Chapter4: Radiation

Example no:4.1,Page no:4.7

In [41]:
#Heat loss by radiaiton

#Variable declaration
e=0.9   #[Emissivity]
sigma=5.67*10**-8    #[W/m**2.K**4]
T1=377  #[K]
T2=283  #[K]

#Calculation
Qr_by_a=e*sigma*(T1**4-T2**4) #[W/sq m]

#Result
print"Heat loss by radiation is",round(Qr_by_a),"W/m^2"

Heat loss by radiation is 704.0 W/m^2

Example no:4.2,Page no:4.7

In [42]:
#Radiation from unlagged steam pipe

#Variable declaration
e=0.9   #Emissivity
T1=393  #[K]
T2=293  #[K]
sigma=5.67*10**-8    #[W/sq m.K]
#Calculation
Qr_by_a=e*sigma*(T1**4-T2**4) #W/sq m
#Result
print"Rate of heat transfer by radiation is",round(Qr_by_a,1),"W/sq m"

Rate of heat transfer by radiation is 841.2 W/sq m

Example no:4.3,Page no:4.7

In [48]:
#Interchange of radiation energy
import math

#Variable declaration
L=1  #[m]
e=0.8    #Emissivity
sigma=5.67*10**-8     #[m**2.K**4]
T1=423.0   #[K]
T2=300.0   #[K]
Do=60.0    #[mm]
Do=Do/1000   #[m]

#Calculation
A=round(math.pi*Do*L,3)  #[sq m]
Qr=e*sigma*A*(T1**4-T2**4)    #[W/m]

#Result
print"Net radiaiton rate per 1 metre length of pipe is",round(Qr),"W/m(approx)"

Net radiaiton rate per 1 metre length of pipe is 204.0 W/m(approx)

Example no:4.4,Page no:4.8

In [49]:
#Heat loss in unlagged steam pipe
import math
#Variable declaration
e=0.9   #Emissivity
L=1.0 #[m]
Do=50.0   #[mm]
Do=Do/1000  #[m]
sigma=5.67*10**-8    #[W/(m**2.K**4)]
T1=415.0  #[K]
T2=290.0  #[K]
dT=T1-T2    #[K]
#Calculation
hc=1.18*(dT/Do)**(0.25)  #[W/sq m.K]
A=math.pi*Do*L  #Area in [sq m]
Qc=hc*A*dT #Heat loss by convection W/m
Qr=e*sigma*A*(T1**4-T2**4)    #Heat loss by radiation per length W/m
Qt=Qc+Qr    #Total heat loss in [W/m]
#Result
print"Total heat loss by convection is",round(Qt,1),"W/m"

Total heat loss by convection is 344.9 W/m

Example no:4.5,Page no:4.8

In [50]:
#Loss from horizontal pipe
import math
#Variable declaration
e=0.85
sigma=5.67*10**-8    #[W/sq m.K]
T1=443.0  #[K]
T2=290.0  #[K]
dT=T1-T2    #[K]
hc=1.64*dT**0.25     #W/sq m.K
Do=60.0   #[mm]
Do=Do/1000  #[m]
L=6 #Length [m]
#Calculation
A=math.pi*Do*L  #Surface area of pipe in [sq m]
Qr=e*sigma*A*(T1**4-T2**4)    # Rate of heat loss by radiaiton W
Qc=hc*A*(T1-T2) # Rate of heat loss by convection [W]
Qt=Qr+Qc    #Total heat loss  [W]
#Result
print"Total heat loss is",round(Qt),"W"

Total heat loss is 2712.0 W

Example no:4.6,Page no:4.11

In [51]:
#Heat loss by radiation in tube
import math
#Variable declaration
sigma=5.67*10**-8    #[W/m**2.K**4]
e1=0.79 
e2=0.93 
T1=500   #[K]
T2=300   #[K]
D=70    #[mm]
D=D/1000    #[m]
L=3 #[m]
W=0.3   #Side of conduit [m]
#Calculation
A1=math.pi*D*L  #[sq m]
A1=0.659        #Approximate calculation in book in [m**2]
A2=4*(L*W)  #[sq m]
Q=sigma*A1*(T1**4-T2**4)/(1/e1+((A1/A2)*(1/e2-1)))    #[W]
#Result
print"Heat lost by radiation is",round(Q,1),"W"

