CHAPTER 5
APPROXIMATE SOLUTIONS:
THE INTEGRAL METHOD
5.1 Introduction
• Why approximate solution?
• When exact solution is:
• Not Available
• Complex
1
• Requires numerical integration
• Implicit
• Approximate solution by integral method:
• Advantages: simple, can deal with
complicating factors
• Used in fluid flow, heat transfer,
mass transfer
2
5.2 Differential vs. Integral Formulation
Differential Formulation
Conservati on laws are applied to
an infinitesm al element dx × dy
• Result: Solutions are exact
3
Integral Formulation
Conservation laws are satisfied
in an average sense
• Result: Solutions are approximate
4
5.3 Integral Method Approximation:
Mathematical Simplification
• Number of independent variables are reduced
• Reduction in order of differential equation
5.4 Procedure
(1) Integral formulation of the basic laws
• Conservation of mass
• Conservation of momentum
• Conservation of energy
5
(2) Assumed velocity and temperature profiles
• Satisfy boundary conditions
• Several possibilities
• Examples: Polynomial, linear, exponential
• Assumed profile has an unknown parameter
or variable
(3) Determination of the unknown parameter
or variable
• Conservation of momentum gives the
unknown variable in the assumed velocity
6
• Conservation of energy gives the unknown variable in
the assumed temperature
5.5 Accuracy of the Integral Method
• Accuracy depends on assumed profiles
• Accuracy is not sensitive to form of profile
• Optimum profile: unknown
7
5.6 Integral Formulation of the Basic Laws
5.6.1 Conservation of Mass
• Boundary layer flow
• Porous curved wall
• Conservation of mass for element δ × dx :
8
dme
mx
δ
dm x
mx +
dx
dx
dmo
Fig. 5.3
dm x
m x + dmo + dme = m x +
dx
dx
dm x
dme =
dx − dmo
dx
(a)
9
dm e = mass from the external flow
dm o = mass through porous wall
m x = mass from boundary layer rate entering
element at x
One-dimensional mass flow rate:
m = ρVA
(b)
Apply (b) to porous side, assume that injected is
the same as external fluid
dmo= ρ vo Pdx
(c)
10
P = porosity
ρ = density
Applying (b) to dy
dm x = ρ udy
Integrating
δ ( x)
mx =
∫ ρ udy
(d)
0
(c) and (d) into (a)
11
d 
dme =

dx 


ρ udy  dx − ρ v o Pdx

δ ( x)
∫0
(5.1)
5.6.2 Conservation of Momentum
Apply momentum theorem to the element δ × dx
∑ Fx = M x (out ) − M x (in)
(a)
∑ Fx = External x-forces on element
12
M x (in) = x-momentum of the fluid entering
M x (out)= x-momentum of the fluid leaving
dp
( p + )dδ
2
pδ
δ
d
pδ + ( pδ )dx
dx
τ o (1 − P )dx
Fig. 5.4 Forces
13
V∞ ( x )dme
Mx
dM x
Mx +
dx
dx
Fig. 5.4 x - Momentum
Fig. 4 and (a):
dp 
d

