# Chapter 10 : Radiative Heat exchange between surfaces¶

## Example 10.1 Page No : 403¶

In [1]:
import math

# Variables
I = 1350;			#Solar Irradiation in W/m**2
L = (1.5*10**8);			#Approximate dismath.tance in km
D = (1.39*10**6);			#Approximate diameter in km

# Calculations
E = (I*(L*1000)**2*3.14)/((3.14/4)*(D*1000)**2);			#Emissive power of Earth
Ts = (E/(5.67*10**-8))**0.25;			#Surface temperature of sun in K

# Results
print 'Surface temperature of sun is %d K'%(Ts)

Surface temperature of sun is 5770 K


## Example 10.4 Page No : 409¶

In [3]:
# Variables
S = 1;			#Side of a square in m
L = 0.4;			#Distance between the plates in m
T1 = 900;			#Temperature of one plate in degree C
T2 = 400;			#Temperature of the other plate in degree C

# Calculations
R = (S/L);			#Ratio of the side of the square to the distance between plates
F12 = 0.415;			#From Fig.10.4 on page no.409
Q = (5.67*10**-8*S*S*F12*((T1+273)**4-(T2+273)**4))/1000;			#The net heat transfer in kW

# Results
print 'The net exchange of energy due to radiation between the plates is %3.1f kW'%(Q)

The net exchange of energy due to radiation between the plates is 39.7 kW


## Example 10.5 Page No : 411¶

In [4]:
# Variables
A51 = 2;			#Ratio of areas A5 and A1
A21 = 1;			#Ratio of areas A2 and A1
F56 = 0.15;			#Shape factor
F53 = 0.11;			#Shape factor
F26 = 0.24;			#Shape Factor
F23 = 0.2;			#Shape Factor

# Calculations
F14 = (A51*(F56-F53))-(A21*(F26-F23));			#Shape factor

# Results
print 'Shape factor F14 is %3.2f'%(F14)

Shape factor F14 is 0.04


## Example 10.8 Page No : 415¶

In [4]:
# Variables
Th = 40.;			#Radiating heating panel in degree C
Tb = 5.;			#Temperature of black plane in degree C
Tc = 31.;			#Temperature of ceiling in degree C
A = (10.*12);		#Area in m**2

# Calculations
F56 = 0.075;			#Using Fig.10.2 on page no. 408
F63 = 0.04;	    		#Using Fig.10.2 on page no. 408
F12 = 0.052;			#Shape factor
F1w = (1-F12);			#Shape factor between the floor and all the walls but the window
Q12 = (A*F12*5.67*10**-8*((Th+273)**4-(Tb+273)**4));			#Heat exchange between the floor and window in W
#Q1 = (5.67*10**-8*A*((Th+273.15)**4-((F12*(Th+273.15)**4)-(F1w*(Tb+273.15)**4))))/1000;			#Net heat given up by the floor in kW
Q1 = (5.67*10**-8*A*((Th+273)**4-((F12*(Th+273)**4)-(F1w*(294)**4))))/1000;			#Net heat given up by the floor in kW

# Results
print 'Heat exchange between the floor and window is %3.0f W \n \
Net heat given up by the floor is %3.1f kW'%(Q12,Q1)

# Note : rounding off error.

Heat exchange between the floor and window is 1283 W
Net heat given up by the floor is 110.1 kW


## Example 10.14 Page No : 424¶

In [5]:
# Variables
A2 = (6.*2);			#Area of windows in m**2
A1 = (10.*12);			#Area of floor in m**2
Th = 40.;			#Radiating heating panel in degree C
Tb = 5.;			#Temperature of black plane in degree C
F12 = 0.052;			#Shape factor

# Calculations
F12a = ((A2-(A1*F12**2))/(A1+A2-(2*A1*F12)));			#Shape factor
Q12 = (A1*F12a*5.67*10**-8*((Th+273)**4-(Tb+273)**4));			#Net heat exchange in W
X = (((A2/A1)-F12)/(1-F12));			#X value for equilibrium temperature
T = (((Th+273)**4+(X*(Tb+273)**4))/(X+1))**0.25;			#Equilibrium temperature in K

