FUNDAMENTALS OF FLUID MECHANICS Chapter 5 Flow Analysis Using …cau.ac.kr/~jjang14/FME/Chap5.pdf · · 2009-09-02Chapter 5 Flow Analysis Using Control Volume. 2 MAIN TOPICS ...

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FUNDAMENTALS OFFLUID MECHANICS

Chapter 5 Flow Analysis Using Control Volume

2

MAIN TOPICS

Conservation of MassNewton’s Second Law – The Linear Momentum

EquationsThe Angular Momentum EquationsFirst Law of Thermodynamics – The Energy EquationSecond Law of Thermodynamics – Irreversible Flow

3

Conservation of Mass –The Continuity Equation 1/4

Basic Law for Conservation of Mass

For the system and a fixed, nondeforming control volume that are coincident at an instant of time, the Reynolds Transport Theorem leads to

( )∫∫∫ ⋅+∂∂

≡CS

dAnVVdt

VdDtD

CVsys

ρρρB=M and b =1

Time rate of change of the mass of the coincident system

Time rate of change of the mass of the content of the coincident control volume

Net rate of flow of mass through the control surface

= +

0dt

dM

system=

VddmM

)system(V)system(Msystem ρ== ∫∫

= 0

= 0

4


System and control volume at three different instances of time. (a) System and control volume at time t – δt. (b) System and control volume at time t, coincident condition. (c) System and control volume at time t + δt.

5


For a fixed, nondeforming control volume, the control volume formulation of the conservation of mass: The continuity equation

0dAnVVdtdt

MdCSCV

system

=⋅ρ+ρ∂∂

≡ ∫∫

∫∫ ⋅ρ−=ρ∂∂

CSCVdAnVVd

t

∑∑ − inout mm

Rate of increaseOf mass in CV

Net influx of mass

6


Incompressible Fluids

For steady flow

0dAnVVdt

0dAnVVdt CSCVCSCV

=⋅+∂∂

→=⋅ρ+∂∂

ρ ∫∫∫∫

The mass flow rate into a control volume must be equal to the mass flow rate out of the control volume.

0dAnVCS

=⋅ρ∫

∑∑ − inout mm

7

Other Definition

Mass flowrate through a section of control surface

The average velocity∑∑∫ −=⋅ρ=ρ= inoutA

mmdAnVQm

AdAnV

V Aρ

∫ ⋅ρ=

8

Fixed, Nondeforming Control Volume 1/2

When the flow is steady

When the flow is steady and incompressible

When the flow is not steady

0Vdt CV

≠ρ∂∂∫

0Vdt CV

=ρ∂∂∫ ∑∑ = inout mm

∑∑ = inout QQ

“+” : the mass of the contents of the control volume is increasing“-” : the mass of the contents of the control volume is decreasing.

9

Fixed, Nondeforming Control Volume 2/2

When the flow is uniformly distributed over the opening in the control surface (one dimensional flow)

When the flow is non-uniformly distributed over the opening in the control surface

AVm ρ=

VAm ρ=

10

Example 5.1 Conservation of Mass –Steady, Incompressible Flow Seawater flows steadily through a simple conical-shaped nozzle at

the end of a fire hose as illustrated in Figure E5.1. If the nozzle exit velocity must be at least 20 m/s, determine the minimum pumping capacity required in m3/s.

Figure E5.1

11

Example 5.1 Solution

Steady flow

( ) ( )

2211

121212CS

QQ

0

ρρ

ρρρ

=

==−=−+=⋅∫ mmormmVAVAdAnV

0dAnVVdt CSCV

=⋅ρ+ρ∂∂

∫∫

The continuity equation

s/m0251.0...AVQQ 3222121 ====⇒ρ=ρ

With incompressible condition

12

Example 5.2 Conservation of Mass –Steady, Compressible Flow

Figure E5.2

Air flows steadily between two sections in a long, straight portion of 4-in. inside diameter as indicated in Figure E5.2. The uniformly distributed temperature and pressure at each section are given. If the average air velocity (Non-uniform velocity distribution) at section (2) is 1000ft/s, calculate the average air velocity at section (1).

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Steady flow

222111

1212CS

VAVA

mm0mmdAnV

ρ=ρ

=⇒=−=⋅ρ∫

0dAnVVdt CSCV

=⋅ρ+ρ∂∂

∫∫


21

21 VV

ρρ

=Since A1=A2s/ft219...V

TpTpV 2

21

121 ==

The ideal gas equation

RTp

=ρ

14

Example 5.3 Conservation of Mass –Two Fluids Moist air (a mixture of dry air and water vapor) enters a

dehumidifier at the rate of 22 slugs/hr. Liquid water drains out of the dehumidifier at a rate of 0.5 slugs/hr. Determine the mass flowrate of the dry air and the water vapor leaving the dehumidifier.

