momentum conservation

IIT-Madras, Momentum Transfer: July 2005-Dec 2005

Momentum Conservation Newton’s Second Law

Force = Mass * Acceleration Alternate method: From Reynold’s theorem

Fluid Flow Force = Momentum flux + Momentum Accumulation rate

Vols

VoldVt

dAnVVF )(.

Flux Accumulation Rate


Example

Straight Pipe

A1

V1

A2

V2

Steady State

22221111 VVAVVAFlux 2211 APAPF

Vols

VoldVt

dAnVVF )(.

0

Assumption: No friction


ExamplePressure Drop for various fluids

1.00

10.00

100.00

1000.00

10000.00

100000.00

Acetone Water 80 C Ammonia(30%)

Water Sulfuric acid(98%)

Phosphoricacid (85%)

Pre

ssur

e D

rop

(Pa)

(Lo

g S

cale

)

10" pipe

2.5" pipe

5" pipe

16 times

38 times

Same flow rate, same length of pipe

Flow in a straight pipe. Realistic case

1 2


Example Problem: 5.32

d2,V2

Steady State

22221111 VVAVVAFlux

FrictionAPAPF 2211

Vols

VoldVt

dAnVVF )(.

0

h=58 cm

d1,V1

d1=8cmd2=5cmV1=5m/s

1=1g/cm3

222111 VAVA

hghgPP waterHg 21

1 2


Example

Example 1 (bent pipe), page 47

1,V1

2,V2

Area=A1 Area=A2=A1

xx BCosAPAPF )(2211

yy BWSinAPF )(0 22

Bx

By

Weight of Fluid

x

y


Example

Steady State d/dt =0

From Eqn of Conservation of Mass

11112222 )( AVVAVCosVFluxX

2222 )(0 AVSinVFluxY

mAVAV 222111

out in

out

in


Example

V1

V2

Area=A1 Area=A2=A1

)()( 212211 CosVVmCosAPAPBx

)sin()( 222 VmWSinAPBy

Bx

By

Vols

VoldVt

dAnVVF )(.


Example

P1,V1

P2,V2

F

2211 APAPFForceTotal

22

212

1 AVAVFlux

Pipe with U turn

Use Gage Pressure!

In case of gas, use absolute pressure to calculate density


Example

P1

P2

F

1 1x xTotal Force F PA

21 1xFlux V A

“L” bend

Assume the force by the pipe on the fluid is in the positive direction

2 2y yTotal Force F P A

22 2yFlux V A

E

N

What will the force be, if the flow is reversed (a) in a straight pipe? (b) in a L bend?


Example

Pbm. 5.13 & 5.19

1122 APAPForce VwV2

V1

12

X Y 02211 VVAVVA ww

Conservation of Mass

P2 P1

dt

dyAV

dt

dxAVVoldV

t Vol

111222)(

Stationary CV yAVxAVCVinMomentum 111222

AYAXM 12 dt

dy

dt

dxVw

2222

2111 VAVAFluxMomentum

Vols

VoldVt

dAnVVF )(.


Example

Pbm. 5.19

VwV2

V1

12

X Y

P1P2

Stationary CV

V1=-3 m/s

Vw=Velocity of Sound in water

V2=0 m/s

Pressure difference?

121112 VVVVPP w

~ 41 atm


Example

Pbm. 5.1,5,3V1=6 m/s

V2=?

0.6 m 1.2 m

2.1

0

111

2.1

0

222 dyyVAdyyVA

21 AA 21

depthmNobjectonDrag /800

Vols

VoldVt

dAnVVF )(.

DragAPAPF 2211 dyyVdyyVFlux22.1

0

11

22.1

0

22

x

y


Momentum Conservation

Angular Momentum In a moving system

Torque = Angular Momentum flux + Angular Momentum Accumulation rate

Vols

VoldVrt

dAnVVr )(.



Example Example 4 in book

Vols

VoldVrt

dAnVVr )(.

Find the torque on the shaft In a moving system

Torque = Angular Momentum flux + Angular Momentum Accumulation rate


Example

Vols

VoldVrt

dAnVVr )(.

Approach-1. Find effective Force in X direction Find the moment of Force Assume: No frictional loss, ignore gravity, steady state,

atmospheric pressure everywhere

Vols

VoldVt

dAnVVF )(.

rmovingVolControl @

..MassCons

rVVin 0

rVVout 0

cos0 rVQmomentumdirectionX out

rVQmomentumdirectionX in 0


Example

cos10 rVQF Fluidon

cos10 rVQF Wheelon

cos10 rVQrT wheelon

Approach-2. Using conservation of angular momentum Stationary CV

S

dAnVVr .

