FLOLAB – FLAT PLATE Anoop Samant Yanyan Zhang Saptarshi Basu Andres Chaparro

FLOLAB – FLAT PLATE

Anoop SamantYanyan ZhangSaptarshi Basu

Andres Chaparro

INTRODUCTION

Model the laminar and turbulent flow over a flat plate

Compare with the theoretical predictions

Study of flow parameters like Nusselt Number, Skin Friction Coefficient, shear layer thickness, velocity and temperature

Effects of change in length and initial flow conditions

THEORY

Boundary layer Assumptions: 1. Steady, Incompressible, 2D, Laminar flow2. The shear layer is thin. This is true if Re>>13. In the boundary layer u i.e. the velocity in the x direction scales

with L4. v the velocity in the y direction scales with 5. u>>v therefore 6. and to simplify the Navier Stokes equations to the

corresponding boundary layer forms7. The term for flow over a flat plate as

constant.

energyy

u

y

Tk

y

Tv

x

TuC

momentumxy

u

dx

dUU

y

uv

x

uu

continuityy

u

x

u

p

ee

.......)()(

......

.....0

22

2

2

2

y

u

x

u

y

v

x

v

01

dx

dp

dx

dUU ee

eU

THEORY

Laminar - Blasius Similarity Solution

Using the similarity variable

we have

Substituing these variables into the x-momentum equation of the boundary layer we will obtain the following ODE as a function of

Assuming no slip conditions we have u(x,0)=v(x,0)=0 and the free stream merge condition u(x,)=Ue

These convert to

x

Uy e

2

)(2

)(

'

'

ffx

Uv

fUu

e

e

0''''' fff

1)(

0)0()0('

'

f

ff

THEORY

Laminar - Blasius Equation Solution

xxRe

xxRe

*

xxRe%99

xfC Re

xDC Re

Parameters Exact from Blasius

0.664

1.721

5

0.664

1.328

THEORY

Laminar - Limitations of B.L.

•Near the edge the boundary layer theory fails as is not valid.

•At Re< 1000 Cf and non-dim pressure becomes high at leading and trailing edges

•At very large x Re gets large and the flow gets separated.

x

THEORY

Definitions

THEORY

Turbulence

Flolab Parameters

Mesh Size Used: Medium Number Iterations: 1000 Fluids: Air and Water Flat Plate Length: 1m and 5m Re # Range: 100 – 2.23*10^9

RESULTS – Velocity Profiles

Laminar – Air

Shear stress and Cf decreases as x increases

becomes thicker as x increases

As x increases we observe that the velocity exceeds the inlet velocity.

In the case of the low Re # (10^2), is much thinner than a higher Re #

Re=10^5 velocity=1.34m/s

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

0.02

-0.4 0.1 0.6 1.1 1.6

velocity

y d

ire

cti

on

inlet

x=0.2

x=0.4

x=0.6

x=0.9

x=0.8

outlet

Re=10^2 velocity=0.00134m/s

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 0.0005 0.001 0.0015 0.002

velocity

y d

ire

cti

on

inlet

x=0.2

x=0.4

x=0.6

x=0.8

x=0.9

outlet

RESULTS – Velocity Profiles

Turbulent – Water

The difference with the laminar case is that the shear layer in turbulence is much thinner than its laminar counterpart.

Re=5*10^6,Vinlet=4.477m/s,T=300-300)

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

3.5 3.7 3.9 4.1 4.3 4.5 4.7

velocity

y

inlet

0.2L

0.4L

0.6L

0.8l

0.9l

outlet

Re=2.23*10^9,Vinlet=2000m/s,T=1000-400)

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

1900 2000

ve locity

y

inlet

0.2l

0.4l

0.6l

0.8l

0.9l

outlet

RESULTS – Skin Friction (Cf)

Laminar –Air/Water

With the increase of x the value of Cf decreases as Rex increases. Cf 1/Rex

As the value of inlet velocity is decreased so that the Re decreases then the skin friction coefficient goes up.

For low Re i.e at 1500 or 100 the Cf does not follow the relation =0.664 as predicted by the Boundary layer solution by Blasius.

Skin friction coefficient Air

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 0.2 0.4 0.6 0.8 1 1.2

x direction

Cf

Re=10^2

Re=10^5

Re=5*10^5

Re=10^6

xfC Re


Turbulent–Air/Water

With the increase of x the value of Cf decreases as Rex increases.

Cf As the value of inlet

velocity is decreased so that the Re decreases the skin friction coefficient goes up.

Cf Vs Reynolds Air

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0 0.2 0.4 0.6 0.8 1 1.2

X

Cf

Re 5*10^6

Re 10^6

Re 10^7

Cf vs Reynolds Water

0

0.001

0.002

0.003

0.004

0.005

0 0.2 0.4 0.6 0.8 1 1.2

x

Cf

5*10 6̂

10 7̂

5.5*e 8̂

2.23*e 9̂

)Re06.0(ln

455.02

x


Low Reynolds #

With the increase of x the value of Cf decreases as Rex increases.

Based on Dennis and Dunwoody Cf show a sharp increase at both the leading and trailing edge.

Based on Flolab the leading edge does match but the trailing edge is drastically different.

Both solutions are numerical solutions full Navier Stokes Eq.

Cf distribution

0.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0 0.5 1 1.5

x direction

1/2*

Cf(R

ex)^

0.5

Creep Re=100

RESULTS – Reynolds Similarity


For two different fluids water and air, varying viscosity and density, Re maintains the same value.

At Re=10^5, Cf gives the same profile at different x as Cf depends on Re only. This shows that Re drives the problem in incompressible fluid flow.