Figure E5.3

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0dAnVVdt CSCV

=⋅ρ+ρ∂∂

∫∫

Steady flow

hr/slugs5.21hr/slugs5.0hr/slugs22mmm

0mmmdAnV

312

321CS

=−=−=

=++−=⋅ρ∫


16

Example 5.4 Conservation of Mass –Nonuniform Velocity Profiles Incompressible, laminar water flow develops in a straight pipe

having radius R as indicated in Figure E5.4. At section (1), the velocity profile is uniform; the velocity is equal to a constant value U and is parallel to the pipe axis everywhere. At section (2), the velocity profile is axisymmetric and parabolic, with zero velocity at the pipe wall and a maximum value of umax at the centerline. How are U and umax related? How are the average velocity at section (2), , and umax related?2V

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Steady flow

2/uVU2u

0rdrRr1u2UA

max2max

R

0

2

max1

==

=

−π+− ∫

0dAnVVdt CSCV

=⋅ρ+ρ∂∂

∫∫

With incompressible condition


0rdr2uUA0dAnVUAR

0 2211A112

=πρ+ρ−=⋅ρ+ρ− ∫∫

21 ρ=ρ

−=

2

max Rr1uu

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Example 5.5 Conservation of Mass –Unsteady Flow A bathtub is being filled with water from a faucet. The rate of flow

from the faucet is steady at 9 gal/min. The tub volume is approximated by a rectangular space as indicate Figure E5.5(a). Estimate the time rate of change of the depth of water in the tub, , in in./min at any instant.t/h ∂∂

Figure E5.5

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Example 5.5 Solution1/2

airvolumewater waterwaterwatervolumeair airair

CSCV

mmVdt

Vdt

0dAnVVdt

+−ρ∂∂

+ρ∂∂

=

=⋅ρ+ρ∂∂

∫∫

∫∫

0mVdt

airFor airvolumeair airair =+ρ∂∂∫


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)ft10)(ft/gal48.7()ft/.in12min)(/gal9(

)Aft10(Q

th

mth)Aft10(

]A)hft5.1()ft5)(ft2(h[Vdt

mVdt

waterFor

23j

2water

waterj2

water

volumewater jwaterwaterwater

volumewater waterwaterwater

=−

=∂∂

=∂∂

−ρ⇒

−+ρ=ρ∂∂

=ρ∂∂

∫

∫

2j ft10A <<

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Moving, Nondeforming Control Volume

When a moving control volume is used, the fluid velocity relative to the moving control is an important variable.W is the relative fluid velocity seen by an observer

moving with the control volume. Vcv is the control volume velocity as seen from a fixed

coordinate system. V is the absolute fluid velocity seen by a stationary

observer in a fixed coordinate system.

CVVVW

−=

0dAnWVdt .S.CCV =⋅ρ+ρ∂∂

∫∫

( )dAnWVdtDt

DMSCCV

sys ⋅+

∂∂

= ∫∫ ρρ ..

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Example 5.6 Conservation of Mass - Compressible Flow with a Moving Control Volume

An airplane moves forward at speed of 971 km/hr as shown in Figure E5.6 (a). The frontal intake area of the jet engine is 0.80m2

and the entering air density is 0.736 kg/m3. A stationary observer determines that relative to the earth, the jet engine exhaust gases move away from the engine with a speed of 1050 km/hr. The engine exhaust area is 0.558 m2, and the exhaust gas density is 0.515 kg/m3. Estimate the mass flowrate of guel into the engine in kg/hr.

Determine the mass flowrate of fuel into the engine in kg/hr

Figure E5.6

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0dAnWVdt .S.CCV =⋅ρ+ρ∂∂

∫∫

hr/kg9100...)km/m1000)(hr/km2021)(m558.0)(m/kg515.0(m

hr/km201hr/km971hr/km1050VVW

WAWAm

0WAWAm

23infuel

plane22

111222infuel

222111infuel

=−=→

=+=−=

ρ−ρ=→

=ρ+ρ−−

=0

The intake velocity, W1, relative to the moving control volume. The exhaust velocity, W2, also needs to be measured relative to the moving control volume.


Assuming one-dimensional flow

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Deforming Control Volume

A deforming control volume involves changing volume size and control surface movement.

The Reynolds transport theorem for a deforming control volume can be used for this case.

CSVWV

+=

( )dAnWVdtDt

DMSCCV

sys ⋅+

∂∂

= ∫∫ ρρ ..

Vcs is the velocity of the control surface as seen by a fixed observer.W is the relative velocity referenced to the control surface.

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Example 5.8 Conservation of Mass –Deforming Control Volume 1/2

A syringe is used to inoculate a cow. The plunger has a face area of 500 mm2. If the liquid in the syringe is to be injected steadily at a rate of 300 cm3/min, at what speed should the plunger be advanced? The leakage rate past the plunger is 0.01 times the volume flowrate out of the needle.

Determine the speed of the plunger be advanced

Figure E5.8

Leakage rate

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Example 5.8 Solutionp1 AA ≅

min/mm660...A

QQV

0QQVA

0QmVAVt

Let

1

leak2p

leak2p1

leak2p1p

==+

=

=ρ+ρ+ρ−

=ρ++ρ−⇒=∂∂

−

22 Qm ρ=

The continuity equation 0dAnWVdt .S.CCV =⋅ρ+ρ∂∂

∫∫

tAVd

t

)VA(Vd0QmVdt

1CV

needle1CVleak2CV

∂∂

ρ=∫ ρ∂∂

→

+ρ=ρ∫=ρ++ρ∫∂∂

→

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The Linear Momentum Equations 1/4

Newton’s second law for a system moving relative to an inertial coordinate system.

systemsysBSsys Dt

PDVdVDtDFFF

=ρ=+= ∫∑∑∑

Time rate of change of the linear momentum of the system = Sum of external forces

acting on the system

VdVdmVP)system(V)system(Msystem ρ== ∫∫

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For the system and a fixed, nondeforming control volume that are coincident at an instant of time, the Reynolds Transport Theorem leads to

( )∫∫∫ ⋅+∂∂

≡CS

dAVVnVdVt

VdVDtD

CVsys

ρρρB=P and Vb

=

( )∫∫∫ ⋅+∂∂

≡CS

dAVnVVdVt

VdVDtD

CVsys

ρρρ

Time rate of change of the linear momentum of the coincident system

Time rate of change of the linear momentum of the content of the coincident control volume

Net rate of flow of linear momentum through the control surface

= +

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External forces acting on system and coincident control volume

∑ ∑= volumecontrolcoincidenttheofcontentssys FF

When a control volume is coincident with a system at an instant of time, the force acting on the system and the force acting on the contents of the coincident control volume are instantaneously identical.

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For a fixed and nondeforming control volume, the control volume formulation of Newton’s second law

( ) ∑∫∫ =⋅+∂∂ FdAnVVVdVt CV

CSρρ Contents of the coincident

control volume

Linear momentum equation

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Example 5.10 Linear Momentum – Change in Flow Direction As shown in Figure E5.10 (a), a horizontal jet of water exits a

nozzle with a uniform speed of V1=10 ft/s, strike a vane, and is turned through an angleθ. Determine the anchoring force needed to hold the vane stationary. Neglect gravity and viscous effects.

Determine the anchoring force needed to hold the vane stationary.

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∑∫∫

∑∫∫

=⋅ρ+ρ∂∂

=⋅ρ+ρ∂∂

zCSCV

xCSCV

FdAnVwVdwt

FdAnVuVdut

The x and z direction components of linear momentum equation

lbsin64.11...sinAVF

lb)cos1(64.11..)cos1(AVFFA)V(sinVA)V()0(FA)V(cosVA)V(V

112

Az

112

Ax

Az21111

Ax211111

θ==θρ=→

θ−−==θ−ρ−=→

=θρ+−ρ=θρ+−ρ

kwiuV

+=

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Example 5.11 Linear Momentum – Weight, pressure, and Change in Speed Determine the anchoring force required to hold in place a conical

nozzle attached to the end of a laboratory sin faucet when the water flowrate is 0.6 liter/s. The nozzle mass is 0.1kg. The nozzle inlet and exit diameters are 16mm and 5mm, respectively. The nozzle axis is vertical and the axial distance between section (1) and (2) is 30mm. The pressure at section (1) is 464 kPa. to hold the vane stationary. Neglect gravity and viscous effects.

34


35


dAwdAnV

ApWApWFdAnVwVdwt 22w11nACSCV

±=⋅

+−−−=⋅ρ+ρ∂∂

∫∫

The z direction component of linear moment equation

With the “+” used for flow out of the control volume and “-” used for flow in.

22w11n21A

22w11n2211

ApWApW)ww(mFApWApW)w(m)w)(m(

−+++−=+−−−=−+−−

mmm 21 == s/kg599.0...QAwmmm 1121 ==ρ=ρ===

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( )

( )

N0278.0...gV)DDDD(h121gVW

N981.0)s/m81.9)(kg1.0(gmW

m6.30...4/D

QAQw

s/m98.2...4/D

QAQw

w2122

12

ww

2nn

222

2

211

1

==

++πρ=ρ=

===

==π

==

==π

==

N8.77...)(...)s/kg599.0(ApWApW)ww(mF 22w11n21A

===−+++−=

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Example 5.12 Linear Momentum – Pressure , Change in Speed, and Friction Water flows through a horizontal, 180° pipe bend. The flow cross-

section area is constant at a value of 0.1ft2 through the bend. The magnitude of the flow velocity everywhere in the bend is axial and 50ft/s. The absolute pressure at the entrance and exit of the bend are 30 psia and 24 psia, respectively. Calculate the horizontal (x and y) components of the anchoring force required to hold the bend in place.

