## Example 13.2 Page 821¶

In [1]:
import math
# (1) Sphere within Cube
F12a = 1                    		;#By Inspection
F21a = (math.pi/6.)*F12a        	; #By Reciprocity
#calculations

# (2) Partition within a Square Duct
F11b = 0                    		;#By Inspection
#By Symmetry F12 = F13
F12b = (1-F11b)/2.      ;		    #By Summation Rule
F21b = math.sqrt(2.)*F12b     ;     #By Reciprocity

# (3) Circular Tube
#From Table 13.2 or 13.5, with r3/L = 0.5 and L/r1 = 2
F13c = .172;
F11c = 0;                    		#By Inspection
F12c = 1 - F11c - F13c      		;#By Summation Rule
F21c = F12c/4.               		;#By Reciprocity
#results

print' %s' %('\n Desired View Factors may be obtained from inspection, the reciprocity rule, the summation rule and/or use of charts')
print '%s %.3f' %('\n (1) Sphere within Cube F21 =',F21a)
print '%s %.3f' %('\n (2) Partition within a Square Duct F21 = ',F21b)
print '%s %.3f' %('\n (3) Circular Tube F21 =',F21c);


Desired View Factors may be obtained from inspection, the reciprocity rule, the summation rule and/or use of charts

(1) Sphere within Cube F21 = 0.524

(2) Partition within a Square Duct F21 =  0.707

(3) Circular Tube F21 = 0.207


## Example 13.3 Page 826¶

In [2]:
import math
import numpy
from numpy import linalg

L = 10        	;#[m] Collector length = Heater Length
T2 = 600      	;#[K] Temperature of curved surface
A2 = 15      	;#[m^2] Area of curved surface
e2 = .5      	;# emissivity of curved surface
stfncnstt = 5.67*math.pow(10,-8);		#[W/m^2.K^4] Stefan-Boltzmann constant
T1 = 1000        ;#[K] Temperature of heater
A1 = 10          ;#[m^2] area of heater
e1 = .9          ;# emissivity of heater
W = 1            ;#[m] Width of heater
H = 1            ;#[m] Height
T3 = 300         ;#[K] Temperature of surrounding
e3 = 1           ;# emissivity of surrounding
#calculations

J3 = stfncnstt*T3*T3*T3*T3; #[W/m^2]
#From Figure 13.4 or Table 13.2, with Y/L = 10 and X/L =1
F12 = .39;
F13 = 1 - F12;   			#By Summation Rule
#For a hypothetical surface A2h
A2h = L*W;
F2h3 = F13;       			#By Symmetry
F23 = A2h/A2*F13;         	#By reciprocity
Eb1 = stfncnstt*T1*T1*T1*T1;#[W/m^2]
Eb2 = stfncnstt*T2*T2*T2*T2;#[W/m^2]
#Radiation network analysis at Node corresponding 1
#-10J1 + 0.39J2 = -510582
#.26J1 - 1.67J2 = -7536
#Solving above equations
A = ([[-10 ,.39],
[.26, -1.67]]);
B = ([[-510582.],
[-7536.]]);

X = numpy.linalg.solve (A,B);

q2 = (Eb2 - X[1])/(1-e2)*(e2*A2);
#results

print '%s %.1f %s' %('\n Net Heat transfer rate to the absorber is = ',q2/1000. ,'kW');

 Net Heat transfer rate to the absorber is =  -77.8 kW


## Example 13.4 Page 830¶

In [3]:
import math

T3 = 300.         					;#[K] Temperature of surrounding
L = .15        						;#[m] Furnace Length
T2 = 1650+273.      				;#[K] Temperature of bottom surface
T1 = 1350+273.      				;#[K] Temperature of sides of furnace
D = .075           					;#[m] Diameter of furnace
stfncnstt = 5.670*math.pow(10,-8);  #[W/m^2.K^4] Stefan Boltzman Constant
#calculations

A2 = math.pi*D*D/4.        			;#[m] Area of bottom surface
A1 = math.pi*D*L        	 		;#[m] Area of curved sides
#From Figure 13.5 or Table 13.2, with ri/L = .25
F23 = .056;
F21 = 1 - F23;        				#By Summation Rule
F12 = A2/A1*F21;         			#By reciprocity
F13 = F12            				;#By Symmetry
#From Equation 13.17 Heat balance
q = A1*F13*stfncnstt*(T1*T1*T1*T1 - T3*T3*T3*T3) + A2*F23*stfncnstt*(T2*T2*T2*T2 - T3*T3*T3*T3);
#results

print '%s %d %s' %('\n Power required to maintain prescribed temperatures is =',q, 'W');

