Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of...

42
57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 1 Chapter 5 Finite Control Volume Analysis 5.1 Continuity Equation RTT can be used to obtain an integral relationship expressing conservation of mass by defining the extensive property B = M such that β = 1. B = M = mass β = dB/dM = 1 General Form of Continuity Equation ρ + ρ = = CV CS dA V V d dt d 0 dt dM or ρ = ρ CV CS V d dt d dA V net rate of outflow rate of decrease of of mass across CS mass within CV Simplifications: 1. Steady flow: 0 V d dt d CV = ρ 2. V = constant over discrete dA (flow sections): ρ = ρ CS CS A V dA V

Transcript of Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of...

Page 1: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 1

Chapter 5 Finite Control Volume Analysis 5.1 Continuity Equation RTT can be used to obtain an integral relationship expressing conservation of mass by defining the extensive property B = M such that β = 1. B = M = mass β = dB/dM = 1 General Form of Continuity Equation

∫ ∫ ⋅ρ+ρ==CV CS

dAVVddtd0

dtdM

or

∫ρ−=⋅∫ρCVCS

VddtddAV

net rate of outflow rate of decrease of of mass across CS mass within CV Simplifications:

1. Steady flow: 0Vddtd

CV=∫ρ−

2. V = constant over discrete dA (flow sections):

∫ ∑ ⋅ρ=⋅ρCS CS

AVdAV

Page 2: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 2

3. Incompressible fluid (ρ = constant)

CS CV

dV dA dVdt

⋅ = −∫ ∫ conservation of volume

4. Steady One-Dimensional Flow in a Conduit:

∑ =⋅ρCS

0AV

−ρ1V1A1 + ρ2V2A2 = 0 for ρ = constant Q1 = Q2

Some useful definitions: Mass flux ∫ ⋅ρ=

AdAVm&

Volume flux ∫ ⋅=

AdAVQ

Average Velocity A/QV =

Average Density ∫ρ=ρ dAA1

Note: m Q≠ ρ& unless ρ = constant

Page 3: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 3

Example

*Steady flow

*V1,2,3 = 50 fps

*@ V varies linearly from zero at wall to Vmax at pipe center *find 4m& , Q4, Vmax

0 *water, ρw = 1.94 slug/ft3

∫ρ−=∫ =⋅ρCVCS

Vddtd0dAV

i.e., -ρ1V1A1 - ρ2V2A2 + ρ3V3A3 + ρ ∫4A

44dAV = 0

ρ = const. = 1.94 lb-s2 /ft4 = 1.94 slug/ft3

∫ρ= 444 dAVm& = ρV(A1 + A2 – A3) V1=V2=V3=V=50f/s

= ( )222 5.1214

50144

94.1 −+π××

= 1.45 slugs/s

4m&

Page 4: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 4

Q4 = 75.m4 =ρ& ft3/s = ∫

4A44dAV

velocity profile

Q4 = ∫ ∫ θ⎟⎟⎠

⎞⎜⎜⎝

⎛−

πor

0

2

0 omax drrd

rr1V

max

2o

2omax

r

0o

3r

0

2

max

r

0 o

2

max

r

0 omax

Vr31

31

21rV2

r3r

2rV2

drrrrV2

rdrrr1V2

o0

o

o

π=⎥⎦⎤

⎢⎣⎡ −π=

⎥⎥⎦

⎢⎢⎣

⎡−π=

∫ ⎥⎦

⎤⎢⎣

⎡−π=

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛−π=

Vmax = 86.2r

31

Q2

o

4 =π

fps

V4 ≠ V4(θ) dA4

2o

max2o

4r

Vr31

AQV

π

π==

= maxV31

Page 5: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 5

5.2 Momentum Equation Derivation of the Momentum Equation Newton’s second law of motion for a system is

time rate of change of the momentum of the system

= sum of external forces acting on the system

Since momentum is mass times velocity, the momentum of a small particle of mass is and the momentum of the entire system is . Thus,

Recall RTT:

With and ,

Page 6: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 6

Thus, the Newton’s second law becomes

where, is fluid velocity referenced to an inertial frame (non-

accelerating) is the velocity of CS referenced to the inertial frame is the relative velocity referenced to CV ∑ ∑ ∑ is vector sum of all forces acting on

the CV

is body force such as gravity that acts on the entire mass/volume of CV

is surface force such as normal (pressure and

viscous) and tangential (viscous) stresses acting on the CS

Page 7: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 7

Note that, when CS cuts through solids, may also include reaction force, (e.g., the reaction force required to hold nozzle or bend when CS cuts through the bolts that are holding the nozzle/bend in place)

∑ ∑ = resultant force on fluid in CV due to and , i.e. reaction force on fluid

Free body diagram

Page 8: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 8

Important Features (to be remembered) 1) Vector equation to get component in any direction must use

dot product x equation

∑ ∫ ⋅ρ+∫ρ=CS

RCV

x dAVuVuddtdF

y equation

∑ ∫ ⋅ρ+∫ρ=CS

RCV

y dAVvVvddtdF

z equation

∑ ∫ ⋅ρ+∫ρ=CS

RCV

z dAVwVwddtdF

2) Carefully define control volume and be sure to include all

external body and surface faces acting on it. For example,

(Rx,Ry) = reaction force on fluid

(Rx,Ry) = reaction force on nozzle

Carefully define coordinate system with forces positive in positive direction of coordinate axes

Page 9: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 9

3) Velocity V and Vs must be referenced to a non-accelerating

inertial reference frame. Sometimes it is advantageous to use a moving (at constant velocity) reference frame: relative inertial coordinate. Note VR = V – Vs is always relative to CS.

4) Steady vs. Unsteady Flow

Steady flow ⇒ ∫ =ρCV

0VdVdtd

5) Uniform vs. Nonuniform Flow

∫ ⋅ρ

CSR dAVV = change in flow of momentum across CS

= ΣVρVR⋅A uniform flow across A 6) Fpres = − ∫ dAnp ∫ ∫=∇

V SdsnfVfd

f = constant, ∇f = 0 = 0 for p = constant and for a closed surface i.e., always use gage pressure 7) Pressure condition at a jet exit

at an exit into the atmosphere jet pressure must be pa

Page 10: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10

Applications of the Momentum Equation Initial Setup and Signs

1. Jet deflected by a plate or a vane 2. Flow through a nozzle 3. Forces on bends 4. Problems involving non-uniform velocity distribution 5. Motion of a rocket 6. Force on rectangular sluice gate 7. Water hammer 8. Steady and unsteady developing and fully developed pipe

flow 9. Empting and filling tanks 10. Forces on transitions 11. Hydraulic jump 12. Boundary layer and bluff body drag 13. Rocket or jet propulsion 14. Propeller

Page 11: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 11

1. Jet deflected by a plate or vane Consider a jet of water turned through a horizontal angle

x-equation: ∑ ∫ ∫ ⋅ρ+ρ==

CSxx dAVuVud

dtdFF

steady flow = )AV(V)AV(V 22x211x1 ρ+−ρ continuity equation: ρA1V1 = ρA2V2 = ρQ Fx = ρQ(V2x – V1x) y-equation: ∑ ∑ ⋅ρ==

CSyy AVvFF

Fy = ρV1y(– A1V1) + ρV2y(– A2V2) = ρQ(V2y – V1y) for above geometry only where: V1x = V1 V2x = -V2cosθ V2y = -V2sinθ V1y = 0 note: Fx and Fy are force on fluid - Fx and -Fy are force on vane due to fluid

∑ ⋅ρ=CS

x AVuF

CV and CS are for jet so that Fx and Fy are vane reactions forces on fluid

for A1 = A2 V1 = V2

Page 12: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 12

If the vane is moving with velocity Vv, then it is convenient to choose CV moving with the vane i.e., VR = V - Vv and V used for B also moving with vane x-equation: ∫ ⋅ρ=

