Chapter 5 - Forced Convection Heat Transfer

Example No : 5.1 - Page : 152

In [3]:
from __future__ import division
from math import sqrt
# given data
rho=1.14   # in kg/m**3
k=2.73*10**-2   # in W/mK
Cp=1.005   # in kg/kgK
v= 16*10**-6   # in m**2/s
Pr=0.67#
# Other data given in the problem are
V=2   # in m/s
w=20*10**-2   # in m
t_infinite= 10   # in degree C
t_s=65   # in degree C
x=0.25   # in m from leading edge
# Re= rho*Vx/miu = V*x/v
Re= V*x/v#
#Since Re<5*10**5 , hence the flow is a laminar flow
#(a) Boundary layer thickness
delta= 5*x/(sqrt(Re))   # in m
delta=delta*10**2   # in cm
print "Boundary layer thickness = %0.3f cm" %delta

#(b) Thermal boundary layer thickness
delta_t= delta/Pr**(1/3)   # in cm
print "Thermal boundary layer thickness = %0.3f cm" %delta_t

#(c) Local friction coefficient
Cfx= 0.664/sqrt(Re)#
print "Local friction coefficient = %0.3f" %Cfx
Cf=2*Cfx#
print "Average friction coefficient %0.3f" %Cf

#(d) Total drag force
A=.25*.2   # in m**2
toh_o=Cf*(rho*V**2/2)#
F=toh_o*A#
print "Total drag force = %0.3f N" %F

#(e) 
# Formula Nux= hx*x/k = 0.332*Re**(1/2)*Pr**(1/3)
hx= 0.332*k/x*Re**(1/2)*Pr**(1/3)   # in W/m**2K
print "Local heat transfer coefficient = %0.3f W/m**2K" %hx
h=2*hx#
print "Average heat transfer coefficient = %0.3f W/m**2K" %h
#(f)
q=h*A*(t_s-t_infinite)#
print "Rate of heat transfer = %0.3f W/m**2K" %q

#Note: In the book, they calculated wrong value of Re so all the answer in the book is wrong
Boundary layer thickness = 0.707 cm
Thermal boundary layer thickness = 0.808 cm
Local friction coefficient = 0.004
Average friction coefficient 0.008
Total drag force = 0.001 N
Local heat transfer coefficient = 5.608 W/m**2K
Average heat transfer coefficient = 11.216 W/m**2K
Rate of heat transfer = 30.844 W/m**2K

Example No : 5.2 - Page : 153

In [2]:
from __future__ import division
# given data
rho=998   # in kg/m**3
k=.648   # in W/mK
v= 0.556*10**-6   # in m**2/s
Pr=3.54#
V=2   # in m/s
t_infinite= 10   # in degree C
t_s=90   # in degree C
Re=5*10**5#
A=1*1   # in m**2
# Re= rho*Vx/miu = V*x/v
x=Re*v/V   # in m
print "Length of the plate = %0.3f m" %x
# Nu = h*x/k =Pr**(1/3)*(0.037*Re**0.8-872)
x=1#
Re= V*x/v#
h= Pr**(1/3)*(0.037*Re**0.8-873)*k/x   # in W/m**2
q=h*A*(t_s-t_infinite)#
print "Heat transfer from entire plate = %0.f kW" %int(q*10**-3)
Length of the plate = 0.139 m
Heat transfer from entire plate = 444 kW

Example No : 5.3 - Page : 154

In [3]:
# given data
rho=1.06   # in kg/m**3
K=.0289#
v= 18.97*10**-6   # in m**2/s
Pr=0.696#
V=2.2   # in m/s
L=0.9   # in m
B=0.45   # in m
t_infinite= 30   # in degree C
t_s=90   # in degree C
#(a) For first half of the plate
x=L/2   # in m
Re=V*x/v#
# Nu = h*x/K = 0.664*Re**(1/2)*Pr**(1/3)
h= 0.664*Re**(1/2)*Pr**(1/3)*K/x   # in W/m**2 degree C
A=x*B#
Q1=h*A*(t_s-t_infinite)   # in watt
print "Heat transfer rate from first half of the plate = %0.f watt" %Q1

#(b) Heat transfer from entire plate
x=L   # in m
Re=V*x/v#
# Nu = h*x/K = 0.664*Re**(1/2)*Pr**(1/3)
h= 0.664*Re**(1/2)*Pr**(1/3)*K/x   # in W/m**2 degree C
A=L*B#
Q2=h*A*(t_s-t_infinite)   # in watt
print "Heat transfer rate from entire plate = %0.3f watt" %Q2

