Laminar Flow Across a Flat Plate
Numerical Analysis for Engineering
April 10, 2001
MEAE-4960
Anya Brunschwig
Carrie-Anne Rondi
Table of Contents
List of Symbols ……………………………………………………………………...…3
Introduction ………………………………………………………………………….…4
Problem Description and Background ……………………………………………...5
Numerical Approaches …………………………………………………………….…6
Runge-Kutta Numerical Method ………………………………………….…6
ANSYS Method …………………………………………………………….…9
Results and Error Analysis ………………………………………………………….10
Results of Runge-Kutta Method …………………………………………….10
Comparison of Runge-Kutta and Howarth ………………………………...12
Results of ANSYS ……………………………………………………………13
Comparison of ANSYS and Howarth ………………………………………14
Conclusion ……………………………………………………………………………15
References ……………………………………………………………………………15
Appendix 1 ……………………………………………………………………………. 16
Appendix 2 ……………………………………………………………………………17
2
List of Symbols
x = Length of the plate
U  = Free-stream velocity
u = Velocity measured
y = Distance away from plate
 = Boundary layer thickness
 = Viscosity
 = Kinematic viscosity
t = Time fluid travels
 = Density
f() = Dimensionless stream function
du/dy = Velocity gradient
= Stream function
 = Constant
3
Introduction
Laminar flow over a flat plate is a fundamental problem of fluid mechanics. The boundary
layer is defined as the thin layer between the transition from zero velocity at the flat plat to full
magnitude at some distance from it. Boundary layer equations for flow along a flat plat have
small viscosity or large Reynolds numbers. Boundary layer thickness decreases with viscosity
and an increasing Reynolds number. For two-dimensional flow of a fluid with small viscosity
adheres to the flat plate. A simplified Navier-Stokes equations yields approximate solutions of
viscosity at high Reynolds numbers. The original 3 Navier-Stokes equations find u, v, and p. The
simplified version, which has been derived by Blasius, only uses 2 unknowns u and v. This
problem can not be solved analytically so a numerical method must be used to get a solution. In
this project, two different numerical methods will be used to solve the Blasius equation for laminar
flow across a flat plate. These solutions will be compared with the results of Howarth.
Problem Description and Background
The boundary-layer flow across a flat plate can be expressed using boundary-layer
equations defined by Blasius. These equations are :
u
u v

0
x y
1
u
v
 2u
v
v 2
x
y
y
2
The boundary conditions for these equations are: y=0, u=v=0, y=  , and u= U  , assuming that
the leading edge of the plate is x=0 and the plate is infinity long. This system of equations can be
simplified further to an ordinary differential equation. To do this we use an equation that reads:
 ~
vx
U
3
To make this quantity dimensionless it can be divided by y to obtain:
y
U
vx
4
The stream function can be found from the equation of continuity as:
  vxU f ()
5
4
where f(n) is the dimensionless stream function. The velocity component u which is equal to

can be expressed as follows:
y
u
since
  

y  y
6
U



. So,
 vxU f ' ( ) , and
y
vx

u  U  f ' ( )
7
Also the transverse velocity component can be expressed as:
v
 1 vU 

(f ' f )
x 2
x
8
Now inserting these equations into the second boundary layer flow equation:

U 2
U2
U2
f ' f ' '  (f ' f )  v  f ' ' '
2x
2x
xv
9
and with furthur simplification we have the Blasius ordinary differential equation.
ff ' '2 f ' ' '  0
10
where the boundary conditions are: when   0 , f  0 and f '  0 ; and when    , f '  1 .
Accuracy in boundary layer solutions can be obtained using numerical analysis methods.
One method that we have learned in class is the Runge Kutta method for high order systems.
Another method is solutions in series. This particular method is broken down into different
subcatigories. Many different people adapted this method to form their own solutions, they
included Howarth (Power series), Falkners (transformation followed by series expansion), Piercy
and Preston (Iterative method). Other methods such as finite element methods can be used to
solve these types of problems.
In this project, the Runge-Kutta method for higher order systems and a finite element
method using ANSYS will be used to solve for the velocity profile of laminar flow across a flat
plate. These solutions will be compared with Howarth's solutions to assess their accuracy.
5
Numerical Approachs
Runge-Kutta Numerical Method
The Runge-Kutta method for systems of differential equations was used to solve the
Blasuis equations for laminar flow over a flat plate. This method is used to solve higher order
initial value problems by transforming the higher order equations into systems of first order
differential equations. The first step in using this method was to formulate the first order
differential equations.
The third order Blasius equation is:
ff ' '2 f ' ' '  0
First the new parameters are defined:
u1  f
du
df
 1
d d
12
d 2 f du 2 d 2 u1


 f ''
d
d 2
d 2
13
u2 
u3 
11
so,
d 3 f du 3
f ''

d
d 3
14
Substituting these new parameters into the original Blasius equation,
du 3
u1u3  2
0
d
15
And splitting it up into three first order equations,
du1
 u2
d
16
du2
 u3
d
17
du 3
uu
 1 3
d
2
18
This yeilds the three first order equations that will be input into the Runge-Kutta program in
Maple. Each of the equations must have an initial condition in order for the program to be able to
find a solution. The Blasius equation gives three boundary conditions which are as follows.
6
19
f(0)=0, f'(0)=0, and f'()=1
The first two boundary conditions can be directed placed into the program as the initial values of
the first and second (du1/dn and du2/dn) first order equations. The initial condition for the last
equation must be computed from the third boundary condition of f' at infinity. By expanding the
Blasius equation about the origin, it is found that,
f  F ()
20
 1 and when   1 , F ( ) is the solution. Taking the limit of f'
where  is determined from f '
as n goes to infinity it is found,
lim f ' ()   2 lim F ' ()   2 lim F ' ( )  1
solving for
21
,
 f ' ( ) 

