CHAPTER 4
BOUNDARY LAYER FLOW:
APPLICATION TO EXTERNAL FLOW
4.1 Introduction
Navier-Stokes equations and the energy equation are simplified using the boundary layer
concept.
x Under special conditions certain terms in the equations can be neglected.
x Two key questions:
(1) What are the conditions under which terms in the governing equations can be
dropped?
(2) What terms can be dropped?
4.2 The Boundary Layer Concept: Simplification of the Governing Equations
4.2.1 Qualitative Description
x Consider convection over a semiinfinite plate (Fig. 4.1).
x Under certain conditions the effect of
viscosity is confined to a thin region
near the surface called the velocity or
viscous boundary layer, G .
x
Under certain conditions the effect of thermal interaction is confined to a thin region
near the surface called the thermal boundary layer G t .
x
Conditions for the formation of the two boundary layers:
x
Velocity boundary layer conditions:
(1) Slender body
(2) High Reynolds number (Re > 100)
x
Thermal boundary layer conditions:
(1) Slender body
(2) High product of Reynolds and Prandtl numbers (Re Pr > 100)
Peclet Number
Pe
RePr
U Vf L c p P
P
k
U c p Vf L
k
(4.1)
2
NOTE:
x Fluid velocity at the surface vanishes.
x Large changes in velocity across G .
x Large changes in temperature across G t .
x Viscosity plays no role outside G .
4.2.2 The Governing Equations
Assumptions: (1) Steady, (2) two-dimensional, (3) laminar, (4) uniform properties, (5) no
dissipation, and (6) no gravity.
Governing equations:
wu wv
wx wy
0
(2.3)
u
wu
wu
v
wx
wy
§ w 2u w 2u ·
1 wp
ǎ ¨ 2 2 ¸
¨ wx
U wx
wy ¸¹
©
(2.10x)
u
wv
wv
v
wx
wy
§ w 2v w 2v ·
1 wp
ǎ ¨ 2 2 ¸
¨ wx
U wy
wy ¸¹
©
(2.10y)
§ wT
wT ·
¸
v
wy ¸¹
© wx
U c p ¨¨ u
§ w 2T w 2T ·
k¨ 2 2 ¸
¨ wx
wy ¸¹
©
(2.19)
4.2.3 Mathematical Simplification
x Simplify the above equations based on boundary layer approximations.
4.2.4 Simplification of the Momentum Equations
(i) Intuitive Argument
x Follow the intuitive argument leading to:
w 2u
w 2u
wy 2
wx 2
(4.2)
wp
|0
wy
(4.3)
and
Therefore
p | p(x)
It follows that
wp dp dp f
|
|
wx dx
dx
x (2.10x) simplifies to the following boundary layer x-momentum equation
(4.4)
3
u
wu
wu
v
wx
wy
w 2u
1 dp f
ǎ 2
U wx
wy
(4.5)
(ii) Scale Analysis
x We start by assuming that
G
L
1
(4.6)
x Follow scale analysis leading to equations (4.2)-(4.4)
x Follow scale analysis leading to:
G
L
1
a
(4.14b)
Re L
x (4.14b) shows that (4.6) is valid when Re L !! 1.
x (4.14b) is generalized as
G
x
1
a
(4.16)
Re x
4.2.5 Simplification of the Energy Equation
x The energy equation for two-dimensional constant properties flow is
§ w 2T w 2T ·
§ wT
wT ·
¸
¸¸ k ¨
v
¨ wx 2 wy 2 ¸
wy ¹
© wx
©
¹
U c 5 ¨¨ u
(2.19)
x (2.19) is simplified for boundary layer flow using two arguments:
(i) Intuitive Argument
x Follow the intuitive argument leading to:
w 2T
w 2T
wx 2
wy 2
x
(4.17)
(2.19) simplifies to the following boundary layer energy
u
wT
wT
v
wx
wy
D
w 2T
wy 2
(4.18)
(ii) Scale Analysis
x We start by assuming that
Gt
L
1
x Follow scale analysis leading to equations (4.17) and (4.18)
x Follow scale analysis for the validity of (4.19). Two cases are considered:
(4.19)
4
Case (1): G t ! G . Follow the argument leading to:
Gt
1
a
L
(4.24)
PrRe L
Thus
Gt
L
1 when
PrRe L !! 1
(4.25)
x The criterion for G t ! G : Taking the ratio of (4.24) to (4.14b) gives
Gt
1
a
G
Pr
(4.27)
Thus
G t ! G when
Pr 1
(4.28)
Case (2): G t G . Follow the argument leading to:
Gt
L
1
a
Pr
1/3
(4.31)
Re L
Thus
Gt
L
1 when Pr 1/3 Re L !! 1
(4.32)
x The criterion for G t G : Taking the ratio of (4.31) to (4.14b)
Gt
1
a 1/3
G Pr
(4.33)
Thus
G t G when Pr 1/3 !! 1
(4.34)
4.3 Summary of Boundary Layer Equations for Steady Laminar Flow
x
Review all assumptions leading to the following boundary layer equations:
Continuity:
wu wv
wx wy
0
(2.3)
x-Momentum:
u
wu
wu
v
wx
wy
w 2u
1 dp f
ǎ 2
U dx
wy
(4.13)
Energy:
u
wT
wT
v
wx
wy
Note the following:
x The continuity is not simplified.