Heat lost by radiation is 1588.5 W

Example no:4.7,Page no:4.11

In [5]:
#Net radiant interchange
#Variable declaration
sigma=5.67*10**-8    #[W/sq m.K**4]
T1=703  #[K]
T2=513  #[K]
e1=0.85 
e2=0.75
#Calculation
Q_by_Ar=sigma*(T1**4-T2**4)/(1/e1+1/e2-1) #[W/sq m]
#Result
print"Net radiant interchange per square metre is",round(Q_by_Ar),"W/sq m"

Net radiant interchange per square metre is 6571.0 W/sq m

Example no:4.8,Page no:4.12

In [31]:
#Radiant interchange between plates
#Variable declaration
L=3  #[m]
A=L**2   #Area in [sq m]
sigma=5.67*10**-8     #[W/sq m.K**4]
T1=373   #[K]
T2=313   #[K]
e1=0.736 
e2=e1 
#Calculation
F12=1.0/((1.0/e1)+(1.0/e2)-1)
Q=sigma*A*F12*(T1**4-T2**4)   #[W]
#Result
print"Net radiant interchange is",round(Q),"W"

Net radiant interchange is 2900.0 W

Example no:4.9,Page no:4.12

In [1]:
#Heat loss from thermofask  
#Variable declaration  
sigma=5.67*10**-8    #[W/sq m.K**4]
e1=0.05 
e2=0.05
#A1=A2=1 (let)
A1=1 
A2=A1 

#Calculation
F12=1.0/(1.0/e1+(A1/A2)*(1.0/e2-1))   
T1=368  #[K]
T2=293  #[K]
Q_by_A=sigma*F12*(T1**4-T2**4)    #Heat loss per unit Area  [W/sq m]
print"Rate of heat loss when of silvered surface is",round(Q_by_A,2),"W/sq m"
#When both the surfaces are black
e1=1 
e2=1 
F12=1/(1/e1+(A1/A2)*(1/e2-1))   
Q_by_A=sigma*F12*(T1**4-T2**4)    #[W/sq m]

#Result
print"\n When both surfaces are black,Rate of heat loss  is",round(Q_by_A),"W/sq m"

Rate of heat loss when of silvered surface is 15.95 W/sq m

 When both surfaces are black,Rate of heat loss  is 622.0 W/sq m

Example no:4.10,Page no:4.13

In [53]:
#Diwar flask
import math

#Variable declaration
e1=0.05
e2=e1
A1=0.6944 
A2=1 
T1=293  #[K]
T2=90   #[K]
sigma=5.67*10**-8    #[W/m**2.K**4]
D=0.3   #Diameter in [m]

#Calculation
F12=1.0/(1.0/e1+(A1/A2)*(1.0/e2-1))
Q_by_A=sigma*F12*(T1**4-T2**4)    #[W/sq m]
Q=Q_by_A*math.pi*(D**2)  #[kJ/h]
Q=Q*3600/1000   #[kJ/h]
lamda=21.44    #Latent heat in [kJ/kg]
m_dot=Q/lamda    #kg/h

#Result
print"The liquid oxygen will evaporate at",round(m_dot,2),"kg/h"

The liquid oxygen will evaporate at 0.59 kg/h

Example no:4.11,Page no:4.13

In [55]:
#Heat flow due to radiation
import math

#Variable declaration
sigma=5.67*10**-8    #W/(m**2.K**4)
e1=0.3 
e2=e1 
D1=0.3  #[m]
D2=0.5  #[m]
T1=90   #[K]
T2=313  #[K]