38


The x direction component of linear moment equation

AxCSCVFdAnVuVdu

t=⋅ρ+ρ

∂∂

∫∫

2211AyCSCVApApFdAnVvVdv

t++=⋅ρ+ρ

∂∂

∫∫

At section (1) and (2), the flow is in the y direction and therefore u=0 at both sections.

0FAx =The y direction component of linear moment equation

∫= AS -F pdAn

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For one-dimensional flow

2211Ay2211 ApApF)m)(v()m)(v( ++=+−+−+

2211Ay21 ApApF)vv(m ++=+−

s/slugs70.9...Avmmm 1121 ==ρ===

lb1324...ApAp)vv(mF 221121Ay −==−−+−=

40

Example 5.13 Linear Momentum – Weight, pressure, and Change in Speed Air flows steadily between two cross sections in a long, straight

portion of 4-in. inside diameter pipe as indicated in Figure E5.13, where the uniformly distributed temperature and pressure at each cross section are given, If the average air velocity at section (2) is 1000 ft/s, we found in Example 5.2 that the average air velocity at section (1) must be 219 ft/s. Assuming uniform velocity distributions at sections (1) and (2), determine the frictional force exerted by the pipe wall on the air flow between sections (1) and (2).

41


The axial component of linear moment equation

2211xCSCVApApRdAnVuVdu

t−+−=⋅ρ+ρ

∂∂

∫∫

2211x2211 ApApR)m)(u()m)(u( −+−=+++−+

)pp(AR)uu(m 212x12 −+−=−

s/slugs297.0...u4D

RTpmmm 2

22

2

221 ==

π

===

)uu(m)pp(AR 12212x −−−=

42


4DA

RTp

22

2

2

22

π=

=ρ

)uu(m)pp(AR 12212x −−−=

lb793...)uu(m)pp(AR 12212x ==−−−=

43

Example 5.14 Linear Momentum –Weight, Pressure,… If the flow of Example 5.4 is

vertically upward, develop an expression for the fluid pressure drop that occurs between sections (1) and (2).

44


22z11CS 22211

22z11CSCV

ApWRAp)dAw()w()m)(w(

ApWRApdAnVwVdwt

−−−=+ρ++−+⇒

−−−=⋅ρ+ρ∂∂

∫∫∫

The axial component of linear moment equation

−=

2

12 Rr1w2w[ ]

3Rw4rdr2w)dAw()w(

)R/r(1w2w2

21

R

0

22CS 222

212

πρ=πρ=+ρ+

−=

∫∫

11

z2

121

22z112

122

1

AW

AR

3wpp

ApWRApRw34Rw

++ρ

=−⇒

−−−=ρπ+ρπ−

45

Example 5.15 Linear Momentum - Trust

A static thrust as sketched in Figure E5.15 is to be designed for testing a jet engine. The following conditions are known for a typical test: Intake air velocity = 200 m/s; exhaust gas velocity= = 500 m/s; intake cross-section area = 1m2; intake static pressure = -22.5 kPa=78.5 kPa (abs); intake static temperature = 268K; exhaust static pressure =0 kPa=101 kPa (abs). Estimate the normal trust for which to design.

46


N83700...)uu(mAAFFApAp)uu(m

uAmuAmmFA)pp(A)pp()m)(u()m)(u(

)AA(pApFApdAnVuVdut

122211th

th221112

22221111

th2atm21atm12211

21atm22th11CSCV

==−+ρ+ρ−=+−=−⇒

ρ==ρ==+−−−=+++−+⇒

−−−+=⋅ρ+ρ∂∂

∫∫


s/kg204...uAmRTp

1111

11 ==ρ==ρ

∫= AS -F pdAn

47

Example 5.16 Linear Momentum –Nomuniform Pressure A sluice gate across a

channel of width b is shown in the closed and open position in Figure E5.16(a) and (b). Is the anchoring force required to hold the gate in place larger when the gate is closed or when it is open?

