# Chapter 16: HEAT TRANSFER BY CONDUCTION

## Example 16.01, page: 367

In [2]:
from __future__ import division
import math

# Initialization  of  Variable
Tsi = 385 #degC
Ka = 0.15 #W/m.K
Kb = 0.08 #W/m2.K
Tinf = 25 #degC
ho = 25 #W/m2.K
Tso = 50 #degC
x = 2

#calculations:
#La
La = (Tsi - Tso)/(Tso - Tinf)/ho*Ka*Kb*2/(2*Kb + Ka)
#Lb
Lb = La/2
#required thickness for the composite window
L = La + Lb

#Results
print  "required  thickness  of  composite window is", round(L,4),"m"

required  thickness  of  composite window is 0.0622 m


## Example 16.02, page: 368

In [3]:
from __future__ import division
import math

# Initialization  of  Variable
Tinf = 25 #degC
h = 100 #W/m2.K
Rtc2 = 0.9E-4 #m2.K/W
Pe2 = 10000 #W/m2
L = 8 #mm
k = 238 #W/m.K

#calculations:
Tc = Tinf + Pe2/(h + 1/(Rtc2 + L*1E-3/k + 1/h))

#Results
print  "maximum allowable temperature is",round(Tc,1),"degC"

maximum allowable temperature is 75.3 degC


## Example 16.03, page: 372

In [4]:
from __future__ import division
import math

# Initialization  of  Variable
Tinf2 = 300 #K
h = 20 #W/m2.K
Tinf1 = 77 #K
p = 804 #kg/m3
hfg = 2E5 #J/kg
r1 = 0.25 #m
r2 = 0.275 #m
k = 0.0017 #W/m.K

#calculations:
#resistances due to conduction
Rtcond = 1/(4*math.pi*k)*(1/r1 - 1/r2)
#resistances due to convection
Rtconv = 1/(h*4*math.pi*r2**2)
#rate of heat transfer to the liquid nitrogen
q = (Tinf2 - Tinf1)/(Rtcond - Rtconv)
#volumetric mass rate of nitrogen boil-off per day
mdotp = q/hfg*3600*24/p*1000

#Results
print  "a) heat  transfer  is", round(q,2),"W"
print  "b) rate of nitrogen boil-off is", round(mdotp,0),"liters/day"

a) heat  transfer  is 13.14 W
b) rate of nitrogen boil-off is 7.0 liters/day


## Example 16.04, page: 375

In [5]:
from __future__ import division
import math

# Initialization  of  Variable
ka = 75 #W/m.K
La = 50 #mm
Lb = 20 #mm
kb = 150 #W/m.K
qbdot = 0
Tinf = 30 #degC
h = 1000 #W/m2.K

#calculations:
#outer surface temperature
#resistance (conduction and converction):
RcondB2 = Lb*1E-3/kb
Rconv2 = 1/h
#temperature at the composite interface
T1 = Tinf + (RcondB2 + Rconv2)*q2
#the inner surface temperature of the composite

#Results
print "Inner and outer surface temperatures of the composite are", To,"degC and", T2,"degC respectively"

Inner and outer surface temperatures of the composite are 140.0 degC and 105.0 degC respectively


## Example 16.05, page: 378

In [6]:
from __future__ import division
import math

# Initialization  of  Variable
Tinf = 25 #degC
r2 = 250 #mm
r1 = 200 #mm
kss = 15 #W/m.K
krw = 20 #W/m.K
qdot = 1E5 #W/m3
h = 500 #W/m2.K

#calculations:
#the tube wall conduction and convection resistances
Rcond1 = math.log(r2/r1)/(2*math.pi*kss)
Rconv1 = 1/(h*2*math.pi*r2*1E-3)
#heat rate per unit length
q1 = qdot*math.pi*(r2*1E-3)**2
#waste surface temperature
Ts1 = Tinf + q1*(Rcond1 + Rconv1)
#the centerline temperature
To = qdot*(r1*1E-3)**2/(4*krw) + Ts1

#Results
print  "Maximum temperature in the system is", round(To,0),"degC"

Maximum temperature in the system is 146.0 degC


## Example 16.06, page: 381

In [1]:
from __future__ import division
import math
#from pylab import *
%pylab inline

# Initialization  of  Variable
Tinf = 25 #degC
h = 100 #W/m2.K
D = 5 #mm
Tb = 100 #degC

#calculations:
#from Table HT-1
kcu = 398  #W/m.K
kal = 180 #W/m.K
kss = 14 #W/m.K
#m
mcu = (4*h/(kcu*D*1E-3))**0.5
mal = (4*h/(kal*D*1E-3))**0.5
mss = (4*h/(kss*D*1E-3))**0.5

