Chapter 6 - Free Convection

Example No : 6.1 - Page : 177

In [10]:
from __future__ import division
# given data
# (i) when
x=.3   # in m
T_s=100   # in degree C
T_infinite=30   # in degree C
T_f=(T_s+T_infinite)/2   # in degree C
T_f=T_f+273   # in K
Bita=1/T_f#
# Other fluid properties at film temperature
Pr=0.703#
K=0.0301   # in W/mK
T=1.8*10**-5       # in m**2/s
g=9.81#
del_T=T_s-T_infinite#
Gr=(g*Bita*del_T*x**3)/T**2#
Ra=Gr*Pr#
print "Rayleigh Number is = %0.2e" %Ra
#Since Ra<10**9, hence flow is laminar, then correlation for vertical plate in laminar flow
# Formula Nu=0.59*Ra**(1/4)=h*x/K
h=0.59*Ra**(1/4)*K/x   # in W/m**2K
A=2*.3*.5#
q1=h*A*(T_s-T_infinite)#
print "Heat transfer rate from the plate, when the vertical height is 0.3 m :",round(q1,3),"W"

#(ii) when
x=0.5   # in m
Gr=(g*Bita*del_T*x**3)/T**2#
Ra=Gr*Pr#
# Formula Nu=0.59*Ra**(1/4)=h*x/K
h=0.59*Ra**(1/4)*K/x   # in W/m**2K
q2=h*A*(T_s-T_infinite)#
print "Heat transfer rate from the plate, when the vertical height is 0.5 m :",round(q2,3),"W"
PercentageDecrease=(q1-q2)/q1*100#
print "Percentage decreases in heat transfer rate when x=0.5 m as compared to when x=0.3 m is :",round(PercentageDecrease,3),"%"

#Note : In the book ,In part (b), calculation of getting the value of h is wrong 
Rayleigh Number is = 1.19e+08
Heat transfer rate from the plate, when the vertical height is 0.3 m : 129.844 W
Heat transfer rate from the plate, when the vertical height is 0.5 m : 114.277 W
Percentage decreases in heat transfer rate when x=0.5 m as compared to when x=0.3 m is : 11.989 %

Example No : 6.2 - Page : 178

In [11]:
# given data
Pr=0.694#
K=0.0296   # in W/mK
rho=1.029   # in kg/m**3
miu=20.6*10**-6   # in poise
x=.2   # in m
T_s=110   # in degree C
T_infinite=30   # in degree C
T_f=(T_s+T_infinite)/2   # in degree C
T_f=T_f+273   # in K
Bita=1/T_f#
g=9.81#
del_T=T_s-T_infinite#
Gr=(rho**2*g*Bita*del_T*x**3)/miu**2#
Ra=Gr*Pr#
#since Rayleigh number is less than 10**10, hence
Nu=0.68*Pr**(1/2)*Gr**(1/4)/((.952+Pr)**(1/4))#
h=Nu*K/x#
A=2*0.2*1#
q=h*A*(T_s-T_infinite)#
print "Heat transfer rate is :",round(q,1),"W"
Heat transfer rate is : 194.7 W

Example No : 6.3 - Page : 179

In [12]:
from numpy import pi
# given data
d=7.5*10**-2   # in m
x=2   # in m
T_s=70   # in degree C
T_infinite=10   # in degree C
del_T=T_s-T_infinite#
g=9.81#
calculation=4.5*10**10# # value of g*Bita*rho**2*C_p/(miu*k)
K=2.75*10**-2   # in W/mK
# g*Bita*rho**2*C_p/(miu*k) = g*Bita*rho**2/miu**2 * miu*C_p/k = (g*Bita*del_T*x**3/T**2 * miu*C_p/k)/(del_T*x**3)
GrxPr= calculation*del_T*x**3# # value of Gr*Pr
Nu= 0.13*(GrxPr)**(1/3)#
# Formula Nu = h*x/k
h= Nu*K/x   # in W/m**2K
A=2*pi*d#
q=h*A*(del_T)   # in W
q=q*60*60   # in J/h
print "Heat transfer rate = %0.3e J/h" %q
Heat transfer rate = 5.067e+06 J/h

