 = 1-F21-F23
print "view factor F22  = %.3f"%(F22)

view factor F11  = 0
view factor F33  = 0.5
view factor F14  = 0.056
view factor F13  = 0.056
view factor F12  = 0.944
view factor F31  = 0.028
view factor F32  = 0.472
view factor F21  = 0.118
view factor F23  = 0.118
view factor F22  = 0.764


## Example 7.10 Page No : 235¶

In :
import math
# Variables
ds = 0.3         			#m, diameter of shell
r1 = 0.1         			#m, dismath.tance from the  centre

#Calculation and Results
#by the defination of view factor
F12 = 1.
print "The view factor from surface 1 to 2 is %.0f"%(F12)
#F21
R = ds/2.          			#m, radius of sphere
r2 = math.sqrt(R**2-r1**2)
A1 = math.pi*r2**2      			#m**2 area
A2 = 2*math.pi*R**2+2*math.pi*R*math.sqrt(R**2-r2**2)
#from reciprocity relation
F21 = (A1/A2)*F12
print "The view factor from surface 2 to 1 is %.3f"%(F21)

The view factor from surface 1 to 2 is 1
The view factor from surface 2 to 1 is 0.167


## Example 7.12 Page No : 237¶

In :
import math
# Variables
d = 0.3     			#m, diameter of steel sphere
Ti = 800.    			#K, initial temp. of sphere
T2 = 303.    			#C,ambient temp.
T1 = 343.     			#C, final tempreture
rho = 7801.  			#kg/m**3, density of steel
cp = 0.473  			#kj/kg C, specific heat of steel
#calculation
R = d/2    			#m, radius of sphere
A1 = 4*math.pi*R**2   			#m**2, area of sphere
m = 4./3*math.pi*R**3*rho  			#m**3, mass of sphere
F12 = 1.              			#view factor
s = 5.669*10**-8      			#stephen Boltzman's consmath.tant
#-dT1/dt = A1*F12*s*(T**4-T2**4)/(m*cp)

def f1(T1):
return (1/(T1**4-T2**4))

t = I/(A1*F12*s/(m*cp*10**3))

# Results
print "The time required for the ball to cool is %.1f h"%(t/3600)

The time required for the ball to cool is 10.3 h


## Example 7.13 Page No : 247¶

In :
import math
#Variables
d = 0.114          			#m, dia.o f pipe
l = 1.                 			#m,  length of pipe
A = (math.pi)*d*l  			#m sq,  area
e1 = 1.             			#emmisivity of black body
F12 =  1.           			#view factor, 1:pipe surface, 2:room walls
s = 5.67*10**-8    			#stephen boltzman consmath.tant
T1 =  440.          			#K, steam temp.
T2 = 300.           			#K, wall temp.
#Caluclation
Q12 = A*e1*F12*s*(T1**4-T2**4)  			#net rate of radiative heat loss

#Results
print "a) Net rate of radiative heat loss Q12  = %.1f W "%(Q12)
#Part-b
e2 = 0.74
Q12 = A*e2*F12*s*(T1**4-T2**4)  			#net rate of radiative heat loss
print "b) Net rate of radiative heat loss Q12  = %.1f W"%(Q12)

a) Net rate of radiative heat loss Q12  = 596.6 W
b) Net rate of radiative heat loss Q12  = 441.5 W


## Example 7.14 Page No : 247¶

In :
import math
#Variable declaration
F12 = 1.                			#view factor
r1 = 0.15              			#m inner radius of phere
r2 = 0.155            			#m , outer radius

#Calculation
A1 = 4.*(math.pi)*r1**2  			#sq m inner area
A2 = 4.*(math.pi)*r2**2  			#sq m,outer area
F21 = A1/A2
h = 200.                 			#J/g, heat of vaporization of nitrogen
s = 5.669*10**-8     			# boltzman consmath.tant
T2 = 298.               			#K, temp. of outer wall
T1 = 77.                			#K, Temp. of inner wall
e1 = 0.06              			#emmisivity
e2 = 0.06              			#emmisivity
x = ((1-e1)/(e1*A1))+(1/(A1*F12))+((1-e2)/(e2*A2))
Q1net = (s*(T2**4-T1**4))/(x)

