CHAPTER 7
FREE CONVECTION
7.1 Introduction
7.2 Features and Parameters of Free Convection
(1) Driving Force.
Requirements
(i) Gravitational field
(ii) Density change with temperature
(2) Governing Parameters. Two parameters:
(i) Grashof number
T ) L3
g (Ts
Grashof number = GrL
2
(7.1)
(ii) Prandtl number
is the Coefficient of thermal expansion, also known as compressibility factor.
For ideal gases it is given by
1 , for ideal gas
T
(2.21)
Rayleigh number
Ra L
GrL Pr
g (Ts T ) L3
ν2
Pr =
g (Ts T ) L3
ν
(7.2)
(3) Boundary Layer.
Flow: Laminar, turbulent, or mixed.
Boundary layer approximations are valid for Ra x
10 4.
(4) Transition from Laminar to Turbulent Flow.
For vertical plates: transition Rayleigh number, Rax t , is
Rax t
109
(7.3)
(5) External vs. Enclosure Free Convection.
(i) External free convection: surface is immersed in infinite medium.
(ii) Enclosure free convection. Free convection takes place inside closed volumetric regions.
(6) Analytic Solutions.
Velocity and temperature fields are coupled.
Momentum and energy equation must be solved simultaneously.
2
7.3 Governing Equations
Approximations:
(1) Constant density, except in evaluating gravity forces.
(2) The Boussinesq approximation (relates density change to temperature change).
(3) No dissipation.
Continuity, momentum, and energy equations are obtained from equations (2.2),
(2.29) and (2.19), respectively
u
x
u
x
u
u
y
v
u
v
x
v
y
u
x
1
y
T
x
v
(p
(7.4)
0
1
g (T T )
v
v
y
(p
p )
v(
2
T
y
v(
p )
T
x2
2
v
x2
2
u
2
x
2
y2
2
v
y2
u
)
)
T
y2
(7.7)
Continuity equation (7.4) is unchanged
x-component of the Navier-Stokes equations simplifies to
u
x
v
u
y
v
βg T T
2
u
y2
(7.8)
Energy equation (7.7)
u
T
x
v
T
y
2
α
T
y2
(7.9)
(7.4), (7.8), and (7.9) contain three unknowns: u, v, and T.
Momentum and energy are coupled.
7.4 Laminar Free Convection over a Vertical Plate: Uniform Surface Temperature
Uniform temperature T s (Fig. 7.1).
Infinite fluid at temperature T .
Determine: velocity and temperature distribution.
7.4.1 Assumptions. Note all assumptions listed in this section.
7.4.2 Governing Equations
(7.6)
2
7.3.1 Boundary Layer Equations
u
(7.5)
3
u
x
u
u
x
v
v
y
u
y
(7.4)
0
2
v
βg T T
u
y2
(7.8)
2
u
where
v
x
α
y
(7.10)
y2
is defined as
T T
Ts T
(7.11)
7.4.3 Boundary Conditions.
Velocity:
(1)
(2)
(3)
(4)
u(x,0) 0
v(x,0) 0
u(x, ) 0
u(0, y) 0
Temperature:
(5) (x,0) 1
(6) (x, ) 0
(7) (0, y) 0
7.4.4 Similarity Transformation. Introduce the similarity variable
Grx
4
1/ 4
y
x
(7.14)
where
g (Ts T ) x 3
Grx
(7.15)
ν2
Let
( )
(7.16)
Grx d
x d
(7.20)
( x, y)
u
2v
v
(Grx )1/ 4
x
Continuity gives
v
1/ 4
(4)
d
d
(7.21)
3
(7.20) and (7.21) into (7.8) and (7.10) and using (7.11) and (7.16), gives
d3
d
3
3
d2
d
2
d
2
d
2
0
(7.22)
4
d2
d
2
3Pr
d
d
0
(7.23)
Transformation of boundary conditions:
Velocity:
d (0)
0
d
(2) (0) 0
d ( )
(3)
0
d
d ( )
(4)
0
d
(1)
Temperature:
(1) (0) 1
(2) ( ) 0
(3) ( ) 0
The problem is characterized by a
single parameter which is the Prandtl
number.
7.4.5 Solution.
(7.22) and (7.23) and their
five boundary conditions are
solved numerically.
