1

MA3006 Fluid MechanicsSemester-2 (13/14)

Dr. YEO Joon HockN3-02b-576790-5500

E-mail: mjhyeo@ntu.edu.sg

2

Prof Yeo JH1. Momentum and Energy equation (7 hr) 2. Dimensional analysis and similitude (4 hr)

Prof Chan WK3. Internal flows and piping systems (8 hr)4. Principles and applications of fluid machines (7 hr)

Textbook BR Munson, DF Young and TH Okiishi, Fundamentals of Fluid Mechanics, 7th edition, John Wiley & Sons,

3

Or along a streamline.

constant2

1 2 gzVp

constant2

2

zg

V

g

p

RevisionsBernoulli’s equation•Applies to steady, incompressible (density ρ = constant), inviscid, and irrotational flow along a streamline.

where p = pressure, z = elevation and V = velocity

System vs. control volumeA system (sys) is a collection of fluid particles. It moves, flows and interacts with its surroundings by exerting a force.

4

5

A mass of air drawn into the compressor of a jet engine is considered a ‘system’. The air changes its shape, size and temperature as it is compressed and expelled through the outlet.

In the system, we (the observers) follow the fluid and observe its behavior as it moves.

6

A control volume (C.V.) is a volume (not mass) in space through which fluid may or may not flow.

C.V. is similar to the free-body diagram in dynamics.

(a) A fluid flowing through a pipe: (1) entrance, (2) exit

(b) A deflating balloon

7

A control surface (C.S.) is a surface of a control volume.

Sections (1) and (2) in (a) are C.S. with flow passing through. The pipe wall is a C.S. without flow passing through.Case (a) is a fixed C.V. but case (b) is a deforming C.V.

In C.V., we (the observers) are usually stationary and observe the fluid behavior.Governing equations of fluid motion are stated in terms of fluid systems, not control volumes.

8

Continuity: mass flow rate is conserved or

time rate of change of Msys (system mass) = 0,

0sysMDt

Dwhere V

sys

sys ρdM

In terms of C.V. and C.S.,

0)ˆ(.. ..

VC SC

dAnVVdt

9

If the flow is steady, then

&0..

VC

Vdt

0)ˆ(..

SC

dAnV

: sum over C.S.

dA: differential area A on C.S. : outward normal vector : component C.S.⊥

..SC

)ˆ( nV

n̂nV ˆ

V

10

11

5.2 Linear momentum equation

Newton’s 2nd law: )( Vmdt

dF

Fluids: continuity equation

.. ..

)ˆ(VC SC

sys dAnVVdt

MDt

D

Inserting into each term,V

.. ..

)ˆ()(VC SC

sys dAnVVVdVt

VMDt

D

(Pg 211)

12

.. ..

)ˆ(VC SC

dAnVVVdVt

F

If the flow is steady, then properties like ρ & are independent of time in C.V.

V

0..

VC

VdVt

Then, ..

)ˆ(SC

dAnVVF

in out

dAnVVdAnVVF )ˆ()ˆ(

13

If ρ & are constant at each inlet, thenV

inin

VAVdAnVV

)ˆ(

Similarly, at the outlet out

out

VAVdAnVV

)ˆ(

Therefore, inout VAVVAVF

inout VmVmF )()(

where is the mass flow rate through A.

AnVm )ˆ(

14

If there are many inlets and outlets, then

inout VmVmF )()(

Example 5.10 : A jet of water exits a nozzle (area A1) with a uniform speed V1. It strikes a vane and is deflected through an angle θ . Find the force needed to hold the vane stationary.

Ignore gravity.

Flow is steady.

15

Select C.V. and assign axes.

Continuity: mmm inout

16

Bernoulli’s equation: (1)→(2)

2

22

21

21

1 22gz

Vpgz

Vp

Ignore gravity: z1 = z2 p1 = p2 = patm

Outcome: V1 = V2 and A1 = A2

Force to hold vane:

Momentum eqn. in x-dir. inxoutxx VmVmF )()(

12 cos VmVmFAx

17

)cos1(1 VmFAx Momentum eqn. in z-dir.

inzoutzz VmVmF )()(

sinsin 12 VmVmFAz

Example 5.15 : A jet engine is mounted on a thrust stand. Velocity at inlet V1, at outlet V2 inlet area A1, outlet area A2 air density ρ , inlet pressure p1 (abs), outlet pressure p2 (abs). Find thrust Fth.