Vols

VoldVrt

dAnVVr )(.

0VrQIn

cos0 rVrrQOut


Example

Consider a jet hitting a moving plate

NozV

PlateV

Vnoz water has entered into the CV

Plate has moved by Vplate

In a control volume which moves with the plate, Vnoz-Vplate water has entered the CV (and exited at the bottom)

PlateV PlateV

After 1 second


Example

Platenozzlein VVVREFERENCEMOVING

Platenozzleout VVVREFERENCEMOVING

)cos(PlatenozzleComponentX

out VVVREFERENCEMOVING

)sin(PlatenozzleComponentY

out VVVREFERENCEMOVING

jPlatenozzleiPlatenozzleout VVVVV

REFERENCEMOVING)sin()cos(

jPlatenozzle

iPlatenozzleiplateout

VV

VVVVREFERENCEStationary

)sin(

)cos(


Example

2nozzle

Plate

VV

0

Platenozzleplateout VVVVREFERENCEStationary

Consider plate moving @ half Vnoz, alpha = 180 degrees

1)(,0)( CosSin

jPlatenozzle

iPlatenozzleiplateout

VV

VVVVREFERENCEStationary

)sin(

)cos(

PlateV


Example

Pbm 5.24 Thickness of slit =t, vol flow rate

=Q, dia of pipe=d, density given Ignore gravity effects

Vols

VoldVrt

dAnVVr )(.


3ft

6ft

0In

drrtVdrtVVrOut 9

3

29

0


Example

Pbm 5.31 P1, P2, density, dia, vol

flow rate given

3ft

1

2

Calculate velocity at 1 (=2)

Vols

VoldVrt

dAnVVr )(.

0In

22222 VArArPT


Energy Conservation

Vols

Voldet

dAnVeDt

DE)(.

dt

dWdAnV

P

dt

dW

dt

dQ

Dt

DE

s

).(

HeatMechanicalWorkdone bythe system

Work doneby pressure force

Friction Loss (Viscous)


Energy Conservation

Vols

Voldet

dAnVP

edt

dW

dt

dW

dt

dQ)(.

energyernalu

ghv

u

MassUnitperEnergye

int2

2

ConstghPv

EqnsBernoulli

2

:'2

No Frictional losses Incompressible Steady No heat, work No internal energy change


Example Flow from a tank

Dia = d1

Dia = d2

22

22

11

21

22gh

Pvgh

Pv

Pressure = atm at the top and at the outlet

1

2

3 Velocity at 1 ~ 0

12 2ghV

h1

h3

0

Toricelli’s Law

3322 VAVA Sections 2 and 3

)(2 313 hhgV

12 2ghAQ )(2

)(2

31

123

hhg

hgAA


Example How long does it take to empty the tank? What if you had a pipe all the way upto level 3?

Dia = d1

Dia = d2

1

2

3gh

A

A2

1

2

h1

h3

02

1

21 V

A

AV

dt

dh

inithht ,0

0, htt

Pressure @ section 2 != atm

Pressure @ section 3 = atm


Example What if you had a pipe all the way upto level 3?

Dia = d1

Dia = d2

1

2

3

3233 VAVAQ

h1

h3

0

)(2 313 hhgV

More flow with the pipe

12 2:.. ghAQpipeow

332 ghPP Turbulence, friction Unsteady flow Vortex formation


150 km/h

Example Moving reference; Aircraft

0, 11 VatmP

60 km/h

Height is known

Find P and r (eg from tables)

2 31

Flight as Reference

150, 11 VatmP

0, 22 VunknownP

60, 33 VunknownP

ConstghPv

2

2


Flow through a siphon vs constriction

h1h2

h3

22

22

11

21

22gh

Pvgh

Pv

atmPPPVV 2132 ,

)( 233 hhgPP atm vaporPPif 3

1 2221121 , VAVAhh

1

2

2

12112 A

AVPP vaporPPif 2


Example Pbm. 6.4 Steady flow through pipe , with friction Friction loss head = 10 psi Area, vol flow rate given Find temp increase

Assume no heat transfer

headLossFrictionghPv

ghPv

22

2

11

21

12 uulossFriction

)( 12 TTCv

Co1


Example Pbm. 6.10 Fluid entering from bottom, exiting at radial direction Steady, no friction

P1=10 psig

P2=atmtD2

D1 Find Q, F on the top plate

22

22

11

21

22gh

Pvgh

Pv

h2

2211 AVAV

tDAD

A 22

21

1 ,4

Vols

VoldVt

dAnVVF )(.