#(c) From next half of the plate
Q3= Q2-Q1#
print "Heat transfer rate from next half of the plate %0.3f" %Q3
Heat transfer rate from first half of the plate = 105 watt
Heat transfer rate from entire plate = 148.342 watt
Heat transfer rate from next half of the plate 43.448

Example No : 5.4 - Page : 155

In [12]:
from math import log
from numpy import pi
# given data
rho=985   # in kg/m**3
k=.654   # in W/mK
Cp=4.18   # in kgJ/kgK
Cp=Cp*10**3   # in J/kgK
v= 0.517*10**-6   # in m**2/s
Pr=3.26#
V=1.2   # in m/s
t_s=85   # in degree C
t_i=40   # in degree C
t_o=70   # in degree C
Ax=15*35   # in mm
P=15+35#
de=4*Ax/(2*P)   # in mm
de=de*10**-3   # in m
Re=V*de/v#
# Formula Nu= h*de/k = 0.023Re**0.8*Pr**0.4
h=0.023*Re**0.8*Pr**0.4*k/de   # in W/m**2K
m=pi*de**2*V*rho/4#
d=de#
L=m*Cp*log((t_s-t_i)/(t_s-t_o))/(pi*d*h)#
print "The length of tube = %0.3f meter" %L
The length of tube = 4.407 meter

Example No : 5.5 - Page : 156

In [4]:
# given data
k=.026   # in W/mK
v= 16.8*10**-6   # in m**2/s
miu=2*10**-5   # in kg/ms
Pr=0.708#
V=15   # in m/s
x=2   # in m
A=2*1   # in m**2
Re=V*x/v#
del_t=40-10   # in degree C
# since Re > 3 *10**5, hence turbulent flow at x=2 m length of laminar flow region is x_L then
Re_1=3*10**5#
# Re_1 = 3*10**5 = V*x_L/v
x_L= Re_1*v/V#

# Part (a)
#Nu= h*x_L/k = 0.664*Re_1**(1/2)*Pr**(1/3)#
h= 0.664*Re_1**(1/2)*Pr**(1/3)*k/x_L   # in W/m**2
print "The average heat transfer coefficient over the laminar boundary layer = %0.3f W/m**2 " %h

# Part(b)
#Nu= h*x/k = (0.037*Re**0.8-872)*Pr**(1/3)#
h= (0.037*Re**0.8-872)*Pr**(1/3)*k/x   # in W/m**2
print "The average heat transfer coefficient over entire plate = %0.3f W/m**2" %h

# Part (c)
q=h*A*del_t#
print "Total heat transfer rate = %0.3f watt" %q

# Note: Calculation of the part(a) in this book is wrong, so answer of the part(a) in the book is wrong
The average heat transfer coefficient over the laminar boundary layer = 25.083 W/m**2 
The average heat transfer coefficient over entire plate = 32.910 W/m**2
Total heat transfer rate = 1974.607 watt

Example No : 5.6 - Page : 158

In [14]:
from numpy import pi
# given data
rho=997   # in kg/m**3
k=0.608   # in W/mK
Cp= 4180   # in J/kg K
miu=910*10**-6   # in Ns/m**2
d=30*10**-3   # in m
m=0.02   # in kg/s
t_o=30   # in degree C
t_i=20   # in degree C
Re= 4*m/(pi*d*miu)#
q_desh=12*10**3   # in W/m**2
# since Re < 2300, flow is laminar one

# Part(a)
# Nu = h*d/k = 4.36
h=4.36*k/d#
print "Heat transfer coefficient = %0.3f W/m**2K" %h

# Part (b)
L=m*Cp*(t_o-t_i)/(q_desh*pi*d)#
print "Length of pipe = %0.3f meter" %L

# Part(c)
# q_desh= h*(t_infinite-t_o)
t_infinite = q_desh/h+t_o#
print "The inner tube surface temperature at the outlet = %0.3f degree C" %t_infinite

# Part(d)
f=64/Re#
print "Friction Factor = %0.3f " %f

# Part(e)
V=4*m/(pi*d**2*rho)   # in m/s  ( because m= rho*V*A , m= rho*V*pi*d**2/4 )
del_P= f*L*rho*V**2/(d*2)   # in N/m**2
print "The pressure drop in the pipe = %0.3f N/m**2" %del_P

