## Example 3.1 Page 104¶

In [1]:
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 = 10+273.; 			#[K] - Temperature of surrounding
Tf = 283.; 					#[K] - Temperature of Fluid Flow
e=.95; 						#  Emissivity of Surface
Lst=.003;    				#[m] - Thickness of Skin
kst=.3; 		 			#  [W/m.K] Effective Thermal Conductivity of Body
kins = .014;  				#  [W/m.K] Effective Thermal Conductivity of Aerogel Insulation
hr = 5.9;    				#[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
q = 100;          			#[W] Given Heat rate
#calculations

#Using Conducion Basic Eq 3.19
Rtot = (Ti-Tsurr)/q;
#Also
#Rtot=Lst/(kst*A) + Lins/(kins*A)+(h*A + hr*A)^-1
#Rtot = 1/A*(Lst/kst + Lins/kins +(1/(h+hr)))

#Thus
#For Air,
h=2.;    					#[W/m^2.k] - Natural Thermal Convectivity from body to air
Lins1 =  kins * (A*Rtot - Lst/kst - 1/(h+hr));

#For Water,
h=200.;    					#[W/m^2.k] - Natural Thermal Convectivity from body to air
Lins2 =  kins * (A*Rtot - Lst/kst - 1/(h+hr));

Tsa=305.;            		#[K]  Body Temperature Assumed

#Temperature of Skin is same in both cases as Heat Rate is same
#q=(kst*A*(Ti-Ts))/Lst
Ts = Ti - q*Lst/(kst*A);
#results

#Also from eqn of effective resistance Rtot F
print '%s %.1f %s' %("\n\n (I) In presence of Air, Insulation Thickness = ",Lins1*1000," mm")
print '%s %.1f %s' %("\n (II) In presence of Water, Insulation Thickness =",Lins2*1000.," mm");
print '%s %.2f %s' %("\n\n  Temperature of Skin =",Ts-273," degC");
#END


(I) In presence of Air, Insulation Thickness =  4.4  mm

(II) In presence of Water, Insulation Thickness = 6.1  mm

Temperature of Skin = 34.44  degC


## Example 3.2 Page 107¶

In [2]:
import math
Tf = 25+273.; 			#[K] - Temperature of Fluid Flow
L=.008;    				#[m] - Thickness of Aluminium
k=239;  				#  [W/m.K] Effective Thermal Conductivity of Aluminium
Rc=.9*math.pow(10,-4);  #[K.m^2/W]    Maximum permeasible Resistane of Epoxy Joint
q=10000.;    			#[W/m^2]    Heat dissipated by Chip
h=100.;    				#[W/m^2.k] - Thermal Convectivity from chip to air
#calculations

#Temperature of Chip

Tc = Tf + q/(h+1/(Rc+(L/k)+(1/h)));
q=(Tc-Tf)/(1/h)+(Tc-Tf)/(Rc+(L/k)+(1/h))
#results

print '%s %.2f %s' %("\n\n Temperature of Chip =",Tc-273,"degC");
print '%s' %("\n Chip will Work well below its maximum allowable Temperature ie 85 degC")
#END


Temperature of Chip = 75.31 degC

Chip will Work well below its maximum allowable Temperature ie 85 degC


## Example 3.3 Page 109¶

In [3]:
import math
D = 14 * math.pow(10,-9);    			# [m]Dia of Nanotube
s = 5*math.pow(10,-6);       			# [m]Distance between the islands
Ts = 308.4;        						#[K] Temp of sensing island
Tsurr = 300;       						#[K] Temp of surrounding
q = 11.3*math.pow(10,-6);    			#[W] Total Rate of Heat flow

#Dimension of platinum line
wpt =math.pow(10,-6);       			#[m]
tpt = 0.2*math.pow(10,-6);   			#[m]
Lpt = 250*math.pow(10,-6);   			#[m]
#Dimension of Silicon nitride line
wsn = 3*math.pow(10,-6);       			#[m]
tsn = 0.5*math.pow(10,-6);  	 		#[m]
Lsn = 250*math.pow(10,-6);   			#[m]
#From Table A.1 Platinum Temp Assumed = 325K
kpt = 71.6;    							#[W/m.K]
#From Table A.2, Silicon Nitride Temp Assumed = 325K
ksn = 15.5;   	 						#[W/m.K]
#calculations