  
 F ' ( ) 
1/ 2

22
since f ' ( )  1 as    so,
  F ' ( )
1
23
2
Taking the third derivative of the equation [20] and putting into the same format as before,
when
lim f ' ' ( )   3 lim F ' ' ()   3 lim F ' ' ( )
24
f ' (0)   3 F ' ' (0)
25
 0,
F(0)=0, F'(0)=0 and F''(0) = 1 are the initial conditions of the modified differential equation
2F'''+FF''=0. So, solving for f''(0) with
  F ' ( )
1
2
as determined above yields,
f''(0)=F'()-3/2
To find f''(0), the third initial condition needed to solve the Blasius equation, first
found. To find
26
F '  must be
F '  , the Runge-Kutta Maple program will be run using the initial conditions for
the modified differential equation, F(0)=0, F'(0)=0, and F''(0)=1. When f'' goes to zero, the
resulting f' is substituted back into equation [26] and the f''(0) initial conditions is found.
Using the Maple program, with the boundary conditions stated above, F(0)=0, F'(0)=0,
and F''(0)=1, and the three first order equations, it was found at time = 7.676 , f'' is equal to zero.
The corresponding f' is equal to 2.08540388. The results of this program are as follows.
7
> alg057();
This is the Runge-Kutta Method for Systems of m equations
Input the number of equations
>3
Input the function F[1](t,y1 ... y3) in terms of t and y1 ... y3 (For example: y1-t^2+1)
> y2
Input the function F[2](t,y1 ... y3) in terms of t and y1 ... y3 (For example: y1-t^2+1)
> y3
Input the function F[3](t,y1 ... y3) in terms of t and y1 ... y3 (For example: y1-t^2+1)
> -1*y1*y3/2
Input left and right endpoints separated by blank
>09
Input the initial condition alpha[1]
>0
Input the initial condition alpha[2]
>0
Input the initial condition alpha[3]
>1
Input a positive integer for the number of subintervals
 34
The output is as follows:
RUNGE-KUTTA METHOD FOR SYSTEMS OF DIFFERENTIAL EQUATIONS
T
W1
W2
W3
0.000 0.00000000 0.00000000 1.00000000
.265 .03503460 .26460360 .99845615
.529 .13997628 .52778104 .98771985
.794 .31402186 .78593809 .95921206
1.059 .55513367 1.03342601 .90625589
1.324 .85958233 1.26327257 .82582072
1.588 1.22174369 1.46835850 .71999825
1.853 1.63424761 1.64284872 .59631439
2.118 2.08849678 1.78348089 .46630306
2.382 2.57545686 1.89026355 .34259093
2.647 3.08652078 1.96632249 .23560399
2.912 3.61422530 2.01698087 .15127017
3.176 4.15266396 2.04846221 .09052412
3.441 4.69756048 2.06669118 .05044592
3.706 5.24607646 2.07651920 .02617063
3.971 5.79648118 2.08145178 .01264218
4.235 6.34780342 2.08375684 .00569038
4.500 6.89954067 2.08476043 .00238930
4.765 7.45145279 2.08516796 .00093744
5.029 8.00343362 2.08532255 .00034450
5.294 8.55543967 2.08537744 .00011895
5.559 9.10745437 2.08539575 .00003875
5.824 9.65947187 2.08540150 .00001198
6.088 10.21149021 2.08540322 .00000353
6.353 10.76350880 2.08540370 .00000100
6.618 11.31552744 2.08540383 .00000028
6.882 11.86754610 2.08540387 .00000008
7.147 12.41956476 2.08540388 .00000002
8
7.412 12.97158342 2.08540388 .00000001
7.676 13.52360208 2.08540388 .00000000
For this solution, W1=f, W2=f'' (the velocity component, u/Uinfinity) and W3=f'''. The final value
for W2 is plugged into the equation [26] to get f''(0)= 0. 332058372. This is the third initial
condition for the Blasius equation.
ANSYS Method
Finite difference methods, such as ANSYS, also can be used to solve the original
problem of laminar flow across a flat plate. With this program, a two dimensional grid was set up
on a rectangle that depicted the flat plate and its immediate surroundings. The program then
solved using the defined boundary conditions. The grid, plate area and boundary conditions are
all set in the input file shown below. The grid was chosen so that the divisions corresponded to
the divisions used in Howarths soluion for n.
/PREP7
!!
!! This program models laminar flow over a flat plate.
! This defines the density and kinematic viscosity of air.
DENS=1.21
VISC= .0000179
!!
!! VARY the VISCOSITY to change the Reynolds number...
!! Number of elements on a side is NSIDE
!! Element spacing ratio is RATIO
NSIDE = 25
RATIO = 1
! This defines the computational domain which is a rectangle in which
! the boundary layer exits
rect,0,1,0,0.25
! This defines the number of subdivisions on each side of domain
! We have defined a mesh that contains 50 divisions on the x-axis and
! 297 divisions on the y-axis. The last number in each line defines the
! distance between the gridlines. The y-axis is divided so that the grid is
! symetric
lesize,1,,,50,1
lesize,2,,,297,1
lesize,3,,,50,1
lesize,4,,,297,1
et,1,141
! FLOTRAN element type
!This creates the mesh
amesh,all
! This sets the boundary conditions for the y-axis at the beginning of the plate
! (the LHS). This defines the velocity to be 1 at the furthest distance from
! the plate.
lsels,s,,,4
nsll
d,all,vx,1.0
d,all,vy,0.0
! This sets the boundary conditions for the plate (x-axis).
lsel,s,,,1,
nsll,,1
d,all,vx,0
d,all,vy,0
9
! This defines the upper boundary condition far away from the plate where the
! velocity is equal to Uinfinity which is defined as 1.
lsel,s,,,3,
nsll,,1
d,all,vx,1
d,all,vy,0
! This defines the boundary conditions for downstream side of the plate.
lsel,s,,,2,
nsll,,1
d,all,pres,0
alls
! These commands assign properties and the number of iterations which ANSYS will
! use to compute the solution.
flda,nomi,dens,DENS
flda,nomi,visc,VISC
flda,iter,exec,50
save
fini.
It is necessary to check the values picked for plate length, freestream velocity and
kinematic viscosity to make sure that the fluid is in the laminar flow region. To do this the
Reynolds number must be calculated. Laminar flow occurs for Reynolds numbers less than
350,000. The equation for Reynolds number is as follows,
Re 
U x

