# Chapter 13 : Radiation Heat Transfer¶

### Example 13.1¶

In :
import math
# View Factors Associated with two Concentric Spheres

# Calculations and Results
#The outer surface of the smaller sphere and inner surface of the larger sphere form a two surface enclosure
N = 2;
print "This enclosure involves ",N**2,"View Factors"
x = (1./2)*N*(N-1)
print "W need to determine only",x,"view factor directly"
F11 = 0
F12 = 1
print ("The Two view Factors")
print "Since no radiation leaving surface 1 strikes itself..... F11 = ",F11
print "Since all radiation leaving surface 1 strikes surface 2    F12 = ",F12
print ("F12 =  ((r1/r2)**2)")
print ("F22 =  1-((r1/r2)**2)")
print ("where r1 and r2 are radius of surface 1 and surface 2")

This enclosure involves  4 View Factors
W need to determine only 1.0 view factor directly
The Two view Factors
Since no radiation leaving surface 1 strikes itself..... F11 =  0
Since all radiation leaving surface 1 strikes surface 2    F12 =  1
F12 =  ((r1/r2)**2)
F22 =  1-((r1/r2)**2)
where r1 and r2 are radius of surface 1 and surface 2


### Example 13.2¶

In :
import math
# Fraction of Radiation Leaving through an Opening

# Variables
r1 = 0.1			#Radius of enclosure[m]
L = 0.1 			#Length of Enclosure[m]
r2 = 0.05
r3 = 0.08			#Inner and outer radii of the ring[m]

# Calculations
#Using Chart in Fig 13.7
F12 = 0.11;
F13 = 0.28;
F1_ring = F13-F12;

# Results
print "The fraction of the radiation leaving the base cyllinder enclosure that escapes \
through coaxial ring opening at its top surface is",F1_ring

The fraction of the radiation leaving the base cyllinder enclosure that escapes through coaxial ring opening at its top surface is 0.17


### Example 13.3¶

In :
import math
# View Factors Associated with a Tetragon

# Variables
#A pyramid with square base and it's sides being isoceles triangle
F11 = 0			#Since base is a flat surface

# Calculations
#F12 = F13 = F14 = F15 = x
x = (1-F11)/4.;

# Results
print "Each side pf the four surfaces of the pyramid recieves",x,"of total radiation"

Each side pf the four surfaces of the pyramid recieves 0.25 of total radiation


### Example 13.5¶

In :
import math

# Variables
# The Crossed-Strings Method for View Factors
a = 12.
b = 5.			#With od long parallel plates[cm]
c = 6.			#Distance between the plates
L1 = a
L2 = b
L3 = c;

# Calculations
L4 = math.sqrt((7**2)+(6**2));
L5 = math.sqrt((5**2)+(6**2));
L6 = math.sqrt((12**2)+(6**2));
F12_1 = ((L5+L6)-(L3+L4))/(2*L1);
F13 = (L1+L3-L6)/(2*L1);
F14 = (L1+L4+L5)/(2*L1);
F12_2 = 1-F13-F14;

# Results
print "F13 = ",F13,"F14 = ",F14,"Therefore from two different methods F12_1 = F12_2 = ",F12_1,

F13 =  0.190983005625 F14 =  1.20957475555 Therefore from two different methods F12_1 = F12_2 =  0.250296378484


### Example 13.6¶

In :
import math
# Radiation Heat Transfer in a Black Furnace

# Variables
F12 = 0.2;
A = 5.*5			#Area of 1 surface of cube[m**2]
Tb = 800
Tt = 1500
Ts = 500			#Temperature of base top and the side surfaces of the furbace[K]

# Calculations and Results
F11 = 0;
Q11 = 0;
F13 = 1-F11-F12;
Q13 = A*F13*(5.67*10**(-8))*((Tb**4)-(Ts**4))			#[kW]
print "The net rate of heat transfer from surface1 to surface3 is",round(Q13/1000),"kW"

Q12 = A*F12*(5.67*10**(-8))*((Tb**4)-(Tt**4))			#[kW]
print "The net rate of radiation heat transfer from siurface1 to surface2 is",round(Q12/1000),"kW"

