CHAPTER 7
FREE CONVECTION
7.1 Introduction
 Applications:
Solar collectors
Pipes
Ducts
Electronic packages
Walls and windows
7.2 Features and Parameters of Free Convection
(1) Driving Force: Natural
1
Requirements: (i) Acceleration field, (ii) Density gradient
Te
g
d
g
e
(2) Governing Parameters: Two:
(i) Grashof number GrL
GrL 
β g(Ts  T )L3
ν
2
(7.1)
 = coefficient of thermal expansion (compressibility), a property

1
T
for ideal gas
(2.21)
2
(ii) Prandtl number Pr
Pr 
c p
k
Rayleigh number Ra L
 g(Ts  T ) L3
 g(Ts  T ) L3
Ra L  GrL Pr 
Pr =
2


(7.2)
(3) Boundary Layer: laminar, turbulent or mixed
Criterion:
Ra x  104
(4) Transition: Laminar to Turbulent Flow
Criterion for vertical plates: Transition Rayleigh number
3
Ra x t  109
(5) External vs. Enclosure free convection:
(i) External: over:
Vertical surfaces
Inclined surfaces
Horizontal cylinders
Spheres
(ii) Enclosure: in:
4
Rectangular confines
Concentric cylinders
Concentric spheres
(6) Analytic Solution
Rs
u
i
n
o
e
c
,
ma
e
o
i
7.3 Governing Equations
Approximations:
(1) Density is constant except in gravity forces
5
(2) Boussinesq approximation: relate density change to temperature
change
(3) Negligible dissipation
Assume:
Steady state
Two-dimensional
Laminar
Continuity:
u v

0
x  y
(7.1)
x-momentum:
6
u
u
1 
 2u  2u
u v
  g (T  T ) 
( p  p )  v ( 2  2 )
x
y
  x
x
y
(7.2)
y-momentum:
v
v
1 
 2v  2v
u v

( p  p )  v ( 2  2 )
x
y
  y
x
y
(7.3)
Energy:
  2T  2T 
T
T
u
v
   2  2 
x
y
y 
 x
(7.4)
NOTE:
(1) Gravity points in the negative x-direction
(2) Flow and temperature fields are coupled
7
7.3.1 Boundary Layer Equations
• Velocity and temperature boundary layers
• Apply approximation used in forced convection
y-component of the Navier-Stokes equations reduces
to

( p  p )  0
y
(a)
External flow: Neglect ambient pressure variation in x

( p  p )  0
x
(b)
Furthermore, for boundary layer flow
8
 2u
x
2

 2u
y
2
(c)
x-momentum:
u
u
 2u
u v
  βg T  T   v 2
x
y
y
(7.5)
Neglect axial conduction
 2T
x
2

 2T
y
2
(d)
Energy: (7.4) becomes
T
T
 2T
u
v
α 2
x
y
y
(7.6)
9
7.4 Laminar Free Convection over a Vertical Plate:
Uniform Surface Temperature
 Vertical plate
 Uniform temperature Ts
 Infinite fluid at temperatureT
Determine: velocity and temperature distribution
7.4.1 Assumptions
(1) Steady state
(2) Laminar flow
(3) Two-dimensional
10
(1)
(2)
(3)
(4)
(5)
(9)
Constant properties
Boussinesq approximation
Uniform surface temperature
Uniform ambient temperature
Vertical plate
Negligible dissipation
7.4.2 Governing Equations
Continuity:
u  v

0
x y
(7.1)
x-momentum:
11
u
u
 2u
u
v
  βg T  T   v 2
x
y
y
(7.5)
Energy:


 2
u
v
α 2
x
y
y
(7.7)
where  is a dimensionless temperature defined as
T  T

Ts  T
(7.8)
7.4.3 Boundary Conditions
Velocity:
(1) u( x ,0)  0
(2) v ( x ,0)  0
12
(3) u( x ,  )  0
(4) u( 0, y )  0
Temperature:
(5)  ( x ,0)  1
(6)  ( x ,  )  0
(7)  ( 0, y )  0
7.4.4 Similarity Transformation
 Transform three PDE to two ODE
 Introduce similarity variable  ( x , y )
y
( x, y)  C
x
1/ 4
(7.8)
13
 βg Ts  T 
C

