# Chapter 8: Natural Convection System¶

### Example 8.1 Page No. 413¶

In [5]:
rou= 0.0551             # density in Ibm/cu.ft
cp=0.2420               # specific heat BTU/(lbm-degree Rankine)
v= 27.88e-5             # viscosity in sq.ft/s
kf = 0.01944             # thermal conductivity in BTU/(hr.ft.degree Rankine)
a = 1.457               # diffusivity in sq.ft/hr
Pr = 0.689              # Prandtl Number
T_inf=120.0+460.0           # wall temperature in degree R
Tw=400.0+460.0              # inside wall temperature in degree R
Beta=1/T_inf

Beta_=0.00116
gc=32.2
L=1.0                     # length of wall in ft
W=2.0                     # width in ft
Gr=(gc*Beta_*(Tw-T_inf)*L**3)/v**2    # Grashof Number
temperature_slope=0.505              #temperature slope from table 8.1
hL=(kf/L)*(4/3.0)*(Gr/4.0)**(1/4.0)*temperature_slope
A=L*W                   # cross sectional area in sq.ft
qw=hL*A*(Tw-T_inf)

print"The heat transferred is",round(qw,0),"BTU/hr"

The heat transferred is 558.0 BTU/hr


### Example 8.2 Page No. 414¶

In [15]:
rou1=1.295            # density in kg/cu.m
cp1=1005.5            # specific heat in J/(kg*K)
v1=12.59e-6           #  viscosity in sq.m/s
Pr1=0.713             # Prandtl Number
kf1=0.02426           # thermal conductivity in W/(m.K)
a1=0.17661e-4         # diffusivity in sq.m/s
T_inf1=0           # inside and outside temperature in K
Beta1=1/(T_inf1+273.0)  # volumetric thermal expansion coefficient at 295 K and 273 K

rou2=1.177            # density in kg/cu.m
cp2=1005              # specific heat in J/(kg*K)
v2=15.68e-6           # viscosity in sq.m/s
Pr2=0.708             # Prandtl Number
kf2=0.02624           # thermal conductivity in W/(m.K)
a2=0.22160e-4         # diffusivity in sq.m/s
T_inf2=22.0           # inside and outside temperature in K
Beta2=1/(T_inf2+273.0)          # volumetric thermal expansion coefficient at 295 K and 273 K

g=9.81
t=0.005               # thickness of glass
L=0.60                # window length in m
k=0.81                # thermal conductivity of glass from appendix table B3
Tw1=18
Tw2=4
Ra1=(g*Beta1*(Tw2-T_inf1)*L**3)/(v1*a1)
hL1=(kf1/L)*(0.68+((0.67*((abs(Ra1)))**(1/4.0))/(1+(0.492/Pr1)**(9/16.0))**(4/9.0)))
Ra2=(g*Beta2*(Tw1-T_inf2)*L**3)/(v2*a2)
hL2=(kf2/L)*(0.68+((0.67*(abs(Ra2))**(1/4.0))/(1+(0.492/Pr2)**(9/16.0))**(4/9.0)))
q1=(T_inf1-T_inf2)/((1/hL2)+(t/k)+(1/hL1))
Tw2_=T_inf2-(q1/hL2)
Tw1_=q1/hL1+T_inf1

Ra1_=3.7*10**8
hL1_=2.92
Ra2_=2.31*10**8
hL2_=2.80
q2=(T_inf2-T_inf1)/((1/hL2_)+(t/k)+(1/hL1_))

Tw2final=q2-T_inf2
Tw1final=10.7

print"The heat loss is ",round(q2,1)," W/sq.m"

The heat loss is  31.2  W/sq.m


### Example 8.3 Page No.419¶

In [30]:
rou= 0.0735     # density in Ibm/cu.ft
cp=0.240        # specific heat BTU/(lbm-degree Rankine)
v= 16.88e-5     # viscosity in sq.ft/s
kf = 0.01516    # thermal conductivity in BTU/(hr.ft.degree Rankine)
a = 0.859       # diffusivity in sq.ft/hr
Pr = 0.708      # Prandtl Number
Tw=90
T_inf=70
g=32.2
L=5.5           # length in ft
W=2+(4/12.0)      # width in ft
Beta=1/(Tw+460.0) # volumetric thermal expansion coefficient in per degree Rankine
Ra=(g*Beta*(Tw-T_inf)*L**3)/(v*a/3600)
hc=(kf/L)*(0.825+((0.387*(Ra)**(1/6.0))/(1+(0.492/Pr)**(9/16.0))**(8/27.0)))**2
q=hc*L*W*(Tw-T_inf)

print"The heat gained is %d BTU/hr",round(q,0),"BTU/hr"

