Finite Control Volume Analysis

Finite Control Volume AnalysisFinite Control Volume Analysis

CVEN 311CVEN 311

Application of Reynolds Transport Theorem

Moving from a System to a Finite Control Volume

Moving from a System to a Finite Control Volume

MassLinear MomentumMoment of MomentumEnergyPutting it all together!

MassLinear MomentumMoment of MomentumEnergyPutting it all together!

Conservation of MassConservation of Mass

BB = Total amount of ____ in the system = Total amount of ____ in the systembb = ____ per unit mass = __ = ____ per unit mass = __ BB = Total amount of ____ in the system = Total amount of ____ in the systembb = ____ per unit mass = __ = ____ per unit mass = __

ˆsys

cv cs

DMdV dA

Dt tr r

¶= + ×¶ ò ò V n̂sys

cv cs

DMdV dA

Dt tr r

¶= + ×¶ ò ò V n

ˆcs cv

dA dVt

r r¶

× =-¶ò òV n̂

cs cv

dA dVt

r r¶

× =-¶ò òV n

massmass11massmass

But DMsys/Dt = 0!But DMsys/Dt = 0!

cv equationcv equation

mass leaving - mass entering = - rate of increase of mass in cvmass leaving - mass entering = - rate of increase of mass in cv

ˆsys

cv cs

DBbdV b dA

Dt tr r

¶= + ×¶ ò ò V n̂sys

cv cs

DBbdV b dA

Dt tr r

¶= + ×¶ ò ò V n

Continuity Equation

Conservation of MassConservation of Mass

1 2

1 1 1 2 2 2ˆ ˆ 0cs cs

dA dAr r× + × =ò òV n V n1 2

1 1 1 2 2 2ˆ ˆ 0cs cs

dA dAr r× + × =ò òV n V n

1122

VV11AA11

If mass in cv is constantIf mass in cv is constant

Unit vector is ______ to surface and pointed ____ of cv

Unit vector is ______ to surface and pointed ____ of cv

ˆcs cv

dA dVt

r r¶

× =-¶ò òV n̂

cs cv

dA dVt

r r¶

× =-¶ò òV n

n̂ normal

out

ˆcs

dAr × =ò V n̂cs

dAr × =ò V n mm

ˆcs

dA

VA

×

=òV n̂cs

dA

VA

×

=òV n

VAr =

n̂

We assumed uniform ___ on the control surface

is the spatially averaged velocity normal to the csVV

[M/T][M/T]

Continuity Equation for Constant Density and Uniform Velocity

Continuity Equation for Constant Density and Uniform Velocity

1 21 1 2 2 0V A V Ar r- + =1 21 1 2 2 0V A V Ar r- + =

1 21 2V A V A Q= =1 21 2V A V A Q= =

1 2

1 1 1 2 2 2ˆ ˆ 0cs cs


1 1 1 2 2 2ˆ ˆ 0cs cs


Density is constant across cs

Density is the same at cs1 and cs2[L3/T][L3/T]

Simple version of the continuity equation for conditions of constant density. It is understood that the velocities are either ________ or _______ ________.

1 1 2 2V A V A Q= =1 1 2 2V A V A Q= =

uniform spatially averaged

Example: Conservation of Mass?Example: Conservation of Mass?

The flow out of a reservoir is 2 L/s. The flow out of a reservoir is 2 L/s. The reservoir surface is 5 m x 5 m. The reservoir surface is 5 m x 5 m. How fast is the reservoir surface How fast is the reservoir surface dropping?dropping?

resAQ

dtdh

resAQ

dtdh

hh

dt

dhAQ res

out dt

dhAQ res

out ExampleExample

Constant densityConstant density

ˆcs cv

dA dVt

r r¶

× =-¶ò òV n̂

cs cv

dA dVt

r r¶

× =-¶ò òV n

ˆcs

VdA

t¶

× =-¶òV n̂

cs

VdA

t¶

× =-¶òV n

out in

dVQ Q

dt- =-out in

dVQ Q

dt- =- Velocity of the reservoir surface

Linear Momentum EquationLinear Momentum Equation

m=B Vm=B Vmm

=V

bmm

=V

b

R.T.T.R.T.T.