D
w 2T
wy 2
(4.18)
5
x Solution to inviscid flow outside the boundary layer gives pressure gradient needed
in (4.13).
x To include buoyancy effect, add [ U E g (T Tf ) ] to (4.13).
4.4 Solutions: External Flow
x For constant properties, velocity distribution is independent of temperature.
x First obtain the flow field solution and then use it to determine temperature
distribution.
4.4.1 Laminar Boundary Layer Flow over Semi-infinite Flat Plate: Uniform Surface
Temperature
The basic problem is shown in Fig. 4.5.
x The plate is at uniform temperature Ts .
x Upstream temperature is Tf .
x Apply all the assumptions summarized in
Section 4.3.
x Governing equations: (continuity, momentum, and energy) are given in (2.3), (4.13),
and (4.18).
(i) Velocity Distribution.
Determine:
x Velocity and pressure distribution.
x Boundary layer thickness G (x) .
x Wall shear W o (x).
(a) Governing equations and boundary conditions
wu wv
wx wy
u
wu
wu
v
wx
wy
0
w 2u
1 dp f
ǎ 2
U dx
wy
(2.3)
(4.13)
The velocity boundary conditions are:
u ( x,0)
0
(4.35a)
v( x,0)
0
(4.35b)
u ( x, f ) Vf
(4.35c)
u (0, y ) Vf
(4.35d)
(b) Scale analysis: boundary layer thickness, wall shear and friction coefficient.
We showed that
6
G
x
a
1
(4.16)
Re x
Define Darcy friction coefficient C f
Cf
Wo
(1 / 2) U Vf2
(4.37a)
Follow scale analysis leading to
Cf a
1
(4.37b)
Re x
(c) Blasius solution: similarity method
x Equations (2.3) and (4.13) are solved analytically by Blasius.
x For inviscid flow over flat plate
u = Vf , v = 0, p = pf = constant
(4.38)
Thus the pressure gradient is
dp f
dx
(4.39)
0
(4.39) into (4.13)
u
wu
wu
v
wx
wy
ǎw
2
u
(4.40)
wy 2
x (2.3) and (4.40) are solved by the method of similarity transformation.
x The basic approach is combining the two independent variables x and y into a single
variable K(x, y) and postulate that u/Vf depends on K only.
x For this problem the correct form of the transformation variable K is
K ( x, y )
x
y
Vf
ǎx
(4.41)
Assume
u
Vf
df
dK
(4.42)
x Using (4.41) and (4.42), integration of the continuity gives
v
Vf
1 ǎ
2 Vf x
·
§ df
¨¨K
f ¸¸
¹
© dK
(4.43)
x Transform all derivatives in terms of f and K , substitute (4.42), (4.43)
2
d3 f
d2 f
f
K
(
)
dK 3
dK 2
x Boundary conditions (4.35a-4.35d) transform to
0
(4.44)
7
df (0)
0
dK
f (0) 0
(4.45a)
(4.45b)
df (f)
dK
1
(4.45c)
df (f)
dK
1
(4.45d)
NOTE:
x The momentum is transformed into
an ordinary differential equation.
x Boundary conditions (4.35c) and
(4.35d) coalesce into a single
condition.
x (4.44) is solved by power series.
The solution is presented in Table
4.1.
x From Table 4.1 we obtain
G
5.2
Re x
x
(4.46)
Scaling gives
G
x
a
1
(4.16)
Re x
x From Table 4.1 we obtain
Cf
0.664
Re x
(4.48)
Scaling gives
Cf a
1
Re x
Table 4.1
Blasius solution [1]
Vf
y
Qx
K
0.0
0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6
4.0
4.4
4.8
5.0
5.2
5.4
5.6
6.0
7.0
8.0
(4.37b)
(ii) Temperature Distribution.
Determine:
x Temperature distribution.
x Thermal boundary layer thickness G t .
x Heat transfer coefficient h(x).
x Nusselt number Nu(x).
