Aula Prática 6&7

Balanços integrais

83

85

423

2

41

23

22

3

0

00

03

420

0

3

0

0

UQ

UUQ

yyUdy

yy

UQ

UQ

QQQ

up

S

S

E

upSE

00

000

0

2 QVFFVVQ

VVVV

E

ES

Resolution

dVolgxpdAnunuudVu

t iisurface

iivc

i

..

dAnuubFsurface

i.

Exit Surface

LUbL

yLyyUb

dyLyU

LyUUbdy

LyUUbdAnuub

L

S

LL

surfaceiS

37

22

322

222

22

222.

2

0

322

2

0

22222

0

Momentum flux at the entrance• The fluid leaving through the area of lower momentum entered through an

surface shorter than 2L. Using the mass conservation one gets:

LL

LUbULb

LUbLUbL

yyUb

dyLyUUb

dAu

E

E

L

m

L

m

surfaceim

23

23

23

23

22

222

222

0

2

0

• The Momentum entering is:

LUbE 232

And the force and Cd

LUbLUbLUbLUbF 2222

31

23

67

23

37

22

313

1

2

2

2 LUb

LUb

LUbFCD

• Let us consider a large number of plates, as in a real turbine.

Resolution

• Maximum power is obtained when the derivative in relation to the angular velocity is null:

4

2DVQ

RVV

RVVVV

j

jrS

jrE

rErS

• In this case there is movement of the plate. The linear speed is:RVc

• The relative velocity has the same modulus and opposite sign. The discharge is the jet discharge, if we assume many plates.

RVQF

FRVRVQ

jj

jjj

2

RRVQPot jj 2

• In this case

RV

RRV

RQRRRVQd

dPot

RRVQPot

j

j

jjj

jj

2

022

2

0

RRVRVabsolute

RRVV

rSS

jrS

• Whole jet the kinetic energy would be extracted. Efficiency would be 100%.

• Mass balance• Momentum Balance• Energy Balance

Resolution

21

222

221

221

2

2

1

2

112

21

121

21

21

21

AAUUUP

UPUP

FPAQUQUQUAUA

Resolution

• Bernoulli

1212

22

22

21

2

2112

2121

hUhUhhgF

ghP

UUhQUFPAPAQUQU

1212

22 2

121 ghUghU

• Mass conservation

1112 hUhU

2

1

2

212

1

2

1

2222

22

1

2

21

21

hh

hhgU

ghhhUghU

Resolution considering

2

1

2

21

1

22

22

21

2

1

21

222

22

22

21

1212

22

22

21

1

2121

21

21

hh

hhghhhhhgF

hhhUhUhhgF

hUhUhhgF

21 UU

22122

21 2

21 hhhghhgF

This equation states that the force must balance the hydrostatic pressure plus the momentum exiting from under the sluice gate. The force tends to zero when h2 goes to zero. If the momentum entering into the system was relevant one had to deduce it from the force:

Description of the problem

• Vj is the jet velocity relative to the boat and Vin is the inlet velocity also relative to the boat. Qj is the jet discharge (equal to the inlet discharge).

• The boat “sees” the water entering into the control volume at its own velocity and leaving it at the relative velocity (Vj).

VVjVinV

k

VQkQQV

VQVQkV

kVVVQF

jjjj

jjj

jj

2

4

02

2

2

The boat velocity depends on the jet velocity, on the discharge and on friction. If there was no friction there would be no force and the maximum boat speed would be the jet speed.

• If we had considered a control volume without the green area, we should consider Vin instead of V, but in that case we should have computed the pressure force in front of the boat, computed as:

VVjVin

ininj FVVQ

• This is the force applied over the fluid inside the green volume. The pressure reduction at the interface between both is the symmetrical of that force.

• This force must appear in the application to the blue control volume:

VVQVVVVQF

VVQFVVQ

FFVVQ

jjininjj

injinjj

ininjj

)(

• Which is the same result as before.

Resolution

0

21

21

21

1

222

211

22

222

211

a

a

pghpDVDV

VpVpVp

2

22

21

1

2

22

212

121

2

22

212

122

222

211

22

222

21

1

2

21

21

21

21

21

DD

ghV

DDVghV

DDVVDVDV

VpVpVghp aa

Resolution

apppVVV

pgHVp

pgLVp

12

32

1222

1233

2121

gHV

gHV

2

021

2

22

LHggLgHp

gLVp

pgLVp

22121

021

3

233

1233