momentummomentum momentum/unit massmomentum/unit mass

Steady stateSteady state

ˆsys

cv cs

DBbdV b dA

Dt tr r

¶= + ×¶ ò ò V n̂sys

cv cs

DBbdV b dA

Dt tr r

¶= + ×¶ ò ò V n

ˆcv cs

DmdV dA

Dt tr r

¶= + ×¶ ò ò

VV V V n̂

cv cs

DmdV dA

Dt tr r

¶= + ×¶ ò ò

VV V V n

ˆcs

DmdA

Dtr= ×ò

VV V n̂

cs

DmdA

Dtr= ×ò

VV V n

0F ¹å 0F ¹å


( ) ( )1 1 1 1 2 2 2 2

DmV A V A

Dtr r=- +

VV V( ) ( )1 1 1 1 2 2 2 2

DmV A V A

Dtr r=- +

VV V

111111 VVM QAV 111111 VVM QAV

222222 VVM QAV 222222 VVM QAV

AssumptionsAssumptionsAssumptionsAssumptions

Vectors!!!Vectors!!!

Uniform densityUniform densityUniform velocityUniform velocity

V AV ASteadySteady

ˆcs

DmdA

Dtr= ×ò

VV V n̂

cs

DmdA

Dtr= ×ò

VV V n

1 2

1 1 1 1 2 2 2 2ˆ ˆcs cs

DmdA dA

Dtr r= × + ×ò ò

VV V n V V n

1 2

1 1 1 1 2 2 2 2ˆ ˆcs cs

DmdA dA

Dtr r= × + ×ò ò

VV V n V V n

V fluid velocity relative to cv

V fluid velocity relative to cv

( )1 2

D m

Dt= = +å

VF M M

( )1 2

D m

Dt= = +å

VF M M

Steady Control Volume Form of Newton’s Second Law

Steady Control Volume Form of Newton’s Second Law

What are the forces acting on the fluid in the control volume?

What are the forces acting on the fluid in the control volume?

21 MMF 21 MMF

1 2 wall wallp p p t= + + + +å F F F F FW1 2 wall wallp p p t= + + + +å F F F F FW

GravityGravityShear forces at the wallsShear forces at the wallsPressure forces at the wallsPressure forces at the wallsPressure forces on the endsPressure forces on the ends

Why no shear on control surfaces? _______________________________No velocity gradient normal to surface

Resultant Force on the Solid Surfaces

Resultant Force on the Solid Surfaces

The shear forces on the walls and the pressure forces on the walls are generally the unknowns

Often the problem is to calculate the total force exerted by the fluid on the solid surfaces

The magnitude and direction of the force determines size of _____________needed to keep pipe in

place force on the vane of a pump or turbine...

The shear forces on the walls and the pressure forces on the walls are generally the unknowns

Often the problem is to calculate the total force exerted by the fluid on the solid surfaces

The magnitude and direction of the force determines size of _____________needed to keep pipe in

place force on the vane of a pump or turbine...

1 2p p ss= + + +å F F F FW1 2p p ss= + + +å F F F FW wallwall

FFpss Fwallwall

FFpss F

=force applied by solid surfaces=force applied by solid surfaces

thrust blocksthrust blocks


1 2p p ss= + + +å F F F FW1 2p p ss= + + +å F F F FW

21 MMF 21 MMF 1 21 2 p p ss+ = + + +M M F F FW1 21 2 p p ss+ = + + +M M F F FW

The momentum vectors have the same direction as the velocity vectors

The momentum vectors have the same direction as the velocity vectors

Fp1Fp1

Fp2Fp2

M1M1

M2M2

FssyFssy

FssxFssx

W

Reducing elbow in vertical plane with water flow Reducing elbow in vertical plane with water flow of 300 L/s. The volume of water in the elbow is 200 of 300 L/s. The volume of water in the elbow is 200 L.L.Energy loss is negligible.Energy loss is negligible.Calculate the force of the elbow on the fluid.Calculate the force of the elbow on the fluid.WW = _________ = _________

section 1section 1 section 2section 2DD 50 cm50 cm 30 cm 30 cmAA __________________ __________________V __________________ __________________pp 150 kPa150 kPa __________________MM __________________ __________________FFpp __________________ __________________

