# Chapter 12 : Fundamentals of Thermal Radiation¶

### Example 12.1¶

In [1]:
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]
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 [2]:
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 [5]:
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
r = 0.75			        #[m]

# Calculations and Results
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"\

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 [6]:
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]
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 [11]:
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

#(b)
ab_b = 0.1
e_b = 0.1			#Grey reflector surface

#(c)
ab_c = 0.9
e_c = 0.1			#Selective Absorber surface

#(d)
ab_d = 0.1
e_d = 0.9			#Selective reflector surface

(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 [12]:
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