(a) Governing equation and boundary conditions
f
0.0
0.02656
0.10611
0.23795
0.42032
0.65003
0.92230
1.23099
1.56911
1.92954
2.30576
2.69238
3.08534
3.28329
3.48189
3.68094
3.88031
4.27964
5.27926
6.27923
df
dK
u
Vf
0.0
0.13277
0.26471
0.39378
0.51676
0.62977
0.72899
0.81152
0.87609
0.92333
0.95552
0.97587
0.98779
0.99155
0.99425
0.99616
0.99748
0.99898
0.99992
1.00000
d2 f
dK 2
0.33206
0.33147
0.32739
0.31659
0.29667
0.26675
0.22809
0.18401
0.13913
0.09809
0.06424
0.03897
0.02187
0.01591
0.01134
0.00793
0.00543
0.00240
0.00022
0.00001
8
u
w 2T
wT
wT
v
wx
wy
D
T ( x,0)
Ts
(4.49a)
T ( x, f )
Tf
(4.49b)
T (0, y )
Tf
(4.49c)
wy 2
(4.18)
The boundary conditions are:
(b) Scale analysis: Thermal boundary layer thickness, heat transfer coefficient and
Nusselt number
x Return to the results of Section 4.2.5:
Case (1): G t ! G (Pr <<1)
Gt
x
a
1
PrRe x
(4.50)
Case (2): G t G (Pr >>1)
Gt
1
a
x
Pr
1/3
(4.51)
Re x
x Scale analysis for h. Begin with
h
wT ( x,0)
wy
k
Ts Tf
(1.10)
Using the scales, the above gives
ha
k
(4.52)
Gt
Case (1): G t ! G (Pr <<1). Substituting (4.50) into (4.52)
ha
k
PrRe x ,
x
for Pr <<1
(4.53)
Defining the local Nusselt number Nu x as
Nu x
hx
k
(4.54)
Substituting (4.53) into (4.54)
Nu x a Pr 1/2 Re x ,
for Pr <<1
(4.55)
Case (2): G t G (Pr >>1). Substituting (4.51) into (4.52)
k 1/3
h a Pr
Re x , for Pr >>1
x
The corresponding Nusselt number is
(4.56)
9
Nu x a Pr
1/3
Re x , for Pr >>1
(4.57)
(c) Pohlhausen’s solution: Temperature distribution, thermal boundary layer
thickness, heat transfer coefficient, and Nusselt number
x Equation (4.18) is solved analytically by Pohlhausen using similarity transformation.
Define
T Ts
Tf Ts
T
(4.58)
(4.58) into (4.18)
u
wT
wT
v
wx
wy
D
w 2T
(4.59)
wy 2
Boundary conditions (4.49) become
T ( x,0) 0
(4.60a)
T ( x, f ) 1
(4.60b)
T (0, y ) 1
(4.60c)
Combine x and y into a single variable K(x, y) given by
K ( x, y )
y
Vf
ǎx
(4.41)
Assume
T ( x, y ) T (K )
Velocity components u and v in (4.59) are given by Blasius solution
u
Vf
v
Vf
1 ǎ
2 Vf x
df
dK
(4.42)
·
§ df
¨¨K
f ¸¸
¹
© dK
(4.43)
(4.41)-(4.43) into (4.59)
d 2T Pr
dT
f (K )
2
2
dK
dK
0
(4.61)
Using (4.41), the three boundary conditions (4.60a-4.60c) transform to
T (0) 0
(4.62a)
T (f ) 1
(4.62b)
T (f ) 1
(4.62c)
Integration details of (4.61) are found in Appendix B. The temperature solution is
10
f
T (K ) 1 ³
K
ªd 2 f )º
«
2 »
¬ dK ¼
f
³
0
2
ªd f º
« 2»
¬ dK ¼
x Surface temperature gradient is
dT (0)
dK
>0.332@
f
³
K
Pr
dK
(4.63)
Pr
dK
1.0
Pr
ªd 2 f º
« 2»
¬ dK ¼
1 0.7( air )
0.8
(4.64)
Pr
100 10
dK
0.1
0.6
T Ts
Tf Ts 0.4
x The integrals in (4.63) and (4.64)
are evaluated numerically.
