Fundamental Concepts of Convection :
Flow and Thermal Considerations
Chapter Six and Appendix D
Sections 6.1 through 6.8
and D.1 through D.3
6.1 Boundary Layers: Physical Features
•
Velocity Boundary Layer
– A consequence of viscous effects associated with relative motion between a
fluid and a surface.
– A region of the flow characterized by shear stresses and velocity gradients.
– A region between the surface and the free stream
whose thickness δ increases in the flow direction.
δ→
u( y)
u∞
= 0.99
– Why does δ increase in the flow direction?
– Manifested by a surface shear stress τ s
that provides a drag force, F D .
– How does τ s vary in the flow
direction? Why?
∂u
τs = µ
∂y
y =0
FD = ∫ τ s dAs
As
• Thermal Boundary Layer
– A consequence of heat transfer between the surface and fluid.
– A region of the flow characterized by temperature gradients and heat fluxes.
– A region between the surface and the free stream whose thickness δ t
increases in the flow direction.
– Why does δ t increase in the
flow direction?
– Manifested by a surface heat flux q′′s and
δt →
Ts − T ( y )
Ts − T∞
q′′s = − k f
a convection heat transfer coefficient h .
– If (Ts − T∞ ) is constant, how do q′′s and
h vary in the flow direction?
h≡
= 0.99
∂T
∂y
y =0
− k f ∂T / ∂y
Ts − T∞
y =0
6.2 Local andAverage Heat Transfer Coefficients
• Local Heat Flux and Coefficient:
q′′s = h (Ts − T∞ )
• Average Heat Flux and Coefficient for a Uniform Surface Temperature:
q = hAs (Ts − T∞ )
q = ∫As q′′dAs = (Ts − T∞ ) ∫A hdAs
s
1
h=
∫As hdAs
As
• For a flat plate in parallel flow:
1 L
h =
∫o h d x
L
6.3 Boundary Layer Transition
• What conditions are associated with transition from laminar to turbulent flow?
• Why is the Reynolds number an appropriate parameter for quantifying transition
from laminar to turbulent flow?
• Transition criterion for a flat plate in parallel flow:
ρu x
Re x ,c ≡ ∞ c → critical Reynolds number
µ
xc → location at which transition to turbulence begins
105 < Re x ,c < 3 x 106
~
~
What may be said about transition if ReL < Rex,c?
If ReL > Rex,c?
• Effect of transition on boundary layer thickness and local convection coefficient:
Why does transition provide a significant increase in the boundary layer thickness?
Why does the convection coefficient decay in the laminar region?
Why does the convection coefficient decay in the turbulent region? Why does it
increase significantly with transition to turbulence, despite the increase in the
boundary layer thickness?
6.4 The Boundary Layer Equations
• Consider concurrent velocity and thermal boundary layer development for steady,
two-dimensional, incompressible flow with constant fluid properties ( µ , c p , k )
and negligible body forces.
• Apply conservation of mass, Newton’s 2nd Law of Motion and conservation of
energy to a differential control volume and invoke the boundary layer approximations.
Velocity Boundary Layer:
Thermal Boundary Layer:
∂ 2u ∂ 2 u
2,
∂x 2
∂y
∂p dp∞
≈
∂x dx
∂ 2T
∂ 2T
2
2
∂x
∂y
• Conservation of Mass:
∂u ∂v
+
=0
∂x ∂y
In the context of flow through a differential control volume, what is the physical
significance of the foregoing terms, if each is multiplied by the mass density of
the fluid?
• Newton’s Second Law of Motion:
x-direction :
∂u
∂u
∂ 2u
1 dp∞
=−
+ν 2
u +v
∂x
∂y
∂y
ρ dx
What is the physical significance of each term in the foregoing equation?
Why can we express the pressure gradient as dp∞/dx instead of∂p / ∂x ?
• Conservation of Energy:
∂T
∂T
∂ T ν ⎛ ∂u ⎞
+v
=α 2 + ⎜ ⎟
u
∂x
∂y
∂y
c p ⎝ ∂y ⎠
2
2
What is the physical significance of each term in the foregoing equation?
What is the second term on the right-hand side called and under what conditions
may it be neglected?
6.5 Boundary Layer Similarity
• As applied to the boundary layers, the principle of similarity is based on
determining similarity parameters that facilitate application of results obtained
for a surface experiencing one set of conditions to geometrically similar surfaces
experiencing different conditions.
• Dependent boundary layer variables of interest are:
τ s and q′′ or h
• For a prescribed geometry, the corresponding independent variables are:
Geometrical: Size (L), Location (x,y)
Hydrodynamic: Velocity (V)
Fluid Properties:
Hence,
Hydrodynamic: ρ , µ
Thermal : c p , k
u = f ( x , y , L, V , ρ , µ )
τ s = f ( x , L, V , ρ , µ )
and
T = f ( x , y , L, V , ρ , µ , c p , k )
;
h = f ( x , L, V , ρ , µ , c p , k )
• Key similarity parameters may be inferred by non-dimensionalizing the momentum
and energy equations.