Since we have chosen, U  =1 m/s,the plate length x=1 m, and the kinematic viscosity, v=14.8e-6
PA-s, then substituting back into the formula for Reynolds number, we obtain a Re = 67,567.
This confirms that the flow is in laminar regime with the chosen velocity and plate length.
After the above command file is read into ANSYS the FLOTRAN solver is run. Once the
solution is finished, the postprocessor is used to obtain the boundary layer thickness and velocity
profile.
Results and Error Analysis
Results of Runge-Kutta Method
Maple was used with the three first order equations derived previously and the three initial
conditions, f(0)=0, f'(0)=0 and f''(0)= .332058372. The results are as follows.
> alg057();
This is the Runge-Kutta Method for Systems of m equations
Input the number of equations
>3
Input the function F[1](t,y1 ... y3) in terms of t and y1 ... y3 (For example: y1-t^2+1)
> y2
Input the function F[2](t,y1 ... y3) in terms of t and y1 ... y3 (For example: y1-t^2+1)
> y3
Input the function F[3](t,y1 ... y3) in terms of t and y1 ... y3 (For example: y1-t^2+1)
> (-1*y1*y3)/2
Input left and right endpoints separated by blank
> 0 11
10
Input the initial condition alpha[1]
>0
Input the initial condition alpha[2]
>0
Input the initial condition alpha[3]
> .332058372
Input a positive integer for the number of subintervals
> 44
The solution is output as follows:
RUNGE-KUTTA METHOD FOR SYSTEMS OF DIFFERENTIAL EQUATIONS
T
W1
W2
W3
0.000 0.00000000 0.00000000 .33205837
.250 .01037682 .08300562 .33191485
.500 .04149384 .16588576 .33091202
.750 .09328397 .24831956 .32820678
1.000 .16557398 .32978094 .32300820
1.250 .25803537 .40955830 .31463409
1.500 .37014208 .48679040 .30258158
1.750 .50113950 .56052036 .28660027
2.000 .65002910 .62976683 .26675256
2.250 .81557252 .69360674 .24344437
2.500 .99631687 .75126065 .21741250
2.750 1.19064044 .80216870 .18966236
3.000 1.39681501 .84604517 .16136123
3.250 1.61307803 .88290251 .13370369
3.500 1.83770659 .91304074 .10777391
3.750 2.06908455 .93700475 .08443083
4.000 2.30575583 .95551803 .06423635
4.250 2.54645949 .96940482 .04743541
4.500 2.79014514 .97951355 .03398442
4.750 3.03597061 .98665203 .02361446
5.000 3.28328553 .99154097 .01591128
5.250 3.53160594 .99478769 .01039443
5.500 3.78058451 .99687818 .00658309
5.750 4.02998079 .99818312 .00404180
6.000 4.27963409 .99897284 .00240566
6.250 4.52944072 .99943615 .00138808
6.500 4.77933603 .99969966 .00077649
6.750 5.02928105 .99984498 .00042113
7.000 5.27925311 .99992267 .00022146
7.250 5.52923942 .99996294 .00011293
7.500 5.77923300 .99998319 .00005585
7.750 6.02923020 .99999306 .00002679
8.000 6.27922913 .99999773 .00001247
8.250 6.52922887 .99999987 .00000563
8.500 6.77922897 1.00000083 .00000247
8.750 7.02922924 1.00000124 .00000105
9.000 7.27922957 1.00000141 .00000043
9.250 7.52922994 1.00000148 .00000017
9.500 7.77923031 1.00000151 .00000007
9.750 8.02923069 1.00000152 .00000003
10.000 8.27923107 1.00000152 .00000001
10.250 8.52923145 1.00000152 .00000000
11
10.500 8.77923183 1.00000152 .00000000
10.750 9.02923221 1.00000152 .00000000
11.000 9.27923259 1.00000152 .00000000
For this solution, W1=f, W2=f'' (the velocity component, u/Uinfinity) and W3=f'''.
Comparison of Runge-Kutta and Howarth
L. Howarth solved the laminar flow over a flat plate problem. In the table below,
Howarth's solution is compared with the solution obtained using the Runge-Kutta method. This
table shows that the two solutions are almost identical. The Howarth values converge to the
solution of f''=1 and f'''=0 at 8.4 while Maple converges at 10.25. The differences between
the solutions becomes less as incrases. One of the reasons for the difference between Maple
and Howarth is round off error. The maple program is accurate to the 8th decimal place where as
Howarth is accurate to the 5th decimal place.

