Chapter 02 - General Heat Conduction Equation

Example No : 2.5 - Page : 51

In [16]:
from __future__ import division
from math import log, pi
# given data
r1=5   # in cm
r2=5+4   # in cm
r3= 9+2.5   # in cm
k1=0.0701   # in W/mK
k2=0.1   # in W/mK
L=20   # in m
T1=234.36   # in degree C
T3=24   # in degree C
sigmaR= (log(r2/r1)/(2*pi*k1*L) + log(r3/r2)/(2*pi*k2*L))#

# Part (i)
q=(T1-T3)/sigmaR   # in watt
print "Heat transfer rate = %0.3f watt" %q

# Part(ii)
# Formula q= (T1-T2)/(log(r2/r1)/(2*pi*k1*L))
T2 =T1- (q*(log(r2/r1)/(2*pi*k1*L)))#
print "Interface temperature of insulation = %0.3f degree" %T2

Heat transfer rate = 2439.473 watt
Interface temperature of insulation = 71.585 degree

Example No : 2.6 - Page : 52

In [18]:
import math
from numpy import pi
# given data
k_brick=0.93   # in W/mK
k_insulation=0.12   # in W/mK
k_wood=0.175   # in W/mK
k_Al=204   # in W/mK
k1=k_brick#
k2=k_insulation#
k3=k_wood#
T1=200   # in degree C
T4=10   # in degree C
x1=10*10**-2   # in m
x2=25*10**-2   # in m
x3=1*10**-2   # in m
A=0.1   # in m**2
sigmaR= x1/(k1*A)+x2/(k2*A)+x3/(k3*A)#
q1=(T1-T4)/sigmaR#
print "Heat transfer rate without rivet = %0.2f Watt" %q1

# Heat transfer rate with rivet
d=3*10**-2   # in meter
x=x1+x2+x3#
k_rivet=k_Al#
A_rivet=pi*d**2/4   # in m**2
R_rivet= x/(k_rivet*A_rivet)#
A_eff=A-A_rivet   # in m**2
sigmaRw= 1/A_eff*(x1/k1+x2/k2+x3/k3)   # in k/W
R_eq= R_rivet*sigmaRw/(R_rivet+sigmaRw)   # in k/W
q2=(T1-T4)/R_eq   # in watt
print "Heat transfer rate with rivet = %0.3f Watt" %q2
percentIncrease=(q2-q1)*100/q1   # percent increase in heat flow due to rivet
print "Percentage increase in heat flow due to rivet = %0.f %%" %math.ceil(percentIncrease)

Heat transfer rate without rivet = 8.45 Watt
Heat transfer rate with rivet = 84.497 Watt
Percentage increase in heat flow due to rivet = 900 %

Example No : 2.7 - Page : 54

In [16]:
from numpy import pi
# given data
k_cu=384   # in W/mK
k_s=1.75   # in W/mK
k1=k_cu#
k2=k_s#
hi=221   #  in W/m**2K
ho=3605   #  in W/m**2K
Ti=100   # in degree C
To=125   # in degree C
r1=0.2   # in m
r2=0.02+0.006   # in m
r3=0.026+0.003   # in m
ri=0.02   # in m
L=1   # in m
# Part(i)
Ao= 2*pi*r3*L#
Ai= 2*pi*r1*L#
# Formula Uo= 1/Ao*sigmaR
Uo= 1/(r3/(ri*hi) + r3/k1*log(r2/r1) + r3/k2*log(r3/r2) + 1/ho)  # in w/m**2K
print "Overall heat transfer coefficient based on outer area = %0.3f W/m**2K" %Uo

#Part(ii)
del_T= To-Ti#
q=Uo*Ao*del_T#
print "Water to air heat transfer rate = %0.3f W/m" %q

#Part (iii)
# Formula q= T/(log(r3/r2)/(2*pi*k*L)) , where T=T2-T3 and k=k_s
k=k_s#
T= q*log(r3/r2)/(2*pi*k*L)#
print "Temperature drop across the scale deposited = %.3f degree C" %T

# Note: In Part (i), they put wrong value of r2 and r1 in log(r2/r1) to calculate the value of Uo. So there is some difference in answer of coding and book

Overall heat transfer coefficient based on outer area = 117.730 W/m**2K
Water to air heat transfer rate = 536.298 W/m
Temperature drop across the scale deposited = 5.326 degree C

Example No : 2.8 - Page : 58

In [21]:
from math import sqrt
# given data
k=0.175   # in W/mK
h_infinite=9.3   #  in W/m**2K
T_infinite=30   # in degree C
T_s=70   # in degree C
d=10*10**-3   # in m
r=d/2#
L=1   # in m
rc=k/h_infinite   # in m
CriticalThickness = rc-r   # in meter
CriticalThickness=CriticalThickness*10**3#
print "Critical thickness = %0.1f mm" %CriticalThickness

q1=2*pi*r*L*h_infinite*(T_s-T_infinite)   # in W/m
q2= (T_s-T_infinite)/(log(rc/r)/(2*pi*k*L)+1/(2*pi*rc*h_infinite))   # in W/m
PerIncHeatDiss= (q2-q1)*100/q1#
print "Percentage increase in heat dissipation rate = %0.3f %%" %PerIncHeatDiss
#Also q1=I1**2*R with bare cable
#     q2=I2**2*R with insulated cable
I2_by_I1 = sqrt(q2/q1)#
# ( I2-I1 ) / I1 = (I2_by_I1 -1) / 1
# Percentage increase in current carrying capacity
PerIncCurrent = (I2_by_I1 -1) / 1 *100#
print "Increase in current carrying capacity = %0.f %%" %math.floor(PerIncCurrent)

