# Chapter 10: Radiative heat transfer¶

### Example 10.1, Page number: 539¶

In [3]:
from __future__ import division
import math

#Variables
T1=2273;             #temp.  of  liquid  air,K
T2=303;              #temp.  of  room,K
T3=973;              #temp.  of  shield,K
D1=0.003;            #diameter  of  crucible,m
D2=0.05;             #diameter  of  shield,m
theta1=330;          #surrounding  angle  of  jet,degree
theta2=30            #  angle  of  slit,degree
Fjr=theta2/360;      #fraction  of  energy  of  view  of  jet  occupied  by  room
Fjs=theta1/360  ;    #fraction  of  energy  of  view  of  jet  occupied  by  shield
sigma=5.67*10**-8;   #Stefen-Boltzman constant

#Calculations
Qnjr=math.pi*D1*Fjr*sigma*(T1**4-T2**4);  #net  heat  transfer  from  jet  to  room,W/m
Qnjs=math.pi*D1*Fjs*sigma*(T1**4-T3**4);  #net  heat  transfer  from  jet  to  shield,W/m
#to  find  the  radiation  from  the  inside  of  the  shield  to  the  room,  we  need  Fshield-room.since  any  radiation  passing  out  of  the  slit  goes  to  the  room,we  can  find  this  view  factor  equating  view  factors  to  the  room  with  view  factors  to  the  slit.
Aslit=math.pi*D2*Fjr;                            #Slit's area, m^2
Fsj=math.pi*D1/Aslit*Fjr;                        #fraction  of  energy  of  view  of  slit  occupied  by  jet
Fss=1-Fsj;                                       #fraction  of  energy  of  view  of  slit    occupied  by  shield.
Fsr=Aslit*Fss/(math.pi*D2*Fjs);                  #fraction  of  energy  of  view  of  shield  occupied  by  room
Qnsr=math.pi*D2*Fjs*sigma*Fsr*(T3**4-T2**4);     #net  heat  transfer  from  shield  to  room,  W/m

#Result
print "Heat transfer from jet to room through the slit is :",round(Qnjr,2),"W/m\n"
print "Heat transfer from the jet to shield is :",round(Qnjs,2)," W/m\n"
print "Heat transfer from inside of shield to the room is :",round(Qnsr,2),"W/m\n"
print "Both the jet and the inside of the shield have relatively small view factors to the room, so that comparatively little heat is lost through the silt \n"

Heat transfer from jet to room through the slit is : 1188.32 W/m

Heat transfer from the jet to shield is : 12636.6  W/m

Heat transfer from inside of shield to the room is : 619.44 W/m

Both the jet and the inside of the shield have relatively small view factors to the room, so that comparatively little heat is lost through the silt



### Example 10.2, Page number: 542¶

In [5]:
from __future__ import division
import math

#Variables
T1=373;                               #temp.  of  shield,K
T2=1473;                              #temp  of  heater,K
h=0.2  ;                              #height  of  disc  heater,m
R1=r1/h  ;                            #factors  necessary  for  finding  view  factor
R2=r2/h  ;                            #factors  necessary  for  finding  view  factor
sigma=5.67*10**-8;                    #Stefen-Boltzman constant

#Calculations
X=1+(1+R2**2)/R1**2;                                                  #factors  necessary  for  finding  view  factor
Fht=0.5*(X-math.sqrt(X**2-4*(R2**2/R1**2)));                          #view  factor
Fhs=1-Fht;                                                            #view  factor  of  heater    occupied  by  shield
Qnhs=math.pi*r2**2*Fhs*sigma*(T2**4-T1**4)/4;                         #Net heat transfer from the heater to shield

#Result
print "Net heat transfer from the heater to shield is : ",round(Qnhs)," W\n"

Net heat transfer from the heater to shield is :  1686.0  W



### Example 10.3, Page number: 547¶

In [6]:
from __future__ import division
import math

#Variables
h=0.2  ;                       #height  of  disc  heater,m
Fhs=0.808;                     #view  factor  of  heater  occupied  by  shield

#Calculations
As=math.pi*(r1+r2)*math.sqrt(h**2+(r2-r1)**2);      #area  of  frustrum  shaped  shield,m**2
Ah=math.pi/4*r2**2;                                 #heater  area,m**2
Fsh=Ah/As*Fhs;                                      #view  factor  of  shield  occupied  by  heater

#Result
print "View factor of shield occupied by heater is :",round(Fsh,3),"\n"

View factor of shield occupied by heater is : 0.065



### Example 10.4, Page number: 548¶

In [4]:
from __future__ import division

#Variables
F1342=0.245;               #view  factor  of  1and  3  occupied  by  2  and  4
F14=0.2;                   #view  factor  of  1  occupied  by  4