# Results
print 'Net heat exchange is %3.0f W Equilibrium temperature is %3.2f K'%(Q12,T)

Net heat exchange is 2409 W Equilibrium temperature is 311.57 K


## Example 10.15 Page No : 430¶

In [8]:
# Variables
D = 0.2;			            #Diameter of each disc in m
L = 2;			                #Distance between the plates in m
T = 800+273,300+273;			#Temperatures of the plates in K
e = [0.3,0.5]       ;			#Emissivities of plates

# Calculations
e1 = (e[0]*e[1]);			#Equivalent emissivity
R = (D/L);			#Ratio between diameter and distance between the plates
F = 0.014;			#F value from Fig.10.4 from page no. 409
Q = (e1*(3.14/4)*D**2*F*5.67*10**-8*((T[0]**4-(T[1]**4))));			#Radiant heat exchange for the plates in W

# Results
print 'Radiant heat exchange for the plates is %3.2f W'%(Q)

Radiant heat exchange for the plates is 4.55 W


## Example 10.16 Page No : 430¶

In [9]:
# Variables
e = 0.8;			#Emissivity of brick wall
D1 = [6,4];			#Width and Height in m
L = 0.04;			#Distance from the wall in m
D2 = [0.2,0.2];			#Dimensions of the furnace wall in m
D3 = [1,1];			#Dimensions at lower and left of the centre of the wall in m
T = [1523+273,37+273];			#Furnace temperature and wall temperature in degree C

# Calculations
F12 = 0.033;			#Shape factor from Fig.10.3 on page no. 409
F13 = 0.05;			#Shape factor from Fig.10.3 on page no. 409
F14 = 0.12;			#Shape factor from Fig.10.3 on page no. 409
F15 = 0.08;			#Shape factor from Fig.10.3 on page no. 409
Fow = (F12+F13+F14+F15);			#Shape factor between opening and wall
Q = (e*L*Fow*5.67*10**-8*(T[0]**4-T[1]**4))/1000;			#Net radiation exchange in kW

# Results
print 'Net radiation exchange between the opening and the wall is %3.1f kW'%(Q)

Net radiation exchange between the opening and the wall is 5.3 kW


## Example 10.17 Page No : 431¶

In [10]:
# Variables
D = [2,1,1];			#Dimensions of the math.tank in m
A = 8;			#Area of the tank in m**2
e = 0.9;			#Surface emissivity
Ts = 25+273;			#Surface temperature in K
Ta = 2+273;			#Ambient temperature in K
e1 = 0.5;			#Emissivity of aluminium

# Calculations
Q = (e*A*5.67*10**-8*(Ts**4-Ta**4))/1000;			#Heat lost by radiation in kW
r = ((e-e1)/e)*Q;			#Reduction in heat loss if the tank is coated with an aluminium paint in kW

# Results
print 'Heat lost by radiation is %3.2f kW \n \
Reduction in heat loss if the tank is coated with an aluminium paint is %3.3f kW'%(Q,r)

Heat lost by radiation is 0.88 kW
Reduction in heat loss if the tank is coated with an aluminium paint is 0.393 kW


## Example 10.18 Page No : 432¶

In [5]:
# Variables
D = 0.2;			#Outer diameter of the pipe in m
Ta = 30+273;			#Temperature of the air in K
Ts = 400+273;			#Surface temperature in K
e = 0.8;			#Emissivity of the pipe surface
D1 = 0.4;			#Diamter of brick in m
e1 = 0.91;			#Emissivity of brick

# Calculations
Q = (e*3.14*D*5.67*10**-8*(Ts**4-Ta**4))/1000;			#Loss of heat by thermal radiation in kW/m
e2 = (1./((1./e)+((D/D1)*((1./e1)-1))));			#Equivalent emissivity
Q1 = (e2*3.14*D*5.67*10**-8*(Ts**4-Ta**4))/1000;			#Heat loss when brick is used in kW/m
r = (round(Q,2)-round(Q1,2))*1000;			#Reduction in heat loss in W/m