# Note: In part(b) value of L is miss printed actual value is .739 m
Heat transfer coefficient = 88.363 W/m**2K
Length of pipe = 0.739 meter
The inner tube surface temperature at the outlet = 165.804 degree C
Friction Factor = 0.069 
The pressure drop in the pipe = 0.679 N/m**2

Example No : 5.7 - Page : 159

In [16]:
from numpy import pi
from math import log
# given data
rho=977.3   # in kg/m**3
kf=0.665   # in W/mK
Cp= 4186   # in J/kg K
miu=4.01*10**-4   # in kg/m-s
Pr=2.524#
d=0.02   # in m
m=0.5   # in kg/s
t_o=70   # in degree C
t_i=20   # in degree C
t_s=100   # in degree C
Re= 4*m/(pi*d*miu)#

# Since Re > 2300, flow is turbulent flow. Then Nusselt Number
# Nu = h*d/k = 0.023*Re**0.8*Pr**0.4
h=0.023*Re**0.8*Pr**0.4*kf/d   # in W/m**2
print "Average heat transfer coefficient = %0.3f W/m**2" %h
L=m*Cp*log((t_s-t_i)/(t_s-t_o))/(pi*d*h)   # in meter
print "Length of tube = %0.3f meter" %L
# Note: Calculation of Re is wrong so the answer in the book is wrong
Average heat transfer coefficient = 9207.036 W/m**2
Length of tube = 3.549 meter

Example No : 5.8 - Page : 160

In [17]:
from numpy import pi
# given data
rho=977   # in kg/m**3
k=0.608   # in W/mK
Cp= 4180   # in J/kg K
miu=910*10**-6   # in poise
d=0.02   # in m
m=0.02   # in kg/s
t_o=40   # in degree C
t_i=10   # in degree C
q_desh= 20*10**3   # in W/m**2

# Part (a)
Re= 4*m/(pi*d*miu)#
print "Reynold number = %0.3f" %Re

# Part(b)
# Nu = h*d/k = 4.364
h=4.364*k/d#
print "Heat transfer coefficient = %0.2f W/m**2K" %h

# Part (c)
# q= q_desh*A = m*Cp*(t_o-t_i)
# q_desh *( pi*d*l) = m*Cp*(t_o-t_i)
l=m*Cp*(t_o-t_i)/(q_desh*pi*d)#
print "Length of pipe = %0.3f meter" %l
Reynold number = 1399.164
Heat transfer coefficient = 132.67 W/m**2K
Length of pipe = 1.996 meter

Example No : 5.9 - Page : 161

In [18]:
from numpy import pi
# given data
rho=7.7*10**3   # in kg/m**3
k=12   # in W/mK
Cp= 130   # in J/kg degree C
Pr=0.011#
delta=8*10**-8   # in m**2/s
d=0.06   # in m
m=4   # in kg/s
t_i=200   # in degree C
del_t=25   # in degree C
miu=rho*delta#
Re= 4*m/(pi*d*miu)#
# From correlation  Nu =h*d/k = 4.82+0.0185*Pe**0.827
Pe=Re*Pr#
h=(4.82+0.0185*Pe**0.827)*k/d   # in W/m**2K
# Length of tube required by doing every balance
#  m*Cp*del_t = h*A*(t_s-t_b) = h*(pi*d*l)*(t_s-t_b)   # its given (t_s-t_b) = 40 degree C
l= m*Cp*del_t/(h*(pi*d)*40)   # in meter
print "Length of tube = %0.3f meter" %l
Length of tube = 0.678 meter

Example No : 5.10 - Page : 162

In [19]:
from numpy import pi
# given data
d=0.058   # in m
t_infinite=30   # in degree C
t_s=155   # in degree C
V=52   # in m/s
T_f=(t_s+t_infinite)/2   # in degree C
T_f=T_f+273   # in K
# Fluid properties at 92.5 degree C and 1 atm
miu= 2.145*10**-5   # in kg/ms
Pr=0.696#
P=1.0132*10**5#
R=287#
k=0.0312   # in W/mK
rho=P/(R*T_f)   # in kg/m**3
Re=rho*V*d/miu#
C=0.0266#
n=0.805#
# Nu = h*d/k = C*(Re)**n*Pr**(1/3)
h=C*(Re)**n*Pr**(1/3)*k/d   # in W/m**2K
#So, heat transfer rate per unit length from cylinder
q_by_L= h*(pi*d)*(t_s-t_infinite)   # in W/m
print "Heat transfer rate per unit length from cylinder = %0.3f W/m" %q_by_L
# Note: Calculation of q_by_L in the book is wrong , so the answer in the book is wrong
Heat transfer rate per unit length from cylinder = 3914.183 W/m