pδ +  p + dδ − pδ − ( pδ )dx − τ o (1 − p )dx =
2
dx

dM x 

M
+
dx  − M x − V∞ ( x )dme
 x
dx


14
p = pressure
V∞ = velocity at the edge of the boundary layer
τ o = wall stress
δ ( x)
∫
ρ u2dy
(b)
∂u( x ,0 )
τo = µ
∂y
(c)
Mx =
0
and
15
(b) and (c) into (a)
∂ u ( x ,0 )
dp
(
)
−δ
=
− µ 1− P
∂y
dx
d
dx
δ (x)
δ (x)
∫
∫
ρ u 2 dy − V ∞ ( x ) d
0
dx
ρ udy − V ∞ ( x )ρ P v o
0
• x-momentum of dmo ?
• Shear force on slanted surface?
• (5.2) applies to laminar and turbulent flow
16
• Curved surface: V∞ ( x ) and p( x )
• (5.2) is the integral formulation of conservation
of momentum and mass
• (5.2) is a first order O.D.E. with x as the
independent variable
Special Cases:
(i) Case 1: Incompressible fluid
dp dp ∞
≈
dx dx
(4.12)
For boundary layer flow
17
∂u
∂u
1 dp∞
∂ 2u
=−
u +v
+ν 2
ρ ∂x
∂x
∂y
∂y
Apply (4.5) at y = δ where
u = V∞
dp dp∞
dV∞
= − ρ V∞ ( x )
≈
dx dx
dx
(5.3) into (5.2), constant
(4.5)
(5.3)
ρ
d V∞
∂ u( x , 0 )
=
− ν (1 − P )
δ V∞ ( x )
dx
∂y
d
dx
δ ( x)
δ ( x)
∫
∫
d
u dy − V∞ ( x )
dx
0
2
0
(5.4)
udy − V∞ ( x ) Pv o
18
(ii) Case 2: Incompressible fluid and impermeable
flat plate
Flat plate, (5.3)
dV ∞ dp dp ∞
=0
=
≈
dx
dx
dx
(d)
Impermeable plate
v o = 0, P = 0
(e)
(d) and (e) into (5.4)
∂u( x ,0 )
d
v
= V∞
∂y
dx
δ (x)
∫0
d
udy −
dx
δ (x)
∫0
u 2dy
(5.5)
19
5.6.3 Conservation of Energy
Neglect:
(1) Changes in kinetic and potential energy
(2) Dissipation
(3) Axial conduction
20
Conservation of energy element δ t × dx
dEe
δt
dE x
Ex +
dx
dx
Ex
dx
dEc
dEo
Fig. 5.6
dE x
E x + dEc + dEo + dEe = E x +
dx
dx
21
Rearranging
dE x
dEc =
dx − dEe − dEo
dx
Fourier’s law
∂T ( x ,0 )
dEc = − k (1 − P )
dx
∂y
(b)
Energy with mass dm e
dEe = c pT∞ dme
22
Use (5.1) for dme
d 