Reducing elbow in vertical plane with water flow Reducing elbow in vertical plane with water flow of 300 L/s. The volume of water in the elbow is 200 of 300 L/s. The volume of water in the elbow is 200 L.L.Energy loss is negligible.Energy loss is negligible.Calculate the force of the elbow on the fluid.Calculate the force of the elbow on the fluid.WW = _________ = _________

section 1section 1 section 2section 2DD 50 cm50 cm 30 cm 30 cmAA __________________ __________________V __________________ __________________pp 150 kPa150 kPa __________________MM __________________ __________________FFpp __________________ __________________

Example: Reducing ElbowExample: Reducing Elbow

1 21 2 p p ss+ = + + +M M F F FW1 21 2 p p ss+ = + + +M M F F FW

1111

22

1 m1 m

y

xDirection of V vectorsDirection of V vectors

0.196 m20.196 m2 0.071 m20.071 m2

1.53 m/s ↑1.53 m/s ↑ 4.23 m/s →4.23 m/s →

-459 N ↑-459 N ↑ 1269 N →1269 N →29,400 N ↑29,400 N ↑

-1961 N ↑-1961 N ↑

??

??

Example: What is p2?Example: What is p2?

2 21 1 2 2

1 21 22 2

p V p Vz z

g gg g+ + = + +

2 21 1 2 2

1 21 22 2

p V p Vz z

g gg g+ + = + +

2 21 2

2 1 1 2 2 2V V

p p z zg g

gé ù

= + - + -ê úë û

2 21 2

2 1 1 2 2 2V V

p p z zg g

gé ù

= + - + -ê úë û

( ) ( ) ( )( )

( )( )

2 23 3

2 2 2

1.53 m/s 4.23 m/s150 x 10 Pa 9810 N/m 0 1 m

2 9.8 m/s 2 9.8 m/sp

é ù= + - + -ê ú

ê úë û( ) ( ) ( )

( )( )( )

2 23 3

2 2 2

1.53 m/s 4.23 m/s150 x 10 Pa 9810 N/m 0 1 m

2 9.8 m/s 2 9.8 m/sp

é ù= + - + -ê ú

ê úë û

P2 = 132 kPaP2 = 132 kPa Fp2 = -9400 NFp2 = -9400 N

Example: Reducing ElbowHorizontal Forces

Example: Reducing ElbowHorizontal Forces

1 21 2 p p ss+ = + + +M M F F FW1 21 2 p p ss+ = + + +M M F F FW

1111

2222

1 21 2ss p p= + - - -F M M F FW1 21 2ss p p= + - - -F M M F FW

xxx pss FMF22

xxx pss FMF22

( ) ( )1269 9400xssF N - N= -( ) ( )1269 9400xssF N - N= -

10.7kNxssF =10.7kNxssF =

Pipe wants to move to the _________leftleft

y

xForce of pipe on fluidForce of pipe on fluid

Fp2Fp2

M2M2

1 21 2x x x x xss x p pF M M F F= + - - -W1 21 2x x x x xss x p pF M M F F= + - - -W

Example: Reducing ElbowVertical Forces

Example: Reducing ElbowVertical Forces

1 21 2y y y y yss y p pF M M F F= + - - -W1 21 2y y y y yss y p pF M M F F= + - - -W

11y y yss y pF M F= - -W11y y yss y pF M F= - -W

( ) ( )459N 1,961N 29,400NyssF =- - - -( ) ( )459N 1,961N 29,400NyssF =- - - -

1111

2222

N9.27 kFyss N9.27 kFyss

Pipe wants to move _________upup

y

x

Fp1Fp1

M1M1

28 kN acting downward on fluid28 kN acting downward on fluid

W

Example: Fire nozzleExample: Fire nozzle

A small fire nozzle is used to create a powerful jet to reach far into a blaze. Estimate the force that the water exerts on the fire nozzle. The pressure at section 1 is 1000 kPa (gage). Ignore frictional losses in the nozzle.

