CHAPTER 5
APPROXIMATE SOLUTIONS:
THE INTEGRAL METHOD
5.1 Introduction

Seek approximate solution when:




Exact solution is unavailable.
Form of exact solution is not suitable or convenient.
Solution requires numerical integration.
The integral method gives approximate solutions.
5.2 Differential vs. Integral Formulation

Example: boundary layer flow, Fig. 5.1.

Differential formulation, Fig. 5.1a: the basic laws are formulated for a differential
element dx  dy.


Solutions satisfy the basic laws exactly (at every point).
Integral formulation, Fig. 5.1b: the basic laws are formulated for the element
dx   .

Solutions satisfy the basic laws in an average sense (for section  ).
5.3 Integral Method Approximation: Mathematical Simplification


Reduction in the number of independent variables.
Reduction of the order of the governing differential equation may result
5.4 Procedure


Integral solutions are obtained for the velocity and temperature fields.
The following procedure is used in obtaining integral solutions:
2
(1) Integral formulation of the basic laws:

Conservation of mass, momentum, and energy.
(2) Assumed velocity and temperature profiles:



Several options. Polynomials are used in Cartesian coordinates.
Assumed velocity and temperature profiles should satisfy known boundary
conditions
Assumed profile contains an unknown parameter or variable.
(3) Determination of the unknown parameter or variable:

Integral form of the basic law gives the unknown parameter or variable.
5.5 Accuracy of the Integral Method




Different assumed profiles give different solutions and accuracy.
Errors are acceptable in many engineering applications.
Accuracy is not very sensitive to the form of an assumed profile.
No procedure is available for identifying assumed profiles that will result in the most
accurate solutions.
5.6 Integral Formulation of the Basic Laws
5.6.1 Conservation of Mass

Boundary layer flow over porous plate of
porosity P with mass injection.

Conservation of mass for element   dx ,
shown in Fig. 5.2 and enlarged in Fig. 5.3,
gives
dm x
(a)
dme 
dx  dmo
dx

d 
dme 
dx 



 ( x)

 u dy  dx   v o Pdx

0


dme
(5.1)
dme is the external mass flow rate into
element.
mx
mx 
dmx
dx
dx
dmo
Fig. 5.3
5.6.2 Conservation of Momentum
 Application of the momentum theorem in the x-direction to the element   dx
 Fx  M x (out )  M x (in )

Axial velocity u varies with x and y.
(a)
3

Pressure p varies with x only (boundary layer approximation).
(p
dp
) d
2
p
V ( x)me

p 
d
( p )dx
dx

Mx
dx
Mx 
dM x
dx
dx
dx
 o (1  P)dx
(b) x  momentum
(a) forces
Fig. 5.4

M x = x-momentum, given by
 ( x)

Mx 
 u 2 dy
(c)
0

 o is wall shear, given by
o  

Equation (a) gives
dp
u x,0 d

  1  P 

dx
y
dx
u  x,0 
y
 ( x)

d
 u 2 dy  V x 
dx
0
(d)
 ( x)
  udy  V

x  Pvo
(5.2)
0
NOTE:
(1)
(2)
(3)
(4)
Equation (5.2) is the integral formulation of conservation of momentum.
Equation (5.2) applies to laminar as well as turbulent flow.
Although u is a function of x and y.
Evaluating the integrals in (5.2) results in a first order ordinary differential equation with
x as the independent variable.
Special Cases:
(i) Case 1: Incompressible fluid

Boundary layer approximation gives
dp dp

dx
dx

The x-momentum equation for boundary layer flow is
(4.12)
4
u
u
u
1 dp
 2u
v

ν 2
x
y
 x
y
(4.5)
dV
dp dp

   V ( x) 
dx
dx
dx
(5.3)
Applying equation (4.5) at y  
Substituting (5.3) into (5.2) and noting that  is constant
 ( x)
dV
u x,0 d
 V ( x)   ν 1  P 

dx
y
dx

 ( x)
u 2 dy  V x 
d
dx
0

u dy  V x  Pvo
(5.4)
0
(ii) Case 2: Incompressible fluid and impermeable flat plate