Q1 = Q11+Q12+Q13			#[kW]
print "Rhe net radiation heat transfer from the base surface is",round(Q1/1000),"kW"

The net rate of heat transfer from surface1 to surface3 is 394.0 kW
The net rate of radiation heat transfer from siurface1 to surface2 is -1319.0 kW
Rhe net radiation heat transfer from the base surface is -925.0 kW


### Example 13.7¶

In :
import math
# Radiation Heat Transfer between Parallel Plates

# Variables
T1 = 800
T2 = 500			#Temp of parallel plates[K]
e1 = 0.2
e2 = 0.7			#Emissivities

# Calculations
q12 = (5.67*10**(-8))*((T1**4)-(T2**4))/((1/e1)+(1/e2)-1);

# Results
print "The net heat at the rate of" \
,round(q12),"W","is transferred from plate 1 to plate 2 by radiation per unit surface area of either plate"

The net heat at the rate of 3625.0 W is transferred from plate 1 to plate 2 by radiation per unit surface area of either plate


### Example 13.8¶

In :
import math
# Radiation Heat Transfer in Cylindrical Furnace

# Variablesa
ro = 1
H = 1			#Radius amd height of cylinder[m]
e1 = 0.8
e2 = 0.4			#Emissivities
T1 = 700
T2 = 500			#Top and base temperatures of furnace[K]
T3 = 400			#Side durface temperature[K]
F11 = 0
F12 = 0.38;

# Calculations and Results
A1 = math.pi*(ro**2)			#[m**2]
A2 = A1			#[m**2]
A3 = 2*math.pi*ro*H			#[m**2]
F13 = 1-F11-F12;
F21 = F12			#Top and Bottom are symmetric
F31 = F13*(A1/A3);
F23 = F13;
F32 = F31;
def rad(J):
i = [0,0,0]
i = J+(((1-e1)/e1)*((F12*(J-(J)))+(F13*((J)-(J)))))-((T1**4)*(5.67*10**(-8)));
i = J+(((1-e2)/e2)*((F21*(J-J))+(F23*(J-J))))-((T2**4)*(5.67*10**(-8)));
i = J-((T3**4)*(5.67*10**(-8)));
#print (J,J,J)
#Q1 = A1*((F12*(J-J))+(F13*(J-J)))			#[kW]
#Q2 = A2*((F21*(J-J))+(F13*(J-J)))			#[kW]
#Q3 = A3*((F31*(J-J))+(F32*(J-J)))			#[kW]
#print "The net rates of radiation heat transfer at the three surfaces are", \
#Q1/1000,Q2/1000,Q3/1000,"kW"



### Example 13.9¶

In :
import math
# Radiation Heat Transfer in a Triangular Furnace

# Variables
A1 = 1.
A2 = 1.
A3 = 1.			#Area of each side[m**2]
T1 = 600.
T2 = 1000.			#[K]
e = 0.7;
F12 = 0.5
F13 = 0.5
F23 = 0.5			#Symmetry

# Calculations
Eb1 = 5.67*10**(-8)*(T1**4)			#[W/m**2]
Eb2 = 5.67*10**(-8)*(T2**4)			#[W/m**2]
Q = (Eb2-Eb1)/(((1-e)/(A1*e))+((((A1*F12)+(1/((1/(A1*F13))+(1/(A2*F23))))))**(-1)))			#[kW]

# Results
print "Heat at the rate of",round(Q/1000),"kW"
print "must be supplied to the heated surface per unit lemgth of the duct \
to maintain steady operation in the furnace"

Heat at the rate of 28.0 kW
must be supplied to the heated surface per unit lemgth of the duct  to maintain steady operation in the furnace


### Example 13.10¶

In :
import math
# Heat Transfer through a Tubular Solar Collector

# Variables
k = 0.02588	    		#[W/m.degree Celcius]
Pr1 = 0.7282
Pr2 = 0.7255			#Prandtl no
nu1 = 1.608*(10**(-5))
nu2 = 1.702*10**(-5)			#[m**2/s]
T1 = 20
T2 = 40     			#[degree Celcius]
Tavg = ((T1+T2)/2)+273			#[K]
Do = 0.1
L = 1			    #Dimensions of glass tube[m]
Di = 0.05			#Inner diameter of tube[m]
Q_glass = 30		#Rate of heat transfer from the outer surface of the glass cover[W]
g = 9.81			#[m**2/s]
eo = 0.9
ei = 0.95			#Emissivity