4v 2

1/ 4
(7.9)
(7.9) into (7.8)7
 Grx 
η

 4 
1/ 4
y
x
(7.10)
Local Grashof number:
Grx 
β g(Ts  T )x 3
ν
2
(7.11)
Let
 ( x , y )   ( )
(7.12)
Stream function  satisfies continuity
14
u

y
(7.14)
Using Blasius solution as a guide,  for this problem is
1
4
 Grx 
ψ  4v 
 ξ η 
 4 
(7.15)
 ( ) is an unknown function
Introduce (7.15) into (7.13) and (7.14)
Grx d
u  2v
x d
(Grx )1 / 4  d

v  1/ 4
  3 

x
( 4)
 d

v
(7.16)
(7.17)
15
Combining (7.5), (7.7), (7.8), (7.12), (7.16), (7.17)
d 
3
d
3
 3
d 2
d
2
2
d 
 d 

2
    0
2
d
 d 
2
 3 Pr
d
0
d
(7.18)
(7.19)
NOTE
 x and y are eliminated (7.18) and (7.19)
 Single independent variable 
Transformation of boundary conditions:
Velocity:
16
d (0)
0
(1)
d
(2)  ( 0)  0
d ( )
(3)
0
d
d ( )
0
(4)
d
Temperature:
(1)  ( 0)  1
(2)  (  )  0
(3)  (  )  0
17
NOTE:
(1) Three PDE are transformed into two ODE
(3) Five BC are needed
(4) Seven BC are transformed into five
(5) One parameter: Prandtl number.
7.4.5 Solution
• (7.18) and (7.19) are solved numerically
• Solution is presented graphically
• Figs. 7.2 gives u(x,y)
• Fig. 7.3 gives T(x,y)
18
19
20
7.4.5 Heat Transfer Coefficient and Nusselt Number
Start with
 T ( x ,0 )
k
y
h
Ts  T
(7.20)
Express in terms of  and 
 k dT d (0) 
h
Ts  T d d y
Use (7.8) and (7.10)
 k  Grx 
h
x  4 
1/ 4
d (0)
d
(7.21)
21
Define: local Nusselt number:
hx
 Grx 
Nu x 
 

k
 4 
1/ 4
d ( 0 )
d
(7.22)
Average heat transfer coefficient
1 L
h   h( x )dx
L 0
(2.50)
(7.21) into (2.50), integrate
4 k  GrL 
h


3 L 4 
1/ 4
d ( 0 )
d
(7.23)
Average Nusselt number
22
hL
4 k  GrL 
Nu L 



k
3 L 4 
1/ 4
d (0)
d
T a b l e 7 . 1 [1 , 2 ]
(7.24)
2
P
d
(0)
d
 Ik f
D
V
O

d 2 (0)
d
2
i s
oP
n
a l
iT 7
f
n
s
d ( 0) d  ( 0)
2
d
d
0.01
0.03
0.09
0.5
0.72
0.733
se
1.0
1.5e
2.0
3.5a
5.0
7.0
10
100
1000
_
r
0.0806
0.9862
0.136
0.219
0.442
0.5045
0.676
0.508
0.6741
a
no
0.5671
0.6421
u
0.6515 n r
0.7165
0.5713
r i
na
0.8558
0.954
u
1.0542 r
1.1649
0.4192
2.191
0.2517
3.9660
0.1450
in Table 7.1 gives surface velocity
gradient and shearing stress
23
Special Cases
Nu x  0.600( PrRa x )1/4 ,
Nu x  0.503( PrGrx )1/4 ,
Pr  0
Pr  
(7.24)
(7.25)
Example 7.1: Vertical Plate at Uniform Surface
Temperature
 8cm  8cm vertical plate in air at 10 oC
 Uniform surface temperature = 70 oC
 Determine the following at x  L = 8 cm:
(1) u at y = 0.2 cm , (2) T at y = 0.2 cm
(3) d , (4) d t
24
(5) Nu L , (6) h(L)
(7) qs (L) , (8) qT
Solution
(1) Observations
 External free convection
 Vertical plate
 Uniform surface temperature
•
Check Rayleigh number for laminar flow
 If laminar, Fig. 7.2 for u(x,y) and Fig. 7.3 T(x,y) and d t
 Determine local Nu and h at x = L
(2) Problem Definition
25
Determine flow and heat transfer characteristics for free convection
over a vertical flat plate at uniform surface temperature.
(3) Solution Plan
 Laminar flow? Compute Rayleigh number
 If laminar, use Figs. 7.2 and 7.3.
 Use solution for Nu and h
(4) Plan Execution
(i) Assumptions
(1) Newtonian fluid
(2) Steady state
(3) Boussinesq approximations
(4) Two-dimensional
26
(5) Laminar flow ( Rax  109 )
(6) Flat plate
(7) Uniform surface temperature
(8) No dissipation
(9) No radiation.
(ii) Analysis and Computation
Compute the Rayleigh number:
 g(Ts  T ) L3
Ra L 

