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
Temperature gradient → density gradient
(2) Governing Parameters: Two:
(i) Grashof number GrL
GrL =
β g(Ts − T∞ )L3
ν
2
(7.1)
β = coefficient of thermal expansion (compressibility), a property
β=
1
T
for ideal gas
(2.21)
2
(ii) Prandtl number Pr
Pr =
c pµ
k
Rayleigh number Ra L
β g(Ts − T∞ ) L3
β g(Ts − T∞ ) L3
Ra L = GrL Pr =
Pr =
2
ν
να
(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
Rax 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
Requires simultaneous integration of
continuity, momentum and energy
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
= β g (T − T∞ ) −
( p − p∞ ) + v ( 2 + 2 )
∂x
∂y
ρ ∞ ∂x
∂x
∂y
(7.2)
y-momentum:
1 ∂
∂v
∂v
∂ 2v ∂ 2v
u +v
=−
( p − p∞ ) + v ( 2 + 2 )
∂x
∂y
ρ ∞ ∂y
∂x
∂y
(7.3)
Energy:
 ∂ 2T ∂ 2T 
∂T
∂T
u
+v
= α  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 temperature T∞
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
∂x
∂y
∂y 2
(7.5)
Energy:
∂θ
∂θ
∂ 2θ
u
+v
=α 2
∂x
∂y
∂y
(7.7)
where θ is a dimensionless temperature defined as
T − T∞
θ=
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) θ ( x ,0) = 1
(6) θ ( x , ∞ ) = 0
(7) θ ( 0, y ) = 0
7.4.4 Similarity Transformation
• Transform three PDE to two ODE
• Introduce similarity variable η ( x , y )
y
η( 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
 Gr 
η= x
 4 
1/ 4
y
x
(7.10)
Local Grashof number:
Grx =
β g(Ts − T∞ )x 3
ν
2
(7.11)
Let
0 ( x , y ) = 0 (η )
(7.12)
Stream function ψ satisfies continuity
14
u=
∂ψ
∂y
(7.14)
Using Blasius solution as a guide, ψ for this problem is
1
4
 Grx 
ψ == 4v 
 ξ (η )
 4 
(7.15)
ξ (η ) is an unknown function
Introduce (7.15) into (7.13) and (7.14)
Grx dξ
u = 2v
x dη
(Grx )1 / 4  dξ

v = 1/ 4
η
−
3
ξ
 dη

x
( 4)


v
(7.16)
(7.17)
15
Combining (7.5), (7.7), (7.8), (7.12), (7.16), (7.17)
3
d ξ
dη
2
2
3
+ 3ξ
d 2θ
dη
2
 dξ 
−
2
  +θ = 0
2
dη
 dη 
d ξ
+ 3 Prξ
dθ
=0
dη
(7.18)
(7.19)
NOTE
• x and y are eliminated (7.18) and (7.19)
• Single independent variable
η
Transformation of boundary conditions:
Velocity:
16
dξ (0)
(1)
=0
dη
(2) ξ (0) = 0
dξ (∞ )
(3)
=0
dη
(4)
dξ (∞ )
=0
dη
Temperature:
(1) θ ( 0) = 1
(2) θ (∞ ) = 0
(3) θ (∞ ) = 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 θ and η
− k dT dθ (0) ∂η
h=
Ts − T∞ dθ dη ∂y
Use (7.8) and (7.10)
− k  Gr x 
h=
x  4 
1/ 4
dθ ( 0 )
dη
(7.21)
21
Define: local Nusselt number:
hx
 Grx 
Nu x =
= −

k
 4 
1/ 4
dθ (0)
dη
(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
dθ (0)
dη
(7.23)
Average Nusselt number
22
hL
4 k  GrL 
Nu L =
=−