# Results
print 'Loss of heat by thermal radiation is %3.1f*10**3 W/m \n \
Reduction in heat loss is %3.0f W/m'%(Q,r)

Loss of heat by thermal radiation is 5.6*10**3 W/m
Reduction in heat loss is 210 W/m


## Example 10.19 Page No : 433¶

In [16]:
# Variables
e = 0.03;			#Emissivity of silver
T2 = -153.+273;			#Temperature of the outer surface of the inner wall in K
T1 = 27.+273;			#Temperature of the inner surface of the outer wall in K
D1 = 0.42;			#Diamter of first sphere in m
D2 = 0.6;			#Diamter of the second sphere in m
V = 220.;			#Rate of vapourization in kJ/kg

# Calculations
e1 = (1./((1./e)+((D1/D2)**2*((1./e)-1))));			#Equivalent emissivity
A = (4*3.14*(D1/2)**2);			#Area in m**2
Q = (e1*A*5.67*10**-8*(T1**4-T2**4))/(1000./3600);			#Radiation heat transfer through walls into the vessel in kJ/h
R = (Q/V);			#Rate of evaporation in kg/h

# Results
print 'Radiation heat transfer through walls into the vessel is %3.3f kJ/h \n \
Rate of evaporation of liqiud oxygen is %3.4f kg/h'%(Q,R)

Radiation heat transfer through walls into the vessel is 18.146 kJ/h
Rate of evaporation of liqiud oxygen is 0.0825 kg/h


## Example 10.20 Page No : 433¶

In [17]:
# Variables
T = [800+273,300+273];			#Temperatures of the plates in K
e = [0.3,0.5];			#Emissivities of the plates

# Calculations
Q = ((5.67*10**-8*(T[0]**4-T[1]**4))/((1./e[0])+((1./e[1]))-1))/1000;			#Net radiant heat exchange in kW/m**2

# Results
print 'Net radiant heat exchange is %3.2f kW/m**2'%(Q)

Net radiant heat exchange is 15.93 kW/m**2


## Example 10.21 Page No : 434¶

In [18]:
# Variables
T1 = 127+273;			#Temperature of the outer side of the brick setting in K
T2 = 50+273;			#Temperature of the inside of the steel plate in K
e1 = 0.6;			#Emissivity of steel
e2 = 0.8;			#Emissivity of fireclay

# Calculations
Q = ((5.67*10**-8*(T1**4-T2**4))/((1./e1)+((1./e2))-1));			#Net radiant heat exchange in W/m**2

# Results
print 'Net radiant heat exchange is %3.0f W/m**2'%(Q)

Net radiant heat exchange is 435 W/m**2


## Example 10.22 Page No : 445¶

In [19]:
# Variables
D = 1;			#Dimension of the plate in m
L = 0.5;			#Distance between the plates in m
Ts = 27+273;			#Surface temperature of the walls in K
T = [900+273,400+273];			#Temperature of the plates in K
e = [0.2,0.5];			#Emissivities of the plates

# Calculations
F12 = 0.415;			#From Fig.10.4 on page no.409
F13 = (1-F12);			#Shape factor
F23 = (1-F12);			#Shape factor
R1 = (1-e[0])/(e[0]*D*D);			#Resistance for 1
R2 = (1-e[1])/(e[1]*D*D);			#Resistance for 2
R3 = 0;			                    #Resistance for 3
A1F12I = (1./(D*D*F12));			#Inverse of the product of area and Shape factor
A1F13I = (1./(D*D*F13));			#Inverse of the product of area and Shape factor
A2F23I = (1./(D*D*F23));			#Inverse of the product of area and Shape factor
Eb1 = (5.67*10**-8*T[0]**4)/1000;			#Emissive power of 1 in kW/m**2
Eb2 = (5.67*10**-8*T[1]**4)/1000;			#Emissive power of 2 in kW/m**2
Eb3 = (5.67*10**-8*Ts**4);			#Emissive power of 3 in W/m**2
J1 = 25;			#Radiosity at node 1 in kW/m**2
J2 = 11.53;			#Radiosity at node 2 in kW/m**2
J3 = 0.46;			#Radiosity at node 3 in kW/m**2
Q1 = ((Eb1-J1)/R1);			#Total heat loss by plate 1 in kW
Q2 = ((Eb2-J2)/R2);			#Total heat loss by plate 2 in kW
Q3 = ((J1-J3)/(A1F13I))+((J2-J3)/(A2F23I));			#Total heat received by the room in kW

