Chapter 6: Momentum PrincipleChapter 6: Momentum Principle

Byy

Dr Ali JawarnehDr Ali JawarnehDepartment of Mechanical Engineering

Hashemite University

1

OutlineIn this lecture we will:• Derive and analyse the momentum

equation.• Discuss each of the terms in the

equation.q• Illustrate the concept with a number of

solved examples.solved examples.

2

6.1: Momentum Equation:Derivation

• Considering Newton’s 2nd law for a fluid particle:

∑ = aF m

• Or:∑ =

dtmd )( vF

• This can also apply for a system comprised of a group of particles:

∑ =dt

d sys )(MomF

3

• Recalling the general form of the C.V equation:

syscv

cs

dB d b dV b VdAdt dt

= ρ + ρ∫ ∫B=Momsys , b=v

• Substituting in Newton’s 2nd law:

dM∫∫ +=cscv

sys dddtd

dtd

AVvvMom

.V ρρ

∫∫∑ +=cscv

dddtd AVvvF .V ρρ

4

V fl id l it l ti t th CS t th l ti• V: fluid velocity relative to the CS at the location where the flow is crossing the surface

• v: the velocity relative to an inertial frame; that is• v: the velocity relative to an inertial frame; that is a frame which does not rotate and can either be fixed or moving at a constant velocity

• The momentum equation states that:The sum of external forces acting on theThe sum of external forces acting on the material in the CV = the rate of momentum change inside the CV + the net rate at which

t fl t f th CVmomentum flows out of the CV

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6.2: Interpretation of the Momentum Eq.

Force Terms ( )F∑( )• These forces can be either:

– Body forces: (gravity, electrostatic,

∑

Body forces: (gravity, electrostatic, magnetic).

– Surface forces: (pressure, shear, (p , ,supports…etc.).

6

Figure 6.1F i t d ith fl i i ( ) iForces associated with flow in a pipe: (a) pipe schematic, (b) control volume situated inside the pipe and (c) control volume surrounding the pipepipe, and (c) control volume surrounding the pipe.

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Momentum accumulation ∫ dd Vρv∫cv

ddt

Vρv

For steady flow this term reduces to zero.

Momentum Flow• For uniform velocity over a flow section theFor uniform velocity over a flow section, the

following is true:• Or:

).(. ∑∫ = AVvAVv ρρcs

d

∑∫ = vAVv md &ρO

• Outward flow is considered positive, inward fl i id d ti h

∑∫ = vAVv mdcs

.ρ

flow is considered negative, hence:

∑∑∫ d vvAVv &&ρ8

∑∑∫ −=cs

iics

oocs

mmd vvAVv .ρ

Momentum Equationq

• The momentum equation can therefore be written as:

∑∑∫∑ ddF &&V

• The momentum equation can be written in

∑∑∫∑ −+=cs

iics

oocv

mmddtd vvvF &&Vρ

• The momentum equation can be written in the form of scalar equations for each of the Cartesian coordinates (x y z) as:the Cartesian coordinates (x, y, z) as:

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dx- direction:

x x xcv cs

d dV .ddt

= ρ + ρ∑ ∫ ∫F v v V Acv cs

dy-direction:

y y ycv cs

d dV .ddt

= ρ + ρ∑ ∫ ∫F v v V Acv cs

d∑ ∫ ∫z- direction:

z z zcv cs

d dV .ddt

= ρ + ρ∑ ∫ ∫F v v V A

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• Example: find the momentum flow ∫cs

d. AVv ρ

• Solution:x-direction:

cs

)(V)(V

)m(v)m(v)m(v xxx 321

00

321

••

•••

++

=+−+−

θ

y-direction:

)m(cosV.)m(V 3311 00 ++− θ

y

)m(v)m(v)m(v yyy 321321

••

•••

=+−+−

)m(sinV)m(V. 332200••

+−+ θ

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• Example: find the momentum flow ∫cs

d. AVv ρ

• Solution:x-direction:

cs

x-direction:

)m(v)m(v)m(v 321

•••

=+−+−

)m(cosV.)m(V

)m(v)m(v)m(v xxx

3311

321

00

321

••

−+−

++

θ

y-direction:•••

)m(sinV)m(V.

)m(v)m(v)m(v yyy

3322

321

00

321

••

−−+

=+−+−

θ

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6.3: Typical Applications

Nozzle

yp pp

Nozzle

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Fluid jetj

Fluid jet striking a flat vane.

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• Example: Steady, uniform flow at each fsection, incompressible, and neglect weight of

90 0 reducing elbow and water. Determine the forced required to hold the elbow in placeforced required to hold the elbow in place.Given:A =0 01 m2A1=0.01 m2

p1=119 kpaA 0 0025 2A2=0.0025 m2

V2=16 m/sp2=patm

ρ=1000 kg/ m3

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• Solution:x-direction:

d dV d= ρ + ρ∑ ∫ ∫F v v V A0.0

x x xcv cs

dV .ddt

= ρ + ρ∑ ∫ ∫F v v V A

)m(v)m(vApR••

+−=+ 2111

0.0V1

)m(v)m(vApR xxx ++ 2111 21

)mVA(pRx

•

+−= 2111

Continuity equation:x 2111

••••

=→+−=→+= ∫∫ 00 mmmmV dAρVdρd 0.0=→+=→+= ∫∫ 212100 mmmmV.dAρVdρ

dt cscv

m/s.AAVVAVρAVρ 4

010002501622

1222111 =×==→=

16

.A 0101

kg/s.AρVm 4001041000111 =××==•

•

y-direction:

N).()mVA(pRx 135040401010119 32111 −=×+××−=+−=

0 0

∫∫∑ +=cs

ycv

yy dA.VvVdvdtdF ρρ

0.0

)()(••

0.0 -V2

)m(v)m(vR yyy +−= 21 21

NVR 6404016•

NmVRy 640401622 −=×−=−=

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• Example: The water leaves the nozzle at 15 m/s (A =0 01 m2) Assuming steadym/s (Anozzle=0.01 m2). Assuming steady, incompressible, and neglect the weight of jet and the plate, change in elevation is also p , gneglected. Determine the reaction forces on the support.

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• Solution:x-direction: ∫∑ =

csxx dA.VvF ρ

0 0 0.0V1

)m(v)m(v)m(vR xxxx

•••

++−= 3321 21

0.0

kg/s.AρVm 150010151000111 =××==•

kg/s.AρVm 150010151000111 ××

kN.mVRx 2521501511 −=×−=−=•

y-direction: ∫∑ =cs

yy dA.VvF ρ

)()()(R•••

++0.0

V2 -V3

From Bernoulli eq. V1=V2=V3

)m(v)m(v)m(vR yyyy ++−= 3321 21

•••••••

From continuity eq. 19

)mmce(sinmmmmm ==→+= 3221321 2

00322 .)mm(VRy =−=••

• Example: Determine the anchoring force d d h ld h i Th blneeded to hold the vane stationary. The problem

is steady, incompressible, neglect the gravity.Given:Given:V1=10 ft/sA 0 06 ft2A1=0.06 ft2

ρfluid=1.94 slug/ft3

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• Solution:x-direction: ∫∑ =

csxx dA.VvF ρ

)()(R••

V1 V2 cosθ

From Bernoulli: V1=V2

)m(v)m(vR xxx +−= 21 21

1 2

From continuity:••

= 21 mm

secslugs/...AρVm 164106010941111 =××==•

gρ 111

Ibf)θ(.)θ(.)θ(VmRx 1cos64111cos1016411cos11 −=−××=−=•

y-direction: ∫∑ =cs

yy dA.VvF ρ

0 0 V2 sin θ

21)m(v)m(vR yyy

••

+−= 21 21

0.0 2

θsin.Ry 6411=

RELATIVE POSITIONThe absolute position of two particles A and B with respect to the fixed x, y, z reference frame are given by rA and rB.The position of B relative to A is represented by rB = rA + rB/A

or

rB/A = rB – rA

Therefore, if rB = (10 i + 2 j ) mand rA = (4 i + 5 j ) m,then rB/A = (6 i – 3 j ) m.

RELATIVE VELOCITYTo determine the relative velocity of B with respect to A, the time derivative of the relative position equation is taken. p q

vB/A = vB – vAor

vB = vA + vB/A

In these equations, vB and vA are called absolute velocities and vB/A is the relative velocity of B with respect to A (The velocity of B as measured by the A)measured by the A).

Note that vB/A = - vA/B .

EXAMPLE

Given: vA = 600 km/hrvB = 700 km/hrB 00 /

Find: vB/A

SOLUTIONvA = 600 cos 35 i – 600 sin 35 j

= (491.5 i – 344.1 j ) km/hrvB = -700 i km/hr

SO U O

B

vB/A = vB – vA = (- 1191.5 i + 344.1 j ) km/hr

khrkmv

AB2.1240)1.344()5.1191( 22

/=+=

where °== − 1.16)1.344(tan 1θ θwhere )5.1191

(

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6.4: Additional Applicationspp• Moving Control Volumes.

example:r c.vV V V= −r r r

The relative velocity is the

Vr1=25 i-10 i=15 i

In case if the car velocity

yfluid velocity relative to themoving control volume-thefluid velocity seen by anobserver riding along on theIn case if the car velocity

is to the left

V =25 i-(-10 i)=35 i

observer riding along on thecontrol volume.The absolute velocity is thefluid velocity as seen by a

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Vr1=25 i-(-10 i)=35 i y ystationary observer in a fixedcoordinate system.

Example:A jet of air tra eling at 12 m/s is directed at aA jet of air traveling at 12 m/s is directed at a90-degree curved passage in a cart that ismoving at constant speed V =5 m/s Themoving at constant speed Vc 5 m/s. Thecurved passage has an inlet diameter of 5cm and outlet diameter of 1.5 cm. The jetjdiameter is 5 cm, what is the jet velocity(m/s) of air at the outlet of the curved

?passage?

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Solution: r c.vV V V= −r r r

221122211121 AVAVAVAVmm =⇒=⇒=••

ρρ 221122211121 AVAVAVAVmm rrrr ⇒⇒ ρρ21

)()( AVVAVVrr VV 4847648476

222111 )()( AVVAVV cjcj −=−

22 01000)12( ππ 22

2 015.04

05.04

)512( ππrV=−

smVr /78.772 =

jiV 78775 +=28

jiVj 78.7752 +=cvjr VVV −= 22

• Example: Determine the reaction forces f i (W 20 ft/ ) Th j tfor a moving vane (W=20 ft/s). The jet velocity is U=100 ft/s. The problem is t d i ibl l t th itsteady, incompressible, neglect the gravity

effect.

Given:2A1=0.006 ft2

ρfluid=1.94 slug/ft3

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• Solution:V1=U1- W1=100-20=80 ft/sFrom Bernoulli: V1=V2=80 ft/sFrom Continuity: ••

= 21 mmslugs/s...AρVm 93120006080941111 =××==

•

x-direction:)m(v)m(vR xxx

••

+−=− 21 21

V2 cosθV1

y-direction:Ibf.)cos(.θ)cosV(mθ)cosV(VmRx 821451809312011211 =−××=−=−=

••

••0.0 V2 sinθy direction: )m(v)m(vR yyy +−= 21 21

Ibf.sin.θsinVmRy 675245809312022 =××==•

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y 2

IbfRRR yx 5722 =+= o5671 .)RR(tan

x

y == −α

• Force on a Rectangular Sluice Gate

x xcs

1 21 1 2 2 viscous G 1 2

F v V.dA

p A p A F F v ( m ) v (m )• •

= ρ

− − − = − +

∑ ∫

1 21 1 2 2 viscous G 1 2

1 1 2 2

1 1 2 2

p p ( ) ( )p (y / 2), p (y / 2)A y b, A y b= γ = γ= =

F F d d h ld h l iFG= Force needed to hold the sluice gate

b = width of the sluice gate and channel

If we neglect the viscous force then:

2 2y b y b

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1 2G 1 2

y b y bF ( ) Q(v v )2 2

= γ − γ +ρ −

6.5: Moment-of-Momentum Equation( )sysd H

Mdt

=∑Where M: momentum

Hsys: total angular momentum of all mass forming the system

H

Control Volume Eqn.:

sysH r x mv=

dB.sys

cv cs

dB d b dV b V dAdt dt

ρ ρ= +∫ ∫

sys sysB H r x mv= = sysHb r x v

m= =

32

( ) ( ) .sys

cv cs

dH d r x v dV r x v V dAdt dt

ρ ρ= +∫ ∫

( ) ( ) .dM r x v dV r x v V dAdt

ρ ρ= +∑ ∫ ∫cv cs

Where:

r: a position vector that extends from the moment center

V: flow velocity relative to the CS

v: flow velocity relative to the inertial framev: flow velocity relative to the inertial frame

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For uniform flow across a CS:

( ) ( ) ( )idM r x v dV r x m v r x m vρ

• •= + −∑ ∑ ∑∫ ( ) ( ) ( )o io o i i

cs cscv

M r x v dV r x m v r x m vdt

ρ= + −∑ ∑ ∑∫

The momentum of momentum equation has the following physical interpretation:

The sum of moments acting on the material within theThe sum of moments acting on the material within the control volume equals the rates of change of angular momentum within the CV plus the net rate of which angular momentums flows out of the CV.

34

35

36

• Example: water enters a rotating sprinkler (ω=500 rev/min) through its base at the steady rate(ω=500 rev/min) through its base at the steady rate of 1x10-3 m3/s. The exit cross section area of each of the two nozzles is 2.6x10-5 m2. The radius of the sprinkler is 0.2 m. Water density is 1000 kg/m3 and pressure at the two exits is atmospheric. Determine the torsional moment (T ) and po er prod ced bthe torsional moment (Tshaft) and power produced by the sprinkler-like turbine.

37

• Solution: dA.V)vr(Mcs

ρ∫∑ ×=

2222211111 AV)vr(AV)vr()k(Mo ρρ ×+×=−

••

−×−+×=− )]()([)]()([)( mivjrmivjrkM ×+×= 2222111 )]()([)]()([)( mivjrmivjrkMo

))(()()()( 222111222111 kmvrmvrmkvrmkvrkMo −+=−+−=−••••

From contin it

•••••

=+=+= mvrmvrmvrmvrmvrMo 2)( 222111222111

From continuity:

kg/s.xρQmmmmmmmmm oooo 50

21011000

222

3

112121 =×

===→=→=→+=−

•••••••••

V1=V2

goo 222112121

rVvrrrr

×+= ω

)()()()( rVvjrkiViv ωω →×+

v: fluid velocity, V: relative velocity, ωr: velocity of nozzle.where V is flow velocity relative to the

t l f d i fl l it

38

111111 )()()()( rVvjrkiViv ωω −=→×+=

222222 )()()()( rVvjrkiViv ωω −=→−×+−=−

control surface, and v is flow velocity relative to an inertial (nonaccelerating) reference frame.

AVQ

Q

QQQQ 2 121

0

0=+=

smxQVV

AVQ

Q

/2.192/1012

5

31

21

1110

====

==

−

rad/s.)π(minv/Reω 352602500500 ===

xA 106.2 51

21 −

)(60

m/s.)..(.ωrVvv 7482035221921 =×−=−==

N.m....mvr)mvrmv(rTM shafto 7481507482022 111222111 =×××==+==•••

Watt...ωTP shaft 4913527481 =×==

39

6.6: Navier-Stokes EquationsqThey are a differential form equations of momentum based on a control volume of infinitesimal size.

For incompressible and constant viscosity flow:

X-direction2 2 2

2 2 2. xu u u u p u u uu v w gt x y z x x y z

ρ ρ μ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

2 2 2

2 2 2. yv v v v p v v vu v w gt x y z y x y z

ρ ρ μ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

Y-direction

y y x y z∂ ∂ ∂⎝ ⎠ ⎝ ⎠

2 2 2w w w w p w w wu v w gρ ρ μ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + = − + + + +⎜ ⎟⎜ ⎟ ⎜ ⎟

Z-direction

40

2 2 2. zu v w gt x y z z x y z

ρ ρ μ+ + + = − + + + +⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

• Example: Find the velocity distribution for steady, L i Fl B Fi d P ll l PlLaminar Flow Between Fixed Parallel Plates.

S l ti• Solution:continuity:

00wvu=

∂+

∂+

∂0.0 0.0

)()x(fuu 100 ≠→=∂

x-direction:

00.zyx=

∂+

∂+

∂)()x(fu.

x n 100 ≠→=∂

0.0 0.0 0.0 0.0 0.0 0.0 0.0

)zu

yu

xuμ(

xpρg)

zuw

yuv

xuu

tuρ( x 2

2

2

2

2

2

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

2

41)(

yuμ

xp. 200 2

2

∂∂

+∂∂

−=

y-direction:0.0 0.0 0.0 0.0 -g 0.0 0.0 0.0

)zv

yv

xvμ(

ypρg)

zvw

yvv

xvu

tvρ( y 2

2

2

2

2

2

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

∂ )(ypρg. 300∂∂

+=

z direction:z-direction:

)zw

yw

xwμ(

zpρg)

zww

ywv

xwu

twρ( z 2

2

2

2

2

2

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

0.0 0.00.00.00.00.0 0.00.0

zyxzzyxt ∂∂∂∂∂∂∂∂

)(zp. 400∂∂

=z∂

From Eq.(4): )z(fp n≠

F E (3) p∂

42

From Eq.(3): )x(fyρgpρgyp

1+−=→−=∂∂

From Eq.(2):2

212

12

2

2111 cyc)y(

xp

μuc)y(

xp

μdydu

xp

μdyud

++∂∂

=→+∂∂

=→∂∂

=

Boundary Conditions:@y=+h u=0.0 )a(chc)h(

xp

μ. 21

2

2100 ++

∂∂

=

@y=-h u=0.0

xμ2 ∂

)b(chc)h(p. 212100 +−

∂= )b(chc)h(

xμ. 21200 +

∂

00c = )h(pc 21 ∂−=001 .c = )h(

xμc2 2 ∂=

]hy[pu 221−

∂=

43

]hy[xμ

u2 ∂

=