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 .dydx

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

ox

e dmdxdx

dmdm (a)

Pdxvdxdyudx

ddm

x

e o

)(

0

(5.1)

edm is the external mass flow rate into

element.

5.6.2 Conservation of Momentum

Application of the momentum theorem in the x-direction to the element dx

)in()out( xxx MMF (a)

Axial velocity u varies with x and y.

xm

edm

5.3 Fig.

dxdx

dmm x

x

odm

3

Pressure p varies with x only (boundary layer approximation).

xM = x-momentum, given by

)(

0

2

x

x dyuM (c)

o is wall shear, given by

y

xuo

0, (d)

Equation (a) gives

oPvxVdyudx

dxVdyu

dx

d

y

xuP

dx

dpxx )(

0

)(

0

20,1 (5.2)

NOTE:

(1) Equation (5.2) is the integral formulation of conservation of momentum.

(2) Equation (5.2) applies to laminar as well as turbulent flow.

(3) Although u is a function of x and y.

(4) 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

dx

dp

dx

dp (4.12)

The x-momentum equation for boundary layer flow is

p dxpdx

dp )(

dxPo )1(

ddp

p )2

(

momentum (b) x

dxdx

dMM x

xxM

emxV )(

5.4 Fig.forces (a)

dxdx

4

2

21

y

u

x

dp

y

uv

x

uu ν (4.5)

Applying equation (4.5) at y

dx

dVxV

dx

dp

dx

dp)( (5.3)

Substituting (5.3) into (5.2) and noting that is constant

oPvxVdyudx

dxVdyu

dx

d

y

xuP

dx

VdxV

xx )(

0

)(

0

20,1)( ν (5.4)

(ii) Case 2: Incompressible fluid and impermeable flat plate

For boundary layer flow over a flat plate

0dx

dp

dx

dp

dx

dV (e)

For an impermeable plate

,0ov 0P (f)

(e) and (f) into (5.4)

dyudx

dudy

dx

dV

y

xu

xx

0

2

0

0,v (5.5)

5.6.3 Conservation of Energy

Application of conservation of energy to the

element dxt , neglecting changes

in kinetic and potential energy, axial

conduction, and dissipation:

Based on these assumptions,

conservation of energy for the

element gives

oex

c dEdEdxdx

dEdE (a)

edE = energy added by external mass

odE = energy added by injected mass

xE = energy convected with boundary layer flow

Formulating each term in (a)

dxy

xTPkdEc

0,)1( (b)

t

cdE

dxdx

dEE x

xxE

edE

5.6 Fig.

dx

odE

5

PdxvTcdxdyudx

dTcdE p

t

pe

x

o

)(

0

(c)

PdxvTcdE opo o (d)

)(

0

xt

px dyTucE (e)

Substituting (b)-(e) into (a)

TTPvcudydx

dTcuTdyc

dx

d

y

xTPk oop

t

p

t

p

xx )()(

00

0,1 (5.6)

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 1P in (5.6)

)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)

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:

dyudx

dudy

dx

dV

y

xu

xx

0

2

0

0,v (5.5)

Assumed velocity profile

33

2210 )()()()(),( yxayxayxaxayxu (a)

Boundary conditions

6

(1) 0)0,(xu

(2) Vxu ),(

(3) 0),(

y

xu

(4) 0)0,(

2

2

y

xu

NOTE: The fourth boundary condition is obtained by setting y = 0 in (2.10x)

Boundary conditions give the four coefficients. Thus

3

2

1

2

3 yy

V

u (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)

dx

dVV 2

280

391

2

3v (b)

Separating variables, integrating and noting that 0)0(

x

dxV

d

0013

140 v

Evaluating the integrals and rearranging

xx ReRex

64.413/280 (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 fC is obtained using (4.36) and

(4.37a)

xVV

xy

u

VC o

f

v32/

0,

2/ 22

Using (5.10) to eliminate )(x

x

fRe

C646.0

(5.11)

Compare with Blasius solution:

7

xRex

2.5, Blasius solution (4.46)

and

x

fRe

C664.0

, Blasius solution (4.48)

Note the small error in prediction .fC

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 ox

is insulated and the remaining part is at

uniform temperature .sT

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 ),( yxu 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

)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7

Assumed temperature profile

33

2210 )()()()(),( yxbyxbyxbxbyxT (a)

Boundary conditions

(1) sTxT )0,(

(2) TxT t ),(

(3) 0),(

y

xT t

(4) 0)0,(

2

2

y

xT

8

NOTE: The fourth condition is obtained by setting 0y in the energy equation (2.19).