The solution is presented
graphically in Figs. 7.2 and
7.3. Fig. 7.2 gives the
5
7.4.6 Heat Transfer Coefficient and Nusselt Number.
Fourier’s law and Newton’s law:
T ( x,0)
y
Ts T
k
h
Express in terms of
(7.24)
and
h
k dT d (0)
T d
d
y
Ts
Use(7.11) and (7.14)
h
1/ 4
k Grx
x
4
d (0)
d
(7.25)
Local Nusselt number
Nu x
Grx
4
hx
k
1/ 4
d (0)
d
(7.26)
Average h
1
L
h
Table 7.1 [1,2]
L
h( x)dx
(2.50)
0
(7.25) into (2.50), and performing the integration
h
4 k GrL
3L 4
1/ 4
d (0)
d
(7.27)
Average Nusselt number is
Nu L
hL
k
4 GrL
3 4
1/ 4
d (0)
d
(7.28)
Solution depends on a single parameter which is the
Prandtl number.
d (0)
Numerical solution gives
, listed in Table 7.1.
d
Special Cases
Very small and very large Prandtl numbers:
Nu x
Nu x
0.600 ( PrRa x )1/4 ,
0.503( PrGrx )1/4 ,
Pr
0
Pr
(7.29a)
(7.29b)
7.5 Laminar Free Convection over a Vertical Plate: Uniform
Surface Heat Flux
Assumptions: Same as constant temperature plate.
Pr
_ d (0)
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
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
d
d 2 ( 0)
d 2
0.9862
0.676
0.6741
0.6421
0.5713
0.4192
0.2517
0.1450
6
Surface boundary conditions
k
T ( x,0)
y
(7.30)
qs
Surface flux is specified.
Determine: Surface temperature Ts (x) and local Nusselt number Nu x .
Solution by similarity transformation.
Solution:
Surface temperature
Ts ( x) T
5
ν 2 (q s ) 4
gk 4
1/ 5
x
( 0)
(7.31)
Local Nusselt number
g qs
Nu x
5ν 2 k
1/ 5
x
4
1
(0)
(7.32)
Table 7.2 [4]
(0) is a dimensionless parameter which depends on the
Prandtl number and is given in Table 7.2 [4].
Correlation equation for (0)
(0)
4 9 Pr
1/ 2
5Pr
10Pr
10
1/ 5
2
, 0.001 Pr 1000
(7.33)
Properties at the film temperature T f
Tf
T
Pr
0.1
1.0
7.34)
Ts (L / 2) / 2
7.6 Inclined Plates
Vertical plate solutions of Sections 7.4 and 7.5 apply
to inclined plates, with g replaced by g cos .
This approach is recommended for
60o .
7.7 Integral Method
7.7.1 Integral Formulation of Conservation of Momentum
Assume:
t
(a)
100
(0)
- 2.7507
- 1.3574
- 0.76746
- 0.46566
7
Application of the momentum theorem in the x-direction to the element
7.6
Fx
dx , Fig.
(b)
M x (out ) M x (in)
dx is enlarged in Fig. 7.7
dM x
dx
dx
Mx
p
d
( p ) dx
dx
dy
o dx
dx
gdxdy
M x dW
( p dp / 2)d
p
Fig. 7.7
p
p
dp
d
2
d
p dx
dx
p
o dx
Mx
dM x
dx
dx
Mx
(c)
Simplify
dp
o dx
dM x
dx
dx
dW
(d)
Wall shearing stress
u x,0
y
o
(e)
Weight of element
dW
dx
g dy
(f)
0
The x-momentum of the fluid entering element
( x)
u 2 dy
Mx
(g)
0
(e), (f) and (g) into (d)
u x,0
y
dp
dx
d
dx
gdy
0
u 2 dy
(h)
0
Combine pressure and gravity terms
dp
dx
dp
dx
g
(i)
8
Multiply by
and rewrite as integral
dp
dx
g
(j)
gdy
0
(j) into (h)
u x ,0
y
g
(
d
dx
)dy
0
u 2 dy
(k)
0
Express density difference in terms of temperature change
(T
(2.28)
T )
(2.28) into (k)
ν
u ( x,0)
y
g (T T ) dy
0
d
dx
u 2 dy
(7.35)
0
(7.35) applies to laminar as well as turbulent flow.
7.7.2 Integral Formulation of Conservation of Energy
Assume:
(1)
(2)
(3)
(4)
No changes in kinetic and potential energy
Negligible axial conduction
Negligible dissipation
Properties are constant
Forced convection formulation of conservation of energy, (5.7), is applicable to free
convection
T x,0
y
( x)
d
dx
u (T
T )dy
(7.36)
0
7.7.3 Integral Solution
Vertical plate, Fig. 7.6.
Uniform surface temperature T s .
We assumed
t . Thus we have two equations, (7.35) and (7.36) for the
determination of a single unknown .