48


When the gate is open, the horizontal forces acting on the contents of the control volume are identified in Figure E5.16 (d).

hbuFbh21bH

21RuuandhHFor

Fbh21RbH

21hbuHbu

Fbh21RbH

21dAnVu

22f

22x21

f2

x22

22

1

f2

x2

CS

ρ−−γ−γ=⇒<<>>

−γ−−γ=ρ+ρ−

−γ−−γ=⋅ρ∫

When the gate is closed, the horizontal forces acting on the contents of the control volume are identified in Figure E5.16 (c).

bH21RRbH

21dAnVu 2

xx2

CSγ=⇒−γ=⋅ρ∫

49

Moving, Nondeforming Control Volume1/3

CVVWV

+=

∫∫∫ ⋅ρ+ρ∂∂

≡ρCSCVsys

dAnWVVdVt

VdVDtD

∑∫∫ =⋅ρ+ρ∂∂ FdAnWVVdVt CSCV

Contents of the coincidentcontrol volume

∑∫∫ =⋅ρ++ρ+∂∂ FdAnW)VW(Vd)VW(t CS CVCV CV


dAnWbVbdtDt

DBCSCV

sys ⋅ρ+ρ

∂∂

= ∫∫Chapter 4: Reynolds transport equation for a control volume moving with constant velocity is

50


( ) ∫∫∫ ⋅ρ+⋅ρ=⋅ρ+CSCVCSCS CV dAnWVdAnWWdAnWVW

( ) 0VdVWt CV CV =ρ+∂∂∫

For a constant control volume velocity, Vcv, and steady flow in the control volume reference frame

For steady flow, continuity equation

=0

0dAnWVdtDt

DM.S.CCV

sys =⋅ρ+ρ∂∂

= ∫∫

0dAnW.S.C =⋅ρ∫

51


( ) ∑∫ =⋅ FdAnWW

CSρ

For an inertial, moving, nondeforming control volume, the linear momentum equation of steady flow


52

Vector Form of Momentum Equation

The sum of all forces (surface and body forces) acting on a Non-accelerating control volume is equal to the sum of the rate of change of momentum inside the control volume and the net rate of flux of momentum out through the control surface.

∫∫∫

=

==

AS

B

-F

BBF

pdAn

VddmCV

ρ

Where the velocities are measuredRelative to the control volume.

( )∫∫

∑∑∑⋅+

∂∂

=

+=

CS

volumecontrolcoincidenttheofcontents

dAnVVVdVt

FFF

CV

BS

ρρn

53

..

Example 5.17 Linear Momentum -Moving Control Volume 1/2

A vane on wheels move with a constant velocity V0 when a stream of water having a nozzle exit velocity of V1 is turned 45° by the vane as indicated in Figure E5.17(a). Note that this is the same moving vane considered in Section 4.4.6 earlier . Determine the magnitude and direction of the force, F, exerted by the stream of water on the vane surface. The speed of the water jet leaving the nozzle is 100ft/s, and the vane is moving to the right with a constant speed of 20 ft/s.

54

..

Example 5.17 Linear Momentum -Moving Control Volume 2/2

55

..


x2211

xCS x

R)m)(45cosW()m)(W(

RdAnWW

−=+°++−+⇒

−=⋅ρ∫

22221111 AWmAWm ρ=ρ=


wz22

WzCS z

WR)m)(45sinW(

WRdAnWW

−=+°+⇒

−=⋅ρ∫

...VVWWAWmAWm

0121

22221111

=−==ρ==ρ=

The z direction component of linear moment equation

56

..


x

z1

2z

2x

w12

1z

12

1x

RRtan

lb3.57...RRR

lb53...W45sinAWR

lb8.21...)45cos1(AWR

−=α

==+=

==+°ρ=

==°−ρ=

1w gAW ρ=

57

First Law of Thermodynamics –The Energy Equation1/4

The first law of thermodynamics for a system isTime rate of increase of the total stored energy of the system

Net time rate of energy addition by heat transfer into the system

Net time rate of energy addition by work transfer into the system

= +

( ) ( ) ( )

( )

gzVue

WQVdeDtDor

WQWWQQVdeDtD

sysinnetinnetsys

sysinnetinnetsysoutinsysoutinsys

++=

+=

+=−+−=

∫

∑ ∑∑ ∑∫

2ˆ

2

//

ρ

ρ

Total stored energy per unit mass for each particle in the system

“+” going into system“-” coming out

The net rate of heat transfer into the system

The net rate of work transfer into the system

58


For the system and the contents of the coincident control volume that is fixed and nondeforming -- Reynolds Transport Theorem leads to

dAnVeVdet

VdeDtD

.S.CCVsys⋅ρ+ρ

∂∂

=ρ ∫∫∫

Time rate of increase of the total stored energy of the system

Net time rate of increase of the total stored energy of the contents of the control volume

The net rate of flow of the total stored energy out of the control volume through the control surface

= +

59


CVCS innetinnetcv )WQ(dAnVeVdet ∫∫ +=⋅ρ+ρ∂∂

For the control volume that is coincident with the system at an instant of time.