# Calculations and Results
Ao = math.pi*Do*L			#Heat transfer surface area of the glass cover[m**2]
print (Ao,Tavg)

Ra_Do = g*Tavg*(T2-T1)*(Do**3)*Pr1/(nu1);
print "The Rayleigh number is",Ra_Do

Nu = ((0.6+((0.387*(Ra_Do**(1./6)))/((1+((0.559/Pr1)**(9./16)))**(8./27))))**2);
print "The nusselt number is",Nu

ho = k*Nu/Do			#[W/m**2.degree Celcius]
Qo_conv = ho*Ao*(T2-T1)			#[W]
Qo_rad = eo*5.67*10**(-8)*Ao*(((T2+273)**4)-((T1+273)**4))			#[W]
Qo_total = Qo_conv+Qo_rad			#[W]
print "The total rate of heat loss from the glass cover ",Qo_total,"W"

Lc = (Do-Di)/2			#The characteristic length
Ai = math.pi*Di*L			#[m**2]
#Assuming
T_tube = 54
T_cover = 26			#Temperature of tube and glass cover[degree Celcius]
T_avg = ((T_tube+T_cover)/2)+273			#[K]
Ra_L = g*T_avg*(T_tube-T_cover)*(Lc**3)*Pr2/(nu2);
print "The Rayleigh number in this case is",Ra_L

F_cyl = ((math.log(Do/Di))**4)/((Lc**3)*(((Di**(-3./5))+(Do**(-3./5)))**5));
k_eff = 0.386*k*((Pr2/(0.861+Pr2))**(1./4))*((F_cyl*Ra_L)**(1./4));
print "The effective thermal conductivity is",k_eff,"W/m.degree Celcius"

QL_conv = 2*math.pi*k_eff*(T_tube-T_cover)/(math.log(Do/Di));
print "The rate of heat transfer between the cylinders by convection is",QL_conv,"W"

QL_rad = ((5.67*10**(-8))*Ai*(((T_tube+273)**4)-((T_cover+273)**4)))/((1/ei)+(((1-eo)/eo)*(Di/Do)));
print "The radiation rate of heat transfer is",QL_rad,"W"

QL_total = QL_conv+QL_rad			#[W]
print "The total rate of heat loss from the glass cover is",QL_total,"W"

(0.3141592653589793, 303)
The Rayleigh number is 2692193.4403
The nusselt number is 19.3316907216
The total rate of heat loss from the glass cover  67.1513072128 W
The Rayleigh number in this case is 57262.3626799
The effective thermal conductivity is 0.0786342130794 W/m.degree Celcius
The rate of heat transfer between the cylinders by convection is 19.9583200961 W
The radiation rate of heat transfer is 27.6572415779 W
The total rate of heat loss from the glass cover is 47.6155616741 W


### Example 13.11¶

In :
import math
# Radiation Shields

# Variables
e = 0.1			#Emissivity of aluminium sheet
T1 = 800
T2 = 500			#Temperatures of two parallel plates[K]
e1 = 0.2
e2 = 0.7			#Emissivities of plates

# Calculations
q12 = ((5.67*10**(-8))*((T1**4)-(T2**4)))/((1/e1)+(1/e2)-1+(1/e)+(1/e)-1)			#[W/m**2]

# Results
print "Radiation Heat Transfer",round(q12),"W/m**2"

Radiation Heat Transfer 806.0 W/m**2


### Example 13.12¶

In :
import math
# Radiation Effect on Temperature Measurements

# Variables
Tw = 400
Tth = 650			#Temperature of duct wall and hota air flowing in it[K]
e = 0.6			#emissivity
h = 80			#Heat transfer coefficient[W/m**2.K]

