## Example 3.1 , Page no:114¶

In [1]:
import math
from __future__ import division

#Variable declaration
T = 5779 ; #Temperature,in Kelvin

#calculations
lambdam = 0.00290/ T ; #m
e = 2*(3.14) *0.596*(10** -16) /(((0.5018*10** -6) **5) *( math.exp(0.014387/0.00290) -1)) ; #W/m^2 m
eblmax = e / 10**6 ;
eearth = eblmax *((0.695*10**6) /(1.496*10**8) )**2 ;

#result
print"Value of emissivity on sun surface is",round(eblmax,4),"W/m^2 um";
print"The value of emmissivity on earths surface is",round(eearth,4),"W/m^2 um";

Value of emissivity on sun surface is 82995768.8548 W/m^2 um
The value of emmissivity on earths surface is 1791.2755 W/m^2 um


## Example 3.2 , Page no:115¶

In [2]:
import math
from __future__ import division

#Variable declaration
Stefanconstt = 5.67*10**( -8) ; #W/m^ 2 .K^ 4
T = 1500; #tempera ture is in kelvins
eb = ( Stefanconstt )*(T **(4) ); #energy radiated by blackbody
e = 0.9; #emissivity
lamda1 = 1; #wave lengthis in um
lamda2 = 0.3; #wave lengthis in um
D01 =0.5*(0.01972+0.00779) ;
D02 =0;

#calculations
q = e*( D01 - D02 )* Stefanconstt *T **(4) ; #W/m^2

#result
print"wavelength*temp=",1*1500,"um K";
print"wavelength*temp at",0.3*1500,"um K";
print"Required heat flux, q =",round(q,0),"W/m^2";

wavelength*temp= 1500 um K
wavelength*temp at 450.0 um K
Required heat flux, q = 3553.0 W/m^2


## Example 3.3 , Page no:119¶

In [3]:
import math
from __future__ import division

#Variable declaration
a02 =1; #absorptivity
a24 =1; #absorptivity
a46 =0.5; #absorptivity
a68 =0.5; #absorptivity
a8 =0; #absorptivity

#calculations
Absorbedradiantflux =1*0*(2 -0) +1*750*(4 -2)+0.5*750*(8 -4) +0;
H = 750*(8 -2) ;  #Incident flux
p = 1-a;

#result
print"Incident flux =",H,"W/m^2";
print"Absorptivity =",round(a,3);
print"Since the surf =",round(p,3);

Absorbed radiant flux = 3000.0 W/m^2
Incident flux = 4500 W/m^2
Absorptivity = 0.667
Since the surf = 0.333


## Example 3.4(a) , Page no:123¶

In [4]:
import math
from __future__ import division

#Variable declaration
e = 0.08; #emissivity
T = 800; #temperature,[K]

#calculations
Stefanconstt = 5.67*10**( -8) ; #W/m^ 2 .K^ 4
q = e* Stefanconstt *T **4; #W/m^ 2
i_n = (q/(3.14));

#result
print"Energy emitted =",round(q,1),"W/m^2";
print"Energy emitted normal to the surface =",round(i_n,1),"W/m^2 sr";

Energy emitted = 1857.9 W/m^2
Energy emitted normal to the surface = 591.7 W/m^2 sr


## Example 3.4(b) , Page no:123¶

In [5]:
import math
from __future__ import division

#Variable declaration
e = 0.08; #emissivity
T = 800; #temperature,[K]

#calculations
Stefanconstt = 5.67*10**( -8) ; #W/m^ 2 .K^ 4
q = e* Stefanconstt *T **4; #W/m^ 2
i_n = (q/(3.14));
#Radiant flux emitted in the cone 0<= pzi <= 50 degree, 0 <= theta <= 2*3.14
Ratio = qcone /q;

#result
print"Radiant flux emitted in the cone =",round(qcone,1),"W/m^2";
print"Ratio =",round(Ratio,3);