The control volume formula for the first law of thermodynamics:

volumecontrolcoincidentinnetinnetsysinnetinnet )WQ()WQ( +=+

60

Rate of Work done by CV

Shaft work : the rate of work transferred into through the CS by the shaft work ( negative for work transferred out, positive for work input required)

Work done by normal stresses on the CS:

Work done by shear stresses on the CS:

Other work

othershearnormalshaft WWWWW +++=ShaftW

∫ ⋅−=⋅=CSnormal dAnVpVF

δnormalW

dAnVWCSshear

⋅τ+= ∫

∫∫∫ ⋅−+=⋅ρ+ρ∂∂

CSinnetshaftinnetCScv dAnVpWQdAnVeVdet

Negligibly small on a control surface

61


inShaftinnetCSCVWQdAnVgzVpuVde

t //

2

)2

ˆ( +=⋅++++

∂∂

∫∫ ρρ

ρ

∫∫∫ ⋅−+=⋅+∂∂

CSinnetShaftinnetCSCVdAnVpWQdAnVeVde

t

ρρ

Energy equation

gzVue ++=2

ˆ2

62

Application of Energy Equation1/2

∫ =ρ∂∂ 0Vdet CV

mgz2

Vpumgz2

VpudAnVgz2

Vpuin

2

out

22

CS

∑∑∫

++

ρ+−

++

ρ+=⋅ρ

++

ρ+

inin

2

outout

2

2

CS

mgz2

Vpumgz2

Vpu

dAnVgz2

Vpu

++

ρ+−

++

ρ+=

⋅ρ

++

ρ+∫

When the flow is steadyThe integral of

dAnVgz2

Vpu2

CS⋅ρ

++

ρ+∫ ???

Uniformly distribution

Only one stream entering and leaving

63

Application of Energy Equation2/2

( )

innetshaftinnet

inout

2in

2out

inoutinout

WQ

zzg2

VVppuum

+=

−+

−+

ρ

−

ρ

+−

ρ+=

puh

( ) in/netshaftin/netinout

2in

2out

inout WQzzg2

VVhhm +=

−+

−+−

If shaft work is involved….

One-dimensional energy equation for steady-in-the-mean flow

Enthalpy The energy equation is written in terms of enthalpy.

64

Example 5.20 Energy – Pump Power 1/2

A pump delivers water at a steady rate of 300 gal/min as shown in Figure E5.20. Just upstream of the pump [section(1)] where the pipe diameter is 3.5 in., the pressure is 18 psi. Just downstream of the pump [section (2)] where the pipe diameter is 1 in., the pressure is 60 psi. The change in water elevation across the pump is zero. The rise in internal energy of water, u2-u1, associated with a temperature rise across the pump is 3000 ft·lb/slug. If the pumping process is considered to be adiabatic, determine the power (hp) required by the pump.

65

Example 5.20 Energy – Pump Power 2/2

66


( )

in/netshaftin/net

12

21

22

1212

WQ

zzg2

VVppuum

+=

−+

−+

ρ

−

ρ

+−

[ ] hp3.32....)s/slugs30.1(W

s/ft123...AQVs/ft0.10.....

AQV

4/DQ

AQV

s/slugs30.1min)/s60)(ft/gal48.7(

min)/gal300)(ft/slug94.1(Qm

innetshaft

22

112

3

3

==

======π

==

==ρ=

One-dimensional energy equation for steady-in-the-mean flow

=0(Adiabatic flow)

67

Example 5.21 Energy – Turbine Power per Unit Mass of Flow Steam enters a turbine with a velocity of 30m/s and enthalpy, h1, of

3348 kJ/kg. The steam leaves the turbine as a mixture of vapor and liquid having a velocity of 60 m/s and an enthalpy of 2550 kJ/kg. If the flow through the turbine is adiabatic and changes in elevation are negligible, determine the work output involved per unit mass of steam through-flow.

68


( ) in/netshaftin/net12

21

22

12 WQzzg2

VVhhm +=

−+

−+−

kg/kJ797...2

VVhhw

ww2

VVhhm

Ww

22

21

21outnetshaft

innetshaftoutnetshaft

21

22

12innetshaft

innetshaft

==−

+−=

−=

−+−==

The energy equation in terms of enthalpy.=0(Adiabatic flow)

69

Example 5.22 Energy – Temperature Change A 500-ft waterfall involves steady flow from one large body of

water to another. Determine the temperature change associated with this flow.