#Calculation
A1=math.pi*D1**2 #Area in [sq m]
A2=math.pi*D2**2#Area in [sq m]
Q1=sigma*A1*(T1**4-T2**4)/(1/e1+(A1/A2)*(1/e2-1))  #[W]
Q1=abs(Q1)  #Absolute value in [W]
print"Rate of heat flow due to radiation is",round(Q1,2),"W"
#When Aluminium is used
e1=0.05
e2=0.5
Q2=sigma*A1*(T1**4-T2**4)/(1/e1+(A1/A2)*(1/0.3-1))  #[W]
Q2=abs(Q2) #Absolute value in [W]
Red=(Q1-Q2)*100/Q1  #Percent reduction

#Result
print"Reduction in heat flow will be",round(Red,2),"%"

Rate of heat flow due to radiation is 36.62 W
Reduction in heat flow will be 79.97 %

Example no:4.12,Page no:4.14

In [56]:
#Heat exchange between concentric shell
import math

#Variable declaration
sigma=5.67*10**-8    #[W/sq m.K**4]
T1=77.0   #[K]
T2=303.0  #[K]
D1=32.0   #cm
D1=D1/100   #[m]
D2=36.0   #[cm]
D2=D2/100   #[m]

#Calculation
A1=math.pi*D1**2 #[sq m]
A2=math.pi*D2**2 #[sq m]
e1=0.03 
e2=e1 
Q=sigma*A1*(T1**4-T2**4)/(1.0/e1+(A1/A2)*(1.0/e2-1))  #[W]
Q=Q*3600.0/1000   #[kJ/h]
Q=abs(Q)    #[kJ/h]
lamda=201.0  #kJ/kg
m_dot=Q/lamda  #Evaporation rate in [kg/h]
#Result
print"Nitrogen evaporates at",round(m_dot,3),"kg/h"

Nitrogen evaporates at 0.047 kg/h

Example no:4.13,Page no:4.15

In [57]:
#Evaporation in concenric vessels
import math
#Variable declaration
D1=250.0  #Inner sphere idameter[mm]
D1=D1/1000  #Outer diameter [m]
D2=350.0  #[mm]
D2=D2/1000  #[m]
sigma=5.67*10**-8    #W/(sq m.K**4)
#Calculation
A1=math.pi*D1**2 #[sq m]
A2=math.pi*D2**2 #[sq m]
T1=76.0   #[K]
T2=300.0  #[K]
e1=0.04 
e2=e1 
Q=sigma*A1*(T1**4-T2**4)/((1.0/e1)+(A1/A2)*((1.0/e2)-1))   #[W]
Q=-2.45     #Approximate
Q=abs(Q)    #[W]
Q=Q*3600.0/1000   #[kJ/h]
lamda=200.0  #kJ/kg
Rate=Q/lamda   #[kg/h]
#Result
print"Rate of evaporation is",Rate,"kg/h"

-2.4112304452
Rate of evaporation is 0.0441 kg/h

Example no:4.15,Page no:4.19

In [60]:
#infinitely long plates
#Variable declaration
sigma=5.67*10**-8    #[W/(m**2.K**4)]
e1=0.4
e3=0.2
T1=473  #[K]
T3=303  #[K]
#Calculation
Q_by_a=sigma*(T1**4-T3**4)/((1.0/e1)+(1.0/e3)-1)  #[W/sq m]
#Q1_by_a=sigma*(T1**4-T2**4)/((1/e1)+(1/e2)-1)=sigma*A*(T2**4-T3**4)/((1/e2)+(1/e3)-1) #[W/sq m]
e2=0.5
#Solving we get
T2=((6.0/9.5)*((3.5/6)*T3**4+T1**4))**(1.0/4.0)  #[K]
Q1_by_a=sigma*(T1**4-T2**4)/((1.0/e1)+(1.0/e2)-1) #[W/sq m]
red=(Q_by_a-Q1_by_a)*100/Q_by_a
#Result
print"Heat transfer rate per unit area(WITHOUT SHIELD) due to radiation is",round(Q_by_a,1),"W/sq m"
print"\nHeat transfer rate per unit area(WITH SHIELD) due to radiation is",round(Q1_by_a,2),"W/sq m"
print"\nReduction in heat loss is",round(red,2),"%"

Heat transfer rate per unit area(WITHOUT SHIELD) due to radiation is 363.1 W/sq m