#creating empty lists for plotting
x =[]
Tcu =[]
Tal =[]
Tss =[]

for h in range(3000):
x.append((h-1)/10)
k = (h-1)/10
#Temp profile
Tcu.append(Tinf + (Tb - Tinf)*math.e**-(mcu*k/1000))
Tal.append(Tinf + (Tb - Tinf)*math.e**-(mal*k/1000))
Tss.append(Tinf + (Tb - Tinf)*math.e**-(mss*k/1000))

# plots
fig  = plt.figure()
ax.plot(x,Tcu,'.')
ax.plot(x,Tal,'.')
ax.plot(x,Tss,'.')
ax.legend('CAS')
xlabel('x(mm)')
ylabel('T (degC)')
title('Temperature distribution along the fin')
show()

#fin heat rate
qfcu = (h*math.pi*D*1E-3*kcu*math.pi/4*(D*1E-3)**2)**0.5*(Tb - Tinf)
qfal = (h*math.pi*D*1E-3*kal*math.pi/4*(D*1E-3)**2)**0.5*(Tb - Tinf)
qfss = (h*math.pi*D*1E-3*kss*math.pi/4*(D*1E-3)**2)**0.5*(Tb - Tinf)

#length
Lcu = 2.65*((kcu*(math.pi*(D*1E-3)**2)/4)/(h*math.pi*D*1E-3))**0.5
Lal = 2.65*(kal*math.pi/4*(D*1E-3)**2/(h*math.pi*D*1E-3))**0.5
Lss = 2.65*(kss*math.pi/4*(D*1E-3)**2/(h*math.pi*D*1E-3))**0.5

#results
print "a) fin heat rate for copper is", round(qfcu,1),"W, for Aluminium is", round(qfal,1),"W and for Stainless Steel is", round(qfss,1),"W"
print "b) length for copper >=", round(Lcu*1000,3),"mm, for Aluminium >=",round(Lal*1000,0),"mm and for Stainless Stell >=", round(Lss*1000,0),"mm"

Populating the interactive namespace from numpy and matplotlib

a) fin heat rate for copper is 45.5 W, for Aluminium is 30.6 W and for Stainless Steel is 8.5 W
b) length for copper >= 34.131 mm, for Aluminium >= 23.0 mm and for Stainless Stell >= 6.0 mm


## Example 16.07, page: 384

In [8]:
from __future__ import division
import math

# Initialization  of  Variable
t = 0.7 #mm
Tinf = 20 #degC
h = 25 #W/m2.K
T1 = 80 #degC
Rtc2 = 1E-3 #m2.K/W
r1 = 2 #mm
r2 = 3 #mm
r3 = 13 #mm
H = 6 #mm
k  = 200 #W/m.K

#calculations:
#Thermal resistances associated with the contact joint and sleeve
Rtc = Rtc2/(2*math.pi*r1*1E-3*H*1E-3)
Rtsleeve = math.log(r2/r1)/(2*math.pi*H*1E-3*k)
#P, Ac
P = 2*(H+t)*1E-3
Ac = t*H*1E-6
#m, L
m = (h*P/(k*Ac))**0.5
L = r3 - r2
#the thermal resistance for a single fin
Rtf = ((h*P*k*Ac)**0.5*(math.sinh(m*L*1E-3) + h/(m*k)*math.cosh(m*L*1E-3))/(math.cosh(m*L*1E-3) + h/(m*k)*math.sinh(m*L*1E-3)))**-1
#thermal resistance of 12 fins
Rtf12 = Rtf/12
#the thermal resistance due to convection
Rtb = 1/(h*(2*math.pi*r2*1E-3 - 12*t*1E-3)*H*1E-3)
#the equivalent resistance is
Requiv = (1/Rtf12 + 1/Rtb)**-1
#total resistance of the finned sleeve
Rtot = Rtc + Rtsleeve + Requiv
#heat transfer rate from the sleeve
qt = (T1-Tinf)/Rtot

#Results
print  "heat  transfer  rate is", round(qt,2),"W"

heat  transfer  rate is 1.63 W


## Example 16.08, page: 390

In [9]:
from __future__ import division
import math

# Initialization  of  Variable
Tinf = 20 #degC
h = 25 #W/m2.K
Ti = 225 #degC
D = 12.5 #mm
t = 6 #min

#calculations:
#from Table HT-2
p = 2225 #kg/m3
c = 835 #J/kg
k = 1.4 #W/m.K
#the characteristic length of the spherical bead
Lc = D*1E-3/6
#Biot num
Bi = h*Lc/k
#temperature T(t) after 6 min
Tt = Tinf + (Ti - Tinf)*math.e**(-1*h*t*60/(p*Lc*c))