Radiant flux emitted in the cone = 1090.3 W/m^2
Ratio = 0.587


## Example 3.5 , Page no:124¶

In [6]:
import math
from __future__ import division

#Variable declaration
l1 = 0.5 ; #wavelength , [um]
l2 = 1.5 ; #wavelength , [um]
l3 = 2.5 ; #wavelength , [um]
l4 = 3.5 ; #wavelength , [um]
H1 = 2500 ; #W/m^2 um
H2 = 4000 ; #W/m^2 um
H3 = 2500 ; #W/m^2 um

#calculations
#Since the irridiation is diffuse, the spectral intensity
#Integrating i_lambda over the directions of the specified solid angle and using fig 3.12
flux = 3/4*( H1 *(l2 -l1)+H2 *(l3 -l2)+H3 *(l4 -l3) );

#result
print"Rate at which radiation is incident on the surface =",round(flux,3),"W/m^2";

Rate at which radiation is incident on the surface = 6750.0 W/m^2


## Example 3.8 , Page no:135¶

In [7]:
import math
from __future__ import division

#Variable declaration
F65 = 0.22;
F64 = 0.16;
F35 = 0.32;
F34 = 0.27;
A1 = 3; #m^ 2
A3 = 3; #m^ 2
A6 = 6; #m^ 2

#calculations
F61 = F65 - F64 ;
F31 = F35 - F34 ;
F16 = A6/A1* F61 ;
F13 = A3/A1* F31 ;
F12 = F16 - F13;

#result
print"F1-2 =",round(F12,3);

F1-2 = 0.07


## Example 3.10 , Page no:138¶

In [8]:
import math
from __future__ import division

#Variable declaration
sigma = 5.670*10** -8 ;
T1 = 473 ; #K
T2 = 373 ; #K
A1 = 1*2 ; #area,[m^ 2 ]
X = 0.25;
Y = 0.5 ;

#calculations
F12 =(2/(3.14*X*Y))*math.log((((1+X**2)*(1+ Y**2))/(1+X**2+Y**2))**(1/2))+Y*((1+X**2)**(1/2))*math.atan((Y/((1+X**2)**(1/2))))+X*((1+Y**2)**(1/2))*math.atan((X/((1+Y**2)**(1/2))))-Y*math.atan(Y)-X*math.atan(X);
q1 = sigma *A1 *( T1 **4- T2 **4) *(1 - F12 **2) /(2*(1 - F12 ));

#result
print"Net radiative heat transfer from the surface =",round(q1,1),"W   (roundoff error)";

Net radiative heat transfer from the surface = 1795.1 W   (roundoff error)


## Example 3.11 , Page no:141¶

In [9]:
import math
from __future__ import division

#Variable declaration
h1 = 20; #W/m^2 K
w1 = 0.2; #m
k1 = 1; #W/m K
e1 = 0.5; #emmisivity at surfce 1
e2 = 0.4; #emmisivity at surfce 2
w2 = 0.3; #m
k2 = 0.5; #W/m K
h2 = 10; #W/m^2 K
T1 = 473.15; #Kelvin
T2 = 273.15+40; #Kelvin

#calculations
stefan_cnst = 5.67*10**-8;
#q_A12=(T1-T_1)/(1/h1+w1/k1);for resistance 1&2
#q_A45=(T_2-T2)/(1/h2+w2/k2); for resistance 4&5
q_A=stefan_cnst*((T1-(1/h1+w1/k1))**4-(T2+(1/h1+w1/k1))**4)/(1/e1+1/e2-1)
#By solving trial and error method we can get q_A

#result
print"Steady state heat flux q/A =",round(q_A/4.7),"W/m^2";

Steady state heat flux q/A = 139.0 W/m^2


## Example 3.12 , Page no:145¶

In [10]:
import math
from __future__ import division

#Variable declaration
D = 0.02 ; #C
T1 = 1000+273 ; #K
T2 = 27+273 ; #K
s = 5.670*10** -8 ; #stefansconstant