For boundary layer flow over a flat plate
dV dp dp


0
dx
dx
dx
(e)
For an impermeable plate
vo  0,
P0
(f)
(e) and (f) into (5.4)
v
u x,0
d
 V
y
dx
 x 

d
udy 
dx
0
 x 

u 2 dy
(5.5)
0
5.6.3 Conservation of Energy


Application of conservation of energy to the
element  t  dx , neglecting changes
in kinetic and potential energy, axial
conduction, and dissipation:
dE x
dx  dEe  dEo
dx
t
Ex 
Ex
Based on these assumptions,
conservation of energy for the
element gives
dEc 
dEe
dE x
dx
dx
dx
(a)
dEc
dEe = energy added by external mass
dEo = energy added by injected mass
dEo
Fig. 5.6
E x = energy convected with boundary layer flow

Formulating each term in (a)
dEc  k (1  P)
T x,0
dx
y
(b)
5
d 

dEe  c p T
dx 

 t ( x) 
 u dy  dx  c p T  vo Pdx

0


(c)
dEo   c p To vo Pdx
t ( x)
(d)
Ex 
(e)
  c uT dy
p
0
Substituting (b)-(e) into (a)
T x,0
d
 k 1  P 

y
dx
 ( x)

t
 ( x)
d
 c p uTdy  c p T
dx
0

t
 udy   c p vo PTo  T 
(5.6)
0
NOTE:

Equation (5.6) is integral formulation of conservation of mass and energy.

Although u and T are functions of x and y, evaluation of the integrals gives a first
order ordinary differential equation with x as the independent variable.
Special Case: Constant properties and impermeable flat plate

Setting P  1 in (5.6)
T  x,0 
d


y
dx

t ( x)
u (T  T )dy
(5.7)
0
5.7 Integral Solutions


Flow field solution.
Temperature field solution.
5.7.1 Flow Field Solution: Uniform Flow over a
Semi-Infinite Plate

Integral form of governing equation:
v ux,0  V
y
 x 
d
dx

udy 
 x 
d
dx
0


u 2 dy
(5.5)
0
Assumed velocity profile
u ( x, y)  a0 ( x)  a1 ( x) y  a2 ( x) y 2  a3 ( x) y 3

Boundary conditions
(a)
6
(1) u ( x,0)  0
(2) u( x,  )  V
(3)
(4)
u ( x,  )
0
y
 2 u ( x,0)
y 2
0
NOTE: The fourth boundary condition is obtained by setting y = 0 in (2.10x)

Boundary conditions give the four coefficients. Thus
u
3 y 1 y
    
V 2    2   
3
(5.9)

Note that the assumed velocity is in terms of the unknown variable  (x).

Boundary layer thickness. Use (5.5) to determine  (x). Substituting (5.9) into (5.5)
3
vV 1  39 V2 d
2
 280
dx
(b)
Separating variables, integrating and noting that  (0)  0


0
140 v
 d 
13 V

x
dx
0
Evaluating the integrals and rearranging

x

280 / 13

Re x

4.64
Re x
(5.10)
Friction coefficient. (5.10) into (5.9) gives u as a function of x and y. With the
velocity distribution determined, friction coefficient C f is obtained using (4.36) and
(4.37a)
u
  x ,0 
o
y
3v
Cf 


2
2
V   x 
 V / 2
 V / 2
Using (5.10) to eliminate  (x )
Cf 

Compare with Blasius solution:
0.646
Re x
(5.11)
7

x

5 .2
Re x
,
Blasius solution
(4.46)
,
Blasius solution
(4.48)
and
Cf 

0.664
Re x
Note the small error in prediction C f .
5.7.2 Temperature Solution and Nusselt
Number: Flow over a Semi-Infinite Plate
(i) Temperature Distribution

A leading section of the plate of length x o
is insulated and the remaining part is at
uniform temperature Ts .