A small fire nozzle is used to create a powerful jet to reach far into a blaze. Estimate the force that the water exerts on the fire nozzle. The pressure at section 1 is 1000 kPa (gage). Ignore frictional losses in the nozzle.

8 cm8 cm 2.5 cm2.5 cm

Example: Momentum with Complex Geometry

Example: Momentum with Complex Geometry

L/s 101 Q L/s 101 Q

cs1cs1

cs3cs3

0yF 0yF

1

2cs2cs2

Find Q2, Q3 and force on the wedge.

xx

yy

3

m/s 201 V m/s 201 V

3 50 2 130

1000 kg / m3

1 10

Unknown: ________________Q2, Q3, V2, V3, FxQ2, Q3, V2, V3, Fx

5 Unknowns: Need 5 Equations5 Unknowns: Need 5 Equations

cs1cs1

cs3cs3

1

cs2cs2

xx

yy

3

Unknowns: Q2, Q3, V2, V3, FxUnknowns: Q2, Q3, V2, V3, Fx

ContinuityContinuity

Bernoulli (2x)Bernoulli (2x)

Momentum (in x and y)Momentum (in x and y)

Q Q Q1 2 3 Q Q Q1 2 3

2

1 2 31 2 3 p p p ss+ + = + + + +M M M F F F FW1 2 31 2 3 p p p ss+ + = + + + +M M M F F F FW

V V1 2

V V1 3

2 21 1 2 2

1 21 22 2

p V p Vz z

g gg g+ + = + +

Identify the 5 equations!

Solve for Q2 and Q3Solve for Q2 and Q3

xx

yy

1 2 31 2 3 p p p ss+ + = + + + +M M M F F F FW1 2 31 2 3 p p p ss+ + = + + + +M M M F F F FW

F M M My y y y 0 1 2 3F M M My y y y 0 1 2 3

0 1 1 1 2 2 2 3 3 3 QV Q V Q Vsin sin sin0 1 1 1 2 2 2 3 3 3 QV Q V Q Vsin sin sin

V sin V sin Component of velocity in y directionComponent of velocity in y direction

Mass conservationMass conservationQ Q Q1 2 3 Q Q Q1 2 3

V V V1 2 3 V V V1 2 3 Negligible losses – apply BernoulliNegligible losses – apply Bernoulli

atmospheric pressureatmospheric pressure

Solve for Q2 and Q3Solve for Q2 and Q3

0 1 1 2 2 3 3 Q Q Qsin sin sin

Q Q2 11 3

2 3

sin sin

sin sin

a fa f

Q Q2 1

10 50

130 50

sin sin

sin sin

af a fa f a f

Q2 6.133 L / s

Q3 3.867 L / s

Why is Q2 greater than Q3?

Q Q Q3 1 2 Q Q Q3 1 2

0 1 1 1 2 2 2 3 3 3 QV Q V Q Vsin sin sin0 1 1 1 2 2 2 3 3 3 QV Q V Q Vsin sin sin

Solve for FxSolve for Fx

F M M Mx x x x 1 2 3F M M Mx x x x 1 2 3

F QV Q V Q Vx 1 1 1 2 1 2 3 1 3cos cos cosF QV Q V Q Vx 1 1 1 2 1 2 3 1 3cos cos cos