Boundary conditions give the four coefficients. Thus

3

3

2

1

2

3)(),(

ttss

yyTTTyxT (5.13)

(5.9) and (5.13) into (5.7), evaluating the integral, gives

42

280

3

20

3)(

2

3 tts

t

s VTTdx

dTT (5.14)

Simplification of (5.14). Note that

1t

, for 1Pr (5.15)

It follows that 2

20

3

280

3 tt

(5.14) simplifies to 2

10 t

t dx

dV (b)

where

V

x

13

280 (c)

Boundary condition

0)( ot x (h)

Solution to (b) 3/1

4/3

1/21/31

528.4

x

x

RePrx

o

x

t (5.17b)

(ii) Nusselt Number. Local Nusselt number is defined as

k

hxNux (j)

h is the local heat transfer coefficient given by

TT

y

xTk

hs

)0,(

(k)

Using (5.13) into (k)

t

kxh

2

3)( (5.19)

9

Eliminating t by using (5.17b)

1/21/3

3/14/3

1331.0)( xo RePrx

x

x

kxh (5.20)

Substituting into (j)

1/21/3

3/14/3

1331.0 xo

x RePrx

xNu (5.21)

(iii) Special Case: Plate with no Insulated Section

Set 0ox in (5.17b), (5.20) and (5.21)

1/21/3

528.4

xRePrx

t (5.23)

1/21/3331.0)( xRePrx

kxh (5.24)

1/21/3331.0 xRePrNux (5.25)

Examine the accuracy of the local Nusselt number. For 10Pr equation (4.72c)

gives Pohlhausen’s solution

10for,339.0 3/1 PrRePrNu xx (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

yaau 10 (b)

Boundary conditions

(1) 0)0,(xu

(2) Vxu ),(

Velocity solution

10

yVu (c)

Integral solution to

xRex

12 (5.26)

Assumed temperature

ybbT 10 (f)

Boundary conditions

(1) sTxT )0,(

(2) TxT t ),(

Temperature solution

tss

yTTTT )( (g)

Integral solution to t

3/14/3

1/3)/(1

121xxx

VProt

ν (o)

Solution to local Nusselt number

3/14/31/3 )/(1289.0 xxRePrNu oxx (5.27)

Special Case: no insulated section, set 0ox in (5.27) gives

xx RePrNu 1/3289.0 (5.28)

Comments. Table 5.1 compares exact solutions for x/ and 1/23/1/xx RePrNu 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 xRex

2/13/1 eRPr

Nu x

Exact (Blasius/ Pohlhausen) 5.2 0.332

3rd

degree polynomial 4.64 0.339

Linear 3.46 0.289

11

5.7.3 Uniform Surface Flux

Plate with an insulted leading section of

length ox .

Plate is heated with uniform flux sq

along its surface .oxx

Steady, two-dimensional, laminar flow.

Determine surface temperature and the local Nusselt number.

Surface temperature is unknown.

Solution

Newton’s law of cooling gives

TxT

qxh

s

s

)()(

Introducing the definition of the Nusselt number

TxTk

xqNu

s

sx

)( (b)

Need surface temperature ).(xTs Use the integral form of the energy equation to

determine )(xTs

)(

0

)(0,

xt

dyTTudx

d

y

xT (5.7)

),( yxu for a third degree polynomial is given by (5.9)

3

2

1

2

3 yy

V

u (5.9)

Assume temperature ),( yxT

33

2210 ybybybbT (c)

Boundary conditions

(1) sqy

xTk

0,

(2) TxT t,

12

(3) 0,

y

xT t

(4) 00,

2

2

y

xT

Application of boundary conditions give the coefficients in (c)

k

qyyTyxT s

t

t 2

3

3

1

3

2),( (5.29)

Surface temperature. Set 0y in the above

ts

sk

qTxTxT

3

2)0,()( (5.30)

(5.30) into (b)

x

xNu

tx

2

3 (5.31)

Must determine .t

(5.9) and (5.29) into (5.7), evaluate the integral

3

2

140

1

10

1 ttt

dx

d

V (e)

Simplify for Prandtl numbers larger than unity, 1/t

3

10 t

dx

d

V

Thermal boundary layer thickness. Integrating and use boundary condition

0)( ot x , gives

3/1

)(10 ot xxV

(j)

Use (5.10) to eliminate in (j), rearrange

3/1

1/21/31

594.3

x

x

RerPx

o

x

t (5.32)

Surface temperature. (5.32) into (5.30) gives

1/21/3

3/1

1396.2)(

x

oss

RerP

x

x

x

k

qTxT (5.33)

Local Nusselt number. (5.32) into (5.31) gives

13

1/21/3

3/1

1417.0 xo

x RerPx

xNu (5.34)

Special Case: Plate with no insulated section, set 0ox in (5.33) and (5.34)

1/21/3396.2)(

x

ss

RerP

x

k

qTxT (5.35)

1/21/3417.0 xx RerPNu (5.36)

Compare with differential formulation solution:

1/21/3453.0 xx RerPNu (5.37)