Example No : 5.11 - Page : 163

In [20]:
from numpy import pi
# given data
delta=15.68*10**-6   # in m**2/s
t_infinite=25+273   # in K
t_s=80+273   # in K
t_infinite=25+273   # in K
k=0.02625   # in W/m degree C
Pr=0.708#
miu_infinite=1.846*10**-5   #in kg/ms
miu_s= 2.076*10**-5   # in kg/ms
d=10*10**-3   # in m
V=5   # in m/s
A=4*pi*(d/2)**2#
Re=V*d/delta#
Nu= 2+ (0.4*Re**(1/2)+0.06*Re**(2/3))*Pr**0.4*(miu_infinite/miu_s)**(1/4)#
# Nu = h*d/k
h=Nu*k/d   # in W/m**2K
# heat transfer rate
q=h*A*(t_s-t_infinite)   # in watt
print "Heat transfer rate = %0.3f watt" %q
Heat transfer rate = 1.456 watt

Example No : 5.12 - Page : 164

In [21]:
import math
# given data
Cp=4179   # in J/kg-K
rho= 997   # in kg/m**3
V=2   # in m/s
miu= 855*10**-6   # in Ns/m**2
Pr=5.83#
k=0.613#
Do=6   #outer dia in cm
Di=4   #inner dia in cm
# de= 4*A/P = 4*pi/4*(Do**2-Di**2)/(pi*(Do+Di))
# or
de= Do-Di   # in cm
de=de*10**-2   # in m
Re= rho*V*de/miu#
# Since Re > 2300, hence flow is turbulent. Hence using Dittus Boelter equation
# Nu= 0.023*Re**0.8*Pr**0.4 =h*de/k
h= 0.023*Re**0.8*Pr**0.4*k/de   # in W/m**2K
print "Heat transfer coefficient = %0.f W/m**2K" %math.floor(h)
Heat transfer coefficient = 7752 W/m**2K

Example No : 5.13 - Page : 165

In [22]:
from numpy import pi
# given data
Cp=0.138   # in KJ/kg-K
m=8.33   # in kg/sec
Pr=0.0238#
k=8.7   # in W/mk
d=1.5*10**-2   # in m
miu=1.5*10**-3   # in kg/ms
Re=4*m/(pi*miu*d)#
Pe=Re*Pr#
# Nu = h*d/k = 7+0.025*Pe**0.8
h= (7+0.025*Pe**0.8)*k/d   # in W/m**2 degree C
print "Heat transfer coefficient = %0.3f W/m**2 degree C" %h
Heat transfer coefficient = 29255.771 W/m**2 degree C

Example No : 5.14 - Page : 166

In [23]:
# given data
rho=887   # in kg/m**3
Pr=0.026#
k=25.6   # in W/mk
d=2.5*10**-2   # in m
miu=0.58*10**-3   # in kg/ms
V=3   # in m/s
Re=rho*V*d/(miu)#
Pe=Re*Pr#
Nu = 4.8+0.015*Pe**0.85*Pr**0.08
h= Nu*k/d   # in W/m**2 degree C
print "Heat transfer coefficient = %0.3f W/m**2 degree C" %h
#Note: There is some difference in coding and book answer because they did not take aqurate calculation
Heat transfer coefficient = 15217.633 W/m**2 degree C

Example No : 5.15 - Page : 166

In [24]:
from numpy import pi
# given data
delta=38.1*10**-6   # in m**2/s
Pr=501#
Prs=98#
K=0.138   # in W/mk
T_infinite=353   # in K
T_s=423   # in K
V=2   # in m/s
d=12.5*2*10**-3   # in m
Re=V*d/delta#
n=0.36# for Pr >= 10
C=0.26   # for Re between 10**3 and 2*10**5
m=0.6   # for Re between 10**3 and 2*10**5
Nu= C*Re**m*Pr**n*(Pr/Prs)**(1/4)#
h= Nu*K/d   # in W/m**2 degree C
A=pi*25*10**-3#
del_t=T_s-T_infinite#
# Formula q=h*A*del_t
q_by_L = h*A*del_t#
print "Initial rate of heat loss per meter length of cylinder = %0.3f" %q_by_L
# Note: calculation in the book is wrong  so answer in the book is wrong
Initial rate of heat loss per meter length of cylinder = 8260.795