F V Q Q Qx 1 1 1 2 2 3 3cos cos cosF V Q Q Qx 1 1 1 2 2 3 3cos cos cos

F Nx 226F Nx 226

Fx

L

N

MMMM

O

Q

PPPP1000 kg / m 20 m / s

0.01 m / s cos 10

0.006133 m / s cos 130

0.003867 m / s cos 50

3

3

3

3

c ha fc hafc ha fc ha f

Fx

L

N

MMMM

O

Q

PPPP1000 kg / m 20 m / s

0.01 m / s cos 10

0.006133 m / s cos 130

0.003867 m / s cos 50

3

3

3

3

c ha fc hafc ha fc ha f

Force of wedge on fluidForce of wedge on fluid

Vector solutionVector solution

200N111 VQM 200N111 VQM

122.66N222 VQM 122.66N222 VQM

77.34N333 VQM 77.34N333 VQM

sLQ /133.62 sLQ /133.62

sLQ /867.33 sLQ /867.33

sLQ /102 sLQ /102

M M M F1 2 3 ss

Vector AdditionVector Addition

cs1

cs3

1

2cs2

x

y

3

1M1M

2M2M3M3M

ssFssF

M M M F1 2 3 ss

Where is the line of action of Fss?

Moment of Momentum EquationMoment of Momentum Equation

m=B r×Vm=B r×V

mm

=r×V

bm

m=

r×Vb

( ) ( )ˆcs

dAr= ×òT r×V V n( ) ( )ˆcs

dAr= ×òT r×V V n

R.T.T.R.T.T.

Moment of momentumMoment of momentum

Moment of momentum/unit massMoment of momentum/unit mass

Steady stateSteady state

ˆsys

cv cs

DBbdV b dA

Dt tr r

¶= + ×¶ ò ò V n̂sys

cv cs

DBbdV b dA

Dt tr r

¶= + ×¶ ò ò V n

( ) ( ) ( )ˆcv cs

D mdV dA

Dt tr r

¶= + ×¶ ò ò

r×Vr×V r×V V n

( ) ( ) ( )ˆcv cs

D mdV dA

Dt tr r

¶= + ×¶ ò ò

r×Vr×V r×V V n

rr22

rr11

VVnnVVtt

Application to TurbomachineryApplication to Turbomachinery

( ) ( )z t n

cs

T rV V dAr=ò( ) ( )z t n

cs

T rV V dAr=ò

( ) ( )2 12 1z t tT Q r V rVr é ù= -ë û( ) ( )2 12 1z t tT Q r V rVr é ù= -ë û

cs1cs1 cs2cs2

( ) ( )ˆcs

dAr= ×òT r×V V n( ) ( )ˆcs

dAr= ×òT r×V V nVt Vn

Example: SprinklerExample: Sprinkler

( ) ( )2 12 1z t tT Q r V rVr é ù= -ë û( ) ( )2 12 1z t tT Q r V rVr é ù= -ë û

Total flow is 1 L/s.Total flow is 1 L/s.Jet diameter is 0.5 cm.Jet diameter is 0.5 cm.Friction exerts a torque of Friction exerts a torque of 0.1 N-m-s0.1 N-m-s22 22.. = 30º.= 30º.Find the speed of rotation.Find the speed of rotation.

Total flow is 1 L/s.Total flow is 1 L/s.Jet diameter is 0.5 cm.Jet diameter is 0.5 cm.Friction exerts a torque of Friction exerts a torque of 0.1 N-m-s0.1 N-m-s22 22.. = 30º.= 30º.Find the speed of rotation.Find the speed of rotation.

vvttvvtt

10 cm10 cm10 cm10 cm 2

220.1 tQr Vw r- =

2

220.1 tQr Vw r- =

2 2sinjett

jet

QV θ r

Aw=- +

2 2sinjett

jet

QV θ r

Aw=- +

222

2 sin2/4

1.0 rθdQ

Qr

222

2 sin2/4

1.0 rθdQ

Qr

0sin2

1.0 2222

22 θ

drQQr

0sin

21.0 22

222

2 θd

rQQr

Vt and Vn are defined relative to control surfaces.