18

Steady flow:Select C.V. & assign axes.

mmAVm outin 11

Momentum eqn. in x-dir. inxoutxx VmVmF )()(

19

)()()( 122211 VVmAppAppF atmatmth )(

)(

12

212211

VVm

AApApApF atmth

Example 5.12 : Water enters a horizontal 180° pipe bend at speed V1. X-sectional area of bend is uniformly A1 = A2. Internal gauge pressures at (1)& (2) are p1 and p2. Find the force needed to hold the bend in place.Weight of nozzle: Wn

Weight of water in nozzle: Ww

20

Flow is steady.Select C.V. & assign axes. , ,

11AVmin

22 AVmout

As A1 = A2 = A, V1 = V2. Force: Inlet , outlet

Continuity: mmm inout outin AVAV )()(

21

Momentum eqn. in y-dir.

inyoutyy VmVmF )()(

)()( 1121 VmVmApApFF Ayy 121 2)( VmAppFAy

There is no flow in the x and z direction

Momentum eqn. in x-dir.

Momentum eqn. in z-dir.

inxoutxx VmVmF )()( 0AxF

inzoutzz VmVmF )()(

0 nwAz WWF

22

23

Example 5.11 Water at flow rate 0.6 L/s exits a vertical nozzle of mass m = 0.1 kg.

Inlet (1) diameter d1 = 16mm

Outlet (2) diameter d2 = 5mmz1 − z2 = 0.03mVn = 2.84×10−6m3

Find inlet pressure p1 and the force to hold nozzle.

24

25

Steady flow: Q = 0.6 ×10−3m3/s

Select C.V. & assign z −axis.

Continuity: outin mm m/s6.30/ 2112 AAVV

26

Bernoulli’s equation: (1)→(2)

2

22

21

21

1 22gz

Vpgz

Vp

m03.0 ,0 212 zzpp atm

)(2

)(21

21

22

1 zzgVV

p

So, p1 = 463.5 kPa (gauge)

Nozzle weight: Wn = 0.1(9.81) NWater weight: Ww = 1000(9.81)2.84×10−6 N

27

Pressure force at inletF1 = p1A1 = 93.3 NPressure force at outletF2 = p2A2 = 0

Momentum eqn. in z −dir.

N8.77

)()(

)()(

12

21

A

wnAz

inzoutzz

F

VmVm

FFWWFF

VmVmF

28

Example 5.16The width of gate is b. The water depth is H upstream of the gate. Find the force required to hold the gate when h is the waterdepth downstream the gate. Ignore friction.

29

Momentum equation:

inxoutxx VmVmF )()(

Section (1): hydrostatic force

HbAH

hAghF cc ,2

,1

Section (2): hydrostatic force

, ,

2

21

2

hbumHbum

hbh

gF

outin

30

If Rx is the force from the gate onto the water, then

HbuhbuFFR

Hbuhbu

umumVmVm

FFRF

x

inoutinxoutx

xx

21

2221

21

22

12

21

)()(

Continuity:

h

Huu

mm outin

12

31

Bernoulli’s:

Hbuhbh

HuFFR

hH

ghu

ghu

gHu

x2

1

2

121

2

1

22

21

2

22

Continuity equation & linear momentum equation for ‘moving’ C.V. in steady flow

.where CV

SC

SC

VVWdAnWWF

dAnW

)ˆ(

)ˆ(0

..

..

32

Example 5.17Water exits a nozzle (A = 5.6×10−4 m3) with a speed 31 m/s. It strikes a vane moving at speed 7 m/s and is turned through angle θ = 45°.Find the force exerted by the water on the vane. Ignore gravity. Ww = 2N

33

Steady flow: attach C.V. on the vane so

(1):

(2):

Force on C.V. from vane:Mass flow rate: AWm 11

34

Continuity:

..

12

122

)ˆ(

24

SC

dAnWWF

WW

mAWm

m/s

Assume uniform at inlet/outlet.)ˆ( nW

N

N

230sin

sin

)()(

95)cos1(

cos

)()(

11

22

11

1122

WmWR

WmWR

WmWmWR

WmR

WmWmR

WmWmR

wz

wz

inzoutzwz

x

x

inxoutxx

dir.z

dir.x

35

36

Angular momentum equation (Sect. 5.2)

....

)ˆ(SCVC

sys dAnVVdt

MDt

D

V

Continuity into Insert

....

....

)ˆ(

)ˆ()(

SCVC

SCVC

sys

dAnVVVdVt

F

dAnVVVdVt

VMDt

D

momentum linear get to

37

Vr

dAnVVrVdVrt

Fr

r

SCVC

velocity having particle fluidof vector position :

: momentum angular get to equation momentum linear into Inserting

....

)ˆ()()(

38

in

outSC

AnVVr

AnVVrdAnVVr

nVVr

)ˆ()(

)ˆ()()ˆ()(

)ˆ()(

..

then lets, inlets/out at uniform is If

inout

VrmVrmFr )()(

Or,

. to equal is

torque shaft the then involved, is shaft aIf

Fr

T

39

Example 5.7 Water enters a sprinkler from its base and leaves through 2 horizontal nozzles in the tangential direction. (1) inlet, (2) outlet

40

Find the speed of water leaving the nozzle (relative to nozzle) if (a) the arms are stationary, (b) each arm rotates at 600 rpm.

Note the direction of ω.

41

Rotating arms: C.V. (dashed line) attached to arms

..22

2

2

:

:

VCVWV

V

W

ground fixed to relative fluid of velocity

C.V. to relative fluid of velocity

42

..

....

....

0)ˆ(

0)ˆ(

VC

SCVC

SCVC

VVW

dAnWVdt

dAnVVdt

where

to changed is

Continuity

! of tindependen is m/s . , Here

:flow Steady

7.16)2/(

)2(

00)ˆ(

0/

222

2221

12

..

WAQW

WAmQm

m mdAnW

t

SC

43

Example 5.18 For the setting in Example 5.7, find (a) torque to hold sprinkle stationary (i.e. ω = 0), (b) torque when the sprinkle rotates at 500 rpm,(c) angular speed of sprinkle if the torque is zero.

(a) Stationary sprinkle:select a stationary C.V. (i.e. disk) and z−axis.

44

/sm36101000

)()(

ˆ

Q

VrmVrm

FrkT

inout

shaft

At the inlet (1), V1 = Q/A1. But A1 = ?

Velocity vector coincides with the z−axis. Therefore, r1, which is the distance between and z−axis, is zero, i.e. r1 = 0.

1V

1V

45

m/smm

(sec.2), outlet the At

. So,

2 67.16 ,30

2

0)(

22

22

11

VA

A

QV

VrmVrmin

46

(b) Rotating arms: select C.V. attached to arms

47

From Example 5.7,W2 = Q/(2A2) = 16.7m/s

Consider the direction of angular speed ω.

With its direction shown, the C.V. moves at

where U2 = r2ω is the speed of each arm-tip.

48

49

(c) Consider Tshaft = 0.

it is concluded that V2 = 0 because , r2 ≠ 0.

Therefore, W2 = V2−(−r2ω)

is reduced toW2 = 0−(−r2ω)

or

0m

rpm 7972

2 r

W

50

51

Energy equation:e: total energy per unit

in fluid particle : internal energy ( ) per unit

( : heat capacity, T: temperature)V2/2: kinetic energy per unit gz: potential energy ( ) per unit

It is actually “power” (instead) of energy that it dealt with here as

gzV

ue 2

2m

u Tcm p m

pc2/2Vm m

gzm m

.dtdmdAnVm /)ˆ(

52

to get momentum equation

....

)ˆ()(SCVC

sys dAnVVVdVt

VMDt

D

Insert e into continuity

0)ˆ(....

SCVC

dAnVVdt

VdDt

D

to get energy equation

Insert into continuityV

0)ˆ()(....

SCVC

sys dAnVVdt

MDt

D

53

: rate of heat transfer in/out of system : rate of work done (or transfer) into/out

of system

and are associated with the ‘+’ sign as they are into the system. and are associated with the ‘−’ sign as they are out of the system.

outinQ /

outinW /

inW

outW

inQ

outQ

sysoutinsysoutin

SCVCsys

WWQQ

dAnVeVdet

VdeDt

D

....

)ˆ(Equ 5.59Page 235

54

A common example of work done (or transfer) into the system through the rotating shaft in a pump is . )(shaftinW

: shaft torqueω : angular speed of shaft

shaftTshaftshaftin TW )(

55

An example of work transfer out of the system through the rotating shaft in a turbine is . shaftoutshaft TW

56

Another important example of is due to pressure:

inW

CS

pin dAnVpW )ˆ)(()(

Note that multiplying the force (−p)dA by velocity gives the power.

In MA3006, fluid flow is usually adiabatic i.e NO heat transfer, .

)ˆ( nV

sysoutsysin QQ 0

57

For Steady, adiabatic flow with , the energy equation is

)()( shaftinpinin WWW

sysoutinsysoutin

SCVC

WWQQ

dAnVeVdet

....