(7.2)
g  gravitational acceleration = 9.81 m/s 2
L  plate length = 0.08 m
27
Ra L  Rayleigh number at the trailing end x  L
Ts  surface temperature = 70o C
T  ambient temperature =10o C
  thermal diffusivity, m 2 /s
  coefficient of thermal expansion  1 / T f K -1
  kinematic viscosity, m 2 /s
Properties at T f
Ts  T (70  10)o C
Tf 

 40o C
2
2
k  thermal conductivity = 0.0271W/m o C
Pr  0.71
  16.96  106 m 2 /s
28
16.96  106 m 2 /s


 23.89  106 m 2 /s
Pr
0.71


1
40o C  273.13
 0.0031934K -1
Substituting into (7.2)
Ra L 
0.0031934( K 1 )9.81(m/s2 )(70  10)(K)(0.08)3 (m 3 )
16.96 106 (m 2 /s)23.89 106 (m 2 /s)
 2.3752 106
Thus the flow is laminar
(1) Axial velocity u:
Grx d
u  2v
x d
(7.10)
29
d
Fig. 7.2:
vs.
d
 Grx 
η

 4 
1/ 4
y
x
(a)
Ra L 2.3792  10 6
Grx  GrL 

 3.351  10 6
Pr
0.71
Use (a), evaluate at x  0.08 m and y  0.002 m
 3.351  10
η  
4

6




1/ 4
0.0032( m )
 1.21
0.08( m )
Fig. 7.2, at   1.21 and Pr = 0.71, gives
30
d
x
u
 0.27
d
2 Grx
Solve for u
u
2 GrL
0.27
L
 0.27
2(16.96  10
6
2
6
)(m /s) 3.351  10
 0.2096 m/s
0.08(m)
(1) Temperature T: At   1.21 and Pr = 0.71, Fig. 7.3 gives
T  T

 0.43
Ts  T
T  T  0.43(Ts  T )  10(o C)  0.43(70  10)(o C)  35.8o C
(3) Velocity B.L. thickness
d:
31
At y  d , axial velocity u  0, Fig. 7.2 gives
 GrL 
η( x , d )  5  

 4 
Solve for d
1/ 4
d
L
4


d  5(0.08)(m ) 
6
 3.37321  10 
1/ 4
 0.0132 m  1.32 cm
(4) Temperature B.L. thickness d t :
At
y  d t , T  T ,   0. Fig. 7.3 gives
 GrL 
η( x , d t )  4.5  

 4 
1/ 4
d
L
32
Solve for d t
d t  4.5(0.08)(m ) 
4

6
 3.3511  10 
1/4
 0.0119 m  1.19 cm
(5) Local Nusselt number:
hx
Grx 
Nux 
 
k
 4 
1/4
d (0)
d
(7.22)
d (0)
Table 7.1 gives
at Pr = 0.71
d
d (0)
 0.5018
d
Nusselt Number: Use (7.22), evaluate at x = L = 0.08 m
33
NuL 
hL
k
 GrL 

 4 
1/ 4
6 1 / 4
 3.351  10
 0.5018

d
4


d ( 0 )
 15.18
(6) Local heat transfer coefficient: at x = L = 0.08 m
k
0.0271(W/m o C)
h( L)  NuL 
15.18  5.14 W/m 2  o C
L
0.08(m)
(7) Heat flux: Newton’s law gives
qs  h(Ts  T )  5.14( W/m  o C)(70  10)( o C)  308.4 W/m 2
(8) Total heat transfer:
qT  hA(Ts  T )
34
4 k  GrL 
h