Since both (7.35) and (7.36) must be satisfied, we introduce another unknown as
follows:
Assumed Velocity Profile:
u x, y
a0 ( x) a1 ( x) y a2 ( x) y 2
Boundary conditions on the velocity
(1) u(x,0)
0
(2) u(x, ) 0
a3 ( x) y 3
(a)
9
(3)
u( x, )
y
2
(4)
u ( x,0)
y
2
0
g
(Ts
T )
Applying the four boundary conditions gives a n . Equation (a) becomes
g (Ts T )
4ν
u
2
y
2
y
1
(b)
Let
g (Ts T )
4ν
u o ( x)
2
(c)
(b) becomes
u
y
uo ( x)
1
y
2
(7.37)
Treat uo (x) as the second unknown function, independent of .
Assumed Temperature Profile:
T ( x, y)
b0 ( x) b1 ( x) y b2 ( x) y 2
(d)
The boundary conditions are
(1) T ( x,0) Ts
(2) T ( x, ) T
T ( x, )
(3)
0
y
Application of the above boundary conditions gives
T ( x, y) T
(Ts T ) 1
y
2
(7.38)
Heat Transfer Coefficient and Nusselt Number
T ( x,0)
y
Ts T
k
h
(7.24)
(7.38) into (7.24)
h
2k
( x)
(7.39)
Thus the local Nusselt number is
Nu x
Must find and uo (x) and
(x).
hx
k
2
x
( x)
(7.40)
10
Solution
(7.37) and (7.38) into (7.35)
ν
uo
g (Ts T )
2
y
1
d u o2
dx 2
dy
0
y
y2 1
4
dy
(e)
0
Evaluate the integrals
1 d 2
uo δ
105 dx
1
βg Ts T
3
δ
ν
uo
δ
(7.41)
(7.37) and (7.38) into (7.36)
( x)
2 (Ts T )
1
d uo
T )
dx
(Ts
y
y 1
4
dy
(f)
0
Evaluate the integrals
1 d
uo
60 dx
1
(7.42)
(7.41) and (7.42) are two equation for (x) and uo (x).
Assume a solution of the form
u o ( x)
Ax m
(7.43)
( x)
Bx n
(7.44)
A, B, m and n are constants.
substitute (7.43) and (7.44) into (7.41) and (7.42)
2m n 2 2m
A Bx
105
n 1
m n
ABx m
210
1
g To T Bx n
3
1
x
B
n 1
A m
vx
B
n
n
(7.45)
(7.46)
Exponents in each equation must be identical. Thus
2m n 1 n
m n 1
m n
(g)
n
(h)
1
4
(i)
Solve (g) and (h) for m and n gives
m
1
, n
2
(i) into (7.45) and (7.46) gives A and B
A 5.17v Pr
and
20
21
1/ 2
g (Ts T )
v2
1/ 2
(l)
11
20
21
B 3.93 Pr -1/2 Pr
1/ 4
g (Ts T )
1/ 4
(m)
v2
(i) and (m) into (7.44)
20 1
3.93
1
21 Pr
x
1/ 4
( Ra x )
1/ 4
(7.47)
(7.47) into (7.40)
1/ 4
20 1
0.508
1
21 Pr
Nu x
( Ra x )1/ 4
(7.48)
7.7.4 Comparison with Exact Solution for Nusselt Number
(7.26) is the exact solution to the local Nusselt number
1/ 4
Grx
4
Nu x
d (0)
d
(7.26)
Rewrite (7.26) as
Grx
4
1/ 4
d (0)
d
Nu x
(7.49)
Rewrite (7.48)
Grx
4
1/ 4
Nu x
20 1
0.508
1
21 Pr
The right hand side of (7.49) and (7.50) are
compared in Table 7.3.
The exact solution for Pr 0
Nu x
exact
0.600 ( PrRa x )1/4 , Pr
integral
0.514( PrRa x )1 / 4 , Pr
0
Exact and integral solutions for Pr
Nu x
exact
Nu x
integral
0.503( Ra x )1/4 ,
0
(7.51a)
are
Pr
(7.29b)
0.508( Ra x )1 / 4 , Pr
(7.51b)
NOTE: The error ranges from 1% for Pr
14% for Pr 0 .
(4 Pr )1/ 4
(7.50)
Table 7.3
Pr
_ d (0)
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
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
d
0.508
20 1
1
21 Pr
(7.29a)
0
Applying integral solution (7.47) to Pr
Nu x
1/ 4
to
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
2.2665
4.0390
1/ 4
(4 Pr )1/4