Assume laminar, steady, two-dimensional,
constant properties boundary layer flow and
neglect axial conduction and dissipation.
Determine  t , h(x), and Nu(x).

Must determine flow field u ( x, y ) and temperature T(x,y).


Flow field solution of Section 5.7.1 applies to this case, equation (5.9).
Equation (5.7) gives the integral formulation of conservation of energy for this
problem

T x,0
d

y
dx
 t ( x)

u (T  T )dy
(5.7
0

Assumed temperature profile
T ( x, y)  b0 ( x)  b1 ( x) y  b2 ( x) y 2  b3 ( x) y 3

Boundary conditions
(1) T ( x,0)  Ts
(2) T ( x,  t )  T
(3)
T ( x,  t )
0
y
(4)
 2T ( x,0)
0
y 2
(a)
8
NOTE: The fourth condition is obtained by setting y  0 in the energy equation (2.19).

Boundary conditions give the four coefficients. Thus
3 y 1 y3 
T ( x, y )  Ts  (T  Ts ) 


3
 2  t 2  t 
(5.13)
(5.9) and (5.13) into (5.7), evaluating the integral, gives
3 T  Ts
d 


(T  Ts ) V 
2
t
dx 


4
 3   2
3   t   
t
   
  
280     
 20   

(5.14)
Simplification of (5.14). Note that
t

 1 , for Pr  1
(5.15)
It follows that
3 t 
3 t 
  
 
280   
20   
2
(5.14) simplifies to

d  t  
10  V
   
t
dx     
2
(b)
where
280  x
13 V

Boundary condition
(c)
 t ( xo )  0
(h)
Solution to (b)
t
x
1/ 3

4.528
Pr 1/3 Re x 1/2
  x  3 / 4  
o
1     
x
    
(5.17b)
(ii) Nusselt Number. Local Nusselt number is defined as
Nu x 
hx
k
(j)
h is the local heat transfer coefficient given by
T ( x,0)
y
Ts  T
k
h
(k)
Using (5.13) into (k)
h( x ) 
3 k
2 t
(5.19)
9
Eliminating  t by using (5.17b)
k   x 
h( x)  0.331 1   o 
x  x 

3 / 4  1 / 3



Pr 1/3 Re x1/2
(5.20)
Substituting into (j)
  x  3 / 4 
Nu x  0.331 1   o  
  x  
1 / 3
Pr1/3 Re x1/2
(5.21)
Nu x  0.331Pr 1/3 Re x1/2
(5.25)
(iii) Special Case: Plate with no Insulated Section
Set xo  0 in (5.17b), (5.20) and (5.21)
t
x
4.528

Pr
1/3
Re x 1/2
k
h( x)  0.331 Pr 1/3 Re x 1/2
x

(5.23)
(5.24)
Examine the accuracy of the local Nusselt number. For Pr  10 equation (4.72c)
gives Pohlhausen’s solution
Nu x  0.339 Pr 1/ 3 Re x , for Pr  10
(4.72c)
Comparing this result with integral solution (5.25) shows that the error is 2.4%.
Example 5.1: Laminar Boundary Layer Flow over a Flat Plate:
Uniform Surface Temperature




This is a repeat of the Section 5.7.1 and 5.72, using assume linear velocity and
temperature profiles.
A linear profile gives less accurate flow and heat transfer results.
The procedure of the previous sections is repeated in this example.
The following is a summary of the results.