3 L 4 
h
4 (0.0271)(W/m  C)  3.351  10
o
3
0.08(m)


4
1/4
d (0)
d
6  1/4


( 0.5018) 
6.86 W/m2  o C
qT  6.86(W/m2  o C)0.08(m)0.08(m)(70  10)(o C)  2.63 W
(iii) Checking
Dimensional check:
Units for
u, T , d , d t , Nu, h, q and qT are consistent
Quantitative check:
35
(i) h is within the range given in Table 1.1
(ii) d
 d t . This must be the case
(5) Comments
(i) Magnitudes of u and h are relatively small
(ii) h( x )  h ( x )
7.5 Laminar Free Convection over a Vertical
Plate: Uniform Surface Heat Flux
• Vertical Plate
• Uniform surface heat flux
• Infinite fluid at temperature T
• Assumptions: Same as Section 7.4
36
• Governing equations: Same as Section 7.4
• Boundary conditions: Replace uniform surface temperature with
uniform surface flux
 T ( x ,0 )
k
 q s
y
 Surface temperature is variable and unknown: Ts ( x )
 Objective: Determine Ts ( x ) and
Nu x
• Solution: Similarity transformation (Appendix F)
• Results:
(1) Surface temperature
37
  (q s ) 
Ts ( x )  T   5
x
4
  gk

2
 (0) = constant, depends on Pr,
4
1/ 5
 ( 0)
given in Table 7.2
(2) Nusselt number
  gq s 4 
Nu x    2 x 
 5 k 
1/ 5
1
 ( 0)
(7.28)
Correlation equation for  (0) [5]
 4  9 Pr  10 Pr 
 ( 0)   

2
5 Pr


1/2
(7.27)
Table 7.2 [4]
P
0.1
1.0
 ( 0)
10
- 2.7507
- 1.3574
- 0.76746
100
- 0.46566
r
1/ 5
(7.29)
38
Example 7.2: Vertical Plate at Uniform Surface Flux
 8cm  8cm vertical plate in air at 10 oC
 Uniform surface flux = 308.4 oC
 Determine the following at x  2, 4, 6 and 8 cm
(1) Surface temperature
(2) Nusselt number
(3) Heat transfer coefficient
Solution
(1) Observations
• External free convection
• Vertical plate
39
• Uniform surface heat flux
• Check Rayleigh number for laminar flow
• If laminar:
Equation (7.27) gives Ts ( x )
Equation (7.28) gives
Nux
Nux gives h(x)
(2) Problem Definition
Determine Ts ( x ) ,
in free convection
Nux and h(x) for uniformly heated vertical plate
(3) Solution Plan
• Laminar flow? Compute Rayleigh number
40
 Use (7.27) for Ts ( x )
 Use (7.28) for
Nux
 Nux gives h(x)
(4) Plan Execution
(i) Assumptions
(10) Newtonian fluid
(11) Steady state
(12) Boussinesq approximations
(13) Two-dimensional
(14) Laminar flow ( Rax  10 )
9
(15) Flat plate
41
(16) Uniform surface heat flux
(17) No dissipation
(18) No radiation
(ii) Analysis
Rayleigh number:
 g(Ts  T ) L3
Ra L 

(7.2)
Surface temperature:
  (q s ) 
Ts ( x )  T   5
x
4
  gk

2
4
1/ 5
 ( 0)
(7.27)
Nusselt number:
42
  gq s 4 
Nu x    2 x 
 5 k 
1/ 5
1
 ( 0)
(7.28)
Heat transfer coefficient
k
h( x )  Nu x
x
(a)
Determine  (0) : Table 7.2 or equation (7.29)
 4  9 Pr  10 Pr 
 ( 0)   

2
5 Pr


1/ 2
1/ 5
(7.29)
(iii) Computations
Properties at T f
43
Ts  T
Tf 
2
(b)
where
Ts ( L)  Ts (0)
Ts 
2
(c)
Ts (L) is unknown. Use iterative procedure:
(1) Assume Ts (L)
(2) Compute T f
(3) Find properties at T f
(4) Use (7.27) to calculate Ts (L)
(5) Compare with assumed value in step (1)
(6) Repeat until (1) and (5) agree
44
Assume Ts ( L)  130 C
o
Ts ( L)  Ts (0) (130  10)(o C)
Ts 

 70o C
2
2
Ts  T (70  10)o C
Tf 

 40o C
2
2
o
Properties of air at 40 C :
k  0.0271 W/m o C
Pr  0.71
  16.96  10 6 m 2 /s
45
16.96  106 m 2 /s