k
3 L 4 
1/ 4
dθ (0)
dη
Table 7.1 [1,2]
(7.24)
Pr
dθ ( 0 )
dη
•
•
•
•
•
Is key factor in solution
Depends on Prandtl number
Values are listed in Table 7.1
Obtained from numerical solution
d 2ξ (0)
dη
2
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
_ dθ ( 0 )
dη
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 2ξ ( 0)
dη 2
0.9862
0.676
0.6741
0.6421
0.5713
0.4192
0.2517
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) δ , (4) δ 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 δ 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:
β g(Ts − T∞ ) L3
Ra L =
να
(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
α = thermal diffusivity, m 2 /s
β = coefficient of thermal expansion = 1 / T f K -1
ν = kinematic viscosity, m 2 /s
Properties at T f
Ts + T∞ (70 + 10)o C
= 40o C
Tf =
=
2
2
k = thermal conductivity = 0.0271 W/m − o C
Pr = 0.71
ν = 16.96 × 10 −6 m 2 /s
28
16.96 × 10 −6 m 2 /s
=
= 23.89 × 10 −6 m 2 /s
α=
Pr
0.71
ν
β=
1
40o C + 273.13
= 0.0031934 K -1
Substituting into (7.2)
Ra L =
0.0031934 ( K −1 )9.81(m/s 2 )(70 − 10)(K)(0.08) 3 (m 3 )
16.96 × 10 − 6 (m 2 /s)23.89 × 10 − 6 (m 2 /s)
= 2.3752 × 10 6
Thus the flow is laminar
(1) Axial velocity u:
Grx dξ
u = 2v
x dη
(7.10)
29
Fig. 7.2:
dξ
vs. η
dη
 Grx 
η=

 4 
1/ 4
y
x
(a)
Ra L 2.3792 × 106
Grx = GrL =
=
= 3.351 ×
Pr
0.71
Use (a), evaluate η 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 η = 1.21 and Pr = 0.71, gives
30
dξ
x
=u
≈ 0.27
dη
2ν Grx
Solve for u
u=
2ν 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 η = 1.21 and Pr = 0.71, Fig. 7.3 gives
T − T∞
θ =
≈ 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
δ:
31
At y = δ , axial velocity u ≈ 0, Fig. 7.2 gives
 GrL 
η( x , δ ) ≈ 5 = 

 4 
Solve for δ
1/ 4
δ
L
1/ 4
4


δ = 5(0.08)(m ) 
6
 3.37321 × 10 
= 0.0132 m = 1.32 cm
(4) Temperature B.L. thickness δ t :
At
y = δ t , T ≈ T∞ , θ ≈ 0. Fig. 7.3 gives
 GrL 
η( x , δ t ) ≈ 4.5 = 

 4 
1/ 4
δ
L
32
Solve for δ t
δ t = 4.5(0.08)(m ) 
1/4
4

6
 3.3511 × 10 
= 0.0119 m = 1.19 cm
(5) Local Nusselt number:
hx
 Grx 
Nu x =
= −
k
 4 
1/4
dθ ( 0 )
dη
(7.22)
dθ ( 0 )
Table 7.1 gives
at Pr = 0.71
dη
dθ ( 0 )
= −0.5018
dη
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 

dη
4


dθ ( 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 (T s − 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 
4 (0.0271)(W/m − C)  3.351 × 10
o
h=−
3
0.08(m)


4
1/4
dθ (0)
dη
6  1/4


( −0.5018 ) =
6.86 W/m 2 − o C
qT = 6.86(W/m 2 − o C)0.08(m)0.08(m)(70 − 10)( o C) = 2.63 W
(iii) Checking
Dimensional check:
Units for
u, T , δ , δ t , Nu, h, q ′′ and qT are consistent
Quantitative check:
35
(i) h is within the range given in Table 1.1
(ii) δ
> δ 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
 ν (q ′s′ ) 
Ts ( x ) − T∞ = − 5
x
4
 β gk

2
θ (0) = constant, depends on Pr,
4
1/ 5
θ ( 0)
given in Table 7.2
Table 7.2 [4]
(2) Nusselt number
 β gq ′s′ 4 
Nu x = −  2 x 
 5ν k 
1/ 5
(7.27)
1
θ ( 0)
(7.28)
Correlation equation for θ ( 0) [5]
 4 + 9 Pr + 10 Pr 
θ ( 0) = − 