x Boundary layer thickness G t is
determined from Fig. 4.6. The
edge of the thermal layer is
defined as the distance y where
T | Tf . This corresponds to
T
Pr
0.2
0
2
0.01
6
4
K
8
10
12
Fig. 4.6 Pohlhausen ' s solution for temperature
distribution for laminar flow over a
semi - infinte isothermal flat plate
T Ts
| 1 , at y
Tf Ts
Gt
(4.65)
The heat transfer coefficient h is determined using equation (1.10)
h
wT ( x,0)
wy
k
Ts Tf
(1.10)
Using (4.41) and (4.58) into the above
h( x ) k
V f dT (0)
ǎ x dK
(4.66)
Average heat transfer coefficient
L
1
L
h
³
h ( x ) dx
(2.50)
k
dT (0)
Re L
L
dK
(4.67)
0
Substituting (4.66) into (2.50) and integrating
h
14
y Vf Q x
2
The local Nusselt number is obtained by substituting (4.66) into (4.54)
11
dT (0)
Re x
dK
Nu x
(4.68)
The corresponding average Nusselt number is
dT (0)
Re L
dK
Table 4.2 gives dT (0) / dK for various values of Pr.
Nu L
(4.69)
2
dT (0)
dK
dT (0)
dK
0.564 Pr 1 / 2 ,
0.332 Pr 1 / 3 ,
dT (0)
dK
Pr < 0.05
(4.71a)
0.6 < Pr < 10
(4.71b)
Pr >10
(4.71c)
0.339 Pr 1 / 3 ,
Table 4.2
dT (0)
Pr
dK
0.001
0.01
0.1
0.5
0.7
1.0
7.0
10.0
15.0
50
100
1000
0.0173
0.0516
0.140
0.259
0.292
0.332
0.645
0.730
0.835
1.247
1.572
3.387
4.4.2 Applications: Blasius Solution, Pohlhausen’s Solution, and Scaling
x Review Examples 4.1, 4.2, and 4.3. They illustrate the application of Blasius
solution, Pohlhausen’s solution, and scaling to the solution of convection problems.
4.4.3 Laminar Boundary Layer Flow over Semi-infinite Flat Plate: Variable Surface
Temperature
x Surface temperature varies with axial
distance x according to
Ts ( x) Tf
Cx n
(4.74)
x Assumptions: see Section 4.3.
(i) Velocity Distribution. Blasius flow field solution is applicable to this case:
u
Vf
v
Vf
1 ǎ
2 Vf x
where the similarity variable K is defined as
df
dK
(4.42)
§ df
·
¨¨K
f ¸¸
© dK
¹
(4.43)
12
K ( x, y )
Vf
ǎx
y
(4.41)
(ii) Governing Equations for Temperature Distribution. Based on the assumptions
listed in Section 4.3, temperature is governed by energy equation (4.18)
u
wT
wT
v
wx
wy
w 2T
D
(4.18)
wy 2
The boundary conditions for this problem are:
T ( x,0)
Tf Cx n
Ts
(a)
T ( x, f )
Tf
(b)
T (0, y )
Tf
(c)
T Ts
Tf Ts
(4.58)
(iii) Solution. Define T as
T
Assume
T ( x, y ) T (K )
(4.75)
Using (4.41)-(4.43), (4.58), (4.74) and (4.75), energy equation (4.18) transforms to (see
Appendix C for details)
d 2T
df
Pr
dT
n
Pr
T
f
K
(
1
)
(
)
dK
dK
2
dK 2
0
(4.76)
Boundary conditions (a)-(c) become
T (0)
0
(4.77a)
T (f ) 1
(4.77b)
T (f ) 1
(4.77c)
Local heat transfer coefficient and Nusselt number are determined using (1.10)
h
wT ( x,0)
wy
k
Ts Tf
(1.10)
Using (4.41), (4.58) and (4.72) into the above
wT ( x,0)
wy
Cx n
Vf dT (0)
ǎ x dK
Substituting into (1.10)
h( x )
k
Vf dT (0)
ǎ x dK
(4.78)
The average heat transfer coefficient for a plate of length L is defined in equation (2.50)
13
L
1
L
h
³ h( x)dx
(2.50)
0
(4.78) into (2.50)
h
2
k
dT (0)
Re L
L
dK
(4.79)
(4.78) into (4.54) gives the local Nusselt number
dT (0)
Re x
dK
Nu x
(4.80)
The corresponding average Nusselt number is
dT (0)
Re L
dK
x Key factor: surface temperature gradient dT (0) / dK.
Nu L
(4. 81)
2
2.0
(iv) Results. (4.76) is solved
numerically subject to boundary
conditions (4.77). Fig. 4.8 gives
dT (0) / dK for three Prandtl numbers.