The heat gained is %d BTU/hr 142.0 BTU/hr


### Example 8.4 Page no. 421¶

In [2]:
rou= 1123         # density in kg/m^3
cp= 1006.7        # specific heat in J/(kg*K)
v= 17.204e-6      # vismath.cosity in m^2/s
Pr =0.703        # Prandtl Number
kf= 0.02738       # thermal conductivity in W/(m.K)
a = 0.2446e-4     # diffusivity in m^2/s
g=9.81
L=5.0
theta=45
T_inf=20.0           # ambient air temperature in degree C
Tw=65              # roof surface temperature in degree C
Beta=1/(T_inf+273.0) # volumetric thermal math.expansion coefficient in per K

import math
x=((3e5*math.exp(0.1368*math.cos(90-theta))*v*a)/(g*math.cos(theta)*Beta*(Tw-T_inf)))**(1/3.0)
x=0.051
print"The Laminar-turbulent  transition length by Vliet equation is ",round(x,3),"m\n\n"
lists=[0.02,0.04,0.051,0.051,0.1,1.0,3,5]
Ra=[0,0,0,0,0,0,0,0]
hc=[0,0,0,0,0,0,0,0]
print"_______________________________________________________"
print"x(m)\t\tRaL\t\t\thc(W/[m.K])"
print"_______________________________________________________"
for i in range(0,8):
if lists[i]<x:
# Laminar Flow regime exists
Ra[i]=(g*math.cos(math.pi*45.0/180.0)*Beta*(Tw-T_inf)*lists[i]**3)/(v*a)
hc[i]=(kf/lists[i])*(0.68+(0.670*Ra[i]**(1/4.0))/(1+(0.492/Pr)**(9/16.0))**(4.0/9.0))
print lists[i],"\t\t%.4g"%Ra[i],"\t\t%.3g"%hc[i]
else:
# Turbulent Flow regime exists
Ra[i]=(g*Beta*(Tw-T_inf)*lists[i]**3)/(v*a)
hc[i]=(0.02738/lists[i])*(0.825+0.324*Ra[i]**(1/6.0))**2
print lists[i],"\t\t%.4g"%Ra[i],"\t\t%.3g"%hc[i]

print"\n\nNOTE:\nCalculation mistake in book in calculation of Ral and hc,when x=0.04(2nd step in loop)"

The Laminar-turbulent  transition length by Vliet equation is  0.051 m

_______________________________________________________
x(m)		RaL			hc(W/[m.K])
_______________________________________________________
0.02 		2.025e+04 		9.32
0.04 		1.62e+05 		7.52
0.051 		4.749e+05 		7.3
0.051 		4.749e+05 		7.3
0.1 		3.58e+06 		6.39
1.0 		3.58e+09 		4.99
3 		9.667e+10 		4.73
5 		4.475e+11 		4.66

NOTE:
Calculation mistake in book in calculation of Ral and hc,when x=0.04(2nd step in loop)


### Example 8.5 Page No. 424¶

In [17]:
rou= 0.0735           # density in lbm/cu.ft
cp=0.240              # specific heat BTU/(lbm-degree Rankine)
v= 16.88e-5           # viscosity in sq.ft/s
kf = 0.01516          # thermal conductivity in BTU/(hr.ft.degree Rankine)
a = 0.859             # diffusivity in sq.ft/hr
Pr = 0.708            # Prandtl Number
Tw=100.0              # temperature of outside surface temperature of oven in degree F
T_inf=60.0            # ambient temperature  in degree F
g=32.2
L=2.0                  # length in ft
W=2.0                  # width in ft

Beta=1/(T_inf+460.0)      # volumetric thermal expansion coefficient in per degree Rankine
Ra=(g*Beta*(Tw-T_inf)*L**3)/(v*a/3600.0)
hc=(kf/L)*(0.68+(0.670*Ra**(0.25))/(1+(0.492/Pr)**(9/16.0))**(4/9.0))
q1side=hc*L*W*(Tw-T_inf)
Lc=0.5
Ra_L=(g*Beta*(Tw-T_inf)*Lc**3)/(v*a/3600.0) # Rayleigh number based on characteristic length
hc_L=(kf/Lc)*0.54*(Ra_L)**(1/4.0)
qtop=hc_L*L*W*(Tw-T_inf)

print"The heat transferred from one side is ",round(q1side,1),"BTU/hr"
print"The heat transferred from top is ",round(qtop,0),"BTU/hr"

if qtop < q1side:

print"More heat is transfered from side"
else:
print "More heat is transfered top side"

