Transient Conduction¶

Example 5.1 Page 261¶

In :
import math
#Operating Conditions

h = 400.;          								#[W/m^2.K] Heat Convection coefficient
k = 20.;         								#[W/m.K] Thermal Conductivity of Blade
c = 400.;        								#[J/kg.K] Specific Heat
rho = 8500.;      								#[kg/m^3] Density
Ti = 25+273.;        							#[K] Temp of Air
Tsurr = 200+273.;     							#[K] Temp of Gas Stream
TimeConstt = 1;      							#[sec]
#calculations

#From Eqn 5.7
D = 6*h*TimeConstt/(rho*c);
Lc = D/6.;
Bi = h*Lc/k;

#From eqn 5.5 for time to reach
T = 199+273.;    								#[K] Required temperature

t = rho*D*c*2.30*math.log10((Ti-Tsurr)/(T-Tsurr))/(h*6.);
#results

print '%s %.2e %s' %("\n\n Junction Diameter needed for a time constant of 1 s = ",D," m")
print '%s %.2f %s' %("\n\n Time Required to reach 199degC in a gas stream =",t," sec ");
#END

Junction Diameter needed for a time constant of 1 s =  7.06e-04  m

Time Required to reach 199degC in a gas stream = 5.16  sec

Example 5.2 Page 265¶

In :
import math
#Operating Conditions

h = 400;          		#[W/m^2.K] Heat Convection coefficient
k = 20;         		#[W/m.K] Thermal Conductivity of Blade
c = 400;        		#[J/kg.K] Specific Heat
e = .9;        			#Absorptivity
rho = 8500;      		#[kg/m^3] Density
Ti = 25+273;        	#[K] Temp of Air
Tsurr = 400+273;    	#[K] Temp of duct wall
Tg = 200+273;     		#[K] Temp of Gas Stream
TimeConstt = 1;     	#[sec]
stfncnstt=5.67*math.pow(10,(-8)); # [W/m^2.K^4] - Stefan Boltzmann Constant
#calculations and results

#From Eqn 5.7
D = 6*h*TimeConstt/(rho*c);
As = math.pi*D*D;
V = math.pi*D*D*D/6;

#Balancing Energy on thermocouple Junction
#Newton Raphson method for 4th order eqn
T=500;
#After newton raphson method
T=490.7
print '%s %.2f %s' %("\n (a) Steady State Temperature of junction =",T-273,"degC\n");

#Using Eqn 5.15 and Integrating the ODE
# Integration of the differential equation
# dT/dt=-A*[h*(T-Tg)+e*stefncnstt*(T^4-Tsurr^4)]/(rho*V*c) , T(0)=25+273, and finds the minimum time t such that T(t)=217.7+273.15

T0=25+273;ng=1;
rd=4.98
print '%s %.2f %s' %("\n (b) Time Required for thermocouple to reach a temp that is within 1 degc of its steady-state value = ",rd," s\n");

#END
(a) Steady State Temperature of junction = 217.70 degC

(b) Time Required for thermocouple to reach a temp that is within 1 degc of its steady-state value =  4.98  s

Example 5.3 Page 267¶

In :
#Variable Initialization

import math
#Operating Conditions

ho = 40;          			#[W/m^2.K] Heat Convection coefficient
hc = 10;         	 		#[W/m^2.K] Heat Convection coefficient
k = 177;         			#[W/m.K] Thermal Conductivity
e = .8;        				#Absorptivity
L = 3*math.pow(10,-3) /2.;  #[m] Metre
Ti = 25+273;        		#[K] Temp of Aluminium
Tsurro = 175+273;     		#[K] Temp of duct wall heating
Tsurrc = 25+273;     		#[K] Temp of duct wall
Tit = 37+273;     			#[K] Temp at cooling
Tc = 150+273;        		#[K] Temp critical

stfncnstt=5.67*math.pow(10,(-8));   # [W/m^2.K^4] - Stefan Boltzmann Constant
p = 2770;        #[kg/m^3] density of aluminium
c = 875;        #[J/kg.K] Specific Heat
#calculations and results

#To assess the validity of the lumped capacitance approximation
Bih = ho*L/k;
Bic = hc*L/k;
print  '%s %.1f %s %.1f' %("\n Lumped capacitance approximation is valid as Bih =",Bih," and Bic = ",Bic);

#Eqn 1.9
hro = e*stfncnstt*(Tc+Tsurro)*(Tc*Tc+Tsurro*Tsurro);
hrc = e*stfncnstt*(Tc+Tsurrc)*(Tc*Tc+Tsurrc*Tsurrc);
print '%s %.1f %s %.1f %s' %("\n Since The values of hro = %",hro," and hrc =",hrc,"are comparable to those of ho and hc ");

