Chapter 4 - Transient (Unsteady State) Heat Conduction

Example No : 4.1 - Page : 102

In [1]:
from __future__ import division
# given data
L=1   # in m
rho=1600   # in kg/m**3
k=40   # in w/mK
Cp=4*10**3   # in J/kgK
a=900   # in  degree C
b=-300   # in  degree C/m
c=-50   # in  degree C/m**2
Qg=1*10**3   # in kW/m**2
A=10   # area in m**2
#t=a+b*x+c*x**2 at any instant, so
# dtBYdx= b+2*c*x
# d2tBYdx2 = 2*c, then

# Part(a)
#q1= -k*A*dtBYdx , at
x=0#
q1= -k*A*(b+2*c*x)   # in w
#q2= -k*A*dtBYdx , at
x=L#
q2= -k*A*(b+2*c*x)   # in w
E_stored= (q1-q2)+Qg*A*L   # in watt
print "The rate of change of energy storage = %0.1e watt" %E_stored

# Part(b)
alpha= k/(rho*Cp)   # in m**2s
d2tBYdx2 = 2*c#
dtBYdtoh= alpha*(d2tBYdx2+Qg/k )   # in degree C/sec
print "Rate of change of temperature = %0.3e degree C/sec" %dtBYdtoh
print "Since dt by dx is independent of x. Hence time rate of charge of temperature throughout wall will remain same."

The rate of change of energy storage = -3.0e+04 watt
Rate of change of temperature = -4.688e-04 degree C/sec
Since dt by dx is independent of x. Hence time rate of charge of temperature throughout wall will remain same.

Example No : 4.2 - Page : 109

In [5]:
from math import exp
from numpy import log, pi
# given data
k=40   # in W/mK
rho=7800   # in kg/m**3
C=450   # in J/kgK
d=20*10**-3   # in m
r=d/2#
t_i=400   # in degree C
t=85   # in degree C
t_infinite=25   # in degree C
h=80   # in W/m**2K
#l_s=V/A = (4/3*pi*r**3)/(4*pi*r**2) = r/3
l_s=r/3   # in m
Bi= h*l_s/k#
# since Biot number is less than 0.1, hence lumped heat capacity system analysis can be applied

# Part(a)
# Formula (t-t_infinite)/(t_i-t_infinite)= %e**(-h*A*toh/(rho*V*C)) = %e**(-h*toh/(rho*l_s*C))
toh= -log((t-t_infinite)/(t_i-t_infinite))*(rho*l_s*C)/h   # in sec
print "The time require to cool the sphere = %0.3f sec" %toh

# Part(b)
# dtBYdtoh = h*A*(t_i-t_infinite)/(rho*V*C) = h*(t_i-t_infinite)/(rho*l_s*C)
dtBYdtoh = h*(t_i-t_infinite)/(rho*l_s*C)   # in degree C/sec
print "Initial rate of cooling = %0.3f degree C/sec" %dtBYdtoh

# Part(c)
A=4*pi*r**2#
toh=60#
q_in= h*A*(t_i-t_infinite)*exp(-h*toh/(rho*l_s*C))   # in watt
print "Instantaneous heat transfer rate = %0.3f watt" %q_in

# Part(d) Total energy transferred during first one minute
V=4/3*pi*r**3#
TotalEnergy = rho*C*V*(t_i-t_infinite)*(1-exp(-h*toh/(rho*C*l_s)))#
print "Total energy transferred during first one minute = %0.3f watt" %TotalEnergy

# Note: Answer of first and last part in the book is wrong

The time require to cool the sphere = 268.015 sec
Initial rate of cooling = 2.564 degree C/sec
Instantaneous heat transfer rate = 25.013 watt
Total energy transferred during first one minute = 1855.401 watt

Example No : 4.3 - Page : 111

In [13]:
from math import exp
# given data
k=40   # in W/mK
rho=8200   # in kg/m**3
C=400   # in J/kgK
D=6*10**-3   # in m
R=D/2#
t_i=30   # in degree C
t_infinite1=400   # for 10 sec in degree C
t_infinite2=20   # for 10 sec in degree C
h=50   # in W/m**2K