The heat transferred from one side is  93.7 BTU/hr
The heat transferred from top is  138.0 BTU/hr
More heat is transfered top side


### Example 8.6 Page No.427¶

In [52]:
rou= 1.177           # density in kg/cu.m
cp= 1005.7           # specific heat in J/(kg*K)
v= 15.68e-6          # viscosity in sq.m/s
Pr =0.708            #  Prandtl Number
kf=0.02624       # thermal conductivity in W/(m.K)
a=0.22160e-4    # diffusivity in sq.m/s
g=9.81
L=4.0             # length in m
D=15/100.0        # diameter in m
T_inf=5.0         # ambient air temperature in degree C
Tw=50.0            # outside surface temperature in degree C

import math
Beta=1/(T_inf+273.0) # volumetric thermal expansion coefficient in per K
Ra=(g*Beta*(Tw-T_inf)*D**3)/(v*a)
hc_h=(kf/D)*(0.60+(0.387*Ra**(1/6.0))/(1+(0.559/Pr)**(9/16.0))**(8/27.0))**2
As=math.pi*D*L
q_hor=hc_h*As*(Tw-T_inf)
hc_v=(kf/D)*0.6*(Ra*(D/L))**(1/4.0)
q_ver=hc_v*As*(Tw-T_inf)
q=round(q_ver,0)+round(q_hor,0)

print"The heat transferred from the horizontal length of 4 m is ",round(q_hor,0),"W"
print"The heat transferred from the vertical length of 4 m is ",round(q_ver,0),"W"
print"nThe total heat lost from the pipe is ",round(q,2),"W"

The heat transferred from the horizontal length of 4 m is  477.0 W
The heat transferred from the vertical length of 4 m is  246.0 W
nThe total heat lost from the pipe is  723.0 W


### Example 8.7 Page No. 430¶

In [54]:
rou= 0.0809          # density in lbm/cu.ft
cp=0.240             # specific heat BTU/(lbm-degree Rankine)
v= 13.54e-5          # viscosity in sq.ft/s
kf = 0.01402         # thermal conductivity in BTU/(hr.ft.degree Rankine)
a = 0.685            # diffusivity in sq.ft/hr
Pr = 0.712           # Prandtl Number
Tw=0                 # temperature of outside surface temperature of oven in degree F
T_inf=70.0             # ambient temperature  in degree F
g=32.2
Beta=1/(T_inf+460.0)   # volumetric thermal expansion coefficient in per degree Rankine
Lc=1/((1/1)+(1/1.2))
Ra=(g*Beta*abs(Tw-T_inf)*Lc**3)/(v*a/3600.0)
hc=(kf/Lc)*0.6*(Ra)**(1/4.0)

print"The value of convection coefficient is ",round(hc,2),"BTU/(hr.sq.ft.degree R)"

The value of convection coefficient is  1.11 BTU/(hr.sq.ft.degree R)


### Example 8.8 Page No.433¶

In [59]:
rou= 0.998                # density in kg/cu.m
cp= 1009.0                # specific heat in J/(kg*K)
v= 20.76e-6               # viscosity in sq.m/s
Pr =0.697                 # Prandtl Number
kf= 0.03003               # thermal conductivity in W/(m.K)
a = 0.2983e-4             # diffusivity in sq.m/s
g=9.81
T_inf=35                   # ambient air temperature in degree C
Tw=100                     # surface temperature in degree C
Beta=1/(T_inf+273.0)         # volumetric thermal expansion coefficient in per K
rou_Al=2702                 # density in kg/cu.m
k_Al=236                   # thermal conductivity in W/(m.K)
cp_Al=896                  # specific heat in J/(kg*K)
a_Al=97.5e-6               # diffusivity in sq.m/s
b=46/100.0
w=24/100.0

zeta=((w*v**2)/(g*Beta*(Tw-T_inf)*Pr))**(1/4.0)
L=1.54*(k_Al/kf)**(1/2)*zeta
S=2.89*zeta
q=(b*w*(Tw-T_inf)*1.3*(k_Al*kf)**(1/2.0))/(6*zeta)
N=b/(2*S)

print"The heat transfer rate is ",round(q,0),"W"
print"The number of fins can be atmost",round(N,0)

The heat transfer rate is  1423.0 W
The number of fins can be atmost 27.0

In [ ]: