SEH NE U Topic 05

409.319A

Systems Engineering HydrodynamicsDepartment of Nuclear Engineering, Seoul National University

A Q i k L kA Quick Look

Reynolds’ Transport TheoremLinear MomentumLinear Momentum

Newton’s law of motion The linear momentum theorem

Applications of the Linear Momentum Theorem Propulsion, wind turbines, wakes/jets, turbine blades, free surface flow

Angular MomentumAngular Momentum Newton’s law of angular momentum The angular momentum theorem

Applications of the Angular Momentum Theorem Centrifugal pumps/compressors, axial flow turbines/compressors

Momentum Conservation

409.319A


R ld ’ T ThConsider a material volume m containing the same fluid as it

Reynolds’ Transport Theorem

moves and deforms following the motion of a fluid at three successive times t1, t2 and t3.

m mm

t1 t2 t3

Consider a fluid scalar property b/unit mass which might be a thermodynamic t h i t l th l it t fproperty such as internal energy or enthalpy per unit mass or components u,v,w of

velocity . The material volume m, unlike the control volume , moves with the fluid

V


enclosing the same fluid particles.

409.319A


Li M

Reynolds’ transport theorem

Linear Momentum

Reynolds transport theorem

dSnVbdbddB

S

dSnVbdbdtdt

T k hTake , thenVb

dMd

S

dS)nV(VbdVdtd

dtMd


409.319A


E lExample - Liquid Flow from the Tube

dSnVddtdmass

S

Atube

dtdLV

dt

L(t)

momentum

2

S

Ld

dS)nV(bVdVdtd

dtdM

V

2

2

dtLdAL


409.319A


N ’ L f M i (1)Apply Newton’s law of motion to the material volume that

Newton’s Law of Motion (1)Apply Newton s law of motion to the material volume that

coincides with our CV at a particular time Determine the forces that act on the fluid within the CV as

p

SFdS)np(

g

S

Fdg

FdS

gg

Then obtain the linear momentum theorem as

dS)nV(VdVdtd

S


dgdSdS)np(

SS

409.319A


N ’ L f M i (2) When the inflow and outflow streams are identified, the momentum equation takes

Newton’s Law of Motion (2)

the form

dgdSdS)np()Vm()Vm(dVdit

dgdSdS)np()Vm()Vm(dVdt SSinout

So far we assumed that the control volume contained only the fluid of interestSo far we assumed that the control volume contained only the fluid of interest. However, it may simplify the analysis to select a fixed control volume that encloses solid objects, then add the external force to the momentum equation as

inout )Vm()Vm(dVdtd

exSS

FdgdSdS)np(dt


409.319A


E lPatm

Example - Fire Hose Nozzle (1)

PatmFex

steady

Q=150 gpm

ρD=1 inch

inout

FddSdS)(

)Vm()Vm(dVdtd

exSS

FdgdSdS)np(

2

2Q4F


2ex DF

409.319A


E lmomentumHorizontal

Example - Fire Hose Nozzle (2)

Patm

.B.NFA)pp()Vm()Vm( cinoutininout

DdInviscid Flow

Fc Pin

equations'Bernoulli

)AA(pApAp inatmatmatmatminin

Pout = Patm

2Vp

2Vp 2

outout2inin

mAVAV

mass

outoutinin

1

AA

AA

21QVF

in

out

out

inoutc


409.319A


A li i f h Li M Th (1)Rocket Engine Compressible Flow of combustion gases

Application of the Linear Momentum Theorem (1)g

FFuel burns at m = 2kg/s

.Ve =200m/s

Far enough down stream so that the plume pressure =atmosphere pressureatmosphere pressure

)Vm()Vm(dVdtd

inout

mkg

FdgdSdS)np( exSS


N400sm200

skg2VmF e

409.319A


A li i f h Li M Th (2) Jet Engine

Application of the Linear Momentum Theorem (2)g

Vf= 250 m/s

Ain FVe

= 500m/s= 250 m/s F

mf

.

3

ff

s/m500Vm/kg40

s/kg2ms/m250V

2in

ea

m0.1A

s/m500Vm/kg4.0

FddSdS)(

)Vm()Vm(dVdtd

inout

s/m250s/kg100s/m500s/kg102VmVmF

FdgdSdS)np(

finout

exSS

Momentum ConservationN106.2

s/m250s/kg100s/m500s/kg102VmVmF4

fineout

409.319A


A li i f h Li M Th (3) Propellers Dp

Application of the Linear Momentum Theorem (3)p

Patm PoutPin

Vp (propeller)

Patm

VwVf

p (p p )F

CV1(Wake)(Flight)

1C

CV2

A(1) (2)ppoutin mAV)VA()VA(

mass

pppoutin

)Vm()Vm(dVdmomentum

)()(

exSS

inout

FdgdSdSnp

)Vm()Vm(dVdt

Momentum Conservationpinout A)pp(F

409.319A


A li i f h Li M Th (4)

i'B lli2C

Application of the Linear Momentum Theorem (4)

2Vp

2Vp,

2Vp

2Vp

equations'Bernoulli2

wa2

pout2

pin2

fa

F000VmVm0

equationMomentum2222

F000VmVm0 fw

p )())((A

Force

fwppfwpwp

FVPFV)(AVPPower

)VV(VA)VV)(VV(2

F

ff

fvpinoutppp

V2FVPefficiencyopulsivePr

FVP,FV)pp(AVP

Momentum Conservationfw

f

p

f

p

vprop VV

V2FVFV

PP

409.319A


A li i f h Li M Th (5)Wake Vf


F

fVf

VFAw

Vw

Vf

x

inout )Vm()Vm(dVdtd

exSS

FdgdSdS)np(

)VA(/F41VV 2

Momentum Conservation2

)VA(/F41VVV fwwff

w

409.319A


A li i f h Li M Th (6) Jet

Application of the Linear Momentum Theorem (6) Jet

Jet Source

Entrainment

As

Aj VjVs

s

inout )Vm()Vm(dVdtd

exSS

FdgdSdS)np(

A

Momentum Conservationjj

sssj A

AVV

409.319A


A li i f h Li M Th (7) Flow through a sluice gate i

Application of the Linear Momentum Theorem (7) Flow through a sluice gate

Paair

F gUpstream

hout PaVin

hinwater2

in

h

0 a hg21dz)pp(in

Vout

a

inout )Vm()Vm(dVdtd

)W/F(widthperForce

exSS

FdgdSdS)np(dt

M

222

outoutinin

h1F

)Wh(V)Wh(VMass


in

out2outout

2out

2in h

h1Vh)hh(g21

WF

409.319A


A li i f h Li M Th (8)Bernoulli’s equation


gh2

Vpgh2

Vpout

2outa

in

2ina

dhhhg2V

outin

2in2

out

hhgh2

hh1h)hh(g

21

WF

and2inout

out2out

2in

)hh(hh4)hh)(hh()hh(2

ghhh2W

outinoutinoutin2out

2in

outin

outinin

)hh(2)hh(g

outin

3outin

outin


409.319A


A li i f h Li M Th (9)Hydraulic Jump

Application of the Linear Momentum Theorem (9)y p

hh1Vh)hh(

2g0

MomentumMass

out2outout

2out

2in

subcriticalsupercritical

gair

F = 0 11hh

hh

21

hV

h)(

2

inin2out

inoutoutoutin

p

V

hin water Vout hout

11hh

hh

21

ghV

hh2gh

outout2in

outoutout

Vin

gatesluicethebeneathflowtheForjumpichydrodynamacrosshh

hh2gh

inout

ininin

11)h/h(

)h/h(2ghV

gatesluicethebeneathflowtheFor

outin

2outin

out

2out


)(g outinout

409.319A


A l M ThNewton’s Law of Angular Momentum

Angular Momentum Theoremg

Multiply the linear momentum by R

FR)V(dR

FR)VmR(dor

FR)Vm(dt

R

FR)VmR(dt

or

Define the angular momentum as

d)VR(H

A l R ld ’ hApply Reynolds’ transport theorem as

dS)nV)(VR(d)VR(dHd


S

dS)nV)(VR(d)VR(dtdt

409.319A


E lKnown for the inflow at R1 andVDetermine)a(

Example - Steady Swirling Inward Flow (1)

R1

11 cosV 11 sinV

11

1

RtoRfromstreamlineaalongppDetermine)b(

andVDetermine)a(

.

1

R

V1

sinV

cosV r

1

i)sinV(i)cosV(V

havingflowInward

R

V

equationmomentumAngular

)1(d)gR(dS)R(dS))np(R(

dS)nV)(VR(d)VR(dtd

S


)1(d)gR(dS)R(dS))np(R(SS

409.319A


E lequationonconservatiMass

Example - Steady Swirling Inward Flow (2)

)2(0dS)nV(ddtd

S

sinVRsinRV)2(sincosVRsincosVR)1(

111

112

121

22

VRV 11

1

VpVpequations'Bernoulli

VR

V

22

1

RV

2Vp

2Vp

22

11


1RR

2Vpp 11

1

409.319A


E lMoment of external forces

Example - Sprinkler (1)

inout

Td)gR(dS)R(dS))np(R(

)VmR()VmR(d)VR(dtd

exSS


Angular speed of sprinkler and velocity V of fluid stream relative to ground

V

α

Vθ =Vcosα

T i)RcosV(i)sinV(V r

)VR()VR(d)VR(d

RAΩ R

exSS

inout


)VmR()VmR(d)VR(dtd

Ω

VAR2Tcos

RV

2


V VAR2R

409.319A


E l Convert the sprinkler to a reaction turbine by attaching an

Example - Sprinkler (2)

electric generator that applies a restraining torque to the rotor and absorbs an amount of power.

Pfor)b(

turbineofP)a(maxout

out

out )RcosV(RmTP

)RcosV(RmT)a(

vesselpressuretheinfluidtheofp)c( s

2

outmaxout

out

0d

dPwhenoccursP)b(

)cosV(

2maxout

equations'Bernoulli)c(4

)cosV(mP

2a

2in

2in

2)R(p

2Vp

2V

q)(


22as )R(V

2pp

409.319A


A li i f h A l M ThAxial Flow Turbines / Compressors

Application of the Angular Momentum Theoremp

(1)

Stator V2zV2

2zz

equations'Bernoulli)R(miRmiPpowerThe

(2)

Stator

izΩ R

2z

~ 222

21

2221

V)R(Vpp2

V2

Vpp:stator

(3)

rotor V3

Ω R

V3 z2332

equationonconservatiMass2

V2

)R(2

Vpp:rotor

231

3z21

)R(pp

VVV

Angular momentum equation

inout )VmR()VmR(d)VR(dtd


exSS


SEH NE U Topic 05

Documents

momentum equation

seoul national universityn

fluid particles

fluid of interestso

seoul national universitya

fixed control volume

newton s law of motion

thermodynamic t h i