#Result-a-i
print "a-i) View factor F12  =  %.0f"%(F12)
print "view factor F21  =  %.3f"%(F21)
#Result- b
print "ii) The net rate of heat gain Q1net  = %.1f J/s"%(Q1net)
nl = Q1net/h
nl = nl*3600        			#g/h
print "b) Rate of nitrogen loss  =  %.0f g/h"%(nl)

a-i) View factor F12  =  1
view factor F21  =  0.937
ii) The net rate of heat gain Q1net  = 4.0 J/s
b) Rate of nitrogen loss  =  72 g/h


## Example 7.15 Page No : 248¶

In :
import math
# Variables
x = 0.15             			#m, length of opening on a furnace
y = 0.12             			#m, width of opening on a furnace
x1 = 6.               			#m, width of wall
y1 = 5.               			#m, height of wall
e2 = 0.8             			#emissivity of wall
T1 = 1400.            			#C, furnace temp.
T2 = 35.              			#C, wall temp.
T3 = 273.             			#C, smath.radians(numpy.arcmath.tan(ard temp.
s = 5.669*10**-8     			#stephen boltzman's consmath.tant
#in fig. 7.29
l1 = 2.               			#m, l1 = AF
l2 = 1.5             			#m, l2 = AH
h = 3.                			#m, E = dA1

# Calculations
F1 = (1./(2*math.pi))*((l2/(math.sqrt(l2**2+h**2)))*math.tanh(l1/(math.sqrt(l2**2+h**2)))+(l1/(math.sqrt(l1**2+h**2)))*math.tan(l2/(math.sqrt(l1**2+h**2))))
#Similarly
#for the dA1-A3 pair the equation is
F2 = 0.1175
#for the dA1-A4 pair the equation is
F3 = 0.1641
#for the dA1-A5 pair the equation is
F4 = 0.0992
#view factor b/w the opening (dA1)and the wall (W) is
F5 = F1+F2+F3+F4
dA1 = x*y
Aw = x1*y1
Eb1 = s*(T1+T3)**4
Ebw = s*(T2+T3)**4
F6 = dA1*F5/Aw
Q = dA1*F5*e2*(Eb1*(1-(1-e2)*F6)-Ebw)

# Results
print "the net rate of radiant heat transfer to the wall is %.0f W"%(round(Q,-2))

the net rate of radiant heat transfer to the wall is 2900 W


## Example 7.16 Page No : 250¶

In :
#Variable declaration
l = 3.           			#m, length of wall
w = 2.           			#m,  width of, wall
d = 3.           			#m
R1 = l/d
A1 = l*w        			#sq m,area  1: front part
A2 = A1         			#sq m ,  area, 2"back part
e1 = 0.7        			#emmisivity
e2 = 0.7        			#emmisivity
T1 = 673.        			#k
T2 = 523.        			#k
s = 5.669*10**-8  			#stephen boltzman consmath.tant

#Calculation
F12 =  0.148  			#view factor ,from fig. 7.12
x = (A1+A2-2*A1*F12)/(A2-(A1*(F12**2)))+((1/e1)-1)+(A1/A2)*((1/e2)-1)

#Results
Q1net = -1*A1*(s*(T2**4-T1**4))/(x)
print "the net rate of radiant heat loss  = %.1f kW "%(Q1net/1000)
# (b)
F24 = 1.        			#from fig 7.12
T20 = 333.      			#K, outer surface temp. of surface 2
T4 = 303.       			#K, ambient temp
q1 = q/1000     			# Kw
h = q/(A2*(T20-T4))
print "convective heat transfer coeff.  = %.0f W/sq m C"%(h)

the net rate of radiant heat loss  = 17.1 kW
convective heat transfer coeff.  = 90 W/sq m C