# Part(a)
#l_s= V/A = R/3
l_s= R/3   # in m
#toh= rho*V*C/(h*A) = rho*C*l_s/h
toh= rho*C*l_s/h   # in sec
print "Time constance = %0.1f sec" %toh

# Part (b)
Bi= h*l_s/k#
# since Bi < 0.1 , hence lumped heat capacity analysis is valid. Now , temperature attained by junction in 10 seconds when exposed to hot air at 400 degree C
toh=10   # in sec
# (t-t_infinite1)/(t_i-t_infinite1)= %e**(-h*A*toh/(rho*V*C)) = %e**(-h*toh/(rho*l_s*C))
t= exp(-h*toh/(rho*l_s*C))*(t_i-t_infinite1)+t_infinite1   # in degree C

print "The junction is taken out from hot air stream and placed in stream of still air 20 degree C."
print "The initial temperature in this case = ",round(t,3)
t_i=t#
toh=20   # in sec
t= exp(-h*toh/(rho*l_s*C))*(t_i-t_infinite2)+t_infinite2   # in degree C
print "The temperature attained by junction = %0.3f degree C" %t

# Note: In the last, calculation to find the value of t is wrong so Answer in the book is wrong

Time constance = 65.6 sec
The junction is taken out from hot air stream and placed in stream of still air 20 degree C.
The initial temperature in this case =  82.314
The temperature attained by junction = 65.939 degree C

Example No : 4.4 - Page : 112

In [14]:
from math import log
# given data
k=8   # in W/mK
alpha=4*10**-6   # in m**2/s
h=50   # in W/m**2K
D=6*10**-3   # in m
R=D/2#
T=0.5   # where T = (t-t_infinite)/(t_i-t_infinite)
#l_s= V/A = R/3
l_s= R/2   # in m
Bi= h*l_s/k#
# since Bi < 0.1 , hence lumped heat capacity analysis can be applied
# toh= rho*V*C/(h*A) = rho*C*l_s/h = k*l_s/(h*alpha)
toh=  k*l_s/(h*alpha)   # in seconds
print "Time constant = %0.f seconds" %toh
# It is given that (t-t_infinite)/(t_i-t_infinite) = 0.5 =  %e**(-h*A*c /(rho*V*C)) = %e**(-h*c/(rho*l_s*C)) = %e**(-h*alpha*c/(l_s))
# or (t-t_infinite)/(t_i-t_infinite) = %e**(-h*alpha*c/(l_s)#
c= -log(T)*l_s/(h*alpha)   # in sec
print "The time required to temperature change to reach half of its initial value = %0.1f seconds" %c

Time constant = 60 seconds
The time required to temperature change to reach half of its initial value = 5.2 seconds

Example No : 4.5 - Page : 113

In [2]:
# given data
#t=450-500*x+100*x**2+150*x**3 at any instant, so
# dtBYdx= -500+200*x+450*x**2
L=0.5   # thickness of the wall in meter
k=10   # in W/mK
# Rate of heating entering in the wall per unit area, at
x=0#
#q1= -k*dtBYdx
q1= -k*(-500+200*x+450*x**2)   # in W/m**2
# Rate of heat going out of the wall per unit area , at
x=L#
q2= -k*(-500+200*x+450*x**2)   # in W/m**2
E=q1-q2   # in W/m**2
print "Heat energy stored per unit area = %0.0f W/m**2" %E

Heat energy stored per unit area = 2125 W/m**2

Example No : 4.6 - Page : 114

In [14]:
from __future__ import division
# given data
k=385   # in W/mK
h=100   # in W/m**2K
delta =2*10**-3   # thickness of plate in meter
A=25*25   # area of plate in square meter
rho=8800   # kg/m**3
C=400   # J/kg-K
# l_s= V/A= L*B*delta/(2*L*B) = delta/2
l_s= delta/2   # in meter
Bi= h*l_s/k#
# since Bi < 0.1 , hence lumped heat capacity analysis can be applied