Apt = wpt*tpt;        					#Cross sectional area of platinum support beam
Asn = wsn*tsn-Apt;    					#Cross sectional area of Silicon Nitride support beam
Acn = math.pi*D*D/4.;       			#Cross sectional Area of Carbon nanotube

Rtsupp = 1/(kpt*Apt/Lpt + ksn*Asn/Lsn); #[K/W] Thermal Resistance of each support

qs = 2*(Ts-Tsurr)/Rtsupp;    			#[W] Heat loss through sensing island support
qh = q - qs;    						#[W] Heat loss through heating island support

Th = Tsurr + qh*Rtsupp/2.;    			#[K] Temp of Heating island

#For portion Through Carbon Nanotube

kcn = qs*s/(Acn*(Th-Ts));
qs = (Th-Ts)/(s/(kcn*Acn));
#results

print '%s %.2f %s' %("\n\n Thermal Conductivity of Carbon nanotube =",kcn,"W/m.K");
#END


Thermal Conductivity of Carbon nanotube = 3111.86 W/m.K


## Example 3.4 Page 113¶

In [10]:
import math
import numpy
from numpy import linspace
import matplotlib
from matplotlib import pyplot
a = 0.25;
x1 = .05;      #[m] Distance of smaller end
x2 = .25;      #[m] Distance of larger end
T1 = 400;      #[K] Temperature of smaller end
T2 = 600;      #[K] Temperature of larger end
k = 3.46;      #[W/m.K] From Table A.2, Pyroceram at Temp 285K
T=numpy.zeros(100)
#calculations

x = numpy.linspace(0.05,100,num=100);
i=1;
for i in range (0,99):
z=float(x[i]);
T[i]=(T1 + (T1-T2)*((1/z - 1/x1)/(1/x1 - 1/x2)));

pyplot.plot(x,T);
pyplot.xlabel("x (m)");
pyplot.ylabel("T (K)");
pyplot.show()
qx = math.pi*a*a*k*(T1-T2)/(4*(1/x1 - 1/x2));            #[W]
#results

print '%s %.2f %s' %("\n\n Heat Transfer rate =",qx," W");
#END


Heat Transfer rate = -2.12  W


## Example 3.5 Page 119¶

In [2]:
%matplotlib inline
import math
import numpy
from numpy import linspace
import matplotlib
from matplotlib import pyplot
k = .055;    				#[W/m.K] From Table A.3, Cellular glass at Temp 285K
h = 5;       				#[W/m^2.K]
ri = 5*math.pow(10,-3);     #[m]  radius of tube
#calculations

rct = k/h;     				# [m] Critical Thickness of Insulation for maximum Heat loss or minimum resistance

x = numpy.linspace(0,100,num=99);
ycond= numpy.zeros(99);
yconv= numpy.zeros(99);
ytot= numpy.zeros(99);
for i in range (0,99):
z=float(x[i]);
ycond[i]=(2.30*math.log10((z+ri)/ri)/(2*math.pi*k));
yconv[i]=1/(2*math.pi*(z+ri)*h);
ytot[i]=yconv[i]+ycond[i];

pyplot.plot(x,ytot);
pyplot.xlabel("r-ri (m)");
pyplot.ylabel("R (m.K/W)");
pyplot.show();
#results

print '%s %.3f %s' %("\n\n Critical Radius is =",rct,"  m ")
print '%s %.3f %s' %("\n Heat transfer will increase with the addition of insulation up to a thickness of",rct-ri," m");
#END

Populating the interactive namespace from numpy and matplotlib


Critical Radius is = 0.011   m

Heat transfer will increase with the addition of insulation up to a thickness of 0.006  m


## Example 3.6 Page 122¶

In [18]:

import math
k = .0017;    						#[W/m.K] From Table A.3, Silica Powder at Temp 300K
h = 5;       						#[W/m^2.K]
r1 = 25./100.;      				#[m]  Radius of sphere
r2 = .275;          				#[m]  Radius including Insulation thickness