CSRx dAVuF

Fx = ρV1x[-(V – Vv)1A1] + ρV2x[(V – Vv)2A2] Continuity: 0 = ∫ ⋅ρ dAVR

i.e., ρ(V-Vv)1A1 = ρ(V-Vv)2A2 = ρ(V-Vv)A Qrel Fx = ρ(V-Vv)A[V2x – V1x]

Qrel on fluid V2x = (V – Vv)2x V1x = (V – Vv)1x Power = -FxVv i.e., = 0 for Vv = 0 Fy = ρQrel(V2y – V1y)

For coordinate system moving with vane

Page 13: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 13

2. Flow through a nozzle Consider a nozzle at the end of a pipe (or hose). What force is required to hold the nozzle in place? Assume either the pipe velocity or pressure is known. Then, the unknown (velocity or pressure) and the exit velocity V2 can be obtained from combined use of the continuity and Bernoulli equations.

Bernoulli: 2222

2111 V

21zpV

21zp ρ+γ+=ρ+γ+ z1=z2

22

211 V

21V

21p ρ=ρ+

Continuity: A1V1 = A2V2 = Q

1

2

12

12 V

dDV

AAV ⎟

⎠⎞

⎜⎝⎛==

0dD1V

21p

42

11 =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−ρ+

Say p1 known: ( )

2/1

41

1

dD1

p2V⎥⎥⎥

⎢⎢⎢

⎟⎠⎞⎜

⎝⎛ −ρ

−=

To obtain the reaction force Rx apply momentum equation in x-direction

CV = nozzle and fluid ∴ (Rx, Ry) = force required to hold nozzle in place

Page 14: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 14

∫ ⋅ρ+∫ ρ=∑CSCV

x dAVuVdudtdF

=∑ ⋅ρCS

AVu

Rx + p1A1 – p2A2 = ρV1(-V1A1) + ρV2(V2A2) = ρQ(V2 - V1) Rx = ρQ(V2 - V1) - p1A1 To obtain the reaction force Ry apply momentum equation in y-direction ∑ ∑ =⋅ρ=

CSy 0AVvF since no flow in y-direction

Ry – Wf − WN = 0 i.e., Ry = Wf + WN Numerical Example: Oil with S = .85 flows in pipe under pressure of 100 psi. Pipe diameter is 3” and nozzle tip diameter is 1”

V1 = 14.59 ft/s V2 = 131.3 ft/s Rx = 141.48 – 706.86 = −569 lbf Rz = 10 lbf This is force on nozzle

steady flow and uniform flow over CS

65.1g

S =γ=ρ

D/d = 3

Q = 2

2

V121

4⎟⎠⎞

⎜⎝⎛π

= .716 ft3/s

Page 15: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 MeProfessor F

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Page 16: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 MeProfessor F

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Page 17: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 17

⎟⎟⎠

⎞⎜⎜⎝

⎛−ρ

12

2

y1

y1

bQ

5. Application of relative inertial coordinates for a moving but non-deforming control volume (CV) The CV moves at a constant velocity CSV with respect to the absolute inertial coordinates. If RV represents the velocity in the relative inertial coordinates that move together with the CV, then: R CSV V V= − Reynolds transport theorem for an arbitrary moving deforming CV:

SYS

RCV CS

dB d d V n dAdt dt

βρ βρ= ∀ + ⋅∫ ∫

For a non-deforming CV moving at constant velocity, RTT for incompressible flow:

syst

RCV CS

dBd V ndA

dt tβρ ρ β∂= ∀ + ⋅

∂∫ ∫

1) Conservation of mass systB M= , and 1β = :

RCS

dM V ndAdt

ρ= ⋅∫

For steady flow: 0R

CS

V ndA⋅ =∫

Page 18: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 18

2) Conservation of momentum

( )CSsyst RB M V V= + and syst R CSdB dM V Vβ = = +

( ) ( ) ( )[ ]CS CSR RCSR R

CV CS

d M V V V VF d V V V ndA

dt tρ ρ

+ ∂ += = ∀ + + ⋅

∂∑ ∫ ∫

For steady flow with the use of continuity:

( )CSR RCS

CSR R RCS CS

F V V V ndA

V V ndA V V ndA

ρ

ρ ρ

= + ⋅

= ⋅ + ⋅

∑ ∫

∫ ∫0

R RCS

F V V ndAρ= ⋅∑ ∫

Example (use relative inertial coordinates): Ex) A jet strikes a vane which moves to the right at constant velocity on a

frictionless cart. Compute (a) the force required to restrain the cart and (b) the power delivered to the cart. Also find the cart velocity for which (c) the force is a maximum and (d) the power is a maximum.