Cf Air at Re=10^5

0

0.005

0.01

0.015

0.02

0.025

0.03

0 0.5 1x

Cf

Re=10^5


Laminar – Air

With x increasing the value of Cf*(Re)^0.5 becomes independent of x (theoretically)

Flolab results showed a pretty good agreement as far the constant, but the power of Re does not match 0.5

Cf

0.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0 0.5 1

x direction

Cf*

Re^

0.5

flowlab Re=10^5

flowlab Re=5*10^5

theoretical


Turbulent – Water

With x increasing the value of

so Cf*Ln^2(Re) becomes independent of x

Flolab results showed a good agreement as far the constant, but the power of Re does not match Ln^2(Re)

)Re06.0(ln/455.0 2xfC

RESULTS – Nusselt Number

Laminar –Water

With x increasing the value of Nu/(Re)^0.5 becomes independent of x (theoretically for a particular Pr)

With Twall=Te the profile doesn’t obey the Polhaussen Pr1/3 law

3/12/1 PrRe332.0)(/ xeww TTkxqNu


Laminar – Air

for laminar case. Here for a constant Pr as Re increases with x,the nusselt number increases as well as Re, so Nu/Re should be a constant

Flolab results showed a pretty good agreement as far the power of 0.5, but the constant does not match

Nu

0.25

0.29

0.33

0.37

0.41

0.45

0 0.2 0.4 0.6 0.8 1 1.2

x direction

Nu

/(R

e^0.

5*P

r^0.

3)

flowlab Re=10 5̂

theoretical

RESULTS – Reynolds Similarity


Nu varies as Pr1/3.Pr of air is approx=0.72 and Pr for Water=6.So the Nuwater/Nuair=2.02.From the chart it is obvious that the flowlab results do agree with these for a constant Re=10^5

Nu at Re=10^5 with heat transfer, air

0

20

40

60

80

100

120

0 0.5 1 1.5x

Nu

Re=10^5 flowlab


Turbulent –Air/Water

for turbulent case. Here for a constant Pr as Re increases with x, the nusselt number increases as well as Re4/5.

Nu vs Re(with heat transfer)

0500000

10000001500000200000025000003000000350000040000004500000500000055000006000000

0 0.2 0.4 0.6 0.8 1 1.2

x

Nu

10 7̂

5.586e 8̂

2.233e 9̂

Nu Vs Re Temp with heat transfer

0

2000

4000

6000

8000

10000

12000

14000

0 0.2 0.4 0.6 0.8 1 1.2

X

Nu

Re 10 6̂

Re 5*10 6̂

Re 10 7̂

60Pr6.0...........PrRe0296.0 3/15/4 xNu



for turbulent case. Without heat transfer the empirical law is not valid. The flowlab results showed a steep increase of Nu, even when Tw = Te

60Pr6.0...........PrRe0296.0 3/15/4 xNuRe=2.2*10^9,Vinle t=2000m/s,water without heat transfer

0

500000

1000000

1500000

2000000

2500000

3000000

3500000

4000000

4500000

5000000

0 0.2 0.4 0.6 0.8 1 1.2

x

Nu

f low lab

theoretical


Turbulent –Water

With x increasing the value of

so Nu/Re^0.8 becomes independent of x for a Pr

Flolab results showed a good agreement as far the constant, but the power of Re does not match Re^0.8

60Pr6.0...........PrRe0296.0 3/15/4 xNu

RESULTS – Temperature Profiles

Laminar – Water The profiles closely

matches the profile as found by Polhaussen

The temperature is almost constant when Tw=Te

Re=10^5 without heat transfer

299.993

299.994

299.995

299.996

299.997

299.998

299.999

300

300.001

0 0.1 0.2 0.3 0.4 0.5 0.6

y direction

T

inlet x=0.2 x=0.4 x=0.6 x=0.8 x=0.9 outlet

RESULTS – Temperature Profiles


The thickness of the thermal boundary layer in the case of turbulence decreases as compared to the laminar case.

Re=2.233*10^9,Vinlet=2000m/s,T=400-1000

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

395 397 399 401 403 405 407 409

Temp

y

inlet

0.2l

0.4l

0.6l

0.8l

0.9l

outlet

Re= 5*10^6, Vinlet= 67.14, T= 300-400

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

290 300 310 320 330 340 350

Temp

Y

inlet

x=0.2*l

x=0.4*l

x=0.6*l

x=0.8*l

x=0.9*l

outlet

RESULTS – Transition

Comparison of Laminar and Turbulent Model

Turbulent model predicts a much higher Nu and Cf than the laminar counterpart. We believe that at 106 the turbulent model must be the better one to solve the flow problem.

Cf Re=10^6

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0 0.2 0.4 0.6 0.8 1 1.2

x direction

Cf laminar flow

turbulent flow

Nu number Re=10^6

0

200

400

600

800

1000

1200

1400

1600

1800

0 0.2 0.4 0.6 0.8 1 1.2

X direction

Nu lamina flowr

turbulent flow

RESULTS – Change of Length

Change of Length from 1m to 5 m

if the length of the plate and inlet velocity is changed such that Re is constant then Nu and Cf do not change at all as evident from the above graphs.

RESULTS – Velocity Contour

Sample velocity contour as generated by FLOLAB at different X locations

Conclusions

Flowlab shows fairly good agreement in all respects as compared to the theoretical values obtained by the boundary layer theory

However it is not possible to get *, , directly from the flowlab results.

The overshoot in the velocity profiles above the inlet velocity also couldn’t be explained .

Documents

FLOLAB – FLAT PLATE Anoop Samant Yanyan Zhang Saptarshi Basu Andres Chaparro