70


( ) innet12

21

2212

12 Qzzg2

VVppuum =

−+

−+

ρ

−

ρ

+−

waterofheatspecifictheis)Rlbm/(Btu1cwherec

uuTT 1212

°⋅=

−=−

V2=V1

R643.0)]slb/()ftlbm(2.32)][Rlbm/(lbft778[

)ft500)(s/ft2.32(c

)zz(gTT 2

212

12 °=⋅⋅°⋅⋅

=−

=−

The temperature change is related to the change of internal energy of the water

One-dimensional energy equation for steady-in-the-mean flow without shaft work

=0(Adiabatic flow)

71

Energy Equation vs. Bernoulli Equation 1/4

( )innetinoutin

2inin

out

2outout quugz

2Vpgz

2Vp

−−−++ρ

=++ρ

( ) innetinout

2in

2outinout

inout Qzzg2

VVppuum =

−+

−+

ρ

−

ρ

+−

m÷

For steady, incompressible flow…One-dimensional energy equation

m/Qq innetinnet =

in

2in

inout

2out

out z2Vpz

2Vp γ+

ρ+=γ+

ρ+

0quu innetinout =−−

where

For steady, incompressible, frictionless flow…

Bernoulli equation

Frictionless flow…

( )

innetshaftinnet

inout

2in

2out

inoutinout

WQ

zzg2

VVppuum

+=

−+

−+

ρ

−

ρ

+−

72

Energy Equation & Bernoulli Equation 2/4

For steady, incompressible, frictional flow…

0quu innetinout >−−

lossquu innetinout =−−

lossgz2

Vpgz2

Vpin

2inin

out

2outout −++

ρ=++

ρ

Defining “useful or available energy”… gz2

Vp 2

++ρ

Defining “loss of useful or available energy”…

Frictional flow…

73


( ) innetshatfinnetinout

2in

2outinout

inout WQzzg2

VVppuum +=

−+

−+

ρ

−

ρ

+−

m÷ )quu(wgz2

Vpgz2

Vpinnetinoutinnetshaftin

2inin

out

2outout −−−+++

ρ=++

ρ

For steady, incompressible flow with friction and shaft work…

losswgz2

Vpgz2

Vpinnetshaftin

2inin

out

2outout −+++

ρ=++

ρ

g÷Lsin

2inin

out

2outout hhz

g2Vpz

g2Vp

−+++γ

=++γ

Q

W

gm

W

g

wh in/netshaftin/netshaftin/netshaft

S γ=≡=

glosshL =Head lossShaft head

74


For turbineFor pumpThe actual head drop across the turbine

The actual head drop across the pump

)0h(hh TTs >−=

Ps hh = hp is pump headhT is turbine head

TLsT )hh(h +−=

pLsp )hh(h −=

Lsin

2inin

out

2outout hhz

g2Vpz

g2Vp

−+++γ

=++γ

75

Example 5.23 Energy – Effect of Loss of Available Energy Compare the volume flowrates associated with two

different vent configurations, a cylindrical hole in the wall having a diameter of 120 mm and the same diameter cylindrical hole in the wall but with a well-rounded entrance (see Figure E5.23a). The room pressure is held constant at 0.1 kPa above atmospheric pressure. Both vents exhaust into the atmosphere. As discussed in Section 8.4.2. the loss in available energy associated with flow through the cylindrical bent from the room to the vent exit is 0.5V2

2/2 where V2 is the uniformly distributed exit velocity of air. The loss in available energy associated with flow through the rounded entrance vent from the room to the vent exit is 0.05V2

2/2, where V2 is the uniformly distributed exit velocity of air.

76


211

211

2

222 lossgz

2Vpgz

2Vp

−++ρ

=++ρ

For steady, incompressible flow with friction, the energy equationV1=0 No elevation change

[ ]

[ ]2/K1pp

4DVAQ

2/K1ppV

2VKlosslosspp2V

L

212

222

L

212

22

L212121

2

+ρ−π

==

+ρ−

=

=

−

ρ−

=

77

Example 5.24 Energy – Fan Work and Efficiency An axial-flow ventilating fan driven by a motor that delivers 0.4 kW

of power to the fan blades produces a 0.6-m-diameter axial stream of air having a speed of 12 m/s. The flow upstream of the fan involves negligible speed. Determine how much of the work to the air actually produces a useful effects, that is, a rise in available energy and estimate the fluid mechanical efficiency of this fan.

78


++

ρ−

++

ρ=− 1

211

2

222

innetshaft gz2

Vpgz2

Vplossw

For steady, incompressible flow with friction and shaft work…

p1=p2=atmospheric pressure, V1=0, no elevation change

kg/mN0.722

Vlossw2

2innetshaft ⋅==−

innetshaft

innetshaft

wlossw −

=ηEfficiency

kg/mN8.95AV

Wm

Ww innetshaftinnetshaft

innetshaft ⋅=ρ

==

79

Example 5.25 Energy – Head Loss and Power Loss The pump shown in Figure E5.25 adds 10 horsepower to the water

as it pumps water from the lower lake to the upper lake. The elevation difference between the lake surfaces is 30 ft and the head loss is 15 ft. Determine the flowrate and power loss associated with this flow.

