One-dimensional, Steady State Conduction¶
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
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
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
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
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
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
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
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
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
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
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