#Liquid Nitrogen Properties
T = 77;        						#[K] Temperature
rho = 804;     						#[kg/m^3] Density
hfg = 2*100000.;  					#[J/kg] latent heat of vaporisation

#Air Properties
Tsurr = 300;   						#[K] Temperature
h = 20        						;#[W/m^2.K]  convection coefficient
#calculations

Rcond = (1/r1-1/r2)/(4*math.pi*k);  #Using Eq 3.36
Rconv = 1/(h*4*math.pi*r2*r2);
q = (Tsurr-T)/(Rcond+Rconv);

print '%s %.2f %s' %("\n\n (a)Rate of Heat transfer to Liquid Nitrogen",q," W");

#Using Energy Balance q - m*hfg = 0
m=q/hfg;    						#[kg/s] mass of nirtogen lost per second
mc = m/rho*3600*24*1000.;
#results

print '%s %.2f %s' %("\n\n (b)Mass rate of nitrogen boil off ",mc,"Litres/day");
#END


(a)Rate of Heat transfer to Liquid Nitrogen 13.06  W

(b)Mass rate of nitrogen boil off  7.02 Litres/day


## Example 3.7 Page 129¶

In [5]:
import math

Tsurr = 30+273.;   						#[K] Temperature of surrounding Water
h = 1000.;     							#[W/m^2.K] Heat Convection Coeff of Water
kb = 150.;    							#[W/m.K] Material B
Lb = .02;    							#[m] Thickness Material B
ka = 75.;    							#[W/m.K] Material A
La = .05;    							#[m] Thickness Material A
qa = 1.5*math.pow(10,6);				#[W/m^3] Heat generation at wall A
qb = 0;  								#[W/m^3] Heat generation at wall B
#calculations
T2 = Tsurr + qa*La/h;
To = 100+273.15;    				    #[K] Temp of opposite end of rod
Rcondb = Lb/kb;
Rconv = 1/h;
T1 = Tsurr +(Rcondb + Rconv)*(qa*La);
#From Eqn 3.43
T0 = qa*La*La/(2*ka) + T1;

#results

print '%s %d %s' %("\n\n (a) Inner Temperature of Composite To = ",T0-273," degC")
print '%s %d %s' %("\n (b) Outer Temperature of the Composite T2 =",T2-273," degC");
#END


(a) Inner Temperature of Composite To =  140  degC

(b) Outer Temperature of the Composite T2 = 105  degC


## Example 3.9 Page 145¶

In [3]:
#Variable Initialization
%matplotlib inline

import math
import numpy
from numpy import linalg
import matplotlib
from matplotlib import pyplot
%matplotlib inline
kc = 398.;    						#[W/m.K] From Table A.1, Copper at Temp 335K
kal = 180.;   	 					#[W/m.K] From Table A.1, Aluminium at Temp 335K
kst = 14.;    						#[W/m.K] From Table A.1, Stainless Steel at Temp 335K
h = 100.;      						#[W/m^2.K] Heat Convection Coeff of Air
Tsurr = 25+273.;    				#[K] Temperature of surrounding Air
D = 5/1000.;    					#[m] Dia of rod
To = 100+273.15;    				#[K] Temp of opposite end of rod
#calculations

#For infintely long fin m = h*P/(k*A)
mc = math.pow((4*h/(kc*D)),.5);
mal = math.pow((4*h/(kal*D)),.5);
mst = math.pow((4*h/(kst*D)),.5);
x = numpy.linspace(0,0.3,100);
Tc= numpy.zeros(100);
Tal= numpy.zeros(100);
Tst= numpy.zeros(100);
for i in range (0,100):
z=x[i];
Tc[i] =Tsurr + (To - Tsurr)*math.pow(2.73,(-mc*z)) - 273;
Tal[i] = Tsurr + (To - Tsurr)*math.pow(2.73,(-mal*z)) -273;
Tst[i] = Tsurr + (To - Tsurr)*math.pow(2.73,(-mst*z)) -273;

pyplot.plot(x,Tc,label="Cu");
pyplot.plot(x,Tal,label="2024 Al");
pyplot.plot(x,Tst,label="316 SS");
pyplot.legend();
pyplot.xlabel("x (m)");
pyplot.ylabel("T (C)");
pyplot.show();