• Recast the boundary layer equations by introducing dimensionless forms of the
independent and dependent variables.
x
x ≡
L
*
y
y ≡
L
*
u
u ≡
V
*
v
v ≡
V
*
T − Ts
T ≡
T∞ − Ts
*
• Neglecting viscous dissipation, the following normalized forms of the x-momentum
and energy equations are obtained:
*
*
*
2 *
∂
∂
∂
u
u
dp
1
u
*
*
+v
=− * +
u
*
*
∂x
∂y
dx
Re L ∂y*2
*
*
2 *
∂
∂
∂
T
T
1
T
*
*
+v
=
u
*
*
∂x
∂y
Re L Pr ∂y*2
ρVL VL
=
→ the Reynolds Number
µ
v
cpµ v
Pr ≡
=
→ the Prandtl Number
k
α
How may the Reynolds and Prandtl numbers be interpreted physically?
Re L ≡
(
u * = f x* , y* , Re L
• For a prescribed geometry,
∂u
τs = µ
∂y
y =0
)
*
⎛ µV ⎞ ∂u
=⎜
⎟ *
⎝ L ⎠ ∂y
y* = 0
The dimensionless shear stress, or local friction coefficient, is then
τs
2 ∂u *
Cf ≡
=
2
ρV / 2 Re L ∂y*
y* = 0
∂u *
∂y*
(
)
(
)
= f x* , Re L
y* = 0
Cf =
2
f x* , Re L
Re L
What is the functional dependence of the average friction coefficient?
• For a prescribed geometry,
(
T * = f x* , y* , Re L , Pr
h=
−k f ∂T / ∂y
y =0
Ts − T∞
)
=−
k f (T∞ − Ts ) ∂T *
L (Ts − T∞ ) ∂y*
y* = 0
k f ∂T *
=+
L ∂y*
The dimensionless local convection coefficient is then
hL ∂T *
Nu ≡
= *
kf
∂y
(
= f x* , Re L , Pr
)
y* = 0
Nu → local Nusselt number
What is the functional dependence of the average Nusselt number?
How does the Nusselt number differ from the Biot number?
y* = 0
6.6 Physical Significance of the Dimensionless Parameters
ρVL ρV 2
inertial force
Re L =
=
=
µ
µV / L viscous force
δ
≈ Pr n
δt
for a gas: δ t ≈ δ
for a liquid metal: δ t δ
for an oil: δ t δ
6.7
The Reynolds Analogy
• Equivalence of dimensionless momentum and energy equations for
negligible pressure gradient (dp*/dx*~0) and Pr~1:
*
*
2 *
∂
u
∂
u
1
∂
u
*
u*
v
+
=
∂x*
∂y *
R e ∂ y *2
Advection terms
Diffusion
*
*
2 *
∂
T
∂
T
1
∂
T
*
u*
+
v
=
R e ∂ y *2
∂x*
∂y *
• Hence, for equivalent boundary conditions, the solutions are of the same form:
u =T
*
*
∂u *
∂y*
y* = 0
∂T *
= *
∂y
y* = 0
Re
Cf
= Nu
2
or, with the Stanton number defined as,
h
Nu
=
St ≡
ρVc p Re Pr
With Pr = 1, the Reynolds analogy, which relates important parameters of the
velocity and thermal boundary layers, is
Cf
2
= St
• Modified Reynolds (Chilton-Colburn) Analogy:
– An empirical result that extends applicability of the Reynolds analogy:
Cf
2
= St Pr
2
3
≡ jH
0.6 < Pr < 60
Colburn j factor for heat transfer
– Applicable to laminar flow if dp*/dx* ~ 0.
– Generally applicable to turbulent flow without restriction on dp*/dx*.
Problem 6.19: Determination of heat transfer rate for prescribed turbine blade operating
conditions from wind tunnel data obtained for a geometrically similar but
smaller blade. The blade surface area may be assumed to be directly
proportional to its characteristic length ( A ∝ L ) .
s
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Surface area A is
directly proportional to characteristic length L, (4) Negligible radiation, (5) Blade shapes are
geometrically similar.
ANALYSIS: For a prescribed geometry,
Nu =
hL
= f ( ReL , Pr ) .
k
The Reynolds numbers for the blades are
ReL,1 = ( V1L1 /ν1 ) = 15m2 /s ν1
Hence, with constant properties
ReL,2 = ( V2L2 /ν 2 ) = 15m2 /s ν 2.
( v1 = v2 ) , ReL,1 = ReL,2.
Therefore,
Nu 2 = Nu 1
( h 2 L2 / k 2 ) = ( h1L1 / k1 )
L1
L1
q1
h2 =
h1 =
L2
L 2 A1 Ts,1 − T∞
(
)
Also,
Pr1 = Pr2
The heat rate for the second blade is then
(
)
L
q 2 = h 2 A 2 Ts,2 − T∞ = 1
L2
A2
A1
( Ts,2 − T∞ ) q
( Ts,1 − T∞ ) 1
Ts,2 − T∞
( 400 − 35)
q2 =
q1 =
(1500 W )
Ts,1 − T∞
300
−
35
(
)
q 2 = 2066 W.
COMMENTS:
(i) The variation in ν from Case 1 to Case 2 would cause ReL,2 to differ from ReL,1. However
for air and the prescribed temperatures, this non-constant property effect is
small.
(
)
(ii) If the Reynolds numbers were not equal Re L,1 ≠ Re L 2 , knowledge of the specific form of
(
)
f Re L, Pr would be needed to determine h2.