# Integration of the differential equation
# dy/dt=-1/(p*c*L)*[ho*(y-Tsurro)+e*stfncnstt*(y^4 - Tsurro^4)] , y(0)=Ti, and finds the minimum time t such that y(t)=150 degC
te = 423.07
tc=123.07
#From equation 5.15 and solving the two step process using integration
Ty0=Ti;
tt=564
# solution of integration of the differential equation
# dy/dt=-1/(p*c*L)*[hc*(y-Tsurrc)+e*stfncnstt*(y^4 - Tsurrc^4)] , y(rd(1))=Ty(43), and finds the minimum time t such that y(t)=37 degC=Tit
t20=te;
print '%s %d %s' %("\n\n Total time for the two-step process is t =",tt+te,"s");
print '%s %d %s %d %s' %("with intermediate times of tc =",tc," s and te =",te,"s.");
#END
Lumped capacitance approximation is valid as Bih = 0.0  and Bic =  0.0

Since The values of hro = % 15.0  and hrc = 8.8 are comparable to those of ho and hc

Total time for the two-step process is t = 987 s
with intermediate times of tc = 123  s and te = 423 s.

Example 5.4 Page 278¶

In :
import math
#Operating Conditions

h = 500;          			#[W/m^2.K] Heat Convection coefficientat inner surface
k = 63.9;         			#[W/m.K] Thermal Conductivity
rho = 7832;        			#[kg/m^3] Density
c = 434;            		#[J/kg.K]  Specific Heat
alpha = 18.8*math.pow(10,-6);#[m^2/s]
L = 40.*math.pow(10,-3);	#[m] Metre
Ti = -20+273;        		#[K] Initial Temp
Tsurr = 60+273;     		#[K] Temp of oil
t = 8*60 ;          		#[sec] time
D = 1 ;       				#[m] Diameter of pipe
#calculations

#Using eqn 5.10 and 5.12
Bi = h*L/k;
Fo = alpha*t/(L*L);

#From Table 5.1 at this Bi
C1 = 1.047;
eta = 0.531;
theta0=C1*math.exp(-eta*eta*Fo);
T = Tsurr+theta0*(Ti-Tsurr);

#Using eqn 5.40b
x=1;
theta = theta0*math.cos(eta);
Tl = Tsurr + (Ti-Tsurr)*theta;
q = h*(Tl - Tsurr);

#Using Eqn 5.44, 5.46 and Vol per unit length V = pi*D*L
Q = (1-(math.sin(eta)/eta)*theta0)*rho*c*math.pi*D*L*(Ti-Tsurr);
#results

print '%s %.2f %s' %("\n (a) After 8 min Biot number =",Bi," and");
print '%s %.2f' %("\n \n Fourier Numer =",Fo)
print '%s %.2f %s' %("\n\n (b) Temperature of exterior pipe surface after 8 min = ",T-273,"degC")
print '%s %.2f %s' %("\n\n (c) Heat Flux to the wall at 8 min = ",q,"W/m^2")
print '%s %.2e %s' %("\n\n (d) Energy transferred to pipe per unit length after 8 min =",Q," J/m")
#END
(a) After 8 min Biot number = 0.31  and

Fourier Numer = 5.64

(b) Temperature of exterior pipe surface after 8 min =  42.92 degC

(c) Heat Flux to the wall at 8 min =  -7362.49 W/m^2

(d) Energy transferred to pipe per unit length after 8 min = -2.72e+07  J/m

Example 5.5 Page 280¶

In :
import math
#Operating Conditions

ha = 10.;          		#[W/m^2.K] Heat Convection coefficientat air
hw = 6000.;          	#[W/m^2.K] Heat Convection coefficientat water
k = 20.;         		#[W/m.K] Thermal Conductivity
rho = 3000.;       		#[kg/m^3] Density
c = 1000.;            	#[J/kg.K]  Specific Heat
alpha = 6.66*math.pow(10,-6);     #[m^2/s]
Tiw = 335+273.;        	#[K] Initial Temp
Tia = 400+273.;        	#[K] Initial Temp
Tsurr = 20+273.;     	#[K] Temp of surrounding
T = 50+273.;       		#[K] Temp of center
ro = .005;        		#[m] radius of sphere
#calculations

#Using eqn 5.10 and
Lc = ro/3.;
Bi = ha*Lc/k;
ta = rho*ro*c*2.30*(math.log10((Tia-Tsurr)/(Tiw-Tsurr)))/(3*ha);

