# Chapter 12 : Fundamentals of Thermal Radiation¶

### Example 12.1¶

In :
import math
# Radiation Emission from a Black Ball

# Variables
T = 800     		    	#Temperature of suspended ball[K]
D = 0.2		            	#Diameter[m]
C1 = 3.74177*10**8			#[(micrometer**4)/m**2]
C2 = 1.43878*10**4			#[micrometer.K]
lambda_ = 3		        	#[micrometer]

# Calculations and Results
Eb = (5.67*10**(-8))*(T**4)			#[W/m**2]
print "The ball emits",Eb/1000,"kJ","of energy in the form of energy in the form of electromagnetic radiation per second per m**2"

As = math.pi*(D**2)			#[m**2]
print "The total Surface area of the ball is",As,"m**2"

del_t = 5*60.    			#[seconds]
Q_rad = Eb*As*del_t			#[J]
print "The total amount of radiation energy emitted from the entire ball is",Q_rad/1000,"kJ"

#Solution (c)
Eb_lambda = C1/((lambda_**5)*((math.exp(C2/(lambda_*T)))-1))			#[W/m**2.micrometer]
print "The spectral blackbody emissive power",round(Eb_lambda),"W/m**2.micrometer"

The ball emits 23.22432 kJ of energy in the form of energy in the form of electromagnetic radiation per second per m**2
The total Surface area of the ball is 0.125663706144 m**2
The total amount of radiation energy emitted from the entire ball is 875.536237159 kJ
The spectral blackbody emissive power 3846.0 W/m**2.micrometer


### Example 12.2¶

In :
import math
# Emission of Radiation from a Lightbulb

# Variables
T = 2500    			#Temp of the filament[K]
lambda1 = 0.4
lambda2 = 0.76			#Visible ranfe[micrometer]
f1 = 0.000321
f2 = 0.053035			#The black body radiation functions corresponding to lamda1*T and lambda2*T

# Calculations and Results
f3 = f2-f1;
print "Fraction of radiation emitted between the two given wavelengths is",f3
lambda_max = 2897.8/T			#[micrometer]
print "The wavelength at which the emission of radiation from the filament peaks is",lambda_max,"micron"

Fraction of radiation emitted between the two given wavelengths is 0.052714
The wavelength at which the emission of radiation from the filament peaks is 1.15912 micron


### Example 12.3¶

In :
import math
# Radiation Incident on a small surface

# Variables
A1 = 3**10.**(-4)			#[m**2]
T1 = 600.			        #[k]
A2 = 5*10.**(-4)			#[m**2]
theta1 = math.pi*55./180
theta2 = math.pi*40./180			#[Radian]
r = 0.75			        #[m]

# Calculations and Results
w_2_1 = (A2*math.cos(theta2))/(r**2)			#[Steradian]
print "The solid angle subtended by a2 when viewed from A1 is",w_2_1,"sr"

I1 = (5.67*10**(-8))*(T1**4)/(math.pi)			#[W/m**2.sr]
print "The Intensity of radiation emitted by A1 is",I1,"W/m**2.sr"

Q1_2 = I1*(A1*math.cos(theta1))*w_2_1			#[W]
print "The rate of radiation energy emitted by A1 in the direction of"\
,theta1,"radians","through the solid angle",w_2_1,"Steradian","is ",Q1_2,"W"

The solid angle subtended by a2 when viewed from A1 is 0.000680928393884 sr
The Intensity of radiation emitted by A1 is 2339.04290284 W/m**2.sr
The rate of radiation energy emitted by A1 in the direction of 0.959931088597 radians through the solid angle 0.000680928393884 Steradian is  0.913647447646 W


### Example 12.4¶

In :
import math
# Emissivity of a surface and emissive Power

# Variables
e1 = 0.3			#For 0< =  lambda < =  3micron
e2 = 0.8			#3micron< = lambda< = 7micron
e3 = 0.1			#7micron< = lamda<infinity
lambda1 = 3
lambda2 = 7			#[micron]
T = 800 			#[K]

# Calculations and Results
p = lambda1*T			#[micron.K]
q = lambda2*T			#[micron.K]
#Hence blackbody radiation functions are
f1 = 0.140256;
f2 = 0.701046;
f0_1 = f1-0;
f2_inf = 1-f2;
e_T = e1*f1+e2*(f2-f1)+e3*(1-f2);
print "Average emissivity of the surface is",e_T