 Power required to maintain prescribed temperatures is = 1830 W


## Example 13.5 Page 834¶

In [4]:
import math

T2 = 300      	;#[K] Temperature of inner surface
D2 = .05      	;#[m] Diameter of Inner Surface
e2 = .05      	;# emissivity of Inner Surface
T1 = 77      	;#[K] Temperature of Outer Surface
D1 = .02        ;#[m] Diameter of Inner Surface
e1 = .02      	;# emissivity of Outer Surface
D3 = .035       ;#[m] Diameter of Shield
e3 = .02        ;# emissivity of Shield
stfncnstt = 5.670*math.pow(10,-8)        ;#[W/m^2.K^4] Stefan Boltzman Constant
#calculations

#From Equation 13.20 Heat balance
q = stfncnstt*(math.pi*D1)*(T1*T1*T1*T1-T2*T2*T2*T2)/(1/e1 + (1-e2)/e2*D1/D2) ;#[W/m]

RtotL = (1-e1)/(e1*math.pi*D1) + 1/(math.pi*D1*1) + 2*((1-e3)/(e3*math.pi*D3)) + 1/(math.pi*D3*1) + (1-e2)/(e2*math.pi*D2)    ;#[m^-2]
q2 = stfncnstt*(T1*T1*T1*T1 - T2*T2*T2*T2)/RtotL;   #[W/m]
#results

print '%s %.2f %s' %('\n Heat gain by the fluid passing through the inner tube =',q,'W/m')
print '%s %.2f %s' %('\n Percentage change in heat gain with radiation shield inserted midway between inner and outer tubes is =',(q2-q)*100/q,'percent');

 Heat gain by the fluid passing through the inner tube = -0.50 W/m

Percentage change in heat gain with radiation shield inserted midway between inner and outer tubes is = -49.55 percent


## Example 13.6 Page 836¶

In [5]:
import math

T2 = 500      					;#[K] Temperature of Painted surface
e2 = .4     	 				;# emissivity of Painted Surface
T1 = 1200      					;#[K] Temperature of Heated Surface
W = 1          					; #[m] Width of Painted Surface
e1 = .8      					;# emissivity of Heated Surface
er = .8        					;# emissivity of Insulated Surface
stfncnstt = 5.670*math.pow(10,-8);#[W/m^2.K^4] Stefan Boltzman Constant

#By Symmetry Rule
F2R = .5;
F12 = .5;
F1R = .5;
#calculations

#From Equation 13.20 Heat balance
q = stfncnstt*(T1*T1*T1*T1-T2*T2*T2*T2)/((1-e1)/e1*W+ 1/(W*F12+1/((1/W/F1R) + (1/W/F2R))) + (1-e2)/e2*W) ;#[W/m]

#Surface Energy Balance 13.13
J1 = stfncnstt*T1*T1*T1*T1 - (1-e1)*q/(e1*W)		;# [W/m^2] Surface 1
J2 = stfncnstt*T2*T2*T2*T2 - (1-e2)*(-q)/(e2*W)		;# [W/m^2] Surface 2
#From Equation 13.26 Heat balance
JR = (J1+J2)/2.;
TR = math.pow((JR/stfncnstt),.25);
#results

print '%s %.2f %s' %('\n Rate at which heat must be supplied per unit length of duct = ',q/1000.,'kW/m')
print '%s %d %s' %('\n Temperature of the insulated surface = ',TR,'K');

 Rate at which heat must be supplied per unit length of duct =  36.98 kW/m

Temperature of the insulated surface =  1102 K


## Example 13.7 Page 841¶

In [6]:
import math

T1 = 1000.      				;#[K] Temperature of Heated Surface
e1 = .8      					;# emissivity of Heated Surface
e2 = .8       					; # emissivity of Insulated Surface
r = .02        					;#[m] Radius of surface
Tm = 400        				;#[K] Temperature of surrounding air
m = .01        					;#[kg/s] Flow rate of surrounding air
p = 101325     					;#[Pa] Pressure of surrounding air
stfncnstt = 5.670*math.pow(10,-8);#[W/m^2.K^4] Stefan Boltzman Constant
#Table A.4 Air Properties at 1 atm, 400 K
k = .0338        				;#[W/m.K] conductivity
u = 230*math.pow(10,-7)    		;#[kg/s.m] Viscosity
cp = 1014        				;#[J/kg] Specific heat
Pr = .69         				;# Prandtl Number
#calculations and results

#Hydraulic Diameter
Dh = 2*math.pi*r/(math.pi+2.)   ;#[m]
#Reynolds number
Re = m*Dh/(math.pi*r*r/2.)/u;
#View Factor
F12 = 1 ;

print '%s %d %s' %("\n As Reynolds Number is",Re,", Hence it is Turbulent flow inside a cylinder. Hence we will use Dittus-Boelter Equation");

#From Dittus-Boelter Equation
Nu = .023*math.pow(Re,.8) *math.pow(Pr,.4);
h = Nu*k/Dh;            		#[W/m^2.K]

#From Equation 13.18 Heat Energy balance
#Newton Raphson
T2=600;        					#Initial Assumption

 As Reynolds Number is 16912 , Hence it is Turbulent flow inside a cylinder. Hence we will use Dittus-Boelter Equation