#Calculations
F12=F1342-F14;             #view  factor  of  1  occupied  by  2

#Results
print "View factor of 1 occupied by 2  is :",F12,"\n"

View factor of 1 occupied by 2  is : 0.045



### Example 10.8, Page number: 554¶

In [9]:
from __future__ import division
import math
from sympy import *

#Variables
T1=80;             #temp.of  liquid  nitrgen,K
T2=230;             #temp  of  chamber  walls,K
D1=0.00635;         #outer  diameter  of  steel,  m
D2=0.0127;         #diameter  of  2nd  steel  tube,  m
e1=0.2  ;           #emissivity  0f  steel
sigma=5.67*10**-8;  #Stefen-Boltzman constant

#Calculations
x = Symbol('x');
#the  nitrogen  coolant  will  hold  the  surface  of  the  line  at  essentially  80  K,  since  the  thermal  ressistance  of  tube  wall  and  int.  convection  or  boiling  process  are  small.
Qgain=math.pi*D1*e1*sigma*(T2**4-T1**4);  #  net  heat  gain  of  line  per  unit  length,W/m
#with  the  shield  ,  assuming  that  the  chamber  area  is  large  compared  to  the  shielded  line.
Qgain1=math.pi*D1*sigma*(T2**4-T1**4)/(((1-e1)/e1+1)+D1/D2*(2*(1-e1)/e1+1));  #net  heat  gain  with  shield,W/m
s=(Qgain-Qgain1)/Qgain*100;                        									#rate  of  heat  gain  reducton  in  percentage
T = (230**4 -0.328/(3.14*D2*e1*sigma))**(1/4)  						#Temp. of the shield

#Result
print "Net heat gain of line per unit length is :",round(Qgain,2)," W/m\n"
print "Rate of heat gain reducton is :",round(s,2),"  percent \n"
print "Temp. of the shield is : ",round(T,2)," C\n"


Net heat gain of line per unit length is : 0.62  W/m

Rate of heat gain reducton is : 47.37   percent

Temp. of the shield is :  213.38  C



### Example 10.9, Page number: 557¶

In [10]:
from __future__ import division
from numpy import mat
from numpy.linalg import inv
from sympy import solve, symbols

#Variables
T1=250  ;          #temp.of  surrounding,K
l1=1;              #width  of  strips,  m
l2=2.4;            #distance  between  strips,m
F12=0.2;           #view  factor  of  1  occupied  by  2.

#Calculations
A=mat('1  -0.14;-1,  10')  ;    #matrix  representation  for  solving  the  linear  equations,  for  black  surroundings
B=mat('559.6;3182.5');          #matrix  representation  for  solving  the  linear  equations.
X=inv(A)*B;

Qn12=(X.item(0)-X.item(1))/(1/(0.9975*F12));            #net  heat  flow  from  1  to  2  for    black  surroundings.
#since  each  strip  loses  heat  to  the  surrounding,Qnet1,  Qnet2  and  Qnet1-2  are  different.
#  three  equations  will  be
#(1451-B1)/2.33  =  (B1-B2)/(1/0.2)+(B1-B3)/(1/0.8)......(1)
#(459.B2)  =  (B2-B1)/(1/0.2)+(B2-B3)/(1/0.8)............(2)
#0=(B3-B1)/(1/0.8)+(B3-B2)/(1/0.8).....................(3)
#solving  these  equations,  we  get    the  values  of  B1,B2  and  B3.
B1=987.7                                 #heat  flux  by  surface  1.
B2=657.4                                 #heat  flux  by  surface  2.
B3=822.6                                 #heat  flux  by  surface  3.
qn12=(B1-B2)/(1/F12)+(B1-B3)/(1/(1-F12));#  net  heat  transfer  between  1  and  2  if  they  are  connected  by  an  insulated  diffuse  reflector  between  the  edges  on  both  sides.

#Results
print "Net heat transfer between 1 and 2 if the surroundings are black is :",round(Qn12,2),"W/m^2\n"
print "Net heat transfer between 1 and 2 if they are connected by an insulated diffuse reflector between the edges on both sides is : ",qn12," W/m^2\n"

x=symbols('x');
x=solve(sigma*(x**4)-822.6,x);          #Solving for Temp. of the reflector.
print "Temperature of the reflector is : ",round(x[1],2)," K\n"

Net heat transfer between 1 and 2 if the surroundings are black is : 46.53 W/m^2

Net heat transfer between 1 and 2 if they are connected by an insulated diffuse reflector between the edges on both sides is :  198.14  W/m^2

Temperature of the reflector is :  347.06  K



### Example 10.10, Page number: 561¶

In [11]:
from __future__ import division
from numpy import array,dot
from numpy.linalg import inv