#Using eqn 3.80
qfc = math.pow((h*math.pi*D*kc*math.pi/4*D*D),.5)*(To-Tsurr);
qfal = math.pow((h*math.pi*D*kal*math.pi/4*D*D),.5)*(To-Tsurr);
qfst = math.pow((h*math.pi*D*kst*math.pi/4*D*D),.5)*(To-Tsurr);

print '%s %.2f %s %.2f %s %.2f %s' %("\n\n (a) Heat rate \n        For Copper = ",qfc,"W \n        For Aluminium =",qfal," W \n        For Stainless steel = ",qfst," W");

#Using eqn 3.76 for satisfactory approx
Linfc = 2.65/mc;
Linfal = 2.65/mal;
Linfst = 2.65/mst;

print '%s %.2f %s %.2f %s %.2f %s' %("\n\n (a) Rods may be assumed to be infinite Long if it is greater than equal to \n For Copper =",Linfc,"m \n        For Aluminium = ",Linfal," m \n        For Stainless steel =",Linfst,"m");
#END

Populating the interactive namespace from numpy and matplotlib
Populating the interactive namespace from numpy and matplotlib


(a) Heat rate
For Copper =  8.33 W
For Aluminium = 5.60  W
For Stainless steel =  1.56  W

(a) Rods may be assumed to be infinite Long if it is greater than equal to
For Copper = 0.19 m
For Aluminium =  0.13  m
For Stainless steel = 0.04 m


## Example 3.10 Page 156¶

In [21]:
import math
H = .15;   						#[m] height
k = 186;    					#[W/m.K] alumunium at 400K
h = 50;      					#[W/m^2.K] Heat convection coefficient
Tsurr = 300;    				#[K] Temperature of surrounding air
To = 500;    					#[K] Temp inside

#Dimensions of Fin
N = 5;
t = .006;    					#[m] Thickness
L = .020;     					#[m] Length
r2c = .048;        				#[m]
r1 = .025;           			#[m]
#calculations

Af = 2*math.pi*(r2c*r2c-r1*r1);
At = N*Af + 2*math.pi*r1*(H-N*t);

#Using fig 3.19
nf = .95;

qt = h*At*(1-N*Af*(1-nf)/At)*(To-Tsurr);
qwo = h*(2*math.pi*r1*H)*(To-Tsurr);
#results

print '%s %.2f %s' %("\n\n  Heat Transfer Rate with the fins =",qt,"W ")
print '%s %.2f %s' %(" \n  Heat Transfer Rate without the fins =",qwo,"W")
print '%s %.2f %s' %("\n Thus Increase in Heat transfer rate of",qt-qwo," W is observed with fins");
#END


Heat Transfer Rate with the fins = 689.60 W

Heat Transfer Rate without the fins = 235.62 W

Thus Increase in Heat transfer rate of 453.98  W is observed with fins


## Example 3.11 Page 158¶

In [22]:
import math
Wc =.05;    				#[m] width
H = .026;  	 				#[m] height
tc = .006;  				#[m] thickness of cell
V = 9.4;    				#[m/sec] vel of cooling air
P = 9;      				#[W] Power generated
C = 1000;    				#[W/(m^3/s)] Ratio of fan power consumption to vol flow rate
k = 200;    				#[W/m.K] alumunium
Tsurr = 25+273.15;    		#[K] Temperature of surrounding air
Tc = 56.4+273.15;    		#[K] Temp of fuel cell
Rtcy = math.pow(10,-3);     #[K/W]   Contact thermal resistance
tb = .002;           		#[m] thickness of base of heat sink
Lc = .05;        			#[m] length of fuel cell
#Dimensions of Fin
tf = .001;    				#[m] Thickness
Lf = .008;     				#[m] Length
#calculations

Vf = V*(Wc*(H-tc));    		#[m^3/sec] Volumetric flow rate
Pnet = P - C*Vf;