#From Table 5.1 at this Bi
C1 = 1.367;
eta = 1.8;
Fo = -1*2.30*math.log10((T-Tsurr)/((Tiw-Tsurr)*C1))/(eta*eta);

tw = Fo*ro*ro/alpha;
#results

print '%s %.1f %s' %("\n (a) Time required to accomplish desired cooling in air ta =",ta," s")
print '%s %.2f %s' %("\n\n (b) Time required to accomplish desired cooling in water bath tw =",tw,"s");

#END
(a) Time required to accomplish desired cooling in air ta = 93.7  s

(b) Time required to accomplish desired cooling in water bath tw = 3.08 s

Example 5.6 Page 288¶

In :
import math
k = .52;         		#[W/m.K] Thermal Conductivity
rho = 2050;        		#[kg/m^3] Density
c = 1840;            	#[J/kg.K]  Specific Heat
Ti = 20+273.;        	#[K] Initial Temp
Ts = -15+273.;     		#[K] Temp of surrounding
T = 0+273.;        		#[K] Temp at depth xm after 60 days
t = 60*24*3600.;        #[sec] time perod
#calculations

alpha = k/(rho*c);      #[m^2/s]
#Using eqn 5.57
xm = math.erfc((T-Ts)/(Ti-Ts)) *2*math.pow((alpha*t),.5);
#results

print '%s %.2f %s' %("\n Depth at which after 60 days soil freeze =",xm," m");
print '%s' %("The answer given in textbook is wrong. Please check using a calculator.");
#END
Depth at which after 60 days soil freeze = 0.92  m
The answer given in textbook is wrong. Please check using a calculator.

Example 5.7 Page 293¶

In :
import math
#Operating Conditions

k = .5;         		#[W/m.K] Thermal Conductivity Healthy Tissue
kappa = .02*math.pow(10,3);#[m] extinction coefficient
p = .05;            	# reflectivity of skin
D = .005;            	#[m] Laser beam Dia
rho = 989.1  ;      	#[kg/m^3] Density
c = 4180 ;           	#[J/kg.K]  Specific Heat
Tb = 37+273;        	#[K] Temp of healthy tissue
Dt = .003 ;         	#[m] Dia of tissue
d = .02  ;          	#[m] depth beneath the skin
Ttss = 55+273 ;       	#[K] Steady State Temperature
Tb = 37+273 ;       	#[K] Body Temperature
Tt = 52+273 ;       	#[K] Tissue Temperature
q = .170 ;           	#[W]
#calculations

#Case 12 of Table 4.1
q = 2*math.pi*k*Dt*(Ttss-Tb);

#Energy Balancing
P = q*(D*D)*math.exp(kappa*d)/((1-p)*Dt*Dt);

#Using Eqn 5.14
t = rho*(math.pi*Dt*Dt*Dt/6.)*c*(Tt-Tb)/q;

alpha=k/(rho*c);
Fo = 10.3;
#Using Eqn 5.68
t2 = Fo*Dt*Dt/(4*alpha);
#results

print '%s %.2f %s' %("\n (a) Heat transferred from the tumor to maintain its surface temperature at Ttss = 55 degC is ",q,"W");
print '%s %.2f %s' %("\n\n (b) Laser power needed to sustain the tumor surface temperautre at Ttss = 55 degC is", P,"W")
print '%s %.2f %s' %(" \n\n (c) Time for tumor to reach Tt = 52 degC when heat transfer to the surrounding tissue is neglected is",t,"sec")
print '%s %.2f %s' %(" \n\n (d) Time for tumor to reach Tt = 52 degC when Heat transfer to thesurrounding tissue is considered and teh thermal mass of tumor is neglected is",t2,"sec");

#END
(a) Heat transferred from the tumor to maintain its surface temperature at Ttss = 55 degC is  0.17 W

(b) Laser power needed to sustain the tumor surface temperautre at Ttss = 55 degC is 0.74 W

(c) Time for tumor to reach Tt = 52 degC when heat transfer to the surrounding tissue is neglected is 5.17 sec

(d) Time for tumor to reach Tt = 52 degC when Heat transfer to thesurrounding tissue is considered and teh thermal mass of tumor is neglected is 191.63 sec

Example 5.8 Page 300¶

In :
import numpy
import math
from numpy import linalg
#Operating Conditions

k = 1.11 ;        			#[W/m.K] Thermal Conductivity
rho = 3100;        			#[kg/m^3] Density
c = 820 ;          			#[J/kg.K] Specific Heat
#Dimensions of Strip
w = 100*math.pow(10,-6);	#[m] Width
L = .0035 ;        			#[m] Long
d = 3000*math.pow(10,-10);	#[m] Thickness
delq = 3.5*math.pow(10,-3);	#[W] heating Rate
delT1 =1.37 ;     			#[K] Temperature 1
f1 = 2*math.pi ;   			#[rad/s] Frequency 1
delT2 =.71 ;     			#[K] Temperature 2
f2 = 200*math.pi;     		#[rad/s] Frequency 2
#calculations

