Chapter # 9
Free Convection
OBJECTIVES
• Obtain an appreciation for the physical origins and nature of buoyancy-
driven flows.
• Acquire tools for performing heat transfer calculations for natural
convection.
2
GENERAL CONSIDERATIONS
• Free convection refers to fluid motion induced by buoyancy forces.
• Buoyancy forces may arise in a fluid for which there are density gradients
and a body force that is proportional to density.
• In heat transfer, density gradients are due to temperature gradients and the
body force is gravitational.
• Stable and Unstable Temperature Gradients
T2  T1
Buoyancy forces are able to overcome the
retarding influence of viscous forces.
T1  T2
Conduction occurs
3
GENERAL CONSIDERATIONS
• Free Boundary Flows
➢ Occur in an extensive (in principle, infinite), quiescent (motionless at
locations far from the source of buoyancy) fluid.
➢ Plumes and Buoyant Jets:
• Free Convection Boundary Layers
➢ Boundary layer flow on a hot or cold
surface (Ts  T ) induced by buoyancy forces.
4
GENERAL CONSIDERATIONS
• Pertinent Dimensionless Parameters
➢ Grashof Number:
g  (Ts − T ) L3 Buoyancy Force
GrL =

2
Viscous Force
L → characteristic length of surface
 → thermal expansion coefficient (a thermodynamic property of the fluid)

 = − 1  
  T  p
[K-1]
Liquids:  → Tables A.5, A.6
Ideal Gas:  = 1/T ( K )
➢ Rayleigh Number:
RaL = GrL Pr =
g  (Ts − T ) L3

5
GENERAL CONSIDERATIONS
• Mixed Convection
➢ A condition for which forced and free convection effects are comparable.
➢ Exists if
Also called Richardson number
( Gr
L
/ ReL2 )  1
- Free convection → ( GrL / ReL2 )  1
- Forced convection → ( GrL / ReL2 )  1
➢ Heat Transfer Correlations for Mixed Convection:
n
n
Nu n  NuFC
 NuNC
NuFC → Nusselt number for forced convection
NuNC → Nusselt number for natural (free) convection
+ → assisting or transverse flows
- → opposing flows
n3
Although, for transverse flows:
n = 3.5 (horizontal plates),
n = 4 (cylinder or sphere)
6
VERTICAL PLATES
• Free Convection Boundary Layer Development on a Hot Plate:
x-component velocity
temperature
➢ Ascending flow with the maximum velocity occurring in the boundary layer
and zero velocity at both the surface and outer edge.
➢ How do conditions differ from those associated with forced convection?
➢ How do conditions differ for a cold plate (Ts  T ) ?
7
VERTICAL PLATES
• Form of the x-Momentum Equation for Laminar Flow
2

u

u

u +  = g  (T − T ) + u2
x
y
y
Net Momentum Fluxes Buoyancy Force
( Inertia Forces)
Viscous Force
➢ Temperature dependence requires that solution for u (x,y) be obtained
concurrently with solution of the boundary layer energy equation for T (x,y).
2

T

T

u
+v
=  T2
x
y
y
– The solutions are said to be coupled.
8
VERTICAL PLATES
• Similarity Solution
➢ Based on existence of a similarity variable,  , through which the x-momentum
equation may be transformed from a partial differential equation with twoindependent variables (x and y) to an ordinary differential equation expressed
exclusively in terms of  .
1/4
y Gr
   x 
x 4 
➢ Transformed momentum and energy equations:
f  + 3 ff  − 2 ( f  ) + T  = 0
2
T * + 3 Pr fT * = 0
f  ( ) 
df
= x ( Grx−1/2 ) u
d 2
T 
T − T
Ts − T
9
VERTICAL PLATES
➢ Numerical integration of the equations yields following results for f  ( ) and T  :
dimensionless x-component velocity
dimensionless temperature
➢ Velocity boundary layer thickness ( ) →   5 for Pr  0.6
 Grx 
Pr

0.6
:

=
5
x
➢


 4 
−1/4
= 7.07
x
 x1/4
1/4
( Grx )
10
VERTICAL PLATES
➢ Nusselt Numbers ( Nux and Nu L ) :
1/4
 Gr 
Nu x = hx = −  x 
k
 4 
g ( Pr ) =
1/4

dT
d
 =0
 Gr 
= x 
 4 
g ( Pr )
0.75 Pr1/2
( 0.609 + 1.221 Pr
1/2
+ 1.238 Pr )
1/4
( 0  Pr   )
h = 1 oL h dx → Nu L = 4 Nu L
L
3
• Transition to Turbulence
➢ Amplification of disturbances
depends on relative magnitudes
of buoyancy and viscous forces.
➢ Transition occurs at a critical
Rayleigh Number.
Rax ,c = Grx ,c Pr =
g  (Ts − T ) x3

 109
11
VERTICAL PLATES
• Empirical Heat Transfer Correlations
➢ Laminar Flow ( RaL  109 ) :
Nu L = 0.68 +
0.670 Ra1/4
L
1 + ( 0.492 / Pr )

(9.27)
9/16 4/9


➢ All Conditions: Churchill and Chu correlation


0.387 Ra1/6


L
Nu L = 0.825 +

9/16 8/27




1 + 0.492 / Pr )
 (
 

2
(9.26)
12
HORIZONTAL PLATES
• Buoyancy force is normal, instead of parallel, to the plate.
• Flow and heat transfer depend on whether the plate is hot or cold and
whether it is facing upward or downward.
• Hot Surface Facing Upward or Cold Surface Facing Downward
Ts  T
Ts  T
Nu L = 0.54 Ra1/4
L
Nu L = 0.15 Ra1/3
L
(10
(10
4
 RaL  107 )
(9.30)
7
 RaL  1011 )
(9.31)
where L = As/P. For various shapes (e.g. squares, rectangles, or circles).
13
HORIZONTAL PLATES
• Hot Surface Facing Downward or Cold Surface Facing Upward
Ts  T
Nu L = 0.52 Ra1/5
L
Ts  T
(10
4
 RaL  109 )
(9.32)
➢ Why do these conditions yield smaller heat transfer rates than those
for a hot upper surface or cold lower surface?
14
LONG HORIZONTAL CYLINDER
• Boundary Layer Development and Variation of the Local Nusselt Number
for a Hot Cylinder:
• The Average Nusselt Number:


1/6
0.387 RaD


Nu D = 0.60 +

9/16 8/27
1 + ( 0.559 / Pr )  


 

2
RaD  1012
(9.34)
• How do conditions change for a cold cylinder?
15
SPHERES
• The Average Nusselt Number:
Nu D = 2 +
0.589 Ra1/4
D
1 + ( 0.469 / Pr )

9/16 4/9


 RaD  1011 
 Pr  0.7 
(9.35)
➢ In the limit as RaD → 0, how may conditions be characterized?
o Conduction → Case 1 of Table 4.1
Table 9.2 Summary of free convection empirical correlations
16
Problem
A square aluminum plate 5 mm thick and 200 mm on a side is heated while vertically
suspended in quiescent air at 40 oC. Determine the average heat transfer coefficient
for the plate when its temperature is 15 oC by two methods: using results from the
similarity solution to the boundary layer equations, and using results from an
empirical correlation.
similarity
solution:
Nu L = 4 NuL
3
1/4
 Gr 
Nu x = hx =  x 
k  4 
g ( Pr ) ;
g ( Pr ) =
0.75 Pr1/2
( 0.609 + 1.221 Pr1/2 + 1.238 Pr )
1/4
Empirical correlation:
Nu L = 0.68 +
0.670 Ra1/4
L
1 + ( 0.492 / Pr )9/16 


4/9
17
Problem
A circular grill of diameter 0.25 m and emissivity 0.9 is maintained at a constant
surface temperature of 130 oC. What electrical power is required when the room air
and surroundings are at 24 oC?
Horizontal (hot) surface at the top (laminar flow) :
Nu L = 0.54 Ra1/4
L
(10
4
 RaL  107 )
18