30
dT (0)
dK
10
1.0
Pr
0
Fig.4.8
0.5
n
1.0
0.7
1.5
dT (0)
for plate with varying surface temperature,
dK
Ts Tf Cx n [4]
4.4.4 Laminar Boundary Layer Flow over a Wedge: Uniform Surface Temperature
x Assumptions: listed in Section 4.3.
x x-momentum equation:
u
wu
wu
v
wx
wy
w 2u
1 dp f
ǎ 2
U dx
wy
(4.13)
x Inviscid flow solution for Vf ( x) is
Vf ( x )
Cx m
C is a constant and m is defined as
(4.82)
14
m
E
(4.83)
2E
Application of (4.13) to the inviscid flow outside the viscous boundary layer, gives
1 dpf
U dx
Vf
wVf
wx
Substituting into (4.13)
u
wu
wu
v
wx
wy
Vf
wVf
w 2u
ǎ 2
wx
wy
(4.84)
The boundary conditions are
u ( x,0)
0
(4.85a)
v ( x,0)
0
(4.85b)
u ( x, f) Vf ( x)
Cx m
(4.85c)
Solution to the velocity distribution is obtained by the method of similarity. Define K as
K ( x, y )
Vf ( x)
ǎx
y
y
C
ǎ
x ( m1) / 2
(4.86)
Assume
u
Vf ( x )
dF
dK
(4.87)
Continuity equation (2.3), (4.86), and (4.87) give the vertical velocity component v
v
Vf ( x)
ǎ
m 1 ª
1 m dF º
F
K
«
xVf ( x) 2 ¬
1 m dK »¼
(4.88)
(4.82) and (4.86)-(4.88) into (4.84)
2
ª dF º
d 3F m 1 d 2F
F
m« » m
3
2
2
dK
dK
¬ dK ¼
0
(4.89)
This is the transformed momentum equation. Boundary conditions (4.85) transform to
dF (0)
dK
F (0)
dF (f)
dK
0
0
1
(4.90a)
(4.90b)
(4.90c)
x Solution. (4.89) is integrated numerically. The solution gives F (K ) and dF / dK.
These in turn give the velocity components u and v.
x Temperature distribution. Start with the energy
15
u
wT
wT
v
wx
wy
D
w 2T
(4.59)
wy 2
Boundary conditions
T ( x,0) 0
(4.60a)
T ( x, f ) 1
(4.60b)
T (0, y ) 1
(4.60c)
T Ts
Tf Ts
(4.58)
T (K )
(4.75)
Where T is defined as
T
Assume
T
K is defined in (4.86). (4.86)-(4.88) and (4.75) into (4.59) and (4.60)
d 2T Pr
dT
(m 1) F (K )
2
2
dK
dK
T (0) 0
0
(4.91)
(4.92a)
T (f ) 1
(4.92b)
T (f ) 1
(4.92c)
x Solution. Separating variables in (4.91), integrating twice and applying boundary
conditions (4.92), gives the temperature solution as
f
T (K )
1
³
K
f
³
0
ª (m 1) Pr
exp « 2
¬
ª (m 1) Pr
exp « 2
¬
K
³
0
K
³
0
º
F (K ) d K » d K
¼
º
F (K ) d K » d K
¼
(4.93)
(4.93) gives the temperature gradient at the surface
dT ( 0)
dK
­°
®
°̄
f
³
0
ª (m 1) Pr
exp «
2
¬
K
½°
º
F (K )dK » dK ¾
°¿
¼
0
³
x The integrals in (4.93) and (4.94) are evaluated numerically.
x Results for dT (0) / dK are given in Table 4.3
1
(4.94)
16
Table 4.3 Surface temperature gradient
dT (0)
and surface velocity
dK
gradient F cc(0) for flow over an isothermal wedge
m
0
0.111
0.333
1.0
wedge angle SE
dT (0) / dK at five values of Pr
F cc(0)
0
0.3206
0.5120
0.7575
1.2326
o
S / 5 (36 )
S / 2 (90o)
S
(180o)
0.7
0.8
1.0
5.0
10.0
0.292
0.331
0.384
0.496
0.307
0.348
0.403
0.523
0.332
0.378
0.440
0.570
0.585
0.669
0.792
1.043
0.730
0.851
1.013
1.344
x Heat transfer coefficient h and Nusselt number Nu. Equation (1.10) gives h
h
wT ( x,0)
wy
k
Ts Tf
(1.10)
Using (4.58), (4.75) and (4.86) into (1.10) gives
h( x )
k
Vf ( x ) dT ( 0)
ǎ x dK
(4.95)
(4.95) into (4.54) gives the Nusselt number
Nu x
dT (0)
Re x
dK
(4.96)
where Re x is the local Reynolds number defined as
Re x
xVf ( x)
ǎ
(4.97)