3.1 Systems ( 体系 ) versus Control Volumes ( 控制体 )

System ： an arbitrary quantity of mass of fixed identity. Everything external to this system is denoted by the term surroundings, and the system is separated from its surroundings by it‘s boundaries through which no mass across. (Lagrange 拉格朗日 )

Chapter 3 Integral Relations （积分关系式）for a Control Volume in One-dimensional Steady Flows

Control Volume (CV): In the neighborhood of our product the fluid forms the environment whose effect on our product we wish to know. This specific region is called control volume, with open boundaries through which mass, momentum and energy are allowed to across. (Euler 欧拉 )

Fixed CV, moving CV, deforming CV

3.2 Basic Physical Laws of Fluid Mechanics

All the laws of mechanics are written for a system, which state what happens when there is an interaction between the system and it’s surroundings.

If m is the mass of the system

Conservation of mass( 质量守恒 )

0dm

m const ordt

Newton’s second law

F ma dV

mdt

( )d

mVdt

Angular momentum dH

Mdt

( )H r V m

First law of thermodynamic

dQ dW dE

dt dt dt

It is rare that we wish to follow the ultimate path

of a specific particle of fluid. Instead it is likely that the fluid forms the environment whose effect on our product we wish to know, such as how an airplane is affected by the surrounding air, how a ship is affected by the surrounding water. This requires that the basic laws be rewritten to apply to a specific region in the neighbored of our product namely a control volume ( CV). The boundary of the CV is called control surface(CS)

Basic Laws for system for CV

3.3 The Reynolds Transport Theorem (RTT)雷诺输运定理

1122 is CV .

1*1*2*2* is system which occupies the CV at instant t.

d

dm :The amount of per unit mass

CVcv

d cvdm

The total amount of in the CV is :

t+dtt+dt

tt

s

: any property of fluid

( , , , )m mV H E

1[ ( ) ( )]CV CVt dt t

dt

1 1[ ( ) ( ) ( ) ] ( )out in ss t dt d d t

dt dt

1 1[ ( ) ( )] [( ) ( ) ]s out ins t dt t d d

dt dt

1[( ) ( ) ]

sout in

dd d

dt dt

1[( ) ( ) ]

s cvout in

d dd d

dt dt dt

( )CV

d

dt

t+dtt+dt

tt

s

( ) ( ) ( ) ( )in in in ind dm Ads AVdt

In the like manner

( ) ( )out outd AVdt

1[( ) ( ) ]

s cvout in

d dd d

dt dt dt

[( ) ( ) ]cv

out ind

AV AVdt

s

1-D flow : is only the function of s . ( )s

For steady flow :

0cvd

dt

( ) ( )s

out ind

AV AVdt

t+dtt+dt tt

ds

R T T

If there are several one-D inlets and outlets :

( ) ( )souti i i i i i i i in

i i

dAV AV

dt

Steady , 1-D only in inlets and outlets, no matter how the flow is within the CV .

3.3 Conservation of mass ( 质量守恒 ) (Continuity Equation)

=m dm/dm=1

( ) ( ) 0sout ini ii ii i

i i

dmV VA A

dt

( ) ( )out inii ii iii i

AV VA ( ) ( )i in i out

i i

m m

Mass flux ( 质量流量 )m

For incompressible flow:

( ) ( )out ini ii ii i

V VA A i ii Volume fluxQ VA 体积流量

-------Leonardo da Vinci in 15001 21 2V VA A

If only one inlet and one outlet

壶口瀑布是我国著名的第二大瀑布。两百多米宽的黄河河面，突然紧缩为 50 米左右，跌入 30 多米的壶形峡谷。入壶之水，奔腾咆哮，势如奔马，浪声震天，声闻十里。 “黄河之水天上来”之惊心动魄的景观。

Example:

A jet engine working at design condition. At the inlet of the nozzle

At the outlet

Please find the mass flux and velocity at the outlet.