# Calculations
Tf = Tth+((e*5.67*10**(-8)*((Tth**4)-(Tw**4)))/h)			#[K]

# Results
print "The temperature of actual air is",round(Tf),"K"

The temperature of actual air is 715.0 K


### Example 13.13¶

In :
import math
# Effective Emissivity of Combustion Gases
d = 5
H = 5   			#Diameter and height of cylindrical furnace[m]
T = 1200			#Temp of gases[K]
P = 2	    		#Pressure[atm]
yN2 = 0.8
yH2O = 0.08
yO2 = 0.07
yCO2 = 0.05			#Volumetric Composition

# Calculations and Results
Pc = yCO2*P			#[atm]
Pw = yH2O*P			#[atm]
print "The partial pressures of CO2 and H2O are",Pc,"atm","and",Pw,"atm"

L = 0.6*d			#[m]
x = Pc*L
y = Pw*L			#[m.atm]
ec_1 = 0.16
ew_1 = 0.23			#Emissivity of CO2 and H2O at 1 atm pressure
Cc = 1.1
Cw = 1.4			#Pressure Correction Factors are
del_e = 0.048			#Emissivity correction factor at T = 1200K
e_g = Cc*ec_1+Cw*ew_1-del_e;
print "The effectivity of the combustion gases is",e_g

The partial pressures of CO2 and H2O are 0.1 atm and 0.16 atm
The effectivity of the combustion gases is 0.45


### Example 13.14¶

In :
import math
# Radiation Heat Transfer in a Cylindrical Furnace

# Variables
Ts = 600    			#Wall Temperature[K]
d = 5
H = 5	    		#Diameter and Height of cylindrical furnace
Tg = 1200
eg = 0.45			#Average gas temperature and average emissivity of the combustion gases
Pc = 0.10
L = 3
Pw = 0.16			#From Previous s

# Calculations and Results
x = Pc*L*Ts/Tg			#[m.atm]
y = Pw*L*Ts/Tg			#[m.atm]
ec_1 = 0.11
ew_1 = 0.25			#Emissivities of CO2 and H2O corresponding to 600K and 1atm
Cc = 1.1
Cw = 1.4			#Correction Factors
a_c = Cc*((Tg/Ts)**(0.65))*(ec_1);
a_w = Cw*((Tg/Ts)**(0.45))*ew_1;
print "The absorptivities of CO2 and H2O are",a_c,"and",a_w

del_a = 0.027;
a_g = a_c+a_w-del_a;
print "The absorptivity of the combustion gases is",a_g

As = (math.pi*d*H)+(math.pi*(d**2)/2)			#[m**2]
print "the surface area of the cylindrical surface is",round(As),"m**2"

Q_net = round(As)*(5.67*10**(-8))*((eg*(Tg**4))-(a_g*(Ts**4)));
print "The net rate of radiation heat transfer from the combustion gases to walls of the furnace is", \
Q_net,"W"

The absorptivities of CO2 and H2O are 0.189869351691 and 0.478114089864
The absorptivity of the combustion gases is 0.640983441555
the surface area of the cylindrical surface is 118.0 m**2
The net rate of radiation heat transfer from the combustion gases to walls of the furnace is 5687334.8017 W


### Example 13.15¶

In :
import math
# Effect of Clothing on Thermal Comfort

# Variables
h_rad = 4.7
h_conv = 4.0			#The radiation and convection heat transfer coefficient[W/m**2.degree Celcius]
R_clo = 0.6*0.155			#Thermal resismath.tance of clothing[m**2.degree Celcius/W]
T_skin = 33
T_amb = 22			#Skin and Ambient temperature[degree Celcius]
As = 1.8			#Surface area of an average man

# Calculations and Results
h_comb = h_conv+h_rad			#combined heat transfer coefficient[W/m**2.degree Celcius]
Q_sen_clo = As*(T_skin-T_amb)/(R_clo+(1/h_comb))			#[W]
print "The sensible heat loss from this person when clothed is",Q_sen_clo,"W"

#On removing the clothes
#R_clo = 0 Clothing resismath.tance on removing clothes
#Setting both heat transfer rates equal to determine new ambient air temperature
T_amb_new = T_skin-(Q_sen_clo*(1/h_comb)/As)    #[degree Celcius]
print "The ambient temperature now is",T_amb_new,"degree Celcius"

The sensible heat loss from this person when clothed is 95.2186169919 W
The ambient temperature now is 26.9196285446 degree Celcius