#Results
print  "temperature  after  6  min is", round(Tt,0),"degC"

temperature  after  6  min is 40.0 degC


## Example 16.09, page: 390

In [10]:
from __future__ import division
import math

# Initialization  of  Variable
e = 0.8
h = 40 #W/m2.K
Tinf = 175 #degC
Ti = 25 #degC
L = 1.5 #mm
Tsur = 175 #degC
Tc = 150 #degC
t = 5

#calculations:
k = 177 #W/m.K
p = 2770 #kg/m3
c = 875 #J/kg

#biot num
#time required for the panel to reach the cure temperature
#total time to complete the 5-min duration cure
tc = tc + t*60

#Results
print  "Elapsed time for completion of the cure process, tc =", round(tc,0),"sec"

Elapsed time for completion of the cure process, tc = 425.0 sec


## Example 16.10, page: 394

In [11]:
from __future__ import division
import math

# Initialization  of  Variable
L = 25 #mm
Tinf = 175 #degC
h = 100 #W/m2.K
Ti = 25 #degC
p = 2325 #kg/m3
c = 800 #J/kg
k = 1.0 #W/m.K
t = 10 #min

#calculations:
a = k/(p*c)
#a) at t = 10 min
Bi = h*L*1E-3/k
Fo = a*t*60/(L*1E-3)**2
#b)
zeta = 1.1347
C = 1.1949
theta0_s = C*math.e**(-1*Fo*zeta**2)
#midplane temperature
T0_10 = Tinf + theta0_s*(Ti-Tinf)
#surface temperature
theta_s = theta0_s*math.cos(zeta*1)
TL_10 = Tinf + theta_s*(Ti - Tinf)
#c)
#Heat transfer to the slab outer surface
qL2 = h*(TL_10 - Tinf)
#d)
#the energy per unit surface area is
Q2 = (1 - math.sin(zeta)*theta0_s/zeta)*p*c*L*1E-3*(Ti-Tinf)

#Results
print  "a) Biot and Fourier numbers after 10 min are",round(Bi,2),"and", round(Fo,3),"respectively"
print  "b) Slab midplane and surface temperatures after 10 min is", round(T0_10,0),"degC"
print  "c) Heat flux to the slab at 10 min is", round(qL2,0),"W/m2"#answer wrong in book
print  "d) Energy transferred to the slab per unit surface area after 10 min is", round(Q2,0),"J/m2"

a) Biot and Fourier numbers after 10 min are 2.5 and 0.516 respectively
b) Slab midplane and surface temperatures after 10 min is 83.0 degC
c) Heat flux to the slab at 10 min is -3895.0 W/m2
d) Energy transferred to the slab per unit surface area after 10 min is -3549589.0 J/m2


## Example 16.11, page: 395

In [12]:
from __future__ import division
import math

# Initialization  of  Variable
Tinf = 20 #degC
h = 6000 #W/m2.K
Ti = 400 #degC
T0_tc = 50 #degC
r0 = 5 #mm
p = 3000 #kg/m3
c = 1000 #J/kg
k = 20 #W/m.K
a = 6.66E-6 #m2/s

#calculations:
#biot num
Bilcm = h*(r0/3*1E-3)/k
Bi = h*r0*1E-3/k
#from Table 16.6
zeta = 1.8
C = 1.376
#Fourier number
Fo = -1/zeta**2*math.log(1/C*(T0_tc - Tinf)/(Ti - Tinf))
#center termperature
tc = Fo*(r0*1E-3)**2/a

#Results
print  "Time tc required to satisfy the cooling requirement is", round(tc,1),"s"

Time tc required to satisfy the cooling requirement is 3.3 s


## Example 16.12, page: 395

In [13]:
from __future__ import division
import math

# Initialization  of  Variable
Ts = -15 #degC
Ti = 20 #degC
Txm_t = 0#degC
p = 2050 #kg/m3
c = 1840 #J/kg
k = 0.52 #W/m.K
t = 60 #days

#calculations:
a = k/(p*c)
LHS = (Txm_t - Ts)/(Ti - Ts)
#from Appendix HT-6
#if erf(z) = LHS then
z = 0.4
#minimum depth to avoid freezing
xm = 2*(a*t*3600*24)**0.5*z

#Results
print  "Minimum pipe depth to avoid freezing is", round(xm,2),"m"

Minimum pipe depth to avoid freezing is 0.68 m