 23.89  106 m 2 /s
Pr
0.71

1
 o
 0.0031934 K -1
40 C  273.13
Substitute into (7.2)
0.0031934( K 1 )9.81(m/s 2 )(130  10)(K)(0.08) 3 (m 3 )
Ra L 
16.96  10  6 (m 2 /s)23.89 10  6 (m 2 /s)
 4.7504  106
•
Flow is laminar
Use (7.9) to determine  ( 0)
46
 4  9(0.71)1 / 2  10(0.71) 
 ( 0)   

2
5(0.71)


1/ 5
 1.4928
Use (7.27) to compute Ts (L)
o
Ts (L)  10( C) 
1/ 5

6
2
4
2
4
4
8


(16.96  10 ) (m /s )(308.4) (W /m )
o
5
(0.1)(m)
(

1
.
4928
)

63
.
6
C


2
4 4 4 4
 0.0031934(1/K)9.81(m/s )(0.0271) (W /m  K

Computed Ts (L) is lower than assumed value
Repeat procedure until assume Ts (L)  computedTs (L)
Result: Ts ( L)  63.2o C
47
Thus,
T f  23.3o C
o
Properties of air 23.3 C
k  0.02601 W/m  o C
Pr  0.7125 25
  15.55  10 6 m2 /s
  21.825  10 6 m2 /s
  0.003373K -1
(7.29) gives  ( 0)
 (0)  1.4913
(7.27) gives Ts ( x )
48
Ts ( L)  10( o C ) 
6

(15.55  10 ) (m /s )(308.4) (W /m )
5
2
4
4
4
4
0.003373
(1/K)9.81
(m/s
)(0.02601)
(W
/m

K

2
4
2
4
4
8
1/5

(x1)(m) 

( 1.4913)
(7.28) gives Nux
 0.003373(1/K)9.81(m/s )308.4(W/m ) 4 4 
Nux   
(x) (m )
6
2
o
 5(15.55  10 )(m /s)0.02601(W/m  C)

2
2
1/ 5
1
 1.4913
(e)
(a) gives h(x)
49
k
0.02601(W/m o C)
h( L)  NuL 
NuL
L
x (m)
(f)
Use (d)-(f) to tabulate results at x = 0.02, 0.04, 0.06 and 0.08 m
x(m) Ts ( x ) (o C) Nux
0.02
0.04
0.06
0.08
50.3
56.3
60.2
63.2
5.88
10.24
14.16
17.83
h( x )( W/m 2  o C)
qs( W/m 2 )
7.65
6.66
6.14
5.795
308.3
308.4
308.2
308.3
(iii) Checking
50
Dimensional check:
Units for Ts , Nu and h are consistent.
Quantitative check:
(i) h is within the range given in Table 1.1
(ii) Compute q s using Newton’s law gives uniform surface flux
(5) Comments
(i) Magnitude of h is relatively small
(ii) Surface temperature increases with distance along plate
7.6 Inclined Plates
Two cases:
51
•
•
•
Hot side is facing downward, Fig. 7.5a
Cold side is facing upward, Fig. 7.5b
Flow field: Same for both
 Gravity component = gcos 
•
Solution: Replace g by gcos  in
solutions of Sections 7.4 and 7.5
•
Limitations
  60o
7.7 Integral Method
52
• Obtain approximate solutions to free
convection problems
• Example: Vertical plate at uniform surface
temperature
7.7.1 Integral Formulation of
Conservation of Momentum
•
Simplifying assumption
ddt
valid
for
Pr

1
(a)
Apply momentum theorem in x-direction to the elementd  dx
53
 Fx
 M x (out )  M x (in )
 Fx :
•
•
•
d x
M

d
x
d
Wall stress
Pressure
Gravity
(b)
od
d
p
d  ( pd)d
d
d
x
d
M
xd
g
p
d (p
(p


d
d/2
/2
)d
)d
d
d
F 7
dM x 
dp 
d


pd   p  dd  pd   pd dx   o dx   M x 
dx   M x
2
dx
dx



(c)
Simplify
 d dp   o dx  dW 
dM x
dx
dx
(d)
Wall stress:
54
o  
u x ,0 
y
(e)
Weight:
• Variable density
 Integrate weight of element dx  dy
dW  dx
d
0  gdy
(f)
x-momentum:
• Constant density
Mx  
d ( x)
0
u2dy
(g)
Substitute(e), (f) and (g) into (d)
55
d
 u  x ,0 
dp
d