dEe = c pT∞
dx 


ρ udy  dx − c pT∞ ρ v o Pdx

δ ( xt )
∫0
(c)
Energy of injected mass
dE o = ρ c pTo v o pdx
(d)
Energy convected with fluid
Ex =
δ t ( x)
∫0
ρ c p uT dy
(e)
23
(b)-(e) into (a)
∂T ( x ,0 ) d
− k (1 − P )
=
∂y
dx
d
c pT∞
dx
δ t ( x)
∫
ρ c p uTdy −
0
δ t ( x)
∫
(5.6)
ρ udy − ρ c p v o P(To − T∞ )
0
Note
(1) (5.6) is integral formulation of conservation
energy
(2) (5.6) is a first order O.D.E. with x as the
independent variable
24
Special Case: Constant properties and
impermeable flat plate
Simplify (5.6)
−α
∂ T ( x ,0 ) d
=
∂y
dx
δ t ( x)
∫
u(T − T∞ )dy
(5.7)
0
25
5.7 Integral Solutions
5.7.1 Flow Field Solution:
Uniform Flow over a Semi-Infinite Plate
• Integral solution to Blasius problem
• Integral formulation of momentum (5.5)
26
∂u( x ,0 )
d
= V∞
v
∂y
dx
δ (x)
∫0
d
udy −
dx
δ (x)
∫0
u2dy
(5.5)
• Assumed velocity profile u( x , y )
u( x , y ) =
N
∑
an ( x ) y n
(5.8)
n=0
Example, a third degree polynomial
u( x , y ) = a0 ( x ) + a1 ( x ) y + a2 ( x ) y 2 + a3 ( x ) y 3
(a)
27
Boundary conditions give a n
(1) u( x ,0 ) = 0
(3) ∂u( x , δ ) ≅ 0
∂y
(2) u( x , δ ) ≅ V∞
2
∂
u( x ,0 )
(4)
=0
2
∂y
Note:
(1) B.C. (2) and (3) are approximate. Why?
(2) B.C. (4): set
y = 0 in the x-component of the
Navier equations, (2.10x)
(3) 4 B.C. and (a) give:
28
a 0 = a 2 = 0 , a1 =
3 V∞
1 V∞
, a3 =
2 δ
2δ 3
u 3 y 1 y
=  −  
V∞ 2  δ  2  δ 
3
(5.9)
Assumed velocity is in terms of a
single unknown variable δ ( x )
(5.9) into (5.5), evaluate integrals
3
1
39
2 dδ
vV∞ =
V∞
2
δ 280
dx
(b)
29
(b) is first order O.D.E. in δ ( x )
140 v
δ dδ =
dx
13 U ∞
Integrate, use B.C. δ (0) = 0
140 v x
∫0 δ dδ =
∫0 dx
13 U ∞
δ
Evaluate integrals
δ
280 / 13 4.64
=
=
x
Re x
Re x
(5.10)
30
(5.10) into (5.9) gives u( x , y )
Friction coefficient C f :Use (4.36) and (4.37a)
∂u
µ ( x ,0 )
3v
τo
∂y
Cf =
=
=
2
2
V∞δ ( x )
ρV∞ / 2
ρV∞ / 2
Use (5.10) to eliminate δ ( x )
Cf =
0.646
Re x
(5.11)
31
Accuracy: Compare (5.10) and (5.11) with Blasius solution
δ
5.2
=
,
x
Re x
Blasius solution
(4.46)
0.664
,
Cf =
Re x
Blasius solution
(4.48)
Note:
(1) Both solutions have same form
(2) Error in δ ( x ) is 10.8%
32
(3) Error in C f is 2.7%
(4) Accuracy of C f is more important than δ ( x )
5.7.2 Temperature Solution and Nusselt Number:
Flow over a Semi-Infinite Plate
(i) Temperature Distribution
• Flat plate
• Insulated section x o
33
• Surface at T s
• Laminar, steady, two-dimensional,
constant properties boundary layer flow
• Determine: δ t , h( x ) , Nu( x )
• Must determine: u( x , y ) and T ( x , y ) (2)
Integral formulation of conservation of energy
∂T ( x ,0 ) d
−α
=
∂y
dx
δ ( x)
∫
ρ c p u(T − T∞ )dy
(5.7)
0
34
• Integral Solution to u( x , y ) and δ ( x ) :
u 3 y 1 y
=  −  
V∞ 2  δ  2  δ 
3
δ
280 / 13 4.64
=
=
x
Re x
Re x
(5.9)
(5.10)
Assumed temperature
T ( x, y ) =
N
∑
bn ( x ) y n
(5.12)
n=0
35
Let
T ( x , y ) = b0 ( x ) + b1 ( x ) y + b2 ( x ) y 2 + b3 ( x ) y 3
(a)
Boundary conditions give bn ( x )
(1) T ( x ,0 ) = Ts
(2) T ( x , δ t ) ≅ T∞
(3) ∂T ( x , δ t ) ≅ 0
∂y
(4) ∂ 2T ( x ,0 )
=0
2
∂y
36
Note
(1) B.C. (2) and (3) are approximate. Why?
(2) B.C. (4): set y = 0 in the x-component of
the energy equation (2.19)
(3) Four B.C. and (a) give:
3
1 b = 0,
b0 = Ts , b1 = (T∞ − Ts ) , 2
δt
2
1
1
b3 = − (T∞ − Ts ) 3
2
δt
37
Therefore
3 y 1 y3 
T ( x , y ) = Ts + (T∞ − Ts )
−

3
2 δ t 2 δ t 
(5.13)
(5.9) and (5.13) into (5.7) and evaluating the integral
3 T∞ − Ts d 
= (T∞ − Ts )V∞δ
α
2
δt
dx 
4
 3  δ t 2
3  δ t   
   −
  
 20  δ  280  δ   
(5.14)
• (5.14) is simplified for Pr > 1
δt
δ
< 1 , for
Pr > 1
(5.15)
38
Last two term in (5.14):
3  δ t
280  δ