Page 19: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 19

Solution:

Assume relative inertial coordinates with non-deforming CV i.e. CV moves

at constant translational non-accelerating

then R CSV V V= − . Also assume steady flow with and neglect gravity effect. Continuity:

0RCS

V ndA⋅ =∫

Bernoulli without gravity:

1p0 2

1 212 RV pρ+ =

0 22

12 RVρ+

1 2R RV V=

Since 1 1 2 2R RV A V Aρ ρ=

1 2 jA A A= = Momentum: ∑

0 0

Page 20: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 20

cos 1 cos

1 cos

1 cos , 0 0 2 1 cos 2 1 cos

4 3 1 cos 0 3 4 0 4 16 126 4 26

3

3 23 1 cos 427 1 cos

Page 21: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 21

5.3 Energy Equation Derivation of the Energy Equation The First Law of Thermodynamics The difference between the heat added to a system and the work done by a system depends only on the initial and final states of the system; that is, depends only on the change in energy E: principle of conservation of energy ΔE = Q – W ΔE = change in energy Q = heat added to the system W = work done by the system E = Eu + Ek + Ep = total energy of the system potential energy kinetic energy The differential form of the first law of thermodynamics expresses the rate of change of E with respect to time

WQdtdE && −=

rate of work being done by system

rate of heat transfer to system

Internal energy due to molecular motion

Page 22: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 22

Energy Equation for Fluid Flow The energy equation for fluid flow is derived from Reynolds transport theorem with Bsystem = E = total energy of the system (extensive property) β = E/mass = e = energy per unit mass (intensive property) = u + ek + ep

∫ ⋅ρ+∫ ρ= CSCV dAVeVeddtd

dtdE

ˆ ˆ( ) ( )k p k pCV CS

dQ W u e e dV u e e V dAdt

ρ ρ− = + + + + + ⋅∫ ∫& & This can be put in a more useable form by noting the following:

2V

M2/MV

massVvelocitywithmassofKETotale

22

k =Δ

Δ==

gzVzV

ME

e pp =

ΔρΔγ=

Δ= (for Ep due to gravity only)

2 2

ˆ ˆ2 2CV Cs

d V VQ W gz u dV gz u V dAdt

ρ ρ⎛ ⎞ ⎛ ⎞− = + + + + + ⋅⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∫ ∫& &

rate of work rate of change flux of energy done by system of energy in CV out of CV (ie, across CS) rate of heat transfer to sysem

VV2 =

Page 23: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 23

System at time t + Δt

System at time t

CS

Rate of Work Components: fs WWW &&& += For convenience of analysis, work is divided into shaft work Ws and flow work Wf Wf = net work done on the surroundings as a result of

normal and tangential stresses acting at the control surfaces

= Wf pressure + Wf shear Ws = any other work transferred to the surroundings

usually in the form of a shaft which either takes energy out of the system (turbine) or puts energy into the system (pump)

Flow work due to pressure forces Wf p (for system)

Work = force × distance

at 2 W2 = p2A2 × V2Δt rate of work⇒ 2222222 AVpVApW ⋅==&

at 1 W1 = −p1A1 × V1Δt 1111 AVpW ⋅=&

Note: here V uniform over A

(on surroundings)

neg. sign since pressure force on surrounding fluid acts in a direction opposite to the motion of the system boundary

CV

Page 24: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 24

In general, AVpWfp ⋅=& for more than one control surface and V not necessarily uniform over A:

∫ ⋅⎟⎠

⎞⎜⎝

⎛ρ

ρ=∫ ⋅= CSCSfp dAVpdAVpW&

fshearfpf WWW &&& += Basic form of energy equation

2 2

ˆ ˆ2 2

s fshear CS

CV CS

pQ W W V dA

d V Vgz u dV gz u V dAdt

ρρ

ρ ρ

⎛ ⎞− − − ⋅⎜ ⎟⎝ ⎠⎛ ⎞ ⎛ ⎞

= + + + + + ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫ ∫

& & &

2

2

ˆ2

ˆ2

s fshear CV

CS

d VQ W W gz u dVdt

V pgz u V dA

ρ

ρρ

⎛ ⎞− − = + +⎜ ⎟

⎝ ⎠⎛ ⎞

+ + + + ⋅⎜ ⎟⎝ ⎠

& & &

h=enthalpy

Usually this term can be eliminated by proper choice of CV, i.e. CS normal to flow lines. Also, at fixed boundaries the velocity is zero (no slip condition) and no shear stress flow work is done. Not included or discussed in text!

Page 25: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

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s no accehydrosta

Energy

One-Dim

pe system

w)

u V dA⎞

⋅⎟⎠

11 1

2 2 2

AV A

V A

ρ

ρ

+

+

ies acroto be

elerationatically d

Equati

mensiona

m as sho

2

31 1

1

32 2

2

2A

V dA

V

ρ

ρ∫

oss the e straign normaldistribute

ion

l Pipe Fl

wn

2dA

flow seght andl to the ed, i.e.,

Chapter 5 25

low

ections d paralstreamlip/ρ +gz

5

the llel; ines z =

Page 26: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 MeProfessor F

*Furthconstaquanti

2

sQ W

⎛= ⎜⎝

& &

RecallSo tha

Define

i.e.,

Q W−& &

(1 Qm

−&&

Nnote

lamina

echanics of FluFred Stern Fa

hermoreant acroities can

1

2 ˆ

spW

gz u

ρ⎛+ +⎜⎝

+ +

l that at

e:

=αA1

1pW gρ

⎛⎜+ +⎜⎝

) pWρ

− +&

e that:

ar flow α

uids and Transpall 2009

, the inoss the f

then be

2

1 1

2

ˆ

ˆA

gz u

u Vρ

⎞+ +⎠

⎞⎟⎠∫

VQ ∫=dV∫ ⋅ρ

d2

VA

3

∫ρ

K.E. flu

∫ ⎟⎠⎞

⎜⎝⎛

A

3

VV

A1

1 1gz u+ +

11

p gzρ

+ +

α α > 1 i

α = 2

port Processes

nternal flow sectaken ou

11

2 2

AV dA

V dA

ρ

ρ

⎞⎟⎠

+

VdAV =⋅AVdA ρ=

dA ρα=

ux K.3

dA = k

21

1 2V mα

⎞⎟+⎟⎠

1 1ˆ2Vu α+

= 1 if Vf V is no

turbule

energy ctions, iutside th

1

2

1

32

2

A

A

A

V dA

ρ

ρ

+ ∫

AV mA &=

2AV3

α=ρ

.E. flux for

kinetic en

2pmρ

⎛⎜= +⎜⎝

&

21 2

2pVρ

= +

V is constonunifor

ent flow α

u can i.e. T =he integr

31

1

2

2V dA

A

m

m2

V2

r V= V =co

nergy co

2 2ˆgz u+ +

2 2ˆgz u+ +

tant acrorm

α = 1.05

be con= constaral sign t

mass flow

onstant acro

orrection

22

2 2Vα

⎞+

⎠22

2 2 2Vα+

oss the fl

∼ 1 may

Chapter 5 26

nsidered ant. Thto yield

w rate

oss pipe

factor

m⎞⎟⎟⎠&

flow sect

be used

6

as hese

tion

Page 27: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 27

Shaft Work Shaft work is usually the result of a turbine or a pump in the flow system. When a fluid passes through a turbine, the fluid is doing shaft work on the surroundings; on the other hand, a pump does work on the fluid pts WWW &&& −= where tW& and pW& are

magnitudes of power ⎟⎠⎞

⎜⎝⎛

timework

Using this result in the energy equation and deviding by g results in

2 2

1 1 2 2 2 11 1 2 2

ˆ ˆ2 2

p tW Wp V p V u u Qz zmg mg g mg

α αγ γ

−+ + + = + + + + −& &&

& & &

mechanical part thermal part Note: each term has dimensions of length Define the following:

QW

QgW

gmW

h pppp γ

==&&

&

&

gmWh t

t&

&=

2 1ˆ ˆ

Lu u Qh head lossg mg−= − =

&

&

Page 28: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 28

Head Loss In a general fluid system a certain amount of mechanical energy is converted to thermal energy due to viscous action. This effect results in an increase in the fluid internal energy. Also, some heat will be generated through energy dissipation and be lost (i.e. -Q& ). Therefore the term from 2nd law

2 1ˆ ˆ

0Lu u Qhg gm−= − >

&

&

Note that adding Q& to system will not make hL = 0 since this also increases Δu. It can be shown from 2nd law of thermodynamics that hL > 0. Drop ⎯ over V and understand that V in energy equation refers to average velocity. Using the above definitions in the energy equation results in (steady 1-D incompressible flow)

Lt2

22

22

p1

21

11 hhz

g2Vphz

g2Vp +++α+

γ=++α+

γ

form of energy equation used for this course!

represents a loss in mechanical energy due to viscous stresses

Page 29: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 MeProfessor F

Comp Apply

Energ

•If hL conser •Therinterpreffects Summ The en B = E β = e =

echanics of FluFred Stern Fa

parison o

y energy

gy eq :

= 0 (i.e.rvation o

refore, enreted as s (hL) an

mary of th

nergy eq

= total e

= E/M =

uids and Transpall 2009

of Energy

equation

1

2Vp +

γ

., μ = 0) of mecha

nergy eqa modif

nd shaft w

he Energ

quation i

energy o

= energy

port Processes

y Equati

n to a str

1

21 zg

V =+

we get Banical en

quation ffied Bernwork (hp

gy Equa

is derive

of the sy

y per uni

Infi

ion and B

ream tub

222

g2Vp +

γ=

Bernoullnergy alo

for steadynoulli eqp or ht)

tion

d from R

stem

it mass

initesimal s

Bernoull

be witho

2

22 zg

++

li equationg a str

y 1-D piquation t

RTT wit

stream tub

li Equati

out any s

Lh

on and reamline

ipe flow o includ

th

be ⇒ α1=

Chapter 5 29

ion

haft wor

e

can be de viscou

=α2=1

9

rk

us

Page 30: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 30

heat add

Neglected in text presentation

= u + 2V21 +gz

internal KE PE

WQdAVeVeddtd

dtdE

CSCV

&& −=∫ ⋅ρ+∫ρ=

vps WWWW &&&& ++=

( )∫ ⋅ρρ∫ =⋅=CSCV

p dAVpdAVpW&

pts WWW &&& −=

( )∫ ⋅+ρ+∫ρ=+−CSCV

pt dAVepeVeddtdWWQ &&&

21ˆ2

e u V gz= + + For steady 1-D pipe flow (one inlet and one outlet): 1) Streamlines are straight and parallel

⇒ p/ρ +gz = constant across CS

work done

from 1st Law of Thermodynamics

shaft work done on or by system (pump or turbine)

pressure work done

on CS

Viscous stress work on CS

Page 31: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 31

mechanical energy Thermal energy

Note: each term has

units of length

V is average velocity (vector dropped) and corrected by α

2) T = constant ⇒ u = constant across CS

3) define ∫ ⎟⎠⎞

⎜⎝⎛=α

CS

3

dAVV

A1 = KE correction factor

⇒ ∫ α=ρα=ρ m2

VA2VdAV

2

233 &

Lt2

22

22

p1

21

11 hhz

g2Vphz

g2Vp +++α+

γ=++α+

γ

gmWh pp &&=

gmWh tt &&=

2 1ˆ ˆ

Lu u Qhg mg−= − =

&

& head loss

> 0 represents loss in mechanical energy due to viscosity

Page 32: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

EGL1 =for hp =

57:020 MeProfessor F

Conc

1p α+γ

Define

HE

pressu

stagna

EGL1 EGL2

= EGL2 + hL= ht = 0

echanics of FluFred Stern Fa

cept of H

21

1 zg2

V +α

e HGL

EGL

HGL corEGL corr

ure tap:

ation tub

+ hp = E= EGL1

L

uids and Transpall 2009

abrupchange dto hp or

Hydrau

p1 hz =+

L = p +γ

L = zp +γ

rrespondresponds

hp2 =γ

be: p2 +γ

EGL2 + h + hp − h

port Processes

pt due r ht

ulic and

2p α+γ

=

z

g2Vz

2α+

ds to press to stag

g2V2

2 =α+

ht + hL ht − hL

E

DLf

d Energ

22

2 zg2

V +

2

ssure tapgnation tu

h

pagd

EGL = HGL

g2V2

gy Gra

t2 hz ++

p measurube mea

point-byapplicatigraphicadisplayed

h = h ta

if V = 0

hL = f

i.e., liV, an

de Line

Lh+

rement +asuremen

-point ion is ally d

eight of ap/tube

g2

2VDLf

inear variatiod f constant

f = ff = f

Chapter 5 32

es

+ z nt + z

f fluid in

on in L for D

friction factof(Re)

2

D,

or

Page 33: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

1p+

γHGL

Lh =

57:020 MeProfessor F

Helpfu 1. EG

2.&3.

g2

21V

1z =++

L2 = EGL1 - h

g2

2V

= for ab

echanics of FluFred Stern Fa

ful hints

GL = HG

DLfhL =

downwa

2z2p+

γ=

hL

brupt expans

uids and Transpall 2009

for draw

GL + αV

g2V

DL 2

in

ard, exce

Lhg2

22V

++

sion

port Processes

wing HG

V2/2g = H

n pipe me

ept for a

L

L and E

HGL for

eans EG

abrupt ch

GL

V = 0

GL and H

hanges d

HGL will

due to ht

Chapter 5 33

l slope

or hp

3

Page 34: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 MeProfessor F

4. p =

5. for

EG

6. for

i.e.

echanics of FluFred Stern Fa

0 ⇒ HG

DLfhL =

L/HGL

change i

. V1A

V1π

1DV

uids and Transpall 2009

GL = z

g2V

DL 2

=

slope do

in D ⇒

A1 = V2A

V4D2

1 =π

121 DVD =

port Processes

constant

ownward

change i

A2

4DV

22

22D

t × L

d

in V

chHch

i.e., line

increasi

hange inHGL & Ehange du

early incr

ng L wit

n distanceEGL and ue to cha

Chapter 5 34

reased fo

th slope

e betweeslope

ange in h

4

for

g2V

Df 2

en

hL

Page 35: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 MeProfessor F

7. If H

c

gage p

echanics of FluFred Stern Fa

HGL < z

condition

pressure

uids and Transpall 2009

z then p/

n for cav

vapp =

g,vap =

p g,va ≈γ

port Processes

/γ < 0

vitation:

a 2000=

aA pp −=

m10−≈

i.e., cav

2mN0

aatm p−≈

9810 N/

vitation p

atm 10−=

/m3

possible

mN000,00

Chapter 5 35

2mN

5

Page 36: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 MeProfessor F

echanics of FluFred Stern Fa

uids and Transpall 2009

port Processes Chapter 5 36

6

Page 37: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 MeProfessor F

echanics of FluFred Stern Fa

uids and Transpall 2009

port Processes Chapter 5 377

Page 38: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 38

Application of the Energy, Momentum, and Continuity Equations in Combination In general, when solving fluid mechanics problems, one should use all available equations in order to derive as much information as possible about the flow. For example, consistent with the approximation of the energy equation we can also apply the momentum and continuity equations Energy:

Lt2

22

22

p1

21

11 hhz

g2Vphz

g2Vp +++α+

γ=++α+

γ

Momentum: ( )∑ −ρ=ρ−ρ= 121

212

22s VVQAVAVF

Continuity: A1V1 = A2V2 = Q = constant

one inlet and one outlet ρ = constant

Page 39: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 MeProfessor F

AbrupConsito kno

separa Apply

1 zp +γ

Mome ∑ =Fs

next d

echanics of FluFred Stern Fa

pt Expander the f

ow hL = h

ation, i.e

y Energy2

11 g2

Vz =+

entum eq

−= Ap 21

divide m

uids and Transpall 2009

nsion flow fromhL(V1,V

e, p1, an a

y Eq from2 zp +

γ=

q. For CV

−− Ap 22

W siomentum

LA2Δγ

port Processes

m a sma2). Anal

approxim

m 1-2 (α22

2 g2V ++

V shown

αsinW

in α m equati

LzΔ

all pipe tolytic solu

mate solu

α1 = α2 =

Lh+

n (shear

∑ρ= Vu

= (V1 −ρ= 2

2 AVρ

ion by γA

o a largeution to

extremto theflow turbuthe asthat thseparremaiconstvalue

ution for

= 1)

stress ne

⋅ AV

)AV 11 +−2

12 VA ρ−

A2

er pipe. exact prmely dife occurreseparati

ulence. Hssumptiohe pressration regins apprtant and e at the pr hL is po

eglected

V(V 22ρ+

1A

Chapter 5 39

Would lroblem isfficult duence of ons and Howeveron is masure in thgion oximateat the

point of ossible:

d)

)A22

9

like s ue

r, if ade he

ly

Page 40: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 40

÷ γA2 ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=−−

γ−

γ1

AA

AA

gV

AA

gV

gVzzpp

2

1

2

12

1

2

12

122

2121

from energy equation ⇓

2

12

122

L

21

22

AA

gV

gVh

g2V

g2V −=+−

⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

2

12

122

L AA21

g2V

g2Vh

⎥⎦

⎤⎢⎣

⎡−+=

2

121

21

22L A

AV2VVg2

1h

−2V1V2

[ ]212L VV

g21h −=

If 2 1V V ,

2L 1

1h = V2g

continutity eq. V1A1 = V2A2

1

2

2

1

VV

AA =

Page 41: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 MeProfessor F

Forces Examp

First a ∑ =Fx Fx + p Fx = ρ

f

echanics of FluFred Stern Fa

s on Tra

ple 7-6

apply mo

∑ρ= Vu

p1A1 − p2

ρQ(V2 −

force req

uids and Transpall 2009

ansitions

omentum

⋅A

2A2 = ρV

V1) − p1

quired to

port Processes

m theorem

V1(−V1A

1A1 + p2A

hold tra

m

A1) + ρV2

A2

ansition i

Q = .7

head lo

(empir Fluid p1 = 2D1 = 3D2 = 2Fx = ?

2(V2A2)

in place

707 m3/s

oss = V1.

rical equ

= water 250 kPa 30 cm 20 cm ?

Chapter 5 41

g2V2

2

uation)

1

Page 42: Chapter 5 10132009user.engineering.uiowa.edu/.../Chapter_5_10132009.pdf · 57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 10 Applications

57:020 Mechanics of Fluids and Transport Processes Chapter 5 Professor Fred Stern Fall 2009 42

The only unknown in this equation is p2, which can be obtained from the energy equation.

L

222

211 h

g2Vp

g2Vp ++

γ=+

γ note: z1 = z2 and α = 1

⎥⎦

⎤⎢⎣

⎡+−γ−= L

21

22

12 hg2

Vg2

Vpp drop in pressure

⇒ ( ) 11L

21

22

1212x Aphg2

Vg2

VpAVVQF −⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−γ−+−ρ=

p2 In this equation, V1 = Q/A1 = 10 m/s V2 = Q/A2 = 22.5 m/s

m58.2g2

V1.h22

L ==

Fx = −8.15 kN is negative x direction to hold

transition in place

(note: if p2 = 0 same as nozzle)

continuity A1V1 = A2V2

12

12 V

AAV =

i.e. V2 > V1