## Example 7.17 Page No : 251¶

In :
from numpy import array, linalg
import math

# Variables
r1i = 0.1       			#m, inner radius of disk 1
r1o = 0.2       			#m, outer radius of disk 1
r2i = 0.12      			#m, inner radius of disk 2
r2o = 0.25      			#m, outer radius of disk 2
h = 0.08        			#m, dismath.tance between the disks
R2 = r2o/h
R1 = r1o/h
X = 1+(1+R1**2)/R2**2
F23_14 = 1./2*(X-math.sqrt(X**2-4*(R1/R2)**2))

#calculation
R2_ = r2o/h
R1_ = r1i/h
X_ = 1+(1+R1_**2)/R2_**2
F23_4 = 1/2*(X_-math.sqrt(X_**2-4*(R1_/R2_)**2))     			#view factor
#similarly
F3_14 = 0.815          			#view factor
F34 = 0.4              			#view factor
A23 = math.pi*r2o**2        			#area
A3 = math.pi*r2i**2
A1 = math.pi*(r1o**2-r1i**2)
#from eq. 1
F12 = A23*(F23_14-F23_4)/A1-(A3*(F3_14-F34))/A1

#calculation of the rate of radiative heat exchange
# Variables
T1 = 1000.            			#K, temprature of disk 1
T2 = 300.             			#K, temprature of disk 2
s = 5.669*10**-8      			#stephen's Boltzman consmath.tant
e1 = 0.8             			#emissivity
e2 = 0.7
A2 = math.pi*(r2o**2-r2i**2)
F1s = 1-F12
F2s = 1-(A1*F12/A2)
#calculation
#let some quantities equal to
a = (1-e1)/(e1*A1)
b = 1/(A1*F12)
c = (1-e2)/(e2*A2)
d = 1/(A1*F1s)
e = 1/(A2*F2s)
f = s*T1**4
g = s*T2**4
#from eq. 7.42(a)
#(f-J1)/a = (J1-J2)/b+J1/d
#(g-J2)/c = (J2-J1)/b+J1/e
#solving two eqns by matrix
A = array([[-0.0564,0.5036],[0.4712,-0.0564]])
B = array([[161.847],[21376.31]])
X = linalg.solve(A,B)
J1 = X
J2 = X

Q12net = (J1-J2)/17.73

# Results
print "net rate of radiation exchange b/w disk 1 and 2 is %d W/m**2"%(Q12net)

net rate of radiation exchange b/w disk 1 and 2 is 2286 W/m**2


## Example 7.18 Page No : 255¶

In :
from scipy.optimize import fsolve
import math

# Variables
di = 0.0254       			#m, inner diameter of tube
Ti = 77.           			#K, liquid temprature
do = 52.5*10**-3   			#m, pipe internal diameter
To = 270.          			#K, wall temprature
l = 1.             			#m, length of tube
e1 = 0.05         			#emissivity of tube wall
e2 = 0.1          			#emissivity of pipe wall
e3 = 0.02         			#emissivity for inner surface of radiation field
e4 = 0.03         			#emissivity for outer surface of radiation field
s = 5.669*10**-8   			#stephen boltzman math.cosmath.tantl

#Calculation
ds = (do+di)/2    			#m, diameter of radiation shield
Ao = math.pi*do*l     			#m**2, outer pipe area
As = math.pi*ds*l     			#m**2, shield area
Ai = math.pi*di*l     			#m**2, inner pipe area
#View factors
#for the long cylindrical enclosure made up of the outer pipe and the shield
Fso = 1.    			#because outer surface of shield cant see itself
Fos = As/Ao
Fsi = Ai/As
#now assume
#(1-e2)/e2+ 1/Fos +Ao*(1-e4)/(As*e4) = x
#(1-e3)/e3 +1/Fsi +(1/Fsi)*(1-e1)/e1 = y
x = (1-e2)/e2+ 1/Fos +Ao*(1-e4)/(As*e4)
y = (1-e3)/e3 +1/Fsi +(1/Fsi)*(1-e1)/e1
#solving the  equations for heat transfer from the outer pipe and inner pipe
def f(Ts):
return (Ao*(To**4-Ts**4)/x)-(Ai*(Ts**4-Ti**4)/x)
Ts = fsolve(f,1)
Qos = (Ao*s*(To**4-Ts**4))/x