Heat transfer rate per unit area(WITH SHIELD) due to radiation is 248.44 W/sq m

Reduction in heat loss is 31.58 %

Example no:4.16,Page no:4.20

In [2]:
#Heat exchange between parallel plates
from scipy.optimize import fsolve
import math
#In steady state,we can write:
#Qcd=Qdb
#sigma(Tc**4-Td**4)*/(1/ec+1/ed-1)=sigma(Td**4-Tb**4)/(1/ed+1/eb-1)
# i.e Td**4=0.5*(Tc**4-Tb**4)
#Variable declaration
Ta=600  #[K]
eA=0.8 
eC=0.5 
eD=0.4 
sigma=5.67*10**-8        #For air

#Calculation

#(600**4-Tc**4)/2.25=(Tc**4-Td**4)/3.5
#1.56*(600**4-Tc**4)=Tc**4-Td**4
#Putting value of Td in terms of Tc
#1.56*(600**4-Tc**4)=Tc**4-0.5*(Tc**4-300**4)
def f(Tc):
    y=1.56*(600**4-Tc**4)-Tc**4+0.5*(Tc**4-300**4)
    return(y)
Tc=fsolve(f,500)            #[K]
#or
Tc=560.94       #[K] Approximate after solving
Td=math.sqrt(math.sqrt(0.5*(Tc**4-300**4)))         #[K]
Q_by_a=sigma*(Ta**4-Tc**4)/(1/eA+1/eC-1)      #[W/sq m]

#Result

print"Steady state temperatures,Tc=",Tc,"K,and Td=",round(Td,2)," K"
print"Rate of heat exchange per unit area=",round(Q_by_a,2),"W/m**2"

Steady state temperatures,Tc= 560.94 K,and Td= 461.73  K
Rate of heat exchange per unit area= 770.95 W/m**2

Example no:4.17,Page no:4.21

In [14]:
#Thermal radiation in pipe

#Variable declaration
sigma=5.67*10**-8    #[W/(sq m.K**4)]
e=0.8
T1=673   #[K]
T2=303   #[K]
Do=200   #[mm]
Do=Do/1000  #[m]
L=1      #Let [m]

#Calculation
import math
A1=math.pi*Do*L #[m**2/m]
#CAse 1: Pipe to surrundings

Q1=e*A1*sigma*(T1**4-T2**4)   #[W/m]
Q1=5600     #Approximated
#Q1=5600         #[W/m] approximated in book for calculation purpose
#Concentric cylinders
e1=0.8 
e2=0.91 
D1=0.2  #[m]
D2=0.4  #[m]
Q2=sigma*0.628*(T1**4-T2**4)/((1/e1)+(D1/D2)*((1/e2)-1))   #[W/m] length
Red=Q1-Q2   #Reduction in heat loss

#Result
print"Due to thermal radiaiton,Loss of heat to surrounding is",round(Q1),"W/m"
print"When pipe is enclosed in 1 400 mm diameter brick conduit,Loss of heat is",round(Q2),"W/m" 
print" Reduction in heat loss is",round(Red),"W/m"

Due to thermal radiaiton,Loss of heat to surrounding is 5600.0 W/m
When pipe is enclosed in 1 400 mm diameter brick conduit,Loss of heat is 5390.0 W/m
 Reduction in heat loss is 210.0 W/m

Example no:4.18,Page no:4.22

In [64]:
#Heat transfer in concentric tube
#Variable declaration
sigma=5.67*10**-8     #[W/(sq m.K**4)]
T1=813.0   #[K]
T2=473.0   #[K]
e1=0.87 
e2=0.26 
D1=0.25  #[m]
D2=0.3  #[m]
#Calculation
Q_by_a1=sigma*(T1**4-T2**4)/(1.0/e1+(D1/D2)*(1.0/e2-1.0))   #[W/ sqm]
#Result
print"Heat transfer by radiaiton is",round(Q_by_a1),"W/sq m"