Given gas constant

5 21 2.05 10 /p N m

T1 =865K ， V1=288 m/s ， A1=0.19 ㎡；

5 22 1.143 10 /p N m T2 =766K ， A2=0.1538 ㎡

R=287.4 J/kg.K 。 Solution 1 1 1

1

45.1 /p AV

kg sRT

m AVp

AVRT

1 1 1 2 2 2

1 2

p AV p A V

RT RT

According to the conservation of mass

1 1 1 2 2 2m AV A V 1 1 2

2 12 2 1

565.1 /A p T

V V m sA p T

Homework: P185 P3.12, P189P3.36

mV

( )linear momentum

dmVV

dm

momentum perunit mass

( )( ) ( )

sout ini ii ii ii i

i i

d mVV VA AV V

dt

3.4 The Linear Momentum Equation ( 动量方程 )（ Newton’s Second Law ）

( ) ( )out ini ii ii i

m mV V

Newton’s second law

( )sd mVF

dt

( ) ( )out ini ii i

i i

m mV V

F

:Net force on the system or CV ( 体系或控制体受到的合外力 )iim V ： Momentum flux ( 动量流量 )

( ) ( )sout ini ii ii i

i i

dV VA A

dt 1-D in & out

steady RTT

flux

inoutm m m For only one inlet and one outlet

According to continuity

)( )

(s

out ind mV

F m V Vdt

2 － out, 1 － in

Example: A fixed control volume of a streamtube in steady flow has a uniform inlet (,V1 )and a uniform exit (,V2) . Find the net force on the control volume.

1V

1

2V

2

o x

y

2 1( )x x xF m V V

2 1( )y y yF m V V

2 1( )z z zF m V V o x

y

z

2 1( )F m V V Solution

:

2 1( )x x xm V VF 2 1( cos )m V V

2 1( )y y ym V VF 1sinmV

1 21 21 2m V VA A

Given5

21 2 4.19 10 Np pm

78.5 Kgm s 1 210 , 8cm cmd d

3998 Kgm

Neglect the weight of the fluid. Find the force on the water by the elbow pipe.

Example:

12

1

2

Solution:

x

y

o

select coordinate,control volume

2 1( )F m V V

2 1( )x x xm V VF

2mV

2 22sx mp VF A

2 22sx mp VF A

2 2

222

2

478.5998 4

dpd

3696N

22 5934sxs syF F NF 1 1 11 (0 )sy m mp V VF A

-4642N1 11sy mp VF A

In the like manner

1 sy

sx

Ftg

F 141.47

Find the force to fix the elbow.

Solution: coordinate, CV

Net force on the control volume:

2 22( )x L R exa ap p pF A A A A F

Where Fex is the force on the CV by pipe,( on elbow)

2 )(L R L exa a ap p p p FA A A 1

2

x

y

o

22 )(x ex ap pF F A

In like manner 11 )(y ey ap pF F A

Fex

Surface force: (1) Forces exposed by cutting though solid bodies which protrude into the surface.(2)Pressure,viscous stress.

A fixed vane turns a water jet of area A through an angle without changing its velocity magnitude. The flow is steady, pressure pa is everywhere, and friction on the vane is negligible. Find the force F applied to vane.

V

F

V

(cos 1)x mVF siny mVF

A water jet of velocity Vj impinges normal to a flat plate which moves to the right at velocity Vc. Find the force required to keep the plate moving at constant velocity and the power delivered to the cart if the jet density is 1000kg/m3 the jet area is 3cm2, and Vj=20m/s,Vc=15m/s

jVcV

x

Neglect the weight of the jet and plate,and assume steady flow with respect to the moving plate with the jet splitting into an equal upward and downward half-jet.

7.5N

Home work:

P190-p3.46

P191-p3.50

P192-p3.54

P192-p3.58

Derive the thrust( 推力 ) equation for the jet engine. air drag is neglect

:inlet , auniform V p outletouter V ap

inner eV ep

Solution:

( ) ( )out inF mV mV e em V mV ( ) efm m V mV

fm : mass flux of fuel

o

o

e

e

e

em

o

o

V ap

V apxF

right 'e eea ep pA A left ' 'o oaAp

Balance with thrust

ee eea eFx p pA A Fx

eVep

em

V

ap

Fx

)(e e eee aFx m V mV p p A

[ ) ](e e eee aR Fx m V mV p p A [ ( ) ) ](e eee aR m V V p p A

)( eea eFx p p A e em V mV

'' 'o o e eeea a eFx Ap p pA A F

Coordinate, CV

em m

0.02fm m

oom eem

oom eem

Example: In a ground test of a jet engine, pa=1.0133×105N/m2 ,Ae=0.1543m2,Pe=1.141×105N/m2, Ve=542m/s, . 43.4 /m Kg s Find the thrust force.

eVep

emap

Solution:

[ ( ) ]

25493

e e a eR mV P P A

N

0V

F16 R=65.38KNF16 R=65.38KN

[ ( ) ) ](e eee aR m V V p p A

x

coordinate

A rocket moving straight up. Let the initial mass be M0,and assume a steady exhaust mass flow and exhaust velocity ve relative to the rocket. If the flow pattern within the rocket motor is steady and air drag is neglect. Derive the differential equation of vertical rocket motion v(t) and integrate using the initial condition v=0 at t=0 .

Example:

ev, eep A

fm

( )v t

Solution:The CV enclose the rocket,cuts through the exit jet,and accelerates upward at rocket speed v(t).

coordinate

z

v(t)

Z-momentum equation:

( ) ( )out inF mv mv

ef

dvmg m m v

dt

0( ) fm t M m t f

em

dv v dt gdtm

0 0 00

v t tef

f

dtdv m v g dt

M m t

0( ) ln(1 )f

em t

v t v gtM

Am

Am

( ) ( )f e A Am v m v m v f em v

aP A ( )a eP A A e eP A mg dvm

dtF

( )e a e

dvP P A mg m

dt

a eif P P

ev, eep A

fm

( )v tv(t)

z

A

3.5 The Angular-Momentum Equation

( ) ( )s

out ind

AV AVdt

RTT

dr v

dm

2 12 1( )|s

z

dHm v vr r

dt

2 12 1( )zM m v vr r

For turbomachines

( , , ) column coordinate r z

0zif M 2 12 1v vr r 2 12 1( )zM m v vr r

r zo

rvzv

v

( )zH r v m

(Angular-Momentum)

zM

： Net moment( 合力矩 )

Example:Centrifugal ( 离心 )pump

The velocity of the fluid is changed from v1 to v2 and its pressure from p1 to p2.Find (a).an expression for the torque T0 which must be applied those blades to maintain this flow. (b).the power supplied to the pump.

0 2 12 1( ) ( )out inm mv vM r r

blade

2v

2rv

2v

1v

1rv1v

1p

2p

o

1 21 21 2r out rin V m V mm A A

m QFor incompressible flow

( )r 1-D

Continuity :

Solution: The CV is chosen .

0 2 12 1( )m v vM r r

0 2 12 1( )m v vT r r

blade

2v

2rv

2v

1v

1rv1v

1p

2p

o

Pressure has no contribution to the torque

0P T 2 12 1( )m v vr r

2 2 1 1( )t tm v V v V

are blade rotational speeds1 2t tV V

pw

m 2 2 1 1t tv V v V Work on per unit mass

Homework: P192-p3.55; P194-p3.68, p3.78 ; P200-p3.114,p3.116

Brief Review• Basic Physical Laws of Fluid Mechanics:

0dt

dm)( Vm

dt

dF

inoutCVsys AVAVdt

d

dt

d)()()()(

inout VmVmF )()(

inout rVmrVmM )()(

)( rVmdt

dM

dt

dW

dt

dQ

dt

dE

• The Reynolds Transport Theorem:

• The Linear Momentum Equation:

• The Angular-Momentum Theorem:

• Conservation of Mass: outin AVAV )()(

Review of Fluid Statics

• Especially : Cp

zp

z ＝

0

0

QuestionWhen fluid flowing…

Bernoulli(1700~1782)

What relations are there in velocity, height and pressure?

Several Tragedies in History:

• A little railway station in 19th Russia.

• The ‘Olimpic’ shipwreck in the Pacific

• The bumping accident of B-52 bomber of the U.S. air force in 1960s.

3.6 Frictionless Flow:The Bernoulli Equation

1.Differential Form of Linear Momentum Equation Elemental fixed streamtube CV of variable area

A(s),and length ds.

dsAl

ddAA

dVV dpp

2

dpp

pV

A

z

s

Linear momentum relation in the streamwise direction:

inouts VmVmF )()(

dVmVVmRight inout

)( AVm

bodysurs FFLeftF

321 ssssur FFFF

sin)2/(3 ls AdppF

sinsin

)2/(dA

dpp

dAdpp )2/( dsAl

ddAA

dVV dpp

2

dpp

pV

A

z

s

pdApdAAdp

Adp

dAdppdAAdpppA )2/())((

321 ssssur FFFF

cosAdsFbody

Adz

AVdVAdpAdz 0 AdpAdzAVdV

0 VdVgdzdp

one-D,steady,frictionless flow

dsAl

ddAA

dVV dpp

2

dpp

pV

A

z

s

0 VdVgdzdp

For incompressible flow, =const.

Integral between any points 1 and 2 on the streamline:

0)()(2

112

21

22

12

zzgVVpp

czgVpzgVp

2

222

1

211

22

forequation Bernoulli

ssfrictionlesteady

flow ibleincompress.streamline a along

A Question:

Is the Bernoulli equation a

momentum or energy equation?

Hydraulic and energy grade lines for frictionless flow in a duct.

Example 1:

Find a relation between nozzle discharge velocity and tank free-surface height h. Assume steady frictionless flow.

1,2 maximum information is known or desired.

h

1

2

V2

Solution:

h

1

2V2

Continuity: VAVA 2211

Bernoulli: zg

Vpz

gVp

2

222

1

211

22

ppp a 11

)(2 2121

22 zzgVV gh2

AA

ghV 2

122

22

/1

2

AA 21

ghV 22 Torricelli 1644

According to the Bernoulli

equation, the velocity of a fluid

flowing through a hole in the

side of an open tank or reservoir

is proportional to the square root

of the depth of fluid above the

hole.The velocity of a jet of water from an open pop

bottle containing four holes is clearly related to the

depth of water above the hole. The greater the

depth, the higher the velocity.

Review of Bernoulli equation

The dimensions of above three items are the same of length!

czg

Vp

zg

Vp

2

222

1

211

2

2

0 VdVgdzdp

Example 1:

Find a relation between nozzle discharge velocity and tank free-surface height h. Assume steady frictionless flow.

V2

h

1

2

ghV 22

Example 2:

Find velocity in the right tube. h

h

AB

gVp

gVp AABB

22

22

hgV B 2

In like manner:

h

V

)(2

gh

V

Example 3: Find velocity in the Venturi tube.

h

'

12

2 21 1 2 2

1 22 2

p V p V

g g

1 1 2 2AV A V

2V

2 2

2

(1 )

pV

1 2p p p

)β1β

ρρhg2

(

)(2 '

21

21 V

A

AV

As a fluid flows through a Venturi tube, the pressure is reduced in accordance with the continuity and Bernoulli equations.

Example 4: Estimate required to keep the plate in a balance state.

(Assume the flow is steady and frictionless.)

1h

1h

2h

V

A A

Solution:

For plate,

1h

A A

V

2h

1 22gh gh 1 2

1

2h h

12ghV

AghFright 2

VAVFleft by lineal momentum equation,by Bernoulli equation,

Example 5: Fire hose,Q=1.5m3/min Find the force on the bolts.

10cm 3cm ap

1

1

2

2

Solution:1. .F l m e p Bernoulli continuity

By continuity:

1 1 2 2AV A V Q 11

QV

A 2

2

QV

A

By Bernoulli:2 2

1 1 2 2

2 2

p V p V

g g 2 2

1 2 1( )2ap p V V

10cm 3cm ap

1

1

2

2

By momentum :

2 1( )F m V V

1 1 2 1( )gp A F m V V

2 1 1 1( ) gF m V V p A

4067N

4067boltF F N

Example 6: Find the aero-force on the blade (cascade).

A

B

D

C

S

S

1xV1yV

1V2xV

2 yV

2V

Solution:

2 1( )x x xF m V V

2 1( )y y yF m V V

.

1 2 2 1( )x x xp s p s F m V V

.

2 1( )y y yF m V V

.

2 1 2 1( ) ( )x x xF m V V p p s 2 1( )p p s

.

1 2x xm sV sV 1 2x x xV V V

A

B

D

C

S

S

1xV1yV

1V2xV

2 yV

2V

By continuity,

2 22 1

1 2 ( )2

V Vp p

g

2 2 2 22 2 1 1[( ) ( )]

2 x y x yV V V V

2 22 1( )

2 y yV V

2 22 1( )

2x y yF V V s

2 1( )y x y yF sV V V

叶片越弯，做功量越大。

A

B

D

C

S

S

1xV1yV

1V2xV

2 yV

2V

By Bernoulli,

Bernoulli Equation for compressible flow

2 22

2 1

10

2

V Vdp

Specific-heat ratio 1.4 p

v

Ck

C

For isentropic flow:1 2

1 2k k k

p ppC

Gas Weight neglect

1 1

1

2 21 2

1 11 1

[( ) 1]1

kk k

k

p pdp dp kC

k pp

12 2

2 2 11

1

[( ) 1] 01 2

kk

p V VkRT

k p

1

2 22 1 2

11

[1 ( ) ]2 1

kk

V V pkRT

k p

For nozzle: 2 1V V 2 1p p For diffuser:

2 1V V 2 1p p

k

k

k

k

p

C

p

p1

1

11

1

1 11

Extended Bernoulli Equation 2 2

22 1

1 2 s f

V Vdpw w

sw machinefw loss friction

For compressor 多变压缩功

For turbine 多变膨胀功

2 22

2 1

10

2

V Vdp

0sw

0sw

p V

p V

Home work!

• Page 206: P3.158, P3.161

• Page 207: P3.164, P3.165

• 《气体动力学》第二章习题第一部分： Page 20 33 题

Review of examples:

h

V

)(2

gh

V

h

'

12

)1(

)(22

hg

Vhg

ppp

)'(21

)( 11

'12

hhgp

hgghp

h

'

1p 2p

1h

2h

1h

A A

V

2h

1 2

1

2h h

)(

)(

12

.

11

VVm

FAppFFF boltarightleft

boltelsealeft FApApF 11

)( 1 elsearight AApF

NApp a 4872)( 11

NFbolt 4067

•Analysis•Choose your control volumn•Body force and Surface force•Solution

NVVm 805)( 12

.

10cm 3cm ap

1

1

2

2

x

Find the aero-force on the blade (cascade).

2 22 1

1 2 ( )2

V Vp p

g

2 2 2 22 2 1 1[( ) ( )]

2 x y x yV V V V

2 22 1( )

2 y yV V

2 22 1( )

2x y yF V V s

2 1( )y x y yF sV V V

叶片越弯，做功量越大。

A

B

D

C

S

S

1xV1yV

1V2xV

2 yV

2V

By Bernoulli,

3.7 The Energy Equation

• Conservation of Energy

Various types of energy occur in flowing fluids.

Work must be done on the device shown to turn it over because the system gains potential energy as the heavy(dark) liquid is raised above the light(clear) liquid.

This potential energy is converted into kinetic energy which is either dissipated due to friction as the fluid flows down ramp or is converted into power by the turbine and dissipated by friction.

The fluid finally becomes stationary again.

The initial work done in turning it over eventually results in a very slight increase in the system temperature.

sout in

dAV AV

dt

.

( )out inm

,dE

E edm

Energy Per Unit Mass

1

1

2

2

e

First Laws of Thermodynamics

dE dQ dW

dt dt dt

. .

Q W

• Conservation of Energy

internal kinetic potential othere e e e e 2

2

vu gz

2 2.

[( ) ( ) ]2 2out in

dE v vm u gz u gz

dt

1

1

2

2

)(.

inout eemdt

dE

.

W ( )a.

shaftW.

( )SW

( )b.

pressW.

( )pW

only 1-1,2 - 2.

2 2 2 1 1 1pW p A V p AV .

2 1

2 1

( )p p

m

.

viscous stressW( )c

0cV

0vW

2 2. . . .2 1

2 12 1

[ ( )] [( ) ( ) ]2 2

sp p V V

Q W m m u gz u gz

2 2. . .

2 1[( ) ( ) ]2 2

sV p V p

Q m u gz u gz W

2 22 1 2 1 2 1

1( ) ( ) ( )

2sq w V V g z z h h

pu h

ph C T

The energy equation!

Example: A steady flow machine takes in air at section 1 and discharged it at section 2 and 3.The properties at each section are as follows:

section A, Q, T, P, Pa Z,m

1 0.04 2.8 21 1000 0.3

2 0.09 1.1 38 1440 1.2

3 0.02 1.4 100 ? 0.4

2m 3 /m s C

CV

(1)

(2)

(3)

110KW

.

?Q

Work is provided to the machine at the rate of 110kw.

Find the pressure (abs) and the heat transfer .

Assume that air is a perfect gas with R=287, Cp=1005.

3p.

Q

Solution: 3 3 3P RT 3 ?

Mass conservation: 1 1 2 2 3 3Q Q Q 3

2 0.0161 /Kg m 31 0.0119 /Kg m

31 1 2 23

3

0.0112 /Q Q

Kg mQ

3 3 3 1199P RT Pa

11

1

70 /Q

V m sA

2 12.2 /V m s 3 55 /V m s

By energy equation:2 2. . . .

[( )] [( )]2 2

s out in

V VQ W m h gz m h gz

22. . .32

2 32 2 3 3( ) ( )2 2p p

VVQ m C T gz m C T gz

.2.

11 1 1( )

2ps

Vm C T gz W

CV

(1)

(2)

(3)

110KW

.

?Q