)ˆ(

is reduced to

outshaftin

CSSC

WWdAnVpdAnVe )(

..

)ˆ)(()ˆ(

Recall: ,gzV

ue 2

2

58

CSSC

dAnVpdAnVe )ˆ)(()ˆ(..

gz

Vpu

2

2

Let be uniform at inlets/outlets.

inin

outout

SC

dAnVgzVp

u

dAnVgzVp

u

dAnVgzVp

u

)ˆ(2

)ˆ(2

)ˆ(2

2

2

..

2

..

2

)ˆ(2SC

dAnVgzVp

u

59

out

out

in

in

mdAnVmdAnV

)ˆ()ˆ( ,

outshaftin

inout

WW

gzVp

umgzVp

um

)(

22

22

)(

22

22

shaftin

inout

W

gzVp

umgzVp

um

Energy Equation

60

Example 5.20Pump delivers water at Q = 0.02m3/s. Find pump power , when z1 = z2, D1 = 0.09m, D2 = 0.025 m, p1 = 124kPa, p2 = 414kPa, , ρ = 1000kg/m3

)(shaftinW

J/kg27912 uu

61)(

22

22

11

22

7.40/

1.3/

200

shaftin

inout

out

W

gzVp

umgzVp

um

AQV

AQV

QmW

m/sm/s

kg/s ,

outshaftin

inout

WW

gzVp

umgzVp

um

)(

22

22

62

)(

21

2212

12

27882

2

shaftinWLHS

VVppuumLHS

W

)(

22

22

shaftin

inout

W

gzVp

umgzVp

um

Energy Equation

63

Or

m

Wgz

Vp

uugzVp

shaftin

in

inout

out

)(2

2

2

2

: loss of energy per unit mass flow rate to overcome friction in

fluid flowHead loss:

Energy loss:

inout uu

guuh inoutL /)(

)( inoutL uumghm

m

64

outin

L

L

outin

shaftin

zg

V

g

pz

g

V

g

p

h

hzg

V

g

pz

g

V

g

p

W

22

0

22

0

22

22

)(

then ,If

then ,If

Bernoulli’s equation

65

66

Example 5.24A fan inputs 0.4 kW of power to the blades to produce an air stream of diameter D2 = 0.6m at 12m/s. Find hL and η efficiency of fan.

ρair = 1.23kg/m3

67

Using energy equation

m

WW

gzVp

uugzVp

outshaftin

in

inout

out

)(

22

2)(

2

kg/s

,

,

m/s ,

17.44

0

0

12/)(

22

222

121

21)(

2

DVAVm

Vppp

zzW

Vguuh

atm

shaftout

inoutL

68

%7575.0

44.2

17.4

)1000(4.081.9

2

)12(

4.0

)(

)(

2

)(

shaftin

Lshaftin

input

outputL

L

shaftin

W

ghmW

W

W ηh

h

W

m,

,

kW

25% of input power is lost to overcome friction in air flow or dissipated as heat.

69

Example 5.25A pump adds 7.5kW to deliver water from (B) to (A) with head loss hL = 4.6m. Find Q and power loss. Let zA − zB = 9.1m.

70

Using energy equation:

m

VVppp

hW

WW

guuhm

WW

gzVp

uugzVp

atmBA

Lshaftout

pumpshaftin

inoutLoutshaftin

in

inout

out

7500)6.41.9(81.9

00

6.40

5.7

/)(

2)(

2

21

)(

)(

)(

22

, ,

m. ,

kW

where

71

Power loss

Steam enters a turbine at speed 30 m/s with and leaves the turbine at speed 60 m/s with .

kJ/kg 3348)/( pu

kJ/kg2550)/( pu

mW shaftout /)(

For adiabatic flow with z1 = z2, find .

Example 5.22

72

Using energy equation

m

WW

gzVp

uugzVp

outshaftin

in

inout

out

)(

22

2)(

2

kJ/kg

kJ/kg

& With

35.12

3060

22

79833482550

0

2222

21)(

inout

inout

shaftin

VV

pu

pu

zzW

73

kJ/kg 65.79635.1798

22

)(

22)(

m

W

VVpu

pu

m

W

shaftout

inoutinout

shaftout

Example: Steady adiabatic flow through the horizontal connection without & .Find the power lost in this horizontal connection.ρ = 1000 kg/m3

)(shaftinW

outW

74

Continuity:

m/s4.162

332211

321

V

AVAVAV

mmm

Energy equation:

02

22

1

2

1

3

2

3

2

2

2

Vpum

Vpum

Vpum

(2 outlets & 1 inlet)

75

Power lost is

MW 16.1

222

)()()(

3

2

3

2

2

2

1

2

1

113322

Vpm

Vpm

Vpm

umumum

umumghm ininoutoutL

76

77

Flow rate meters (Sec. 8.6, Munson)

Various flow-measuring devices are frequently used to determine the flow rate Q in a pipe experimentally.1. Orifice meter :

Page 457

78

It is a plate with a circular hole of diameter D2 fitted in a pipe of diameter D1 > D2. As the flow is partially blocked by the plate, a pressure drop (p1 − p2 > 0) across the plate is expected. Two pressure taps are drilled upstream and downstream of the plate to measure the pressure drop (Δp = p1 − p2).

2. Nozzle meter: D → d

79

3. Venturi meter : D → d

Generally, a decrease in cross-sectional area A in apipe causes an increase in velocity V because Q=VA.And increasing V leads to a drop in pressure p at a given elevation z, according to Bernoulli’s equation:

constant2

2

zg

V

g

p

80

Consider an incompressible fluid of density ρ flowing through a horizontal pipe of diameter D inserted with a restriction of diameter d.

Let z1 = z2. If the loss of energy hL = 0, then

g

V

g

p

g

V

g

p

22

222

211

81

Cancelling gravity g and rearranging terms,

212

2

212

2

21

pp

V

VV

Continuity (V1A1 = V2A2):44

2

2

2

1

dV

DV

421

2

21422

2

2

1

1

)(2

)(21

ppV

ppV

V

V

D

d

Or,

And, . then , If

82

Define a theoretical or ideal flow rate as Qideal = A2V2.

From V2 derived, 421

2 1

)(2

pp

AQideal

In reality where ,0Lh

Lhg

V

g

p

g

V

g

p

22

222

211

with z1 = z2. In the absence of an accurate ‘theoretical’ formula for hL, the actual flow rate is taken to be Q = CnQideal

83

Cn : discharge coefficient (subscript n for nozzle) found from many experimental measurements

Cn = f (β ,Re) is plotted as a function of β & Re = ρDV /μ where V = V1, and μ is the dynamic viscosity of fluid.

84

Re is the Reynolds number which indicates if the pipe flow (either laminar or turbulent).

Procedure of finding Q when d, D, ρ, μ, Cn are given.1. Measure p1 − p2 = 71.5Pa2. Calculate β = d / D = 0.625,

A1 = πD2/4 = 0.005m3,A2 = πd2/4 = 0.002m3 and

3m022.0

1

)(2421

2

pp

AQideal

85

For an orifice meter of diameter d fitted in a pipe of diameter D > d, Co (subscript ‘o’ for nozzle) is the discharge coefficient in Q = CoQideal.

86

For a Venturi meter of diameter d fitted in a pipe of diameter D > d, Cv (subscript ‘v’ for Venturi) is the discharge coefficient in Q = CvQideal.

87

The orifice meter, although not as accurate as thenozzle meter and the Venturi meter, is widely used because of its simplicity in design and cost involved. Standards are available for values of Co, Cn and Cv for various specific geometric configurations.Example 8.15Alcohol flows through a pipe D = 0.06m at Q = 0.003m3/s. If p1 − p2 = 4kPa across the nozzle is measured, find nozzle d.Given: ρ = 789kg/m3. μ = 1.19×10−3 Ns/m2

88

4

23

421

2

2

221

1

11

21

1

1102.1

)1(

)(2

44

10?

42200Re06.14

dC

ppACQCQ

dApp

dD

d

DV

A

QV

DA

n

nidealn

kPa

, ,

m/s ,

89

Solve this equation for d with the Cn graph.

577.0/

0346.0 102.1

11123

4

Dd

dd

Cn

m

. and Assume

90

568.006.0/0341.0 0341.0

1102.1

577.0972.0

972.0

577.042200Re

4

23

m

. , Letgraph) (from

, & With

d

dC

C

C

n

n

n

With Re = 42200 & β = 0.568, estimate Cn ≈ 0.972 from graph. As this value is near the previous one, the calculation is stopped. Otherwise, repeat the calculation with d = 0.0341m until Cn does not change.

91

In practice, iterative calculations are carried out in a computer, if the lines in the graph are expressed as a function of β and Re.

.

over meters Venturi for74

15.115.42

1.4

10Re107 ,44.03.0

Re

1060033.000175.0

2262.099.0

nC