# Part(i)
# toh= rho*V*C/(h*A) = rho*C*l_s/h
toh= rho*C*l_s/h   # in second
print "Time constant = %0.1f seconds" %toh

# Part(ii)
t_i=400   # in degree C
t=40   # in degree C
t_infinite=25   # in degree C
# (t-t_infinite)/(t_i-t_infinite) =  %e**(-h*A*toh /(rho*V*C)) = %e**(-h*toh/(rho*l_s*C)) 
toh= -log((t-t_infinite)/(t_i-t_infinite))*rho*C*l_s/h   # in sec
print "The time required for the plate to reach the temperature of 40 degree C = %0.1f seconds" %toh

Time constant = 35.2 seconds
The time required for the plate to reach the temperature of 40 degree C = 113.3 seconds

Example No : 4.7 - Page : 114

In [15]:
from math import log
# given data
k=380   # in W/mK
delta =6*10**-2   # thickness of plate in meter
rho=8800   # kg/m**3
C=400   # J/kg-K
# l_s= V/A = delta/2
l_s= delta/2   # in meter
t=80   # in degree C
t_i=200   # in degree C
t_inf=30   # in degree C
hw= 75   # in W/m**2K
ha= 10   # in W/m**2K

# Part(i)
# ha*A*(t-t_inf_a)+ hw*A*(t-t_inf_w) = -rho*V*C*dtBYdtho, since t_ini_a = t_inf_w = t_inf = 30 degree C
# (ha+hw)*A*(t-t_inf)= -rho*V*C*dtBYdtho
# (ha+hw)/(rho*C*V)*A*dtoh = -dt/(t-t_inf)
# integrate('(ha+hw)/(rho*V*C)*A','toh',0,toh) = integrate('1/(t-t_inf)','t',t_i,t)
toh= -rho*l_s*C/(ha+hw)*log((t-t_inf)/(t_i-t_inf))#
print "Time required to cool plate to 80 degree C is :",round(toh,1),"seconds =",round(toh/60,2),"minutes"

# Part (ii)
t= -rho*l_s*C/(2*ha)*log((t-t_inf)/(t_i-t_inf))#
print "Time required to cool plate in only air is :",round(t,1),"seconds =",round(t/60,2),"minutes"

Time required to cool plate to 80 degree C is : 1520.4 seconds = 25.34 minutes
Time required to cool plate in only air is : 6461.5 seconds = 107.69 minutes

Example No : 4.8 - Page : 116

In [16]:
from numpy import pi
# given data
k=45   # in W/m degree C
d =0.1   # in meter
l =0.30   # in meter
t=800   # in degree C
t_i=100   # in degree C
t_infinite=1200   # in degree C
h= 120   # in W/m**2 degree C
alpha=1.2*10**-5   # in meter
rhoC= k/alpha#
V=pi/4*d**2*l   # in m**3
A= pi*d*l + 2*pi/4*d**2   # in m**2
# l_s= V/A = (pi/4*d**2*l)/(pi*d*l + 2*pi/4*d**2) = d*l/(4*l+2*d**2)
l_s = d*l/(4*l+2*d**2)#
Bi= h*l_s/k#
# since Bi < 0.1 , hence lumped heat capacity analysis can be applied
# (t-t_infinite)/(t_i-t_infinite)  =  %e**(-h*A*toh /(rho*V*C)) = %e**(-h*toh/(rho*l_s*C)) = %e**(-h*toh/(rhoC*l_s))
toh = -log((t-t_infinite)/(t_i-t_infinite))*rhoC*l_s/h   # in sec

# So, the velocity of ingot passing through the furnace
FurnaceLength = 8*100   # in cm
time = toh#
Velocity = FurnaceLength/time   # in cm/sec
print "Maximum speed = %0.4f cm/sec" %Velocity

Maximum speed = 1.0291 cm/sec

Example No : 4.9 - Page : 117

In [17]:
from math import log
# given data
rho=8500   # in kg/m**3
C=400   # J/kgK
toh=1   # in sec
h= 400   # in W/m**2 degree C
t=198   # in degree C
t_i=25   # in degree C
t_infinite=200   # in degree C