P = 2*(Lc+tf);
Ac = Lc*tf;
N = 22;
a=(2*Wc - N*tf)/N;
h = 19.1;            		#/[W/m^2.K]
q = 11.25;            		#[W]
m = math.pow((h*P/(k*Ac)),.5);
Rtf = math.pow((h*P*k*Ac),(-.5))/ math.tanh(m*Lf);
Rtc = Rtcy/(2*Lc*Wc);
Rtbase = tb/(2*k*Lc*Wc);
Rtb = 1/(h*(2*Wc-N*tf)*Lc);
Rtfn = Rtf/N;
Requiv = 1/(1/Rtb + 1/Rtfn);
Rtot = Rtc + Rtbase + Requiv;

Tc2 = Tsurr +q*(Rtot);
#results

print '%s %.2f %s' %("\n\n (a) Power consumed by fan is more than the generated power of fuel cell, and hence system cannot produce net power = ",Pnet ,"W ")
print '%s %.2f %s %.2f %s' %("\n\n (b) Actual fuel cell Temp is close enough to ",Tc2-273.," degC for reducing the fan power consumption by half ie Pnet =",C*Vf/2.," W, we require 22 fins, 11 on top and 11 on bottom.");

#END


(a) Power consumed by fan is more than the generated power of fuel cell, and hence system cannot produce net power =  -0.40 W

(b) Actual fuel cell Temp is close enough to  54.47  degC for reducing the fan power consumption by half ie Pnet = 4.70  W, we require 22 fins, 11 on top and 11 on bottom.


## Example 3.12 Page 163¶

In [23]:
import math
hair = 2.;     			#[W/m^2.K] Heat convection coefficient air
hwater = 200.;     		#[W/m^2.K] Heat convection coefficient water
hr = 5.9 ;     			#[W/m^2.K] Heat radiation coefficient
Tsurr = 297.;    		#[K] Temperature of surrounding air
Tc = 37+273.;    		#[K] Temp inside
e = .95;
A = 1.8 ;       		#[m^2] area
#Prop of blood
w = .0005 ;      		#[s^-1] perfusion rate
pb = 1000.;        		#[kg/m^3] blood density
cb = 3600.;        		#[J/kg] specific heat
#Dimensions & properties of muscle & skin/fat
Lm = .03 ;       		#[m]
Lsf = .003 ;     		#[m]
km = .5 ;       		#[W/m.K]
ksf = .3;        		#[W/m.K]
q = 700.;         		#[W/m^3]  Metabolic heat generation rate
#calculations

Rtotair = (Lsf/ksf + 1/(hair + hr))/A;
Rtotwater = (Lsf/ksf + 1/(hwater+hr))/A;
#please correct this in the textbook.
m = math.pow((w*pb*cb/km),.5);
Theta = -q/(w*pb*cb);

Tiair = (Tsurr*math.sinh(m*Lm) + km*A*m*Rtotair*(Theta + (Tc + q/(w*pb*cb))*math.cosh(m*Lm)))/(math.sinh(m*Lm)+km*A*m*Rtotair*math.cosh(m*Lm));
qair = (Tiair - Tsurr)/Rtotair;

Tiwater = (Tsurr*math.sinh(m*Lm) + km*A*m*Rtotwater*(Theta + (Tc + q/(w*pb*cb))*math.cosh(m*Lm)))/(math.sinh(m*Lm)+km*A*m*Rtotwater*math.cosh(m*Lm));
qwater = (Tiwater - Tsurr)/Rtotwater;
#results

print '%s %.2f %s' %("\n\n For Air \n Temp excess Ti = ",Tiair-273," degC ")
print '%s %.2f %s %.2f %s %.2f %s' %("and Heat loss rate =",qair," W \n\n For Water \n Temp excess Ti = ",Tiwater-273," degC and Heat loss rate =",qwater,"W ");
#END


For Air
Temp excess Ti =  34.77  degC
and Heat loss rate = 141.99  W

For Water
Temp excess Ti =  28.25  degC and Heat loss rate = 514.35 W