# Results
print 'Total heat loss by plate 1 is %3.1f kW Total heat loss by plate 2 is %3.1f kW \n \
Total heat received by the room is %3.2f kW'%(Q1,Q2,Q3)

Total heat loss by plate 1 is 20.6 kW Total heat loss by plate 2 is 0.1 kW
Total heat received by the room is 20.83 kW


## Example 10.23 Page No : 447¶

In [20]:
# Variables
T = [800+273,300+273];			#Temperatures of the plates in K
e = [0.3,0.5];			#Emissivities of the plates
e3 = 0.05;			#Emissivity of aluminium

# Calculations
q = ((5.67*10**-8*(T[0]**4-T[1]**4))/((1./e[0])+(1./e[1])-1))/1000;			#Heat transfer without the shield in kW/m**2
R1 = (1-e[0])/e[0];			#Resistance in 1
R2 = (1-e[1])/e[1];			#Resistance in 2
R3 = (1-e3)/e3;			#Resistance in 3
R = (R1+(2*R2)+(2*R3));			#Total resismath.tance
q1 = ((5.67*10**-8*(T[0]**4-T[1]**4))/R)/1000;			#Heat transfer with shield in kW/m**2
r = ((q-q1)*100)/q;			#Reduction in heat transfer
X1 = ((1./e3)+(1./e[1])-1);			#X1 for tempearture T3
X2 = ((1./e[0])+(1./e3)-1);			#X1 for tempearture T3
T3 = (((X1*T[0]**4)+(X2*T[1]**4))/(X2+X1))**0.25;			#Temperature of the sheild in K
T3c = T3-273;			#Temperature of the sheild in degree C

# Results
print 'Percentage reduction in heat transfer is %3.0f percent \n \
Temperature of the sheild is %3.2f degree C'%(r,T3c)

Percentage reduction in heat transfer is  90 percent
Temperature of the sheild is 641.02 degree C


## Example 10.24 Page No : 448¶

In [21]:
# Variables
Q = 79;			#Reduction in net radiation from the surfaces
e1 = 0.05;			#Emissivity of the screen
e2 = 0.8;			#Emissivity of the surface

# Calculations
n = (((Q*((2/e2)-1))-((2/e2)+1))/((2/e1)-1));			#Number of screens to be placed between the two surfaces to achieve the reduction in heat exchange

# Results
print 'Number of screens to be placed between the two surfaces to achieve the reduction in heat exchange is%3.0f'%(n)

Number of screens to be placed between the two surfaces to achieve the reduction in heat exchange is  3


## Example 10.25 Page No : 449¶

In [22]:
# Variables
e = 0.8;			#Emissivity of the pipe
D = 0.275;			#Diameter of the pipe in m
Ts = 500+273;			#Surface temperature in K
Te = 30+273;			#Temperature of enclosure in K
D1 = 0.325;			#Diamter of the steel screen in m
e1 = 0.7;			#Emissivity of steel screen
Tsc = 240+273;			#Temperature of screen in K