 δt 
3
 <<  

δ 
20

 
2
Simplify (5.14)
2

α
d
δ t 
10 = V∞
δ   
δt
dx   δ  
Use (5.10) for
(b)
δ
280 ν x
δ =
13 V ∞
(c)
39
(c) into (b)
3
2
δt 
 δ t  d  δ t  13 1
  + 4 x 
 =
δ 
 δ  dx  δ  14 Pr
δt
Solve (d) for . Let
δ
 δt
r =
δ





(d)
3
(e)
(e) into (d)
4 dr 13 1
r+ x =
3 dx 14 Pr
(f)
40
Separate variables and integrate
3
δt 
− 3 / 4 13 1
r =   = C ( x)
+
14 Pr
δ 
(g)
C = constant. Use boundary condition on δ t
δ t ( xo ) = 0
(h)
13 1 3 / 4
C=−
xo
14 Pr
(i)
Apply (h) to (g)
41
(i) into (h)
δ t 13 1   xo 
=
1 −  
δ 14 Pr   x 
Use (c) to eliminate
δ
3 / 4  1 / 3


 
(5.16)
280 vx
13 V∞
(5.17a)
in (5.16)
13 1   x 
δt = 
1 −  o 
14 Pr   x 
1/ 3
3 / 4 


 
or
   x 
o
1
−
= 1/3




1/2
x Pr Re x    x 
δt
4.528
1/ 3
3/ 4 


 
(5.17b)
42
Re x is the local Reynolds
Re x =
V∞ x
(5.18)
ν
(ii) Nusselt Number
Local Nusselt number:
hx
Nu x =
k
(j)
∂T ( x ,0)
−k
∂y
h=
Ts − T∞
(k)
h is given by
43
Use temperature solution (5.13) in (k)
3k
h( x ) =
2δt
(5.19)
Use (5.17a) to eliminate δ t in (5.19)
3/4
k   xo  
h( x ) = 0.331 1 −   
x   x  
−1 / 3
Pr 1/3 Re x 1/2
(5.20)
1/2
(5.21)
Substitute into (j)
  xo 
Nux = 0.331 1 −  
  x 
3 / 4  −1 / 3



Pr
1/3
Re x
44
(iii) Special Case: Plate with no Insulated Section
set xo = 0 in above solution
1/ 3
δ t 13 1 
=

δ 14 Pr 
δt
x
=
0.975
Pr 1 / 3
4.528
=
Pr
1/3
Re x
1/2
(5.22)
(5.23)
45
k 1/3
h( x ) = 0.331 Pr Re x 1/2
x
(5.24)
Nu = 0.331 Pr 1/3 Re x 1/2
(5.25)
x
Accuracy of integral solution:
(1) for Pr = 1, δ t / δ = 1 . Set
Pr = 1 in (5.22)
δt
= 0.975
δ
Error is 2.5%
(2) Compare with Pohlhausen’s solution. For Pr > 10
Nu x = 0.339 Pr 1 / 3 Re x , for Pr > 10
Error is 2.4%
(4.72c)
46
Example 5.1:
Laminar Boundary Layer Flow over a Flat Plate:
Uniform Surface Temperature
Use a linear profiles. Velocity:
u = a0 + a1 y
(a)
Boundary conditions
(1) u( x ,0) = 0 ,
u = V∞
(2) f u( x , δ ) ≈ V∞
y
(b)
δ
47
Apply integral formulation of momentum, (5.5).
Result:
δ
x
=
12
Re x
(5.26)
Temperature profile:
T = b0 + b1 y
(c)
Boundary conditions
(1) T ( x ,0) = Ts ,
(2) T ( x , δ t ) ≈ T∞
T = Ts + (T∞ − Ts )
y
δt
(d)
48
Apply integral formulation of energy, (5.7).
Result:
Nu x = 0.289 Pr
1/3
[
Re x 1 − ( xo / x )
3/4
]
− 1/3
(5.27)
Special case: x o = 0
Nu x = 0.289 Pr 1/3 Re x
Comments
(i) Linear profiles give less accurate results
than polynomials.
(ii) More accurate prediction of Nusselt number
than viscous boundary layer thickness.
49
5.7.3 Uniform Surface Flux
• Flat plate
• Insulated section of length x o
• Plate is heated with uniform flux q ′s′
• Determine h( x ) and Nux
50
q′′ = h( x ) [Ts ( x ) − T∞ ]
(a)
s
or
h( x ) =
qs′′
Ts ( x ) − T∞
Nusselt number:
q′′ x
Nux =
k [Ts ( x ) − T∞ ]
(b)
s
Apply conservation of energy to determine Ts ( x )
∂T ( x ,0 ) d
−α
=
∂y
dx
∫
δt
u(T − T∞ )dy
(5.7)
0
51
Integral solution to u( x , y ) :
u
3 y  1 y 
=  −  
V∞ 2  δ  2  δ 
3
(5.9)
Assume temperature profile:
T = b0 + b1 y + b2 y 2 + b3 y 3
(c)
Boundary conditions:
∂T ( x ,0 )
= q′s′
(1) − k
∂y
(2) T ( x , δ t ) ≅ T∞
∂T ( x , δ t )
(3)
≅0
∂y
2
∂
T ( x ,0 )
(4)
=0
2
∂y
52
2
1 y 3  q′s′
T ( x , y ) = T∞ +  δ t − y +