# Results
print "The net rate of heat gain of tube is %.2f W"%(Qos)

# note : rounding off error.

The net rate of heat gain of tube is 0.30 W


## Example 7.19 Page No : 258¶

In :
# Variables
T1 = 300 + 273.      #K
Ta = 30. + 273.      # K
w = 0.075            # m
k = 0.08             # W/m c
l = 1.075            # m
delta = 5.669 * 10**-8   # W/m**2 K**4
A1 = 1 * 1.5         # M**2
A2 = A1
A2m = (1.5 + 1.9)/2   # m**2
A3m = (5. + 6.02)/2   # m**2

# Calculations
T2 = 545.1          # k
T2_ = 322.6
T3 = 544.7
T3_ = 328.4
rate_of_heatloss1 = int(A2m*k/w*(T2-T2_))
rate_of_heatloss2 = int(A3m*k/w*(T3-T3_))
total = rate_of_heatloss1 + rate_of_heatloss2

# results
print "Rate of heat loss from the top surface : %d W"%rate_of_heatloss1
print "Rate of heat loss from the side walls : %d W"%rate_of_heatloss2
print "Total rate of heat loss : %d W"%total

Rate of heat loss from the top surface : 403 W
Rate of heat loss from the side walls : 1271 W
Total rate of heat loss : 1674 W


## Example 7.20 Page No : 264¶

In :
import math

# Variables
T = 300.          			#K, temprature
lam = 4.2        			#micrometer, wavelength radiation
L = 0.1          			#m, path length

#calculation
# I2/I1 = f
f = 1-per/100.    			#fraction of incident radiation transmitted
#from  eq. 7.69
a = -math.log(f)/L

# Results
print "the spectral extinction coefficient is %.2f m**-1"%(a)

# note : rounding off error.

the spectral extinction coefficient is 24.08 m**-1


## Example 7.21 Page No : 265¶

In :
# Variables
Ts = 800.         			#C, wall temp.
Tg = 1100.        			#C. burner temprature
CO2 = 8.          			#percent, composition of CO2 in flue gas
M = 15.2         			#percent, composition of moisture in flue gas
a = 0.4          			#m, length of  duct
b = 0.4          			#width of duct
h = 15.           			#W/m**2 C, heat transfer coefficient
P = 1.            			#atm  pressure
#CAlCULATION of Eg(Tg)
pc = CO2/100.*P         			#atm, partial pressure of CO2
pw = M/100.*P          			#atm, partial pressure of moisture
l = 1.                 			#m, length of duct
V = a*b*l             			#m**3, volume of duct
A = 1.6*l             			#m**2 area of duct
Le = 3.6*(V/A)        			#m, mean beam length

pc*Le
pw*Le
Tg_ = Tg+273.
Ts_ = Ts+273.
#from fig 7.38
Ec = 0.06
Eg = 0.048             			#from fig 7.39
#a correction dE need to be calculated
#pw/(pc+pw)
#pc*Le+pw*Le
#from fig. 7.39
dE = 0.003
Eg_Tg = Ec+Eg-dE      			#emissivity at temp. Tg

#Calculation of alpha
#pc*Le*Ts/Tg
#from fig. 7.37
Ec1 = 0.068
#from fig. 7.38
Ew1 = 0.069
Cc = 1                			#correction factor
Cw = 1                			#correction factor
d_alpha = dE          			#AT 1 ATM TOTAL PRESSURE
alpha = Cc*Ec1*(Tg_/Ts_)**0.65+Cw*Ew1*(Tg_/Ts_)**0.45-dE

The total rate of heat transfer from the gas to the wall is 22.5 kW