# Part (1)
# toh =rho*V*C/(h*A) = rho*C*l_s/h
l_s= toh*h/(rho*C)#
# l_s = V/A = r/3 
r=3*l_s   # in m
r=r*10**3   # in mm
d=2*r   # in m
print "Junction diameter needed for the thermocouple = %0.3f mili miter" %d

# Part(ii)
# toh= -rho*V*C/(h*A)*log((t-t_infinite)/(t_i-t_infinite))  
toh = -toh*log((t-t_infinite)/(t_i-t_infinite))#
print "Time required for the thermocouple junction to reach 198 degree C = %0.3f seconds" %toh

Junction diameter needed for the thermocouple = 0.706 mili miter
Time required for the thermocouple junction to reach 198 degree C = 4.472 seconds

Example No : 4.10 - Page : 118

In [3]:
# given data
L=40*10**-2   # in m
k=1.5   # in W/mK
A=4   # in square meter
alpha=1.65*10**-3   # in m**2/h
#T = 50-40*x+10*x**2+20*x**3-15*x**4 , so
# dtBYdx= -40+20*x+60*x**2-60*x**3
# d2tBYdx2 = 20+120*x-180*x**2

# Part (a) Heat entering the slab
#q1= -k*A*dtBYdx , at
x=0#
qi= -k*A*(-40+20*x+60*x**2-60*x**3)   # in w
print "Heat entering the slab = %0.0f watt" %qi
# Heat leaving the slab
#ql= -k*A*dtBYdx , at
x=L#
ql= -k*A*(-40+20*x+60*x**2-60*x**3)   # in w
print "Heat leaving the slab = %0.2f watt" %ql

# Part (b) Rate of heat storage
RateOfHeatStorage = qi-ql   # in watt
print "Rate of heat storage = %0.2f watt" %RateOfHeatStorage

# Part (c) Rate of temperature change
# d2tBYdx2 = 1/alpha*dtBYdtoh
# dtBYdtoh= alpha*d2tBYdx2, at
x=0#
dtBYdtoh = alpha*(20+120*x-180*x**2)   # in degree C/h
print "The rate of temperature change at entering the slab = %0.3f degree C/h" %dtBYdtoh
# dtBYdtoh= alpha*d2tBYdx2, at
x=L
dtBYdtoh = alpha*(20+120*x-180*x**2)   # in degree C/h
print "The rate of temperature change at leaving the slab = %0.3f degree C/h" %dtBYdtoh

# Part (d) for the rate of heating or cooling to be maximum
# dBYdx of dtBYdtoh = 0
# dBYdx of (alpha*d2tBYdx2) =0
# d3tBYdx3 = 0
x=120/360   # in meter
print "The point where rate of heating or cooling is maximum = %0.3f meter" %x

Heat entering the slab = 240 watt
Heat leaving the slab = 157.44 watt
Rate of heat storage = 82.56 watt
The rate of temperature change at entering the slab = 0.033 degree C/h
The rate of temperature change at leaving the slab = 0.065 degree C/h
The point where rate of heating or cooling is maximum = 0.333 meter

Example No : 4.11 - Page : 119

In [18]:
from math import log
# given data
k=40   # in W/m degree C
d =12*10**-3   # in meter
t=127   # in degree C
t_i=877   # in degree C
t_infinite=52   # in degree C
h= 20   # in W/m**2 degree C
rho=7800   # in W/m**2K
C=600   # in J/kg K
r=d/2   # in meter
#l_s = V/A = r/3
l_s =  r/3#
Bi= h*l_s/k#
# since Bi < 0.1 , hence lumped heat capacity analysis can be applied
# (t-t_infinite)/(t_i-t_infinite)  =  %e**(-h*A*toh /(rho*V*C)) = %e**(-h*toh/(rho*l_s*C)) = %e**(-h*toh/(rho*C*l_s))
toh = -log((t-t_infinite)/(t_i-t_infinite))*rho*C*l_s/h   # in sec
print "Time required for cooling process =",round(toh,3),"seconds =",round(toh/60,2),"minutes"