cs2cs2

= 34 rpm= 34 rpm

c = -(1000 kg/m3)(0.001 m3/s)2(0.1m)(2sin30)/3.14/(0.005 m)2c = -(1000 kg/m3)(0.001 m3/s)2(0.1m)(2sin30)/3.14/(0.005 m)2

b = (1000 kg/m3)(0.001 m3/s) (0.1 m) 2 = 0.01 Nmsb = (1000 kg/m3)(0.001 m3/s) (0.1 m) 2 = 0.01 Nms

Example: SprinklerExample: Sprinkler

0sin2

1.0 2222

22 θ

drQQr

0sin

21.0 22

222

2 θd

rQQr

aacbb

242

aacbb

242

22b Qrr= 22b Qrr=

22 2

2sinc Q r θ

dr

p=- 2

2 2

2sinc Q r θ

dr

p=-

c = -1.27 Nmc = -1.27 Nm

= 3.5/s= 3.5/s

a = 0.1Nms2a = 0.1Nms2

ReflectionsReflections

What is the name of the equation that we used to move from a system (Lagrangian) view to the control volume (Eulerian) view?

Explain the analogy to your checking account. The velocities in the linear momentum equation are

relative to …? Why is ma non-zero for a fixed control volume? Under what conditions could you generate power from a

rotating sprinkler? What questions do you have about application of the linear

momentum and momentum of momentum equations?

What is the name of the equation that we used to move from a system (Lagrangian) view to the control volume (Eulerian) view?

Explain the analogy to your checking account. The velocities in the linear momentum equation are

relative to …? Why is ma non-zero for a fixed control volume? Under what conditions could you generate power from a

rotating sprinkler? What questions do you have about application of the linear

momentum and momentum of momentum equations?

RTTRTT

Energy EquationEnergy Equation

First law of thermodynamics: The heat First law of thermodynamics: The heat QQHH added to a system plus added to a system plus

the work the work WW done on the system equals the change in total energy done on the system equals the change in total energy EE of the system. of the system.net net 2 1in in

Q W E E+ = -net net 2 1in in

Q W E E+ = -

net shaftin

ˆcs

DEQ W p dA

Dt= + - ×ò V n

net shaftin

ˆcs

DEQ W p dA

Dt= + - ×ò V n

net pr shaftin

W W W= +net pr shaftin

W W W= +

pr ˆcs

W p dA=- ×ò V npr ˆ

cs

W p dA=- ×ò V n

What is for a system?DE/Dt

ˆsys

cv cs

DBbdV b dA

Dt tr r

¶= + ×¶ ò ò V n̂sys

cv cs

DBbdV b dA

Dt tr r

¶= + ×¶ ò ò V n

ˆcv cs

DEedV e dA

Dt tr r

¶= + ×¶ ò ò V n̂

cv cs

DEedV e dA

Dt tr r

¶= + ×¶ ò ò V n

dE/dt for our System?dE/dt for our System?

netin

DEQ

Dt=

netin

DEQ

Dt=

shaft

DEW

Dt=

shaft

DEW

Dt=

cs

DEp d

Dt=- ×ò V A

cs

DEp d

Dt=- ×ò V A

hp hp

pAF pAF

prW FV=-prW FV=-

Heat transferHeat transfer

Shaft workShaft work

Pressure workPressure work

net shaftin

ˆcs

DEQ W p dA

Dt= + - ×ò V n

net shaftin

ˆcs

DEQ W p dA

Dt= + - ×ò V n

General Energy EquationGeneral Energy Equation

net shaftin

ˆ ˆcs cv cs

DEQ W p dA edV e dA

Dt tr r

¶= + - × = + ×

¶ò ò òV n V n net shaftin

ˆ ˆcs cv cs

DEQ W p dA edV e dA

Dt tr r

¶= + - × = + ×

¶ò ò òV n V n

net shaftin

ˆcv cs

pQ W e d e dA

tr r

ræ ö¶

= + = " + + ×ç ÷è ø¶ ò ò V n net shaftin

ˆcv cs

pQ W e d e dA

tr r

ræ ö¶

= + = " + + ×ç ÷è ø¶ ò ò V n

2

2V

e gz u= + +2

2V

e gz u= + +zz

RTTRTT1st Law of Thermo1st Law of Thermo

PotentialPotential KineticKinetic Internal (molecular spacing and forces)Internal (molecular spacing and forces)