Example No : 6.4 - Page : 180

In [13]:
from math import log
# given data
m=15   # in kg
C_p=420   # in J/kg K
T_s=200   # in degree C
T_infinite=30   # in degree C
T_f=(T_s+T_infinite)/2   # in degree C
T_f=T_f+273   # in K
Pr=0.688#
K=0.0321   # in W/mK
delta=23.18*10**-6   # in m**2/s
Bita=1/T_f#
g=9.81#
x=0.3   # in m
del_T=T_s-T_infinite#
Gr=(g*Bita*del_T*x**3)/delta**2#
Ra=Gr*Pr#
#Since Ra<10**9, hence it is laminar flow using the relation
# Formula Nu=0.59*Ra**(1/4)=h*x/K
h=0.59*Ra**(1/4)*K/x   # in W/m**2K
print "(i) Heat transfer coefficient is :",round(h,3),"W/m**2K"

# (b) Initial rate of cooling
# Formula h*A*(T_s-T_infinite) = m*C_p*dt_by_toh
A=2*0.3*0.5#
dt_by_toh = h*A*(T_s-T_infinite)/(m*C_p)   # in degree C/sec
dt_by_toh=dt_by_toh*60   # in degree C /min
print "(ii) Initial rate of cooling of the plate is :",round(dt_by_toh,3),"degreeC per min"

#(c) Time taken by plate to cool from 200 degree C to 50 degree C
T_i=200   # in degree C
T=50   # in degree C
# Formula (T-T_infinite)/(T_i-T_infinite)= %e**(-h*A*toh/(m*C_p))#
toh= -log((T-T_infinite)/(T_i-T_infinite))*m*C_p/(h*A)   # in sec
toh=toh/60   # in min
print "(iii) Time required to cool plate from 200 degree C to 50 degree C is :",round(toh,2),"minutes"
(i) Heat transfer coefficient is : 6.97 W/m**2K
(ii) Initial rate of cooling of the plate is : 3.385 degreeC per min
(iii) Time required to cool plate from 200 degree C to 50 degree C is : 107.46 minutes

Example No : 6.5 - Page : 182

In [14]:
from numpy import pi
# given data
rho=0.8   # in kg/m**3#
C_p=1.01   # in KJ/kg K
Pr=0.684#
d=15*10**-2   # diameter in meter
K=0.035   # in W/mK
delta=2.78*10**-5   # in m**2/s
g=9.81#
x=2   # in m
T_s=250   # in degree C
T_infinite=30   # in degree C
T_f=(T_s+T_infinite)/2   # in degree C
T_f=T_f+273   # in K
Bita=1/T_f#
del_T=T_s-T_infinite#
#Heat loss from vertical part by free convection
Gr=(g*Bita*del_T*x**3)/delta**2#
Ra=Gr*Pr#
#Since Ra>10**9, hence turbulent flow
# Formula Nu= h*x/K =0.13*Ra**(1/3)
h=0.13*Ra**(1/3)*K/x   # in W/m**2K
A=2*pi*d#
q1=h*A*del_T   # w
q1=q1*10**-3   # in kW
print "Heat loss from vertical part is :",round(q1,3),"kW"
#Heat loss for Horizontal part
# here
x=d#
Gr=(g*Bita*del_T*x**3)/delta**2#
Ra=Gr*Pr#
#Since Ra<10**9, hence laminar fluid flow
# Formula Nu= h*x/K =0.53*Ra**(1/4)
h=0.53*Ra**(1/4)*K/x   # in W/m**2K
A=pi*d*8#
q2=h*A*del_T   # w
q2=q2*10**-3   # in kW
print "Heat loss for horizontal part is :",round(q2,3),"kW"

#Heat loss by radiation
sigma=5.67*10**-8#
epsilon=0.65   # emissivity of steel
A=pi*d*10#
T_s=T_s+273   # in K
T_infinite=T_infinite+273   # in K
q3=sigma*A*epsilon*(T_s**4-T_infinite**4)   # in w
q3=q3*10**-3   # in kW
print "Heat loss by radiation is :",round(q3,3),"kW"
#Total heat loss
theta=q1+q2+q3#
print "Total heat loss is :",round(theta,3),"kW"


#Note : value of q3 and theta in the book is wrong so answer in the book is wrong
Heat loss from vertical part is : 1.572 kW
Heat loss for horizontal part is : 6.447 kW
Heat loss by radiation is : 11.53 kW
Total heat loss is : 19.549 kW