Assumed velocity

Boundary conditions
(1) u ( x,0)  0
(2) u( x,  )  V

Velocity solution
u  a0  a1 y
(b)
10
u  V

Integral solution to 


Assumed temperature
(c)

12
Re x
(5.26)
T  b0  b1 y
(f)
x

y

Boundary conditions
(1) T ( x,0)  Ts
(2) T ( x,  t )  T

Temperature solution
T  Ts  (T  Ts )

(g)
t
Integral solution to  t
t 

y
12 ν
1
1/3
V
Pr

x 1  ( xo / x) 3 / 4

1/ 3
Solution to local Nusselt number

Nu x  0.289 Pr 1/3 Re x 1  ( xo / x) 3 / 4

(o)

 1/ 3
(5.27)
Special Case: no insulated section, set xo  0 in (5.27) gives
Nu x  0.289 Pr 1/3 Re x
(5.28)
Comments. Table 5.1 compares exact solutions for  / x and Nu x / Pr 1 / 3 Rex1/2 with integral
results for the case of a plate with no insulated section based on assumed linear and
polynomial profiles

Note that the integral method gives more accurate prediction of Nusselt number than
of the boundary layer thickness  .
Table 5.1

Solution
Exact (Blasius/ Pohlhausen)
3rd degree polynomial
Linear
x
Re x
5.2
4.64
3.46
Nu x
Pr 1 / 3 Re1 / 2
0.332
0.339
0.289
11
5.7.3 Uniform Surface Flux

Plate with an insulted leading section of
length xo .

Plate is heated with uniform flux q s
along its surface x  xo .

Steady, two-dimensional, laminar flow.

Determine surface temperature and the local Nusselt number.

Surface temperature is unknown.
Solution

Newton’s law of cooling gives
h( x ) 

Introducing the definition of the Nusselt number
Nu x 

q s
Ts ( x)  T
q s x
k Ts ( x)  T 
(b)
Need surface temperature Ts (x). Use the integral form of the energy equation to
determine Ts (x)
T  x,0 
d


y
dx
 t ( x)

u (T  T )dy
(5.7)
0

u ( x, y ) for a third degree polynomial is given by (5.9)
u
3 y 1 y
    
V 2    2   

3
Assume temperature T ( x, y )
T  b0  b1 y  b2 y 2  b3 y 3

(5.9)
Boundary conditions
(1)  k
T x,0
 q s
y
(2) T x,  t   T
(c)
12
(3)
T x,  t 
0
y
(4)
 2 T  x ,0 
0
y 2

Application of boundary conditions give the coefficients in (c)
2
1 y 3  q s
T ( x, y )  T    t  y 

3  t2  k
 3

Surface temperature. Set y  0 in the above
Ts ( x)  T ( x,0)  T 

Must determine  t .

(5.9) and (5.29) into (5.7), evaluate the integral

V

(5.31)
3
d  2  1  t
1   t   



  
 t
dx  10  140     

 

V

d   t3 
 
dx   
Thermal boundary layer thickness. Integrating and use boundary condition
 t ( xo )  0 , gives
 

 t  10
( x  xo ) 
 V

x

3.594
Pr 1/3 Re 1/2
x
(j)
 xo 
1  x 


1/ 3
(5.32)
Surface temperature. (5.32) into (5.30) gives
q 
Ts ( x)  T  2.396 s
k

1/ 3
Use (5.10) to eliminate  in (j), rearrange
t

(e)
Simplify for Prandtl numbers larger than unity,  t /   1
10

(5.30)
3 x
2  t x 


2 q s
t
3 k
(5.30) into (b)
Nu x 

(5.29)
 xo 
1  x 


Local Nusselt number. (5.32) into (5.31) gives
1/ 3
x
Pr
1/3
Re 1/2
x
(5.33)
13
 x 
Nu x  0.417 1  o 
x


1 / 3
Pr 1/3 Re 1/2
x
Special Case: Plate with no insulated section, set xo  0 in (5.33) and (5.34)
q 
x
Ts ( x)  T  2.396 s
1/3
k Pr Re 1/2
x
Nu x  0.417 Pr 1/3 Re 1/2
x

(5.34)
(5.35)
(5.36)
Compare with differential formulation solution:
Nu x  0.453 Pr 1/3 Re 1/2
x
(5.37)