TotalTotal

SteadySteady

Simplify the Energy EquationSimplify the Energy Equation

netin

2

shaft ˆ2

cs

p Vq w m gz u dAr

ræ öæ ö+ = + + + ×ç ÷ ç ÷è ø è øò V n

netin

2

shaft ˆ2

cs

p Vq w m gz u dAr

ræ öæ ö+ = + + + ×ç ÷ ç ÷è ø è øò V n

cgzp

cgzp

2

2V

e gz u= + +2

2V

e gz u= + +

Assume...Assume...Assume...Assume...

But But VV is often ____________ over control surface! is often ____________ over control surface!But But VV is often ____________ over control surface! is often ____________ over control surface!

Hydrostatic pressure distribution at csHydrostatic pressure distribution at cs

ŭ is uniform over csŭ is uniform over cs

not uniformnot uniform

00

net shaftin

ˆcv cs

pQ W e d e dA

tr r

ræ ö¶

+ = " + + ×ç ÷è ø¶ ò ò V n net shaftin

ˆcv cs

pQ W e d e dA

tr r

ræ ö¶

+ = " + + ×ç ÷è ø¶ ò ò V n

netin

q mshaftw m

If V normal to nIf V normal to n

kinetic energy correction termkinetic energy correction term

V = point velocityV = point velocityV = average velocity over csV = average velocity over cs

Energy Equation: Kinetic Energy Term

Energy Equation: Kinetic Energy Term

32

ˆ2 2

cs

V V AdA

rr a

æ ö× =ç ÷è øò V n

32

ˆ2 2

cs

V V AdA

rr a

æ ö× =ç ÷è øò V n

3

3

1

cs

VdA

A Va

æ ö= ç ÷è øò

3

3

1

cs

VdA

A Va

æ ö= ç ÷è øò

3

3

2

2

cs

VdA

V A

r

ar

æ öç ÷è ø

=ò

3

3

2

2

cs

VdA

V A

r

ar

æ öç ÷è ø

=ò

= _________________________= _________________________ = _________________________= _________________________

=___ for uniform velocity=___ for uniform velocity =___ for uniform velocity=___ for uniform velocity11

Energy Equation: steady, one-dimensional, constant densityEnergy Equation: steady, one-dimensional, constant density

netin

2 2

shaft2 2in in out out

in in in out out out

p V p Vgz u q w gz ua a

r r+ + + + + = + + +

netin

2 2




r r+ + + + + = + + +

ˆcs

dA mr × =ò V n ˆcs

dA mr × =ò V n

netin

2 2

shaft 2 2out out in in

out out in in

p V p Vq w m gz u gz u ma a

r ré ùæ ö æ öæ ö+ = + + + - + + +ç ÷ ê úç ÷ ç ÷è ø è ø è øë û

netin

2 2

shaft 2 2out out in in

out out in in

p V p Vq w m gz u gz u ma a

r ré ùæ ö æ öæ ö+ = + + + - + + +ç ÷ ê úç ÷ ç ÷è ø è ø è øë û

mass flux ratemass flux rate

netin

2

shaft ˆ2

cs

p Vq w m gz u dAr

ræ öæ ö+ = + + + ×ç ÷ ç ÷è ø è øò V n

netin

2

shaft ˆ2

cs

p Vq w m gz u dAr

ræ öæ ö+ = + + + ×ç ÷ ç ÷è ø è øò V n

Energy Equation: steady, one-dimensional, constant densityEnergy Equation: steady, one-dimensional, constant density