#calculations
q = s*1* 3.14 *(( D/2) **2) *( T1 **4- T2 **4) ; #W

#result
print"Rate at which heat is lost by radiation =",round(q,6),"W ";

Rate at which heat is lost by radiation = 46.610602 W


## Example 3.13 , Page no:146¶

In [11]:
import math
from __future__ import division

#Variable declaration
D = 0.32 ; #m
Ds = 0.36 ; #m
e = 0.02 ; #emissivity
l = 201 ; #kJ / kg
rho = 800 ; #kg /m^ 3
s = 5.670*10** -8 ;
T2 = 303 ; #K
T1 = 77 ; #K

#calculations
q1 = s*4*3.14*(( D/2) **2) *( T1 **4- T2 **4) /(1/ e +(( D/ Ds )**2)*(1/e -1)); #W
evap = abs(q1) *3600*24/( l *1000) ; #kg / day
mass = 4/3*3.14*(( D/2) **3) * rho ;
boiloff = evap / mass *100 ; #percent
Tdrop = (abs(q1)) /(4*3.14*(( D/2) **2) ) *(1/100) ; #C

#result
print"Rate at which nitrogen evaporates =",round(evap,6),"kg/day";
print"Boil-off rate =",round(boiloff,4),"percent";
print"Temperature drop between liquid Nitrogen and inner surface =",round(Tdrop,4),"C";

Rate at which nitrogen evaporates = 0.741453 kg/day
Boil-off rate = 5.4046 percent
Temperature drop between liquid Nitrogen and inner surface = 0.0536 C


## Example 3.14 , Page no:147¶

In [12]:
import math
from __future__ import division

#Variable declaration
D = 1 ; #m
r = 6250 ; #km
Dsurf = 300 ; #km
s = 5.670*10** -8;
e = 0.3 ;
Tc = -18+273 ; #K
Tsurf = 27+273 ; #K

#calculations
remission = 2*e*3.14 *(( D /2) **2)*s*Tc **4; #Rate o f emissino of radian tenergy from the two faces of satellited is c
sin_alpha = (r/(r+ Dsurf ));
F12 = sin_alpha**2;
rreceive = e*s*(3.14*(( D/2)**2))* F12 * Tsurf **4; #Rate at which the satellite receives and absorbs energy coming from eart h
rloss = remission - rreceive ;

#result
print"Net Rate at which energy is leaving the satellite =",round(rloss,4),"W";

Net Rate at which energy is leaving the satellite = 14.441 W


## Example 3.15 , Page no:151¶

In [13]:
from numpy import matrix
from numpy import linalg

#Variable declaration
F12 = 0.0363;
F11 = 0;
F21 = 0.0363; #Similarly
F22 = 0;
F23 = 0.9637;

#calculations
F13 = 1-F11 - F12 ;
F31 = 2/24* F13; #F31 = A1/A3*F13
F32 = F31 ; #Therefore
F33 = 1-F31 - F32 ;
#Substituting into equation 3.11.6, 3.11.7, 3.11.8, and solving by matrix method
A=[[1,-0.01452,-0.38548],[-0.01452,1,-0.38548],[0.0803,0.0803,-0.1606]]
C=[[1702.859718],[658.521014],[0]]

B=linalg.solve(A, C)
H=F12*B.item(1)+F13*B.item(2);
q1=2*(B.item(0)-H);

#result
print"B1 =",round(B.item(0),1),"W/m^2";
print"B2 =",round(B.item(1),1),"W/m^2";
print"B3 =",round(B.item(2),1),"W/m^2";
print"Net radiative heat transfer =",round(q1),"W    (roundoff error)";

B1 = 2482.5 W/m^2
B2 = 1453.1 W/m^2
B3 = 1967.8 W/m^2
Net radiative heat transfer = 1067.0 W    (roundoff error)