d
 gdy  
u2dy
y
dx 0
dx 0

d

(h)
Combine pressure and gravity terms (B.L. flow)
dp dp

  g
dx
dx
(i)
Rewrite
d
dp
d
    gd      gdy
dx
0
(j)
d
u x ,0
d d 2

 g  (     )dy    u dy
y
dx 0
0
(k)
(j) into (h)
Density change:
56
      (T  T )
(2.28)
(2.28) into (k), assume constant  
d
u( x ,0)
d d 2

  g  (T  T )dy   u dy
y
dx 0
0
(7.30)
NOTE:
(1)
no shearing force on the slanted surface
(2)
(7.30) applies to laminar and turbulent flow
(3)
(7.30) is a first order O.D.E.
7.7.2 Integral Formulation of Conservation of Energy
Assume:
57
•
•
•
•
No changes in kinetic and potential energy
Neglect axial conduction
Neglect dissipation
Properties are constant
 T  x ,0 
d


y
dx
NOTE:
d ( x)
0
u(T  T )dy
(7.31)
I
n
t
e
g
r
a
l
f
o
r
m
u
l
a
t
i
o
n
o
f
e
n
e
r
g
y
i
s
t
h
e
s
a
m
e
f
o
r
f
r
e
e
c
o
n
v
e
c
t
i
o
n
a
n
d
f
o
r
f
o
r
c
e
d
c
o
n
v
e
c
t
i
o
n
7.7.3 Integral Solution
58
• Vertical plate
 Uniform temperature Ts
 Quiescent fluid at uniform temperature T
Solution Procedure for Forced Convection:
(1) Velocity is assumed in terms of d ( x )
59
(1) Momentum gives d ( x )
(2) Temperature is assumed in terms d t ( x )
(3) Energy gives d t ( x )
Solution Procedure for Free Convection:
dd
S
i
n
c
e
w
e
a
s
s
u
m
e
d 
t,
d
e
i
t
h
e
r
m
o
m
e
n
t
u
m
o
r
e
n
e
r
g
y
g
i
v
e
s t.
T
o
s
a
t
i
s
f
y
b
o
t
h
,
i
n
t
r
o
d
u
c
e
a
n
o
t
h
e
r
u
n
k
n
o
w
n
i
n
a
s
s
u
m
e
d
v
e
l
o
c
i
t
y
60
Assumed Velocity Profile
u x , y   a 0 ( x )  a 1 ( x ) y  a 2 ( x ) y 2  a 3 ( x ) y 3
(a)
Boundary conditions:
(1) u( x ,0)  0
(2) u( x , d )  0
(3)
(4)
u( x , d )
0
y
 2 u( x ,0)
y 2

g
(Ts  T )

 Setting y  0 in the x-component, (7.2), to obtain condition (4)
 4 boundary conditions give a n
61
(a) gives
 g(Ts  T ) 
y y2 
u
d y 1  2 

2
2
d d 

4
Rewrite as
2
  g(Ts  T ) 2  y 
y
u
d  1  
2

 d  d 
4
(b)
Introduce a second unknown in (b). Define
  g(Ts  T ) 2 
uo ( x )  
d 
2


4
(c)
(b) becomes
62
y 
y
u  uo ( x ) 1  
d  d
2
(7.32)
Assumed Temperature Profile
T ( x , y )  b0 ( x )  b1 ( x ) y  b2 ( x ) y 2
(d)
Boundary Conditions:
(1) T ( x,0)  Ts
(2) T ( x , d )  T
(3) T ( x , d )  0
y
 3 boundary conditions givebn
63
(d) becomes
y

T ( x , y )  T  (Ts  T ) 1  
 d
2
(7.33)
Heat Transfer Coefficient and Nusselt Number
 T ( x ,0 )
y
Ts  T
k
h
(7.20)
(7.33) into (7.20)
h
2k
d ( x)
(7.34)
Local Nusselt number:
64
Nu x 
hx
x
2
k
d ( x)
(7.35)
Determine d ( x )
Solution
Use momentum: substitute (7.32) and (7.33) into (7.30)

uo
d
  g (Ts  T )
d
0
2
y
d  uo2

1  d  dy  dx  2
d
d
0
4 
y

y 1   dy 
 d

2 
(e)
Evaluating the integrals
 
u
1 d 2
1
uo δ  βg Ts  T  δ  v o
105 dx
3
δ
(7.36)
Use energy: substitute(7.32) and (7.33) into (7.31)
65
d  uo
2 (Ts  T )  (Ts  T ) 
d
dx  d