80


0VV0pp

hhzg2

Vpzg2

Vp

BABA

LsB

2BB

A

2AA

====

−+++γ

=++γ

The energy equation

( )ft03015/1.88/−+===−+= Q

QW

zzhh innetshaftBALs γ

The pump head

Power loss ...QhW Lloss =γ=

81

Application of Energy Equation to Nonuniform Flows 1/2

∫ ⋅ρ dAnV2

V2

.S.C

α−

α=⋅ρ∫ 2

V~

2V~mdAnV

2V 2

inin2outout

2

CS

1

2

22

3

2

2

≥=

⋅

=∫∫

Vm

dAV

Vm

dAnVV

AA

ρρ

α

If the velocity profile at any section where flow crosses the control surface is not uniform…

For one stream of fluid entering and leaving the control volume….

α is the kinetic energy coefficient and V is the instantaneous velocity.For uniform velocity profile, what is the coefficient ?

????

82

Application of Energy Equation to Nonuniform Flows 2/2

losswgz2Vpgz

2Vp

innetshaftin

2ininin

out

2outoutout −++

α+

ρ=+

α+

ρ

For nonuniform velocity profile…….

ρ×

)loss(wz2

Vpz2

Vp innetshaftin

2inin

inout

2outout

out ρ−ρ+γ+ρα

+=γ+ρα

+

g÷

Linnetshaft

in

2ininin

out

2outoutout h

gw

zg2Vpz

g2Vp

−++α

+γ

=+α

+γ

83

Example 5.26 Energy – Effect of Nonuniform Velocity Profile 1/2

The small fan shown in Figure E5.26 moves air at a mass flowrate of 0.1 kh/min. Upstream of the fan, the pipe diameter is 60 mm, the flow is laminar, the velocity distribution is parabolic, and the kinetic energy coefficient, α1, is equal to 2.0. Downstream of the fan, the pipe diameter is 30 mm, the flow is turbulent, the velocity profile is quite uniform, and the kinetic energy coefficient, α2 , is equal to 1.08. If the rise in static pressure across the fan is 0.1 kPa and the fan motor draws 0.14 W, compare the value of loss calculated: (a) assuming uniform velocity distributions, (2) considering actual velocity distribution.

84

Example 5.26 Energy – Effect of Nonuniform Velocity Profile 2/2

85


losswgz2Vpgz

2Vp

in/netshaft1

2111

2

2222 −++

α+

ρ=+

α+

ρ

The energy equation for non-uniform velocity profile…….

s/m92.1...AmVs/m479.0...

AmV

kg/mN0.84min)/s60(min/kg1.0

]W/)s/mN1)[(W14.0(m

motorfantopowerw

22

11

in/netshaft

==ρ

===ρ

=

⋅=⋅

=

=

2V

2Vppwloss

222

21112

innetshaftα

−α

+

ρ−

−=

86


( )1kg/mN975.02V

2Vppwloss

21

222

21112

in/netshaft

=α=α⋅=

α−

α+

ρ−

−=

( )08.1,2kg/mN940.02V

2Vppwloss

21

222

21112

in/netshaft

=α=α⋅=

α−

α+

ρ−

−=

87

Example 5.28 Energy – Fan Performance For the fan of Example 5.19, show that only some of the shaft power

into the air is converted into a useful effect. Develop a meaningful efficiency equation and a practical means for estimating lost shaft energy.

88


losswgz2

Vpgz2

Vpinnetshaft1

211

2

222 −+++

ρ=++

ρ

++

ρ−

++

ρ=

−=

1

211

2

222

in/netshaft

gz2

Vpgz2

Vp

lossweffectuseful

innetshaft

innetshaft

wlossw −

=ηEfficiency

22innetshaft VUw θ+=

(1)

(2)

(3)

(4)

89


)]gz2/V/p()gz2/V/p[(VU 12

1122

2222 ++ρ−++ρ−=η θ

(2)+(3)+(4)

2212

1122

22 VU/]}gz)2/V()/p[(]gz)2/V()/p{[( θ++ρ−++ρ=η

(2)+(4)

FUNDAMENTALS OF FLUID MECHANICS Chapter 5 Flow Analysis Using …cau.ac.kr/~jjang14/FME/Chap5.pdf · · 2009-09-02Chapter 5 Flow Analysis Using Control Volume. 2 MAIN TOPICS ...

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