Heat transfer by radiaiton is 6229.0 W/sq m

Example no:4.19,Page no:4.24

In [3]:
#Heat exchange between black plates
#Variable declaration
sigma=5.67*10**-8    #[W/sq m.K**4]
A1=0.5*1    #[sq m]
F12=0.285
T1=1273 #/[K]
T2=773  #[K]
#Calculation
Q=sigma*A1*F12*(T1**4-T2**4)    #[W]
#Result
print"Net radiant heat exchange between plates is",round(Q),"W"

Net radiant heat exchange between plates is 18334.0 W

Example no:4.20,Page no:4.32

In [66]:
#Radiation shield
import math

#Variable declaration
sigma=5.67*10**-8    #[W/sq m.K**4]
T1=750  #[K]
T2=500  #[K]
e1=0.75 
e2=0.5 

#Calculation
#Heat transfer without shield :
Q_by_a=sigma*(T1**4-T2**4)/((1/e1)+(1/e2)-1)  #[W/sq m]

#Heat transfer with shield:
R1=(1-e1)/e1    #Resistance 1
F13=1 
R2=1/F13        #Resistance 2
e3=0.05
R3=(1-e3)/e3    #Resistance 3
R4=(1-e3)/e3    #Resistance 4
F32=1 
R5=1/F32        #Resistance 5
R6=(1-e2)/e2       #Resistance 6
Total_R=R1+R2+R3+R4+R5+R6   #Total resistance
Q_by_as=sigma*(T1**4-T2**4)/Total_R   #[W/sq m]
Red=(Q_by_a-Q_by_as)*100/Q_by_a     #Reduciton in heat tranfer due to shield 

#Result
print"Reduction in heat transfer rate as a result of radiaiotn shield is",round(Red,2),"percent"

Reduction in heat transfer rate as a result of radiaiotn shield is 94.35 percent

Example no:4.21,Page no:4.33

In [67]:
#Heat transfer with radiaiton shield

#Variable declaration
e1=0.3
e2=0.8

#Calculation
#Let sigma*(T1**4-T2**4)=z=1(const)
z=1     #Let
Q_by_A=z/(1/e1+1/e2-1)  #W/sq m

#Heat transfer with radiation shield 
e3=0.04
F13=1 
F32=1 
#The resistances are:
R1=(1-e1)/e1
R2=1.0/F13
R3=(1-e3)/e3
R4=R3
R5=1.0/F32
R6=(1-e2)/e2
R=R1+R2+R3+R4+R5+R6     #Total resistance
Q_by_As=z/R #where z=sigma*(T1**4-T2**4) #W/sq m
red=(Q_by_A-Q_by_As)*100/Q_by_A    #Percent reduction in heat transfer

#Result
print"The heat transfer is reduced by",round(red,1),"% due to shield"

The heat transfer is reduced by 93.2 % due to shield

Example no:4.22,Page no:4.34

In [2]:
#Radiaition shape factor

#Variable declaration
sigma=5.67*10**-8 
T1=1273 #[K]
T2=773  #[K]
T3=300  #[K]
A1=0.5  #[sq m]
A2=A1   #[sq m]
F12=0.285 
F21=F12 
F13=1-F12 
F23=1-F21 
e1=0.2 
e2=0.5

#Calculation
#Resistance in the network are calculated as:
R1=1-e1/(e1*A1)
R2=1-e2/(e2*A2)
R3=1.0/(A1*F12)
R4=1.0/(A1*F13)
R5=1.0/(A2*F23)
R6=0    #Given (1-e3)/e3*A3=0
#Also
Eb1=sigma*T1**4  #W/sq m
Eb2=sigma*T2**4   #[W/sq m]
Eb3=sigma*T3**4 #[W/sq m]

#Equations are:
#(Eb1-J1)/2+(J2-J1)/7.018+(Eb3-J1)/2.797=0
#(J1-J2)/7.018+(Eb3-J2)/2.797+(Eb2-J2)/2=0

#On solving we get:
J1=33515    #[W/sq m]
J2=15048    #[W/sqm]
J3=Eb3  #[W/sq m]
Q1=(Eb1-J1)/((1.0-e1)/(e1*A1))        #[W/sq m]
Q2=(Eb2-J2)/((1.0-e2)/(e2*A2))        #[W/sq m]
Q3=(J1-J3)/(1.0/(A1*F13))+(J2-J3)/(1.0/(A2*F23))    #[W/sq m]