E = e_T*(5.67*10**(-8))*(T**4)			#[W/m**2]
print "The Emissive Power of the surface is",E,"W/m**2"

Average emissivity of the surface is 0.5206042
The Emissive Power of the surface is 12090.6785341 W/m**2


### Example 12.5¶

In :
import math
# Selective Absorber and Reflective Surfaces

# Variables
G_D = 400
G_d = 300			#Direct and diffuse components of solar radiation[W/m**2]
Ts = 320
T_sky = 260			#[K]
theta = 20*math.pi/180

# Calculations and Results
G_solar = (G_D*math.cos(theta))+G_d
#(a)
ab_a = 0.9
e_a = 0.9			#Grey absorber surface
q_net_rad_a = ab_a*G_solar+e_a*(5.67*10**(-8))*((T_sky**4)-(Ts**4))			#[W/m**2]
print "(a) The net radiation heat transfer is",round(q_net_rad_a),"W/m**2"

#(b)
ab_b = 0.1
e_b = 0.1			#Grey reflector surface
q_net_rad_b = ab_b*G_solar+e_b*(5.67*10**(-8))*((T_sky**4)-(Ts**4))			#[W/m**2]
print "The net radiation heat transfer is",round(q_net_rad_b),"W/m**2"

#(c)
ab_c = 0.9
e_c = 0.1			#Selective Absorber surface
q_net_rad_c = ab_c*G_solar+e_c*(5.67*10**(-8))*((T_sky**4)-(Ts**4))			#[W/m**2]
print "The net radiation heat transfer is",round(q_net_rad_c),"W/m**2"

#(d)
ab_d = 0.1
e_d = 0.9			#Selective reflector surface
q_net_rad_d = ab_d*G_solar+e_d*(5.67*10**(-8))*((T_sky**4)-(Ts**4))			#[W/m**2]
print "The net radiation heat transfer is",round(q_net_rad_d),"W/m**2"

(a) The net radiation heat transfer is 306.0 W/m**2
The net radiation heat transfer is 34.0 W/m**2
The net radiation heat transfer is 575.0 W/m**2
The net radiation heat transfer is -234.0 W/m**2


### Example 12.6¶

In :
import math
# Installing Reflective Films on Windows

# Variables
A_glazing = 40			#[m**2]
SHGC_wof = 0.766
SHGC_wf = 0.261			#[kWh/year]
unit_c_e = 0.08			#[$/kWh] unit_c_f = 0.5 #[$/therm]
COP = 2.5
neta = 0.80;

# Calculations and Results
#For the months of June,July,August and Sepetember
Q_summer = 5.31*30+4.31*31+3.93*31+3.28*30			#[kWh/year]
#For the months oct,Nov,Dec,Jan,Feb,Mar,Apr
Q_winter = 2.80*31+1.84*30+1.54*31+1.86*31+2.66*28+3.43*31+4.00*30			#[kWh/year]
c_l_d = Q_summer*A_glazing*(SHGC_wof-SHGC_wf)			#[kWh/year]
print "The decrease in the annual cooling load is",c_l_d,"kWh/year"

h_l_i = Q_winter*A_glazing*(SHGC_wof-SHGC_wf)			#[kWh/year]
print "The increase in annual heating load is",h_l_i,"kWh/year"

d_c_c = c_l_d*(unit_c_e)/COP			#[$/year] i_h_c = h_l_i*(unit_c_f/29.31)/neta #[$/year]
print "The corresponding decrease in cooling math.costs and the increase in heating math.costs are $"\ ,d_c_c,"and$",i_h_c,"per year"

Cost_s = d_c_c-i_h_c			#[$/year] print "The net annual math.cost savings due to the reflective film is$",Cost_s,"per year"

I_cost = 20*A_glazing			#[$] print "The implementation Cost of installing films is$",I_cost

pp = I_cost/Cost_s			#[years]
print "Payback Period is",pp,"years"

The decrease in the annual cooling load is 10365.428 kWh/year
The increase in annual heating load is 11073.842 kWh/year
The corresponding decrease in cooling math.costs and the increase in heating math.costs are $331.693696 and$ 236.136173661 per year
The net annual math.cost savings due to the reflective film is $95.5575223391 per year The implementation Cost of installing films is$ 800
Payback Period is 8.3719207072 years