Critical thickness = 13.8 mm
Percentage increase in heat dissipation rate = 61.845 %
Increase in current carrying capacity = 27 %

Example No : 2.9 - Page : 59

In [4]:
# given data
k_in=0.3   # in W/mK
k_gw=0.038   # in W/mK
ro=1.5   # in cm
ho=12   #  in W/m**2 degree C
rc=k_in/ho   # in m
rc=rc*10**2   # in cm
print "Critical radius = %0.1f cm" %rc
if ro<rc:
    print "Since radius of insulation (",round(ro,3),"cm) is less than critical radius of insulation (",round(rc,3),"cm)"
    print "so heat transfer rate will increase by adding thsi insulation and hence it is not effective"
ro=ro*10**-2   # in meter
# For effective insulation
# ro>=rc
# Kin/ho<= ro
roho=ro*ho   # in W/mK
# Kin<= ro*ho
print "Maximum value of thermal conductivity = %0.2f W/mK" %roho

Critical radius = 2.5 cm
Since radius of insulation ( 1.5 cm) is less than critical radius of insulation ( 2.5 cm)
so heat transfer rate will increase by adding thsi insulation and hence it is not effective
Maximum value of thermal conductivity = 0.18 W/mK

Example No : 2.10 - Page : 60

In [15]:
from math import log
# given data
d=1.2*10**-3   # in m
r=d/2   # in m
rc=1.8*10**-3   # in m
T1=100   # in degree C
T_infinite=30   # in degree C
k=0.3   # in W/mK
h=10   # in W/m**2K
L=1   # in m
ke=5.1*10**7#
q=(T1-T_infinite)/(log(rc/r)/(2*pi*k)+1/(2*pi*rc*h))   # in W/m
# Volume of wire for one meter length
vol= pi*r**2*L   # in m**3
print "In steady state heat transfer process, the heat produced by the wire is dissipated to surrounding."    
# Heat produced per unit volume of the wire
HeatProduced= q/vol   # in w/m**2
# Formula HeatProduced= I**2*R = I**2/ke
I=sqrt(HeatProduced*ke)   # in amp/m**2
# Area of wire
A= pi*r**2#
# so current carrying capacity of the given wire
Current= I*A#
print "The current carried by the copper wire = %0.3f amphere" %Current

In steady state heat transfer process, the heat produced by the wire is dissipated to surrounding.
The current carried by the copper wire = 20.698 amphere

Example No : 2.11 - Page : 61

In [14]:
from math import log
# given data
d_i=.1   # inner dia in m
r_i=d_i/2   # in m
Ti=473   # in K
T_infinite=293   # in K
k=1   # in W/mK
h=8   # in W/m**2K
rc=k/h   # in m
print "Critical radius = %0.3f meter" %rc
#when
ro=rc#
q_by_L= (Ti-T_infinite)/(log(rc/r_i)/(2*pi*k)+1/(2*pi*rc*h))   # in W/m
print "Heat loss per meter length of pipe = %0.3f W/m" %q_by_L

# Note: To calculate the value of q_by_L the calculation is wrong in the book so answer in the book is wrong

Critical radius = 0.125 meter
Heat loss per meter length of pipe = 590.189 W/m

Example No : 2.12 - Page : 62

In [13]:
from numpy import pi
# given data
r1=100*10**-3   # in m
r2=200*10**-3   # in m
q1=1.16*10**5   # in W/m**2
t2=30   # in degree C
k=50   # in W/mK
L=1   # in m
# Total heat passing through the cylinder q
#q=q1*2*pi*r1*L                     (1)
# and heat conducted through the cylinder
# q= 2*pi*k*L(t1-t2)/log(r2/r1)     (2)
# From (1) and (2)
t1= t2+ q1*2*pi*r1*L*log(r2/r1)/(2*pi*k*L)   # in degree C
print "Temperature of inner surface = %0.2f degree C" %t1

Temperature of inner surface = 190.81 degree C

Example No : 2.13 - Page : 62

In [12]:
from math import log
# given data
d1=1*10**-3   # in m
d2=3*10**-3   # in m
r1=d1/2#
r2=d2/2#
kp=384   # in W/mK
kw=0.35   # in W/mK
rho=1.96*10**-8   # in Wm
t_s=95   # in degree C
t_infinite=40   # in degree C
h=8.75   # in W/m**2K
q_by_L= (t_s-t_infinite)/(log(r2/r1)/(2*pi*kp)+1/(2*pi*r2*h))#
# Also q_by_L = I**2*R/L = I**2*rho/(pi/4*d**2)
I= sqrt(q_by_L*(pi/4*d1**2)/rho)   # in amp
print "The maximum steady state current = %0.3f amphere" %I

The maximum steady state current = 13.481 amphere

Example No : 2.16 - Page : 66

In [11]:
from math import log, sqrt
from math import pi
from __future__ import division
# given data
d1=10*10**-3   # in mm
r1=d1/2#
K=0.2   # in W/mK
T_max=177   # in degree C
T_infinite=27   # in degree C
ho=10   # in W/m**2K
R=10# # in W/m
rc=K/ho   # in m
x=rc-r1   # in m
q_by_L= (T_max-T_infinite)/(log(rc/r1)/(2*pi*K)+1/(2*pi*ho*rc))#
# Also q_by_L = I**2*R
I= sqrt(q_by_L/R)   # in amp
print "The maximum possible current = %0.3f amphere" %I

# Note: Answer in the book is wrong

The maximum possible current = 2.811 amphere