2
3δt  k
3
(5.29)
set y = 0 to obtain Ts ( x )
2 q′s′
Ts ( x ) = T ( x ,0) = T∞ +
δt
3k
(5.30)
5.30 into (b)
3 x
Nu x =
2 δ t (x)
(5.31)
53
Must determine δ t . Substitute (5.9) and (5.29)
into (5.7)
δt

d 
α = V∞ 
dx 
0
∫
3 y 1 y3 2
1 y 3  
−

 δ t − y +
 dy 
3
2
3δt  
2 δ 2 δ 3

(d)
Evaluating the integrals
3
d  2  1 δ t
1  δ t   
= δ t 
−
  
V∞ dx  10 δ 140  δ   
α
(e)
For Pr > 1 , δ t / δ < 1
3
1 δt 
1 δt
<<
 
140  δ 
10 δ
(f)
54
(f) into (e)
d δ t3 
10
=  
V∞ dx  δ 
α
Integrate
α
δ t3
10
x=
+C
V∞
δ
(g)
δ t ( xo ) = 0
(h)
Boundary condition:
Apply (h) to (g)
C = 10
α
V∞
(i)
xo
55
(i) into (g)
 α

δ t = 10 ( x − xo )δ 
 V∞

1/ 3
(j)
Use(5.10) to eliminate δ in (j)
 α
280 280 / 13
δ t = 10 ( x − xo )
13
Re x
 V∞

x

1/ 3
or
3.594  xo 
1− 
= 1/3

x Pr Re x 
x
δt
1/3
(5.32)
Surface temperature: (5.32) into (5.30)
q′′  x 
Ts ( x ) = T∞ + 2.396 s 1 − o 
k 
x
1/3
x
Pr
1/3
Re1/2
x
(5.33)
56
Nusselt number: (5.32) into (5.31)
 x 
Nu x = 0.417 1 − o 
x