1
d ( x)
0
4 
y



y 1   dy 
 d

(f)
Evaluating the integrals
1 d
uod    1
60 dx
d
(7.37)
 The two dependent variables ared ( x ) and uo ( x )
 (7.36) and (7.37) are two simultaneous first order O.D.E.
 Assume a solution of the form
uo ( x )  Ax m
(7.38)
d ( x )  Bx n
(7.39)
 Determine the constants A, B, m and n
66
 Substitute (7.38) and (7.39) into (7.36) and (7.37)
2m  n 2 2m  n1 1
A
A Bx
 g To  T Bx n  vx m  n
105
3
B
mn
1
ABx m  n1   x  n
210
B
(7.40)
(7.41)
T
o
s
a
t
i
s
f
y
(
7
.
4
0
)
a
n
d
(
7
.
4
1
)
a
t
a
l
l
v
a
l
u
e
s
o
f
x
,
x
t
h
e
e
x
p
o
n
e
n
t
s
o
f
i
n
e
a
c
h
t
e
r
m
m
u
s
t
b
e
i
d
e
n
t
i
c
a
l
(7.40) requires
2m  n  1  n  m  n
(g)
Similarly, (7.41) requires that
m  n  1  n
(h)
67
Solve (g) and (h) for m and n gives
1
1
m  , n
2
4
(i)
Introduce (i) into (7.40) and (7.41)
1 2
1
A
A B  g To  T B  v
85
3
B
(j)
1
1
AB  
280
B
(k)
Solve (j) and (k) for A and B
20 

A  5.17  Pr  
21 

1 / 2
  g(Ts  T ) 




2
1/ 2
(l)
and
68
20 

B  3.93 Pr -1/2  Pr  
21 

1/ 4
  g(Ts  T ) 




2
1 / 4
(m)
Substitute (i) and (m) into (7.39)
d
 20 1

 3.93
 1
x
 21 Pr

1/ 4
( Ra x ) 1 / 4
(7.42)
Introduce(7.42) into (7.35)
 20 1

Nu x  0.508
 1
 21 Pr

1 / 4
( Ra x )1 / 4
(7.43)
7.7.4 Comparison with Exact Solution for Nusselt Number
Exact solution:
 Gr 
Nu x    x 
 4 
1/ 4
d (0)
d
(7.22)
69
Rewrite above as
 Grx 
 4 
1 / 4
 20 1

Nux  0.508
 1
 21 Pr 
1 / 4
(4 Pr )1 / 4
(7.45)
Rewrite integral solution (7.43) as
 Grx 
 4 
1 / 4
Nu x  
d (0)
d
(7.44)
 Compare right hand side of (7.45) with  d (0) / d of exact solution
(7.44)
 Comparison depends on the Prandtl number
70
 Table 7.2 gives results
Table 7.3
Limiting Cases
Pr
(1) Pr  0
Nu x exact  0.600 ( PrRa x )1/4 , Pr  0 (7.25a)
Nu x integral  0.514( PrRa x )1 / 4 ,Pr  0 (7.46a)
(2) Pr  
Nu x exact  0.503 ( Ra x )1/4 ,
0.01
0.03
0.09
0.5
0.72
0.733
1.0
1.5
2.0
3.5
5.0
7.0
10
100
1000
_ d (0)
d
0.0806
0.136
0.219
0.442
0.5045
0.508
0.5671
0.6515
0.7165
0.8558
0.954
1.0542
1.1649
2.191
3.9660
Pr  
Nu x integral  0.508( Ra x )1 / 4 , Pr  
 20 1

0.508 
 1
 21 Pr 
1 / 4
(4 Pr )1/4
0.0725
0.1250
0.2133
0.213
0.4627
0.5361
0.5399
0.6078
0.7031
0.7751
0.9253
1.0285
1.1319
1.2488
1.2665
4.0390
(7.25b)
(7.46b)
71
NOTE:
(1) Error ranges from 1% for Pr   to 14% for Pr  0
(2) We assumed d  d t ( Pr  1 ). Solution is reasonable for all Prandtl
numbers
72