#CALCUATIONS
Q = (e*5.67*10**-8*3.14*D*(Ts**4-Te**4))/1000;			#Loss of heat per unit length by radiation in kW/m
e2 = (1./((1./e)+((D/D1)*((1./e1)-1))));			#Equivalent emissivity
Q1 = (e2*5.67*10**-8*3.14*D*(Ts**4-Tsc**4))/1000;			#Radiant heat exchange per unit length of header with screen in kW/m
R = (Q-Q1);			#Reduction in heat by radiation due to the provision of the screen in kW/m

# Results
print 'Loss of heat per unit length by radiation is %3.1f kW/m \n \
Reduction in heat by radiation due to the provision of the screen is %3.2f kW/m'%(Q,R)

Loss of heat per unit length by radiation is 13.7 kW/m
Reduction in heat by radiation due to the provision of the screen is 4.92 kW/m


## Example 10.26 Page No : 451¶

In [23]:
# Variables
e = 0.6;		    	#Emissivity of thermocouple
Ta = 20+273;			#Ambient temperature in K
Tt = 500+273;			#Temperature from the thermocouple in K
e = 0.3;	    		#Emissivity of radiation shield
h = 200;		    	#Convective heat transfer coefficient in W/m**2.K
Ts = 833;			    #Temperature in K

# Calculations
T = ((5.67*10**-8*e*(Tt**4-Ta**4))/(h*1000))+Tt;			#Temperature of the shield in K
T1 = (Ts-T);			#Error between the thermocouple temperature and gas temperature in K
Ts = 825.;  			#Surface temperature with radiation shield in K
Tc = 829.;	    		#Thermocouple temperature with radiation shield in K
e = (Tc-Ts);			#Error between the thermocouple temperature and gas temperature with the shielded thermocouple arrangement in K

# Results
print 'Error between the thermocouple temperature and gas temperature is%3.0f K \n \
Error between the thermocouple temperature and gas temperature with the shielded thermocouple arrangement is%3.0f K'%(T1,e)

Error between the thermocouple temperature and gas temperature is 60 K
Error between the thermocouple temperature and gas temperature with the shielded thermocouple arrangement is  4 K


## Example 10.27 Page No : 452¶

In [24]:
# Variables
D = 0.2;			#Diameter of pipe in m
Ta = 30+273;			#Temperature of air in K
Ts = 200+273;			#Temperature of surface in K
e = 0.8;			#Emissivity of the pipe

# Calculations
Q = (e*5.67*10**-8*3.14*D*(Ts**4-Ta**4));			#Heat lost by thermal radiation in W/m
T = (Ta+Ts)/2;			#Film temperature in degree C
k = 0.03306;			#Thermal conductivity in W/m.K
v1 = (24.93*10**-6);			#Kinematic viscosity in m**2/s
b = (1./388);			#Coefficient of thermal expansion in 1./K
Pr = 0.687;			#Prantl number
Ra = ((9.81*b*D**3*(Ts-Ta)*Pr)/(v1**2));			#Rayleigh number
Nu = (0.53*(Ra)**0.25);			#Nussults number
h = (k*Nu)/D;			#Heat transfer coefficient in W/m**2.K
Q1 = (h*3.14*D*(Ts-Ta));			#Heat lost by convection in W/m
Q2 = (Q+Q1);			#Total heat lost per meter length in W/m

# Results
print 'Heat lost by thermal radiation is %3.0f W/m \n \
Heat lost by convection is %3.1f W/m'%(Q,Q1)

Heat lost by thermal radiation is 1186 W/m
Heat lost by convection is 734.4 W/m


## Example 10.28 Page No : 453¶

In [25]:
# Variables
Ts = 200+273;			#Temperature of stream main in K
Ta = 30+273;			#Rooom temperature in K
h = 17.98;			#Heat transfer coefficient in W/m**2.K
e = 0.8;			#Emissivity of the pipe surface

# Calculations
q = (5.67*10**-8*e*(Ts**4-Ta**4));			#Heat transfer by radiation in W/m**2
hr = (q/(Ts-Ta));			                #Heat transfer coefficient due to radiation in W/m**2.K
hc = (h-hr);			                    #Heat transfer coefficient due to convection in W/m**2.K