−1/3
Pr 1/3 Re 1/2
x
(5.34)
Special case: x o = 0
q′s′
x
Ts ( x ) = T∞ + 2.396
k Pr 1/3 Re1/2
x
(5.35)
Does Ts ( x ) increase or decrease with distance x?
Nusselt number:
Nu x = 0.417 Pr 1/3 Re1/2
x
(5.36)
Exact solution:
Nu x = 0.453 Pr 1/3 Re1/2
x
(5.37)
57
Example 5.2:
Laminar Boundary Layer Flow over a Flat Plate:
Variable Surface Temperature
• Specified surface temperature
Ts ( x ) = T∞ + C x
58
• Determine the local Nusselt number
(1) Observations
• Determine u( x , y ) and T ( x , y )
• Variable Ts ( x )
• Constant properties: T ( x , y ) is independent of u( x , y )
(2) Problem Definition. Determine u( x, y) and T ( x , y )
(3) Solution Plan
• Start with the definition of Nux
• Apply the integral method to determine T ( x , y )
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(4) Plan Execution
(i) Assumptions
(1) Steady state
(2) Constant properties
(3) Two-dimensional
×105)
(4) Laminar flow (Rex < 5×
(5)
(6)
(7)
(8)
(9)
Viscous boundary layer flow (Rex > 100)
Thermal boundary layer (Pe > 100)
Uniform upstream velocity and temperature
Flat plate (9)
Negligible changes in kinetic and potential energy
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(10) Negligible axial conduction
(11) Negligible dissipation
(12) No buoyancy (b = 0 or g = 0)
(ii) Analysis
Nu x =
hx
k
(a)
(1.10) gives h
∂T ( x ,0)
∂y
h=
Ts ( x ) − T∞
−k
(1.10)
To determine T ( x , y ) use (5.7)
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∂T ( x ,0 ) d
−α
=
∂y
dx
δ t ( x)
∫
u(T − T∞ )dy
(5.7)
0
(5.9) gives u(x,y)
u
3 y  1 y 
=  −  
V∞ 2  δ  2  δ 
3
(5.9)
where
280 / 13
280 xν
x=
δ =
13 V∞
Re x
(5.10)
Assume
T ( x , y ) = b0 ( x ) + b1 ( x ) y + b2 ( x ) y 2 + b3 ( x ) y 3
(a)
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Boundary conditions:
(1) T ( x ,0 ) = Ts ( x )
(2) T ( x , δ t ) ≅ T∞
(3) ∂T ( x , δ t )
≅0
∂y
(4) ∂ 2T ( x ,0 )
=0
2
∂y
 3 y 1 y3 

−
T ( x, y ) = Ts ( x) + [T∞ − Ts ( x)] 
3
 2 δ t 2 δ t 
(b) into (1.10)
3k
h( x ) =
2 δt
(b)
(c)
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(c) into (a)
3 x
Nux =
2δt
(d)
• Use (5.7) to determine δ t
• (5.9) and (b) into (5.7), evaluating the integral
3 Ts ( x ) − T∞
=
α
2
δt
d 
[Ts ( x ) − T∞ ]V∞δ
dx 
(e)
4 
 3  δ t 2
3 δt  
−
  
  
 20  δ  280  δ   
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For Pr > 1
δt
δ
(5.15)
for Pr > 1
< 1,
Thus
4
3 δt 
3 δt 
  <<  
280  δ 
20  δ 
2
Simplify (e)
d 
α
10 [Ts ( x ) − T∞ ] = V∞ [Ts ( x ) − T∞ ]δ
dx 
δt
2
δt 
  
 δ  
(f)
However
Ts ( x ) − T∞ = C x
(g)
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(5.10) and (g) into (f)
d 
13 V∞ 2 
α
10 [C x ] = V∞ C x
δt 
dx 
280 νx 
δt
Simplify
3/2
280 α
[ν / V∞ ] x dx = δ t2 dδ t
5
13 ν
Boundary condition on δ t
(h)
:
(i)
δ t ( 0) = 0
Integrate (h) using (i)
1/3
δ t = [10 280 / 13 ] ( Pr )−1/3 (ν x / V∞ )1/2
(j)
(j) into (d)
Nu x = 0.417 Pr 1/3 Re1/2
x
(5.38)
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(3) Checking
• Dimensional check
• Boundary conditions check
(4) Comments
(i) (5.38) is identical with (5.36) for uniform flux
q′s′
x
Ts ( x ) = T∞ + 2.396
k Pr 1/3 Re1/2
x
(5.35)
Rewrite (5.35)
Ts ( x ) = T∞ + C x
(ii) Use same procedure for other specified Ts ( x )
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