Example No : 6.6 - Page : 183

In [15]:
# given data
rho=1.205   # in kg/m**3#
C_p=1006   # in J/kg K
Pr=0.71#
K=0.0256   # in W/mK
delta=1.506*10**-5   # in m**2/s
T_s=35   # in degree C
T_infinite=5   # in degree C
T_f=(T_s+T_infinite)/2   # in degree C
T_f=T_f+273   # in K
Bita=1/T_f#
del_T=T_s-T_infinite#
g=9.81#
# Formula 1/x= 1/Lh + 1/Lv
Lh=50   # in cm
Lv=50   # in cm
x=Lh*Lv/(Lh+Lv)   # in cm
x=x*10**-2   # in m
# Formula Gr=(g*Bita*del_T*x**3)/delta**2#
Gr=(g*Bita*del_T*x**3)/delta**2#
Ra=Gr*Pr#
# Formula Nu= h*x/K =0.53*Ra**(1/4)
h=0.53*Ra**(1/4)*K/x   # in W/m**2K
A=2*(0.5+0.5)#
q=h*A*del_T   # w
print "Heat loss per meter length of pipe is :",round(q,3),"watt"

# Note: In the book, value of h is wrong due to place miss value of x, so the answer in the book is wrong
Heat loss per meter length of pipe is : 272.624 watt

Example No : 6.7 - Page : 184

In [16]:
from scipy.integrate import quad
# given data
L=3   # in m
delta=0#
hx='10*x**(-1/4)'
# (a) Average heat transfer coefficient
def integrand(x):
    return 10*x**(-1/4)

ans, err = quad(integrand, delta, L)

h=1/L*ans
print "(a) Average heat transfer coefficient is :",round(h,2),"W/m**2K"

# (b) Heat transfer rate
A=3*.3   # in m**2
Tp=170   # plate temp. in degree C
Tg=30   # gas temp. in degree C
del_T=Tp-Tg#
q=h*A*del_T   # in W
print "(b) Heat transfer rate is :",round(q,3),"W"

# (c) 
x=2   # in m
qx_by_A= 10*x**(-1/4)*(Tp-Tg)#
print "Local heat flux 2 m from the leading edge is :",round(qx_by_A,1),"W/m**2"
(a) Average heat transfer coefficient is : 10.13 W/m**2K
(b) Heat transfer rate is : 1276.524 W
Local heat flux 2 m from the leading edge is : 1177.3 W/m**2

Example No : 6.8 - Page : 185

In [17]:
from numpy import pi
# given data
Pr=0.712#
K=0.026   # in W/mK
delta=1.57*10**-5   # in m**2/s
T_s=320   # in K
T_infinite=280   # in K
del_T=T_s-T_infinite#
T_f=(T_s+T_infinite)/2   # in K
Bita=1/T_f#
d1=20   # in cm
d2=30   # in cm
x=(d2-d1)/2   # in cm
x=x*10**-2   # in m
g=9.81#
Gr=(g*Bita*del_T*x**3)/delta**2#
Ra=Gr*Pr#
# Formula Nu= h*x/K =0.228*Ra**(0.226)
h=0.228*Ra**(0.226)*K/x   # in W/m**2K
A=pi*(d1*10**-2)**2#
q=h*A*del_T   # w
print "Heat transfer rate is :",round(q,1),"watt"
Heat transfer rate is : 11.4 watt

Example No : 6.9 - Page : 186

In [18]:
from numpy import pi
# given data
K=0.0278   # in W/mK
rho=1.092   # in kg/m**3
miu=19.57*10**-6   # in kg/ms
Cp=1007   # in kg/kg degree C
epsilon=0.9#
sigma=5.67*10**-8#
d=75+2*25   # in mm
d=d*10**-3   # in meter
T_s=80   # in degree C
T_infinite=20   # in degree C
T_f=(T_s+T_infinite)/2   # in degree C
T_f=T_f+273   # in K
Bita=1/T_f#
g=9.81#
del_T=T_s-T_infinite#
Pr=miu*Cp/K#
Gr=(rho**2*g*Bita*del_T*d**3)/miu**2#
# Formula Nu= h*d/K = 0.53*(Gr*Pr)**(1/4)#
h= 0.53*(Gr*Pr)**(1/4)*K/d#
#(a) Heat loss from 6 m length of pipe
A=pi*d*6#
Q_conv=h*A*del_T#
Q_rad=epsilon*sigma*A*((T_s+273)**4-(T_infinite+273)**4)#
#total heat transfer rate
Q=Q_conv+Q_rad#
print "Total heat transfer rate = %0.f W" %Q

# (b) Overall heat transfer coefficient
# Formula Q=U*A*del_T
U=Q/(A*del_T)#
print "Overall heat transfer coefficient is :",round(U,3),"W/m**2 degree C"
Total heat transfer rate = 1863 W
Overall heat transfer coefficient is : 13.178 W/m**2 degree C