2
5 Pr


1/2
Pr
θ ( 0)
0.1
1.0
- 2.7507
10
- 0.76746
100
- 0.46566
- 1.3574
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
9
(14) Laminar flow ( Ra x < 10 )
(15) Flat plate
41
(16) Uniform surface heat flux
(17) No dissipation
(18) No radiation
(ii) Analysis
Rayleigh number:
β g(Ts − T∞ ) L3
Ra L =
να
(7.2)
Surface temperature:
 ν (q ′s′ ) 
Ts ( x ) − T∞ = − 5
x
4
 β gk

2
4
1/ 5
θ ( 0)
(7.27)
Nusselt number:
42
 β gq ′s′ 4 
Nu x = −  2 x 
 5ν k 
1/ 5
1
θ ( 0)
(7.28)
Heat transfer coefficient
k
h( x ) = Nu x
x
(a)
Determine θ ( 0) : Table 7.2 or equation (7.29)
 4 + 9 Pr + 10 Pr 
θ ( 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 ) + T s ( 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
o
Assume Ts ( L) = 130 C
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
ν = 16.96 × 10−6 m 2 /s
45
16.96 × 10 −6 m 2 /s
α=
=
= 23.89 × 10 −6 m 2 /s
Pr
0.71
ν
1
β= 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 × 10 6
•
Flow is laminar
Use (7.9) to determine θ (0)
46
 4 + 9(0.71) + 10(0.71) 
θ ( 0) = − 

2
5(0.71)


1/ 2
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) ≈ computed Ts (L)
Result: Ts ( L) = 63.2o C
47
Thus,
T f = 23.3o C
Properties of air 23.3o C
k = 0.02601 W/m − o C
Pr = 0.7125 25
ν = 15.55 × 10− 6 m 2 /s
α = 21.825 × 10− 6 m 2 /s
β = 0.003373 K -1
(7.29) gives θ ( 0)
θ ( 0) = −1.4913
(7.27) gives Ts ( x )
48
T s ( 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 Nu x
 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)
7.65
6.66
6.14
5.795
q′s′ ( W/m 2 )
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 θ
•
Solution: Replace g by gcos θ in
solutions of Sections 7.4 and 7.5
•
Limitations
θ ≤ 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
δ = δt
(a)
valid for Pr ≈ 1
Apply momentum theorem in x-direction to the element δ × dx
53
∑ Fx
= M x (out ) − M x (in )
∑ Fx :
•
•
•
Mx +
Wall stress
Pressure
Gravity
(b)
dM x
dx
dx
pδ +
d
( pδ )dx
dx
dy
τ odx
dx
M x dW ρ gdxdy
pδ
((pp++dp
dp//22))ddδδ
Fig. 7.7
dM x 
dp 
d


pδ +  p + dδ − pδ − ( pδ )dx − τ o dx − =  M x +
dx  − M x
2
dx
dx



(c)
Simplify
− δ dp − τ o dx − dW =
dM x
dx
dx
(d)
Wall stress:
54
τo = µ
∂u( x ,0 )
∂y
(e)
Weight:
• Variable density
• Integrate weight of element dx × dy
dW = dx
δ
∫0 ρ gdy
(f)
x-momentum:
• Constant density
δ ( x)
Mx = ρ
∫0
u2dy
(g)
Substitute(e), (f) and (g) into (d)
55
δ
δ
∂u( x ,0 )
dp
d
−µ
−δ
− ∫ ρgdy = ρ ∫ u2dy
∂y
dx 0
dx 0
(h)
Combine pressure and gravity terms (B.L. flow)
dp dp∞
= −ρ∞ g
≅
dx
dx
(i)
Rewrite
δ
dp
δ
= − ρ ∞ gδ = − ∫ ρ ∞ gdy
dx
0
(j)
(j) into (h)
∂u( x ,0 )
−µ
+g
∂y
δ
∫0
d
( ρ ∞ − ρ )dy = ρ
dx
δ
∫0
u 2dy
(k)
Density change:
56
ρ ∞ − ρ = ρβ (T − T∞ )
(2.28)
(2.28) into (k), assume constant ρ β
∂u( x ,0)
−ν
+βg
∂y
δ
∫0
d
(T − T∞ )dy =
dx
δ
∫0
u2 dy
(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
−α
=
∂y
dx
NOTE:
δ ( x)
∫0
u(T − T∞ )dy
(7.31)
Integral formulation of energy is the same for
free convection and for forced convection
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 δ ( x )
59
(1) Momentum gives δ ( x )
(2) Temperature is assumed in terms δ t ( x )
(3) Energy gives δ t ( x )
Solution Procedure for Free Convection:
Since we assumed δ = δ t ,
either momentum or energy gives δ t .
To satisfy both,
introduce another unknown in assumed velocity
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 , δ ) ≅ 0
(3)
(4)
∂u( x , δ )
≅0
∂y
∂ 2 u( x ,0)
∂y 2
=−
βg
(Ts − T∞ )
ν
• Setting y = 0 in the x-component, (7.2), to obtain condition (4)
• 4 boundary conditions give an
61
(a) gives
β g(Ts − T∞ ) 
y y2 
δ y 1 − 2 +
u=