Example

P1=10 psigD1

Vols

VoldVt

dAnVVF )(.

11APWeightPlatefromForceF

F

211VAMomentum in

0outMomentum

y

If the velocity distribution just below the top plate is known, then P can be found using Bernoulli’s eqn

..., PlatefromForceHence

A

dAPF


Modifications to Eqn

headLossFrictionghPv

ghPv

22

2

11

21

dst

Vgh

Pvgh

Pv

2

1

2

11

21

22

Unsteady state, for points 1 and 2 along a stream line


Draining of a tank

We can obtain the time it takes to drain a tank (i) Assume no friction in the drain pipe (ii) Assume you know the relationship

between friction and velocity

1 1 2 2V A V A

L

H

R

D

1

2

Le us take that the bottom location is 2 and the top fluid surface is 1

Incompressible fluid


Draining of a tank: Quasi steady state Quasi steady state assumption

Velocity at fluid surface at 1 is very small i.e. R >> D

No friction : L is negligible P1 = P2 = Patm

2 2

2 1

2 2

V Vg H L g H

2 2V g H


Draining of a tank: Quasi steady state

1

( )d HV

dt

2

1

2V

A

dH

t

A

d

2

2

2

1

2

4

D

R

g H

Original level of liquid is at H = H0

Integrating above equation from t=0, H=H0 to t=tfinal, H =0, we can find the efflux time


Draining of a tank: Unsteady state

BSL eg.7.7.1

At any point of time, the kinetic + potential energy of the fluid in tank is converted into kinetic energy of the outgoing fluid

We still neglect friction 2 21

1. .

2VK E R H

42

42

2 1

1

2 6

DR V

RH

Potential Energy of a disk at height z and thickness dz

2. .P E R z g dz 2

0

. .H

Total P E R z g dz


Draining of a tank: Unsteady state

4 2

2 2 22 4

1

2 16 2

d D HR H V R g

dt R

Also, using continuity equation

22

2 2

1

4 2

DV V

2

2 2

4dH RV

dt D

Substituting, you get a 2nd order non linear ODE with two initial conditions. Please refer to BSL for solution


Draining of a tank (accounting for friction)

What if the flow in the tube is laminar and you want to account for friction?

Bernoulli’s eqn is not used (friction present) Continuity

1 1 2 2 3 3V A V A V A

L

H

R

D

1

2

Hagen-Poiseuille’s eqn

43

8

P RQ

L

1

dHV

dt

P g H L


Draining of a tank (accounting for friction)

Substituting and re arranging,

Integrating with limits

21

43

8( ) L Rd H Ldt

H L R g

2

43

1 l 1 )128

n(final

LR HtD Lg

0@ 0,

@ , 0final

t H H

t t H

Note: The answer is given in terms of diameter of tube, so that it is easier to compare with the answer given in the book


Example dst

Vgh

Pvgh

Pv

2

1

2

11

21

22

A1,A2, initial height h1 known A1 >> A2

1

2

Lh1

atmPPP 21

01 V

3

Consider section 3 and 2

23 VV

dt

dV

t

V

Section

2

23

L

dt

dVds

t

V 22

1

Pseudo Steady state ==> Toricelli’s law


Example

22

22 2

dV dt

gh V L

2@ 0, 0t V

2 tanh( 2 )22

V tgh

Lgh

22 2

1 2

dV VL gh

dt

Rearranging and solving, we get

lim tanh( 2 ) 12t

tgh

L

As t increases, the solution approaches the Toricelli’s equation


Appendix:Exampleds

t

Vgh

Pvgh

Pv

2

1

2

11

21

22

Oscillating fluid in a U-tube

atmPPP 21

L3

h1

h2

1

2

321 VVV Let h=h1-h2

)(321

2

1

kConstLhhds dt

dV

t

V s

dt

dxV

2

1 21 hhx

02

2

2

xk

g

dt

xd


Appendix:Example Blood Flow in vessels Minimization of ‘work’

Murray’s Law:

33outin rr

Laminar Flow, negligible friction loss (other than that due to viscous loss in

laminar flow) , steady

Turbulent, pulsating flow

4

8

R

LQp

Assume

4

2

r

QpQWork


Appendix:Example If the ratio of ‘smaller’ to larger capillary is constant

And Metabolic requirement =m= power/volume

inner

outer

rk

r

Work for maintaining blood vessel

Total work

22 21

1m Lr rk

22

4

QW r

r

Optimum radius1

3Q

k

momentum conservation

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