#Variables
T1=773;                              #temp.of  two  sides  of  duct,K
T2=373;                              #temperature  of  the  third  side,K
e1=0.5;                              #emissivity  of  stainless  steel
e2=0.15;                             #emissivity  of  copper
a=5.67*10**-8;                       #stefan  constant
f12=0.4;                             #view  factor  of  1  occupied  by  2.
f21=0.67;                             #view  factor  of    2  occupied  by  1
f13=0.6;                              #  view  factor  of  1  occupied  by  3
f31=0.75;                             #view  factor  of  3  occupied  by  1
f23=0.33;                             #view  factor  of  2  occupied  by  3
f32=0.25;                             #view  factor  of  2  occupied  by  3

#Calculations
A=array(([1,  (-1+e2)*f12,  (e2-1)*f13],[(-1*e1*f21),  1,  (e1*-1*f23)],[(e1*-1*f31),  (e1*-1*f32),  1]));#matrix  method  to  solve  three  equations  to  find  radiosity
B=array(([e2*a*T2**4],[e1*a*T1**4],[e1*a*T1**4]));      #matrix  method  to  solve  three  equations  to  find  radiosity
X=dot(inv(A),B);                                       #solution  of  above  matrix  method
Qn1=0.5*e2/(1-e2)*(a*T2**4-X.item(0));                #net  heat  transfer  to  the  copper  base  per  meter  of  the  length  of  the  duct,W/m
Qn2=Qn1+2.6;

#Result
print "Net heat transfer to the copper base per meter of length of the duct is : ",round(Qn2,2),"W/m ,the -ve sign indicates that the copper base is gaining heat.\n"

Net heat transfer to the copper base per meter of length of the duct is :  -1294.01 W/m ,the -ve sign indicates that the copper base is gaining heat.



### Example 10.11, Page number: 573¶

In [12]:
from __future__ import division

#Variables
T1=1473  ;                                #temp.of  gas,K
T2=573  ;                                 #temp  of  walls,K
D1=0.4;                                   #diameter  of  combustor,  m
a=5.67*10**-8;                            #stefan  boltzman  coefficient,W/(m**2*K**4)
#we  have  Lo=D1=0.4m,  a  total  pressure  of  1  atm.,  pco2=0.2  atm.  ,  using  figure,  we  get  eg=0.098.
eg=0.098;                                 #total  emittance

#Calculations
ag=(T1/T2)**0.5*(0.074);                  #total  absorptance
#now  we  can  calculate  Qnetgas  to  wall.  for  these  problems  with  one  wall  surrounding  one  gas,  the  use  of  the  mean  beam  length  in  finding  eg  and  ag  accounts  for  all  geometric  effects  and  no  view  factor  is  required.
Qngw=math.pi*D1*a*(eg*T1**4-ag*T2**4)/1000;      #net  heat  radiated  to  the  walls,kW/m

#Result
print "Net heat radiated to the walls is : ",round(Qngw,2),"KW/m\n"
#end

Net heat radiated to the walls is :  31.96 KW/m



### Example 10.12, Page number: 577¶

In [13]:
from __future__ import division
from sympy import solve,symbols

#Variables
T1=291;                              #temp.of  sky,K
T2=308;                              #temp  of  air,K
e1=0.9;                              #emissivity  0f  black  paint
h=8;                                 #heat  transfer  coefficient,W/(m**2*K)
P=600  ;                             #Solar radiation of roof, W/m**2
aacr=0.26;                           #heat  flux,W/m**2
ablk=0.9;                            #heat  flux,W/m**2
sigma=5.67*10**-8;                   #Stefen-Boltzman constant

#Calculations-1
#heat  loss  from  the  roof  to  the  inside  of  the  barn  will  lower  the  roof  temp.,  since  we  dont  have  enough  information  to  evaluate  the  loss,  we  can  make  an  upper  bound  on  roof  temp.  by  assuming  that  no  heat  is  transferred  to  the  interior.
T=symbols('T');
T=solve(((e1*sigma*(T**4-T1**4)+h*(T-T2))-ablk*P),T);
Tn=T[1]

#Result-1
print "For non-sensitive black paint, temp. of roof is:",round(Tn)-273,"degree C \n"

#Calculations-2
#for  white  acrylic  paint,  by  using  table,  e=0.9  and  absorptivity  is  0.26,Troof
T=symbols('T');
T=solve(((e1*sigma*(T**4-T1**4)+h*(T-T2))-0.26*P),T);
Tn=T[1]

#Result-2
print "For acrylic paint temp. of the root is :",round(Tn)-273,"degree C \n\nThe white painted roof is only a few degrees warmer than the air.\n"
#end

For non-sensitive black paint, temp. of roof is: 65.0 degree C

For acrylic paint temp. of the root is : 39.0 degree C

The white painted roof is only a few degrees warmer than the air.