#Result
print"Total heat lost by plate 1 is",round(Q1),"W/sq m"
print"Total heat lost by plate 2 is",round(Q2),"W/sq m" 
print"The net energy lost by both plates must be absorbed by the room",round(Q3,1),"=",round(Q1+Q2)

Total heat lost by plate 1 is 14423.0 W/sq m
Total heat lost by plate 2 is 2598.0 W/sq m
The net energy lost by both plates must be absorbed by the room 17032.9 = 17021.0

Example no:4.23,Page no:4.37

In [1]:
#Radiation loss in plates

#Variable declaration
sigma=5.67*10**-8    #[W/sq m.K**4]
e1=0.7 
e2=0.7 
T1=866.5    #[K]
T2=588.8    #[K]

#Calculation
Q_by_A=sigma*(T1**4-T2**4)/((1/e1)+(1/e2)-1)  #[W/sq m]
e1=0.7 
e2=e1 
e3=e1 
e4=e1 
e=e1 
#Q with n shells =1/(n+1)
n=2
Q_shield=1/(n+1) 
es1=e1 
es2=e1 
Q_by_A=sigma*(T1**4-T2**4)/((1/e1)+(1/e2)+2*(1/es1+1/es2)-(n+1))  #[W/sq m]

#Result
print"\n New Radiaiton loss is",round(Q_by_A,1),"W/sq m"

 New Radiaiton loss is 4513.9 W/sq m

Example no:4.24,Page no:4.38

In [77]:
#Radiation in Concentric tube
import math

#Variable declaration
#1.WITHOUT SHIELD
sigma=5.67*10**-8        
e1=0.12 
e2=0.15 
T1=100  #[K]
T2=300  #[K]
r1=0.015    #[m]
r2=0.045    #[m]
L=1         #[m]

#Calculation
A1=2*math.pi*r1*L   #[sq m]
Q_by_L=2*math.pi*r1*sigma*(T1**4-T2**4)/(1/e1+(r1/r2)*(1/e2-1))   #[W/m]
#-ve saign indicates that the net heat flow is in the radial inward direction
print "Without shield,Q=",round(Q_by_L,3),"W/m"
#2.WITH CYLINDRICAL RADIATION SHIELD
e3=0.10 
e4=0.05 
r3=0.0225   #[m]
Qs_by_L=2*math.pi*r1*sigma*(T1**4-T2**4)/(1/e1+r1/r2*(1/e2-1)+(r1/r3)*(1/e3+1/e4-1))  #[W/sq m]
red=(abs(Q_by_L)-abs(Qs_by_L))*100/abs(Q_by_L)  #percent reduction in heat gain

#Radiation network approach
A3=2*math.pi*r3     #[sq m]
A2=2*math.pi*r2     #[sq m]
F13=1 
F32=1 
R1=(1-e1)/(e1*A1)
R2=1.0/(A1*F13)
R3=(1-e3)/(e3*A3)
R4=(1-e4)/(e4*A3)
R5=1.0/(A3*F32)
R6=(1-e2)/(e2*A2)

Qs=sigma*(T1**4-T2**4)/((1.0-e1)/(e1*A1)+1/(A1*F13)+(1.0-e3)/(e3*A3)+(1.0-e4)/(e4*A3)+1/(A3*F32)+(1.0-e2)/(e2*A2))        

#Result
print"With cylindrical radiaiton shield Heat gained by fluid per 1 m lengh of tube is",round(Qs_by_L,3),"W/m" 
print"Percent reduction in heat gain is",round(red,2),"%"
print"With radiaiton network approach",round(Qs,2),"W/sqm "

Without shield,Q= -4.182 W/m
With cylindrical radiaiton shield Heat gained by fluid per 1 m lengh of tube is -1.446 W/m
Percent reduction in heat gain is 65.41 %
With radiaiton network approach -1.45 W/sqm