Introduction¶
Example 1.1 Page 5¶
In [1]:
L=.15; #[m] - Thickness of conducting wall
delT = 1400. - 1150.; #[K] - Temperature Difference across the Wall
A=.5*1.2; #[m^2] - Cross sectional Area of wall = H*W
k=1.7; #[W/m.k] - Thermal Conductivity of Wall Material
#calculations
#Using Fourier's Law eq 1.2
Q = k*delT/L; #[W/m^2] - Heat Flux
q = A*Q; #[W] - Rate of Heat Transfer
#results
print '%s %.2f %s' %("\n \n Heat Loss through the Wall =",q," W");
#END
Example 1.2 Page 11¶
In [2]:
import math
d=.07; #[m] - Outside Diameter of Pipe
Ts = 200+273.15; #[K] - Surface Temperature of Steam
Tsurr = 25+273.15; #[K] - Temperature outside the pipe
e=.8; # Emissivity of Surface
h=15; #[W/m^2.k] - Thermal Convectivity from surface to air
stfncnstt=5.67*math.pow(10,(-8)); # [W/m^2.K^4] - Stefan Boltzmann Constant
#calculations
#Using Eq 1.5
E = e*stfncnstt*Ts*Ts*Ts*Ts; #[W/m^2] - Emissive Power
G = stfncnstt*Tsurr*Tsurr*Tsurr*Tsurr; #[W/m^2] - Irradiation falling on surface
#results
print '%s %.2f %s' %("\n (a) Surface Emissive Power = ",E," W/m^2");
print '%s %.2f %s' %("\n Irradiation Falling on Surface =",G," W/m^2");
#Using Eq 1.10 Total Rate of Heat Transfer Q = Q by convection + Q by radiation
q = h*(math.pi*d)*(Ts-Tsurr)+e*(math.pi*d)*stfncnstt*(Ts*Ts*Ts*Ts-Tsurr*Tsurr*Tsurr*Tsurr); #[W]
print '%s %.2f %s' %("\n\n (b) Total Heat Loss per unit Length of Pipe=",q," W");
#END
Example 1.4 Page 20¶
In [5]:
import math
Ts = 56.4+273.15; #[K] - Surface Temperature of Steam
Tsurr = 25+273.15; #[K] - Temperature of Surroundings
e=.88; # Emissivity of Surface
#As h=(10.9*math.pow(V,.8)[W/m^2.k] - Thermal Convectivity from surface to air
stfncnstt=5.67*math.pow(10,(-8)); # [W/m^2.K^4] - Stefan Boltzmann Constant
A=2*.05*.05; # [m^2] Area for Heat transfer i.e. both surfaces
E = 11.25; #[W] Net heat to be removed by cooling air
#calculations
Qrad = e*stfncnstt*A*(math.pow(Ts,4)-math.pow(Tsurr,4));
#Using Eq 1.10 Total Rate of Heat Transfer Q = Q by convection + Q by radiation
Qconv = E - Qrad; #[W]
#As Qconv = h*A*(Ts-Tsurr) & h=10.9 Ws^(.8)/m^(-.8)K.V^(.8)
V = math.pow(Qconv/(10.9*A*(Ts-Tsurr)),(1/0.8));
#results
print '%s %.2f %s' %("\n\n Velocity of Cooling Air flowing= ", V,"m/s");
#END
Example 1.6 Page 26¶
In [3]:
import math
A=1.8; # [m^2] Area for Heat transfer i.e. both surfaces
Ti = 35+273.; #[K] - Inside Surface Temperature of Body
Tsurr = 297.; #[K] - Temperature of surrounding
Tf = 297.; #[K] - Temperature of Fluid Flow
e=.95; # Emissivity of Surface
L=.003; #[m] - Thickness of Skin
k=.3; # Effective Thermal Conductivity
h=2; #[W/m^2.k] - Natural Thermal Convectivity from body to air
stfncnstt=5.67*math.pow(10,(-8)); # [W/m^2.K^4] - Stefan Boltzmann Constant
#Using Eq 1.5
Tsa=305.; #[K] Body Temperature Assumed