f
Howarth
f, R-K
% error
0
1
2
3
4
5
6
7
8
9
10
10.25
0
0.16557
0.65003
1.39682
2.30576
3.28329
4.27964
5.27926
6.27923
0
.16557398
.65002910
1.39681501
2.30575583
3.28328553
4.27963409
5.27925311
6.27922913
3.98E-06
-9E-07
-5E-06
-4.2E-06
-4.5E-06
-5.9E-06
-6.9E-06
-8.7E-07
f''
Howarth
f'', R-K
0
0
0.32979 .32978094
0.62977 .62976683
0.84605 .84604517
0.95552 .95551803
0.99155 .99154097
0.99898 .99897284
0.99992 .99992267
.99999773
1
1.00000141
1
1.00000152
1
1.00000152
1
% error
f'''
Howarth
f''', R-K
% error
0
-9.06E-06
-3.17E-06
-4.83E-06
-1.97E-06
-9.03E-06
-7.16E-06
2.67E-06
-2.27E-06
1.41E-06
1.52E-06
1.52E-06
0.33206
0.32301
0.26675
0.16136
0.06424
0.01591
0.0024
0.00022
0.00001
0
0
0
.33205837
.32300820
.26675256
.16136123
.06423635
.01591128
.00240566
.00022146
.00001247
.00000043
.00000001
.00000000
-1.63E-06
-1.8E-06
2.56E-06
1.23E-06
-3.65E-06
1.28E-06
5.66E-06
1.46E-06
2.47E-06
4.3E-07
1E-08
0
Results of ANSYS
After running the input file, the post processor is used to obtain a velocity profile plot. To
generate this plot you need to go into the post processor tab in ANSYS. First you hit the 'last set'
button to ensure that the latest solution version is being used. Next you click on 'plot results.
This gives you a few choices from which the 'vector plot' option should be choosen. From the
next menu you pick 'predefined' and then from the resulting window select 'velocity' from the
upper right side. In the space labeled 'scale velocity vectors' you can change the size of the
velocity vectors so that they fit on the plot. This is a vector plot of velocity.
12
The different colors represent the different magnitudes of velocity. The plot shows the
boundary layer thickness grows when traveling down the length of the plate. Also, a plot of
boundary layer thickness and velocity versus distance along the plate was generated. To
generate this plot first a path was defined as a straight line at a certain distance x away from the
beginning of the plate. Once the path is defined (done by using the 'define path' command), the
velocity and boundary layer thickness can be plotted by selecting 'map onto path', choosing the
velcoity vx, and then selecting 'plot item on graph'. Also the results can be listed by selecting 'list
path results'. The results are as follows,
13
This plot shows that velocity increases very sharply with distance until it levels off when it reaches
the freestream velocity at 1. The boundary layer thickness, represented by S, increases slowly in
a linear fashion as you move down the plate.
Through the ANSYS postprocessor it is possible to get a listing of the velocity values
over a given path. To find a velocity profile the path is drawn as a veritcal line at the end of the
plate. This output listing is included in the Appendices.
Comparison of ANSYS and Howarth
In order to compare the solution generated by ANSYS to Howarth's solution some
calculations must be made. Howarth uses all non-dimensional variables while ANSYS uses
actual dimensions. In ANSYS, the user inputs the properties of the working fluid in the desired
dimensions. In this case properties for air at 20 deg C were used (in english units). The first step
in getting comparable values was to re-dimensionalize the value n. This is done using the
formula,
y
