# Two dimensional, Steady State Conduction¶

## Example 4.1 Page 211¶

In [1]:
import math
d = .005;        										#[m] Diameter of wire
k = .35;         										#[W/m.K] Thermal Conductivity
h = 15;          										#[W/m^2.K] Total coeff with Convection n Radiation
#calculations

rcr = k/h;        										# [m] critical insulation radius
tcr = rcr - d/2.;  										# [m] critical insulation Thickness

Rtcond = 2.302*math.log10(rcr/(d/2.))/(2*math.pi*k);    #[K/W] Thermal resistance

#Using Table 4.1 Case 7
z = .5*tcr;
D=2*rcr;
Rtcond2D = (math.acosh((D*D + d*d - 4*z*z)/(2*D*d)))/(2*math.pi*k);
#results

print '%s %.2f %s' %("\n\n The reduction in thermal resistance of the insulation is", Rtcond-Rtcond2D," K/W ");
#END


The reduction in thermal resistance of the insulation is 0.10  K/W


## Example 4.3 Page 224¶

In [2]:
import math
import numpy
from numpy import linalg
Ts = 500.;        	#[K] Temp of surface
Tsurr = 300.;     	#[K] Temp of surrounding Air
h = 10.;          	#[W/m^2.K] Heat Convection soefficient
#Support Column
delx = .25;      	#[m]
dely = .25;      	#[m]
k = 1.;         	#[W/m.K] From Table A.3, Fireclay Brick at T = 478K
#calculations

#Applying Eqn 4.42 and 4.48
A = numpy.array([[-4, 1, 1, 0, 0, 0, 0, 0],
[2, -4, 0, 1, 0, 0, 0, 0],
[1, 0, -4, 1, 1, 0, 0, 0],
[0, 1, 2, -4, 0, 1, 0, 0],
[0,0, 1, 0, -4, 1, 1, 0],
[0, 0, 0, 1, 2, -4, 0, 1],
[0, 0, 0, 0, 2, 0, -9, 1],
[0, 0, 0, 0, 0, 2, 2, -9]]);

C = numpy.array([[-1000], [-500], [-500], [0], [-500], [0], [-2000], [-1500]]);

T = numpy.linalg.solve (A,C);
#results

print '%s' %("\n Temp Distribution in K = ");
print (T);

q = 2*h*((delx/2.)*(Ts-Tsurr)+delx*(T[6]-Tsurr)+delx*(T[7]-Tsurr)/2.);
print '%s %.2f %s' %("\n\n Heat rate from column to the airstream",q," W/m ");
#END

 Temp Distribution in K =
[[ 489.30472333]
[ 485.15381783]
[ 472.06507549]
[ 462.00582466]
[ 436.94975396]
[ 418.73932983]
[ 356.99461052]
[ 339.05198674]]

Heat rate from column to the airstream 882.60  W/m


## Example 4.4 Page 230¶

In [3]:
import math
import numpy
from numpy import linalg
#Operating Conditions

ho = 1000;               #[W/m^2.K] Heat Convection coefficient
hi = 200;                #[W/m^2.K] Heat Convection coefficient
Ti = 400;                #[K] Temp of Air
Tg = 1700;               #[K] Temp of Gas
h = 10 ;                 #[W/m^2.K] Heat Convection coefficient

A = 2*6*math.pow(10,-6) ;#[m^2] Cross section of each Channel
x = .004 ;               #[m] Spacing between joints
t = .006;                #[m] Thickness
k = 25;                  #[W/m.K] Thermal Conductivity of Blade
delx = .001 ;            #[m]
dely = .001 ;            #[m]
#calculations and results

#Applying Eqn 4.42 and 4.48
A = numpy.array([[-(2+ho*delx/k), 1, 0,0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[1,-2*(2+ho*delx/k),1,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0],
[0,1,-2*(2+ho*delx/k),1,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0],
[0,0,1,-2*(2+ho*delx/k),1,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,1,-2*(2+ho*delx/k),1,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,1,-(2+ho*delx/k),0,0,0,0,0,1,0,0,0,0,0,0,0,0,0],
[1,0,0,0,0,0,-4,2,0,0,0,0,1,0,0,0,0,0,0,0,0],
[0,1,0,0,0,0,1,-4,1,0,0,0,0,1,0,0,0,0,0,0,0],
[0,0,1,0,0,0,0,1,-4,1,0,0,0,0,1,0,0,0,0,0,0],
[0,0,0,1,0,0,0,0,1,-4,1,0,0,0,0,1,0,0,0,0,0],
[0,0,0,0,1,0,0,0,0,1,-4,1,0,0,0,0,1,0,0,0,0],
[0,0,0,0,0,1,0,0,0,0,2,-4,0,0,0,0,0,1,0,0,0],
[0,0,0,0,0,0,1,0,0,0,0,0,-4,2,0,0,0,0,1,0,0],
[0,0,0,0,0,0,0,1,0,0,0,0,1,-4,1,0,0,0,0,1,0],
[0,0,0,0,0,0,0,0,2,0,0,0,0,2,-2*(3+hi*delx/k),1,0,0,0,0,1],
[0,0,0,0,0,0,0,0,0,2,0,0,0,0,1,-2*(2+hi*delx/k),1,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,1,-2*(2+hi*delx/k),1,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,-(2+hi*delx/k),0,0,0],
[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,-2,1,0],
[0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,1,-4,1],
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,-(2+hi*delx/k)]]);

C = numpy.array([[-ho*delx*Tg/k],
[-2*ho*delx*Tg/k],
[-2*ho*delx*Tg/k],
[-2*ho*delx*Tg/k],
[-2*ho*delx*Tg/k],
[-ho*delx*Tg/k],
[0],
[0],
[0],
[0],
[0],
[0],
[0],
[0],
[-2*hi*delx*Ti/k],
[-2*hi*delx*Ti/k],
[-2*hi*delx*Ti/k],
[-hi*delx*Ti/k],
[0],
[0],
[-hi*delx*Ti/k]]);

T = numpy.linalg.solve (A,C);
print '%s' %("\n Temp Distribution in K = ");
print (T);
q = 4*ho*((delx/2.)*(Tg-T[0])+delx*(Tg-T[1])+delx*(Tg-T[2])+ delx*(Tg-T[3])+delx*(Tg-T[4])+delx*(Tg-T[5])/2.);
print '%s %.1f %s' %("\n\n Heat rate Transfer = " ,q,"W/m ");
#END

 Temp Distribution in K =
[[ 1525.95413813]
[ 1525.27944565]
[ 1523.59609075]
[ 1521.93574674]
[ 1520.83066847]
[ 1520.45069026]
[ 1519.66699612]
[ 1518.7949547 ]
[ 1516.52842892]
[ 1514.53554374]
[ 1513.30134519]
[ 1512.88873965]
[ 1515.12393697]
[ 1513.70494809]
[ 1509.18712651]
[ 1506.37665411]
[ 1504.9504289 ]
[ 1504.50157796]
[ 1513.41885557]
[ 1511.71377418]
[ 1506.02634497]]

Heat rate Transfer =  3540.6 W/m