Simulation of Turbulent Flows in Channel with Obstructions. Golovnya B. P. ISMEL, Cherkassy, Ukraine.

Simulation of Turbulent Flows in Channel with Obstructions.

Golovnya B. P.

ISMEL, Cherkassy, Ukraine

Briefly about model of turbulence.

It is supposed, then energy stream from mean flow to turbulence is separated on infinite series of parts by turbulence own. So full turbulent energy is separated on infinite series of parts also. Two major parts were named as "primary" and "secondary" turbulence or "big" and "middle" eddies.

Primary turbulence provides interactions with mean flow, secondary is in coincidence with data by coherent structures.

In mathematical sense it is need to introduce adjusting function in a new way.

,Pfy

k

yf

Dt

Dk0

k

t0

201t

0 CPfCkyy

fDt

D

(1)

(2)

2

t

kFC

,Pfy

k

yf

Dt

Dk1110

1

k

1t10

1

1211011

111t10

1 CPfCkyy

fDt

D

1

21

11t

kFC

P)f1(P 01

Secondary turbulence.

2

t y

UP

Primary turbulence.

(3)

(4)

Properties of model.

It can be demonstrated, that this model nothing loses and nothing adds to exact Reynolds equationsOn the base of this model simulations of all main shear turbulent flows were provided. They are - boundary layer, mixing layer, jet and far wake, pipe and channel flows. The models for simulation of full tensors of turbulent stresses and heat fluxes were designed and the numerical simulations of mixed convection from forced flow till natural convection were performed. The numerical simulations of bypass laminar-turbulent transition on cold plate under high free-stream turbulence were performed also. It must be said, that simulations were started from the blade of а plate, i.e. with alone physically valid initial conditions. Calculations of turbulent energy spectra were performed. Results corresponds to "-5/3" law. Dissipative scales in channel flow, which were found by this model are in wonderful accord with Comte‑Bellot experimental data. And it is not complete list of problems, which can be solved on the base of this approach.

Problem for simulation. Model equations.

In given work series of flow in channel with canopies type obstructions were simulated. Presence of obstructions in channel was modeled by additional term introduction into Navier-Stockes equations

hy 0

hy CU

x

p1

y

U

yy

UV

x

UU

2

t

Here coefficient C depends on obstructions form and concentration, h – obstructions height. Turbulence production by obstructions presence was accounted by introduction of additional system for turbulence in wake for obstructions. This system completely coincides with mentioned above turbulence model by form.

First problem.

0 40 80 120

0

1

2

3

U (m /s)

0 40 80 120 0 40 80 120 160

y (m m )

X=150m m X=480m m X=810m m

Obstructions have a cylindrical form.

Fig.1. Comparison of U calculations in three channel sections with experiments.

H channel=320 mm, h obstr.=74 mm, D obstr.=9 mm, Re=10000, C=1.4

init 2/3

initinitchaninit 2initinitinit Lk ,H65.0L,Tu5.1k ,05.0Tu

Second problem.

Obstructions have a tree form – crown of triangle form is placed on a stem.2H chan=1.68 m, h obstr.=h crown+h stem=(0.08+0.04)m, D crown=0.08 m, D stem=0.009 m, Re=10000, C stem=1, C crown=10C stem.

0.0 0.5 1.0

0.00

0.25

0.50

0.75

Y/H

0.5 1.0 0.5 1.0 0.5 1.0 0.5 1.0

X /h=10 X /h=20 X /h=30 X /h=40 X /h=55

U /U m ax

Fig.2. Results of U calculations

0.00 0.25 0.50

0.00

0.25

0.50

0.75

Y/H

0.25 0.50 0.25 0.50 0.25 0.50 0.25 0.50

X /h=10 X /h=20 X /h=30 X /h=40 X /h=55

u'/U m id

Fig.3. Results of velocity fluctuations calculations.

k44.1u0.07Uwwvv ,U36.0uu 2e

2e

0 0.2 0.4 0.6 0.8 1Y/H

0

0.02

0.04

0.06

0.08

0.1

k/U 2m id

big eddies m iddle eddies wake

X/h=40

Fig.4. Components of turbulent energy.

0 0.2 0.4 0.6 0.8 1y/H

0

0.2

0.4

0.6

0.8

1

U/U m ax

crown resistance=A * stem resistance

A=1A=10A=20A=30A=40A=50

X/h=40

0 0.2 0.4 0.6 0.8 1y/H

0

0.1

0.2

0.3

0.4

0.5

u'/U m id

crown resistance=A * stem resistance

A=1A=10A=20A=30A=40A=50

X/h=40

0 0.2 0.4 0.6 0.8 1y/H

0

40

80

120

160

t/crow n resistance=A * stem resistance

A=1A=10A=20A=30A=40A=50

X/h=40

0 0.2 0.4 0.6 0.8 1y/H

0.8

1

1.2

1.4

1.6

L crown resistance=A * stem resistance

A=1A=10A=20A=30A=40A=50X/h=40

Fig.5. Dependence of all parameters of obstructions resistance at X/h=40.

0.00 0.25 0.50 0.75 1.00Y / H

0.00

0.25

0.50k/ U 2

m id

0.00

0.25

W ithout obstructions

W ith obstructions

X /h=10X/h=20X/h=30X/h=40X/h=55

0.00 0.25 0.50 0.75 1.00Y / H

0.0000

0.0025

0.0050-u'v'/ U 2

mid

0.000

0.025

W ithout obstructions

W ith obstructions

X /h=10X/h=20X/h=30X/h=40X/h=55

Fig.6. Development of all main turbulentparameters in flow with and withoutobstructions.

0.00 0.25 0.50 0.75 1.00Y / H

0 . 0

0 . 5

1 . 0

1 . 5L

0 . 0

0 . 5

1 . 0

W ithout obstructions

W ith obstructions

X /h=10X/h=20X/h=30X/h=40X/h=55

0 20 40 60 80 100 120 140X/H

0.01

0.1

1

0 200 400 600 800 1000 1200 1400X/h

ks i w ithout obstructions

ks i w ith obstructions

u* w ithout obstructions

u* w ith obstructions

u ' m ax w ithout obstructions

u ' m ax w ith obstructions

X

P

U5.0

H2ksi

2mid

Fig.7. Development of channel hydraulic resistance, friction on the wall and velocity fluctuations on the way from channel inlet till stabilization point.

Here - channel hydraulic resistance.

0 0.2 0.4 0.6 0.8 1Y/H

0

0.2

0.4

0.6

0.8

1

U/U m ax

X/h=1400 w ithout obstruct.X /h=1400X/h=500X/h=200X/h=40

0 0.2 0.4 0.6 0.8 1Y/H

0

0.1

0.2

0.3

0.4

0.5

u'/U m id

X/h=1400 w ithout obstruct.X /h=1400X/h=500X/h=200X/h=40

0 0.2 0.4 0.6 0.8 1Y/H

0

0.01

0.02

0.03

0.04

-u 'v '/U 2m id

X/h=1400 w ithout obstruct.X /h=1400X/h=500X/h=200X/h=40

0 0.2 0.4 0.6 0.8 1Y/H

0

0.4

0.8

1.2

1.6

2

L X/h=1400 w ithout obstruct.X /h=1400X/h=500X/h=200X/h=40

Fig.8. Comparison of stable state of flows with and without obstructions.