Time required for cooling process = 1122.215 seconds = 18.7 minutes

Example No : 4.12 - Page : 127

In [4]:
# given data
D=10*10**-2   # in m
b=D/2#
h= 100   # in W/m**2 degree C
T_o=418   # in degree C
T_i=30   # in degree C
T_infinite=1000   # in degree C

print " (A) For copper cylinder "
k=350   # in W/mK
alpha=114*10**-7   # in m**2/s
Bi= h*b/k#
theta_0_t = (T_o-T_infinite)/(T_i-T_infinite)#
Fo=18.8#
# Formula Fo= alpha*t/b**2
t=Fo*b**2/alpha#
print "Time required to reach for the cylinder centreline temperature 418 degree C =",round(t,3),"seconds =",round(t/3600,3),"hours"

# (2) Temperature at the radius of 4 cm
theta_0_t = 0.985#
# Formula theta_0_t = (T-T_infinite)/(T_o-T_infinite)
T= theta_0_t*(T_o-T_infinite)+T_infinite   # in degree C
print "Temperature at the radius of 4 cm = %0.2f degree C" %T 
print "It has very less temperature gradients over 4 cm radius"

print " (B) For asbestos cylinder "
k=0.11   # in W/mK
alpha=0.28*10**-7   # in m**2/s
Bi= h*b/k#
theta_0_t = (T_o-T_infinite)/(T_i-T_infinite)#
Fo=0.21#
# Formula Fo= alpha*t/b**2
t=Fo*b**2/alpha#
print "Time required to reach for the cylinder centreline temperature 418 degree C =",round(t,3),"seconds =",round(t/3600,1),"hours"

# (2) Temperature at the radius of 4 cm
theta_x_t = 0.286#
# Formula theta_x_t = (T-T_infinite)/(T_o-T_infinite)
T= theta_x_t*(T_o-T_infinite)+T_infinite   # in degree C
print "Temperature at the radius of 4 cm = %0.3f degree C" %T 
print "It has large temperature gradients"

 (A) For copper cylinder 
Time required to reach for the cylinder centreline temperature 418 degree C = 4122.807 seconds = 1.145 hours
Temperature at the radius of 4 cm = 426.73 degree C
It has very less temperature gradients over 4 cm radius
 (B) For asbestos cylinder 
Time required to reach for the cylinder centreline temperature 418 degree C = 18750.0 seconds = 5.2 hours
Temperature at the radius of 4 cm = 833.548 degree C
It has large temperature gradients

Example No : 4.13 - Page : 133

In [19]:
from numpy import pi
# given data
D=5*10**-2   # in m
b=D/2#
h= 500   # in W/m**2 degree C
k=60   # in W/m**2K
rho=7850   # in kg/m**3
C=460   # in J/kg
alpha=1.6*10**-5   # in m**2/s
T_i=225   # in degree C
T_infinite=25   # in degree C
t=2   # in minute

# Part(i)
Bi= h*b/k#
Fo= alpha*t/b**2#
theta_0_t = 0.18#
# Formula theta_0_t = (T_o-T_infinite)/(T_i-T_infinite)
T_o= theta_0_t*(T_i-T_infinite)+T_infinite   # in degree C
print "Centreline Temperature of the sphere after 2 minutes of exposure = %0.f degree C " %T_o

# Part(2)
depth= 10*10**-3   # in meter
r=b-depth   # in meter
rBYb=r/b#
theta_x_t = 0.95#
# Formula theta_x_t = (T-T_infinite)/(T_o-T_infinite)
T= theta_x_t*(T_o-T_infinite)+T_infinite   # in degree C
print "The Temperature at the depth of 1 cm from the surface after 2 minutes = %0.1f degree C" %T

# Part (3)
BiSquareFo= Bi**2*Fo#
QbyQo= 0.8   # in kJ
A=4/3*pi*b**3#
Qo= rho*A*C*(T_i-T_infinite)   # in J
Qo=Qo*10**-3   # in kJ
# The heat transffered during 2 minute, 
Q= Qo*QbyQo   # in kJ
print "The heat transffered during 2 minutes = %0.3f kJ" %Q

Centreline Temperature of the sphere after 2 minutes of exposure = 61 degree C 
The Temperature at the depth of 1 cm from the surface after 2 minutes = 59.2 degree C
The heat transffered during 2 minutes = 37.814 kJ