#calculations
Ts=307.19
q = k*A*(Ti-Ts)/L; #[W]
print '%s' %("\n\n (I) In presence of Air")
print '%s %.2f %s' %("\n (a) Temperature of Skin = ",Ts,"K");
print '%s %.2f %s' %("\n (b) Total Heat Loss = ",q," W");
#When person is in Water
h = 200; #[W/m^2.k] - Thermal Convectivity from body to water
hr = 0; # As Water is Opaque for Thermal Radiation
Ts = (k*Ti/L + (h+hr)*Tf)/(k/L +(h+hr)); #[K] Body Temperature
q = k*A*(Ti-Ts)/L; #[W]
#results
print '%s' %("\n\n (II) In presence of Water")
print '%s %.2f %s' %("\n (a) Temperature of Skin =",Ts," K");
print '%s %.2f %s' %("\n (b) Total Heat Loss =",q," W");
#END
Example 1.7 Page 30¶
In [2]:
%matplotlib inline
import math
import numpy
from numpy import roots
import matplotlib
from matplotlib import pyplot
Tsurr = 30+273; #[K] - Temperature of surrounding
Tf = 20+273; #[K] - Temperature of Fluid Flow
e=.5; # Emissivity of Surface
a = .8; # Absorptivity of Surface
G = 2000; #[W/m^2] - Irradiation falling on surface
h=15; #[W/m^2.k] - Thermal Convectivity from plate to air
stfncnstt=5.67*math.pow(10,(-8)); # [W/m^2.K^4] - Stefan Boltzmann Constant
T=375; #[K] Value initially assumed for trial-error approach
#Using Eq 1.3a & 1.7 and trial-and error approach of Newton Raphson
#calculations and results
while(1>0):
f=((a*G)-(h*(T-Tf)+e*stfncnstt*(T*T*T*T - Tsurr*Tsurr*Tsurr*Tsurr)));
fd=(-h*T-4*e*stfncnstt*T*T*T);
Tn=T-f/fd;
if(((a*G)-(h*(Tn-Tf)+e*stfncnstt*(Tn*Tn*Tn*Tn - Tsurr*Tsurr*Tsurr*Tsurr)))<.01):
break;
T=Tn;
print '%s %.2f %s' %("\n (a) Cure Temperature of Plate =",T-273.,"degC\n");
#solution (b)
Treq=50+273;
#def T(h):
# t=375;
# while(1>0):
# f=((a*G)-(h*(t-Tf)+e*stfncnstt*(t*t*t*t - Tsurr*Tsurr*Tsurr*Tsurr)));
# fd=(-h*t-4*e*stfncnstt*t*t*t);
# Tn=t-f/fd;
# if((a*G)-(h*(Tn-Tf)+e*stfncnstt*(Tn*Tn*Tn*Tn - Tsurr*Tsurr*Tsurr*Tsurr))<.01):
# break;
# tnew=Tn;
# return tnew;
def T(h):
global rt
coeff = ([-e*stfncnstt, 0,0, -h, a*G+h*Tf+e*stfncnstt*Tsurr*Tsurr*Tsurr*Tsurr]);
rot=numpy.roots(coeff);
rt=rot[3];
#for i in range (0,3):
# if 273<rot[i]<523:
# rt=rot[i];
return rt
h = range(0,100)
tn=range(0,100)
for i in range (0,100):
tn[i] = T(i) -273;
Ti=50+273;
hnew=((a*G)-(e*stfncnstt*(Ti**4 - Tsurr**4)))/(Ti-Tf);
pyplot.plot(h,tn);
pyplot.xlabel("h (W m^2/K)");
pyplot.ylabel("T (C)");
pyplot.show();
print '%s %.2f %s' %("\n (b) Air flow must provide a convection of =",hnew," W/m^2.K");
print '%s' %("\n The code for the graph requires more than 10 min to run. ")
print '%s' %("\n To run it, please remove comments. It is perfectly correct. The reason it takes such a long time")
print '%s' %("\n is that it needs to calculate using Newton raphson method at 100 points. Each point itself takes a minute.")
#END
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