2
2
δ δ 

4ν
Rewrite as
2
 β g(Ts − T∞ ) 2  y 
y
δ  1 − 
u=
2

 δ  δ 
4ν
(b)
Introduce a second unknown in (b). Define
 β g(Ts − T∞ ) 2 
δ 
uo ( x ) = 
2


4ν
(c)
(b) becomes
62
y 
y
u = uo ( x ) 1 − 
δ  δ
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 , δ ) ≅ T∞
(3) ∂T ( x , δ ) ≅ 0
∂y
≅
• 3 boundary conditions give bn
63
(d) becomes
y

T ( x , y ) = T∞ + (Ts − T∞ ) 1 − 
 δ
2
(7.33)
Heat Transfer Coefficient and Nusselt Number
∂ T ( x ,0 )
−k
∂y
h=
Ts − T∞
(7.20)
(7.33) into (7.20)
2k
h=
δ ( x)
(7.34)
Local Nusselt number:
64
Nu x =
hx
x
=2
k
δ ( x)
(7.35)
Determine δ ( x )
Solution
Use momentum: substitute (7.32) and (7.33) into (7.30)
−ν
uo
δ
+ β g (Ts − T∞ )
δ
∫0
2
y
d  uo2

1 − δ  dy = dx  2
δ
δ
∫0
4 
y

y 1 −  dy 
 δ

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
2α (Ts − T∞ ) = (Ts − T∞ ) 
dx  δ
δ

1
δ ( x)
∫0
4 
y



y 1 −  dy 
 δ

(f)
Evaluating the integrals
1 d
[uoδ ] = α 1
60 dx
δ
(7.37)
• The two dependent variables are δ ( 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)
δ ( 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
= βg (To − T∞ )Bx n − vx m − n
105
3
B
m+n
1 −n
m
+
n
−
1
ABx
=α x
210
B
(7.40)
(7.41)
To satisfy (7.40) and (7.41) at all values of x ,
the exponents of x in each term must be identical
(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=
4
2
(i)
Introduce (i) into (7.40) and (7.41)
1 2
1
A
A B = βg (To − T∞ )B − v
85
3
B
(j)
1
1
AB = α
280
B
(k)
Solve (j) and (k) for A and B
−1 / 2
20  −

A = 5.17ν  Pr + 
21 

 β g(Ts − T∞ ) 




ν2
1/ 2
(l)
and
68
1/ 4
20 

B = 3.93 Pr -1/2  Pr + 
21 

 β g(Ts − T∞ ) 




ν2
−1 / 4
(m)
Substitute (i) and (m) into (7.39)
 20 1

= 3.93
+ 1
x
 21 Pr

δ
1/ 4
( Ra x ) −1 / 4
(7.42)
Introduce(7.42) into (7.35)
−1 / 4
−
 20 1

Nu x = 0.508
+ 1
21
Pr


( 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
dθ (0)
dη
(7.22)
69
Rewrite above as
 Grx 
 4 
−1 / 4
−1 / 4
−
 20 1

Nu x = 0.508
+ 1
21
Pr


(4 Pr )1 / 4
(7.45)
Rewrite integral solution (7.43) as
 Grx 
 4 
−1 / 4
Nu x = −
dθ (0)
dη
(7.44)
• Compare right hand side of (7.45) with − dθ ( 0) / dη 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
_ dθ (0)
dη
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 δ = δ t ( Pr = 1 ). Solution is reasonable for all Prandtl
numbers
72