# Results
print 'Heat transfer coefficient due to radiation is %3.1f W/m**2.K \n \
Heat transfer coefficient due to convection is %3.2f W/m**2.K'%(hr,hc)

Heat transfer coefficient due to radiation is 11.1 W/m**2.K
Heat transfer coefficient due to convection is 6.87 W/m**2.K


## Example 10.29 Page No : 461¶

In [26]:
import math
# Variables
t = 0.05;			#Thickness of the gas layer in m
r = 0.1;			#Remaining radiation  intensity

# Calculations
a = (-1./t)*2.3*(math.log(r)/math.log(10));			#Extinction coefficient per m

# Results
print 'Extinction coefficient is %3.2f/m'%(a)

Extinction coefficient is 46.00/m


## Example 10.30 Page No : 462¶

In [27]:
# Variables
A = 30.;			#Total surface area in m**2
V = 10.;			#Volume in m**3
Ts = 1000.;			#Temperature of the furnace in degree C
p = 2.; 			#Total pressure in atm
ph2o = 0.1;			#Partial pressure of water vapour in atm
pco2 = 0.3;			#Partial pressure of CO2

# Calculations
lms = (3.6*V)/A;			#Mean beam length in m
pco2lms = (pco2*lms);			#pco2lms in m.atm
eco2 = 0.16;			#From Fig.10.23 on page no. 458
cco2 = 1.11;			#From Fig.10.23 on page no. 458
cco2eco2 = (cco2*eco2);			#cco2eco2 value
ph2olms = (ph2o*lms);			#ph2olms in m.atm
eh2o = 0.12;			#From Fig.10.24 on page no. 459
P = (p+ph2o)/2;			#P value in atm
ch2o = 1.43;			#From Fig.10.26 on page no. 460
ch2oeh2o = (ch2o*eh2o);			#ch2oeh2o value
P1 = (ph2o/(ph2o+pco2));			#Ratio of pressures
X = (pco2lms+ph2olms);			#X value in m.atm
e = 0.035;			#Error value from Fig. 10.27 on page no.461
et = (cco2eco2+ch2oeh2o-e);			#Total emissivity of the gaseous mixture

# Results
print 'Emissivity of the gaseous mixture is %3.4f'%(et)

Emissivity of the gaseous mixture is 0.3142


## Example 10.31 Page No : 463¶

In [6]:
# Variables
Tg = 950+273;			#Flue gas temperature in K
p = 1;			#Total pressure in atm
pco2 = 0.1;			#Percent of co2
ph2o = 0.04;			#Percent of h2o
D = 0.044;			#Diameter of the tube in m
e = 0.8;			#Emissivity of grey surface
Tw = 500+273.;			#Uniform temperature in K

# Calculations
lms = (3*0.044);			#lms value from Table 10.2 on page no. 457
pco2lms = (pco2*lms);			#pco2lms in m.atm
ph2olms = (ph2o*lms);			#ph2olms in m.atm
eco2 = 0.05;			#From Fig.10.23 on page no. 458
eh2o = 0.005;			#From Fig.10.24 on page no. 459
b = 1.05;			#Correction factor from Fig. 10.28 on page no. 461
eg = 0.061;			#Total emissivity of gaseous mixture
ag = ((0.056*(Tg/Tw)**0.65)+(b*0.021));			#Absorbtivity of the gases
q = (0.5*(e+1)*5.67*10**-8*((eg*Tg**4)-(ag*Tw**4)));			#Heat transfer rate by radiation in W/m**2
hr = (q/(Tg-Tw));			#Radiation heat transfer coefficient in W/m**2.degree C

# Results
print 'Net radiation exchange between the gas and the tube walls is %3.0f W/m**2 \n \
Radiation heat transfer coefficient is %3.2f W/m**2.degree C'%(q,hr)

Net radiation exchange between the gas and the tube walls is 5187 W/m**2
Radiation heat transfer coefficient is 11.53 W/m**2.degree C