U
x
where Uinfinity =1 m/s, x=1 m, and v=17.9 *10e-6 Pa sec. Substituting all these values in and
some chosen values for n, y can be found. These results are found in the table below.
To find the corresponding values from the ANSYS solution, the solution output values are
studied. From the velocity and position values that are listed in the output, as described in the
results section, the velocities are picked out to match as closely as possible to the values taken
from Howarth. The corresponding ANSYS y values can be compared to the calculated y* values
from Howarth's solution. These values are listed below.
n
0.2
0.4
0.6
0.8
1
2
3
4
5
6
7
8
y*,
Velocity y* from Velocity (f'') Difference between
calculated, (f'') from ANSYS, from ANSYS, the velcoity of the
m
Howarth,
m
m/s
two solutions
m/s
0.000846
0.001692
0.002539
0.003385
0.004231
0.008462
0.012693
0.016923
0.021154
0.025385
0.029616
0.033847
0.06641 0.000843
0.13277 0.00168
0.19894 0.00253
0.26471 0.00337
0.32979 0.00421
0.62977 0.00842
0.84605 0.01263
0.95552 0.01684
0.99155 0.0202
0.99898 0.0253
0.99992 0.0295
1
0.0337
0.10107
0.20108
0.29903
0.39378
0.4841
0.83345
0.98967
1.0293
1.0334
1.0316
1.0297
1.0281
0.03466
0.06831
0.10009
0.12907
0.15431
0.20368
0.14362
0.07378
0.04185
0.03262
0.02978
0.0281
This table of results shows that ANSYS has come up with a similar solution to the laminar
flow problem as Howarth. The values are closest farthest away from the plate where the solution
tends toward Uinfinity (1). The solutions are the farthest apart at the intermediate values of n.
The ANSYS solution has a steeper velocity gradient than the Howarth solution. This solution
reaches the freestream velocity at n~3.0 while the Howarth solution does not reach the
freestream velocity until n~8.0. Both the Howarth and ANSYS solutions are accurate to the 5th
decimal place.
14
Conclusion
As shown in this paper, there are a variety of different numerical methods which can be
used to solve a high order differential equation problem such as laminar flow over a flat plate. In
this paper, both Runge-Kutta solution method and ANSYS, a finite element solution method were
used to obtain a solution for this fluid mechanics problem. These solutions were then compared
another numerical solution completed by Howarth. The results for both methods show that the
solutions agree with Howarth's solution. However, the Runge-Kutta method seemed to follow the
Howarth solution more closely. While the ANSYS did predict the same boundary layer and
velocity behavior, the values were not as close to Howarth's as the values from the Runge-Kutta
method. Using two completely different methods gave a well-rounded approach in achieving an
accurate solution for the problem.
References
1.- Schlichting, H., Boundary Layer Theory, 7th ed., McGraw-Hill, New York, 1979, pp. 230-233.
2.- Burden, R. L. and Faires, J.D., Numerical Analysis, 7th ed., Brooks/Cole, Pacific Grove, 2001,
chapter 5.
3. – Curle, N.,The Laminar Boundary Layer Equations, Oxford University Press, London, 1962.
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Appendix 1.
Howarth's Solution
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Appendix 2
ANSYS List Output (found only in print out of paper)
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