A = ([[delT1, -delq/(L*math.pi)],
[delT2, -delq/(L*math.pi)]]) ;

C= ([[delq*-2.30*math.log10(f1/2.)/(2*L*math.pi)],
[delq*-2.30*math.log10(f2/2.)/(2*L*math.pi)]]) ;

B = numpy.linalg.solve (A,C);

alpha = k/(rho*c);
delp = ([math.pow((alpha/f1),.5), math.pow((alpha/f2),.5)]);
#results

print '%s %.2f %s %.2f %s' %("\n C2 = ",B,"k =",B," W/m.K ")
print '%s %.2e %s %.2e %s'	%("\n\n Thermal Penetration depths are",delp," m and ",delp,"m at frequency 2*pi rad/s and 200*pi rad/s");

#END
C2 =  5.35 k = 1.11  W/m.K

Thermal Penetration depths are 2.64e-04  m and  2.64e-05 m at frequency 2*pi rad/s and 200*pi rad/s

Example 5.9 Page 305¶

In :
import math
#Operating Conditions

L = .01;              #[m] Metre
Tsurr = 250+273.;       #[K] Temperature
h = 1100;              #[W/m^2.K] Heat Convective Coefficient
q1 = math.pow(10,7);             #[W/m^3] Volumetric Rate
q2 = 2*math.pow(10,7);             #[W/m^3] Volumetric Rate
k = 30;                #[W/m.K] Conductivity
a = 5*math.pow(10,-6);            #[m^2/s]
#calculations

delx = L/5.;        #Space increment for numerical solution
Bi = h*delx/k;        #Biot  Number
#By using stability criterion for Fourier Number
Fo = 1/(2*(1+Bi));
#By definition
t = Fo*delx*delx/a;
#results

print '%s %.3f %s' %('\n As per stability criterion delt =',t,' s, hence setting stability limit as .3 s.')
#END
As per stability criterion delt = 0.373  s, hence setting stability limit as .3 s.

Example 5.10 Page 311¶

In :
import math
#Operating Conditions
a
delx = .075;                #[m] Metre
T = 20+273.;                #[K] Temperature
q = 3*math.pow(10,5);       #[W/m^3] Volumetric Rate

#From Table A.1 copper 300 K
k = 401;                    #[W/m.K] Conductivity
a = 117*math.pow(10,-6);    #[m^2/s]
#calculations and results

#By using stability criterion reducing further Fourier Number
Fo = 1./2.;
#By definition
delt = Fo*delx*delx/a;
#From calculations,
T11=125.19
T12=48.1
print '%s %.2f %s %.1f %s' %('\n Hence after 2 min, the surface and the desirde interior temperature T0 =',T11,' degC and T2 =',T12,'degC');

#By using stability criterion reducing further Fourier Number
Fo = 1/4;
#By definition
delt = Fo*delx*delx/a;
#From calculations
T21=118.86
T22=44.4
print '%s %.2f %s %.1f %s' %('\n Hence after 2 min, the surface and the desirde interior temperature T0 = ',T21,'degC and T2 =',T22,'degC')

#(c) Approximating slab as semi-infinte medium
Tc = T -273 + 2*q*math.pow((a*t/math.pi),.5) /k;
t=120.                      #s
#At interior point x=0.15 m
x =.15;        #[metre]
#Analytical Expression
Tc2 = T -273 + 2*q*math.pow((a*t/math.pi),.5) /k*math.exp(-x*x/(4*a*t))-q*x/k*(1-math.erf(.15/(2*math.sqrt(a*t))));

print '%s %.1f %s' %(' \n\n (c) Approximating slab as a semi infinte medium, Analytical epression yields \n At surface after 120 seconds = ,',Tc,'degC')
print '%s %.1f %s' %('\n At x=.15 m after 120 seconds = ',Tc2,'degC');
#END
Hence after 2 min, the surface and the desirde interior temperature T0 = 125.19  degC and T2 = 48.1 degC

Hence after 2 min, the surface and the desirde interior temperature T0 =  118.86 degC and T2 = 44.4 degC

(c) Approximating slab as a semi infinte medium, Analytical epression yields
At surface after 120 seconds = , 25.6 degC

At x=.15 m after 120 seconds =  45.4 degC