netin

2 2shaft

2 2

out inin in out out

in in out out

u u qp V w p V

z zg g g g

a ag g

- -+ + + = + + +

netin

2 2shaft

2 2

out inin in out out

in in out out

u u qp V w p V

z zg g g g

a ag g

- -+ + + = + + +

netin

out in

L

u u qh

g

- -=

netin

out in

L

u u qh

g

- -=

shaft

P T

wh h

g= -shaft

P T

wh h

g= -hPhP

Lost mechanical energyLost mechanical energy

divide by gdivide by g

mechanicalmechanical thermalthermal

netin

2 2




r r+ + + + + = + + +

netin

2 2




r r+ + + + + = + + +

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

Thermal Components of the Energy Equation

Thermal Components of the Energy Equation

2

2V

e gz u= + +2

2V

e gz u= + +

v pu c T c T= @v pu c T c T= @

Example

Water specific heat = 4184 J/(kg*K)Water specific heat = 4184 J/(kg*K)

For incompressible liquidsFor incompressible liquids

Change in temperatureChange in temperature

Heat transferred to fluidHeat transferred to fluid

netin

out in

L

u u qh

g

- -=

netin

out in

L

u u qh

g

- -=

( )netin

p out in

L

c T T qh

g

- -=

( )netin

p out in

L

c T T qh

g

- -=

Example: Energy Equation(energy loss)

Example: Energy Equation(energy loss)

datumdatum

2 m2 m4 m4 m

An irrigation pump lifts 50 L/s of water from a reservoir and An irrigation pump lifts 50 L/s of water from a reservoir and discharges it into a farmer’s irrigation channel. The pump discharges it into a farmer’s irrigation channel. The pump supplies a total head of 10 m. How much supplies a total head of 10 m. How much mechanical energy energy is lost?is lost?

p out Lh z h= +p out Lh z h= + L p outh h z= -L p outh h z= -

2.4 m2.4 m

cs1cs1

cs2cs2

hL = 10 m - 4 mhL = 10 m - 4 m

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

We need _______ in the pipe, __, and ____ ____. We need _______ in the pipe, __, and ____ ____. We need _______ in the pipe, __, and ____ ____. We need _______ in the pipe, __, and ____ ____.

Example: Energy Equation(pressure at pump outlet)

Example: Energy Equation(pressure at pump outlet)

datumdatum

2 m2 m4 m4 m

50 L/s50 L/s50 L/s50 L/shhPP = 10 m = 10 mhhPP = 10 m = 10 m

The total pipe length is 50 m and is 20 cm in diameter. The The total pipe length is 50 m and is 20 cm in diameter. The pipe length to the pump is 12 m. What is the pressure in the pipe length to the pump is 12 m. What is the pressure in the pipe at the pump outlet? You may assume (for now) that the pipe at the pump outlet? You may assume (for now) that the only losses are frictional losses in the pipeline. only losses are frictional losses in the pipeline.

The total pipe length is 50 m and is 20 cm in diameter. The The total pipe length is 50 m and is 20 cm in diameter. The pipe length to the pump is 12 m. What is the pressure in the pipe length to the pump is 12 m. What is the pressure in the pipe at the pump outlet? You may assume (for now) that the pipe at the pump outlet? You may assume (for now) that the only losses are frictional losses in the pipeline. only losses are frictional losses in the pipeline.

2.4 m2.4 m2.4 m2.4 m

cs1cs1

cs2cs2

0

/ 0

/ 0

/ 0

/ 0

/ 0

/ 0

/ 0

/velocityvelocity head losshead loss

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

How do we get the velocity in the pipe?

How do we get the frictional losses?

What about ?

How do we get the velocity in the pipe?

How do we get the frictional losses?

What about ?

Expect losses to be proportional to length of the pipeExpect losses to be proportional to length of the pipe

hl = (6 m)(12 m)/(50 m) = 1.44 mhl = (6 m)(12 m)/(50 m) = 1.44 m

V = 4(0.05 m3/s)/[ 0.2 m)2] = 1.6 m/sV = 4(0.05 m3/s)/[ 0.2 m)2] = 1.6 m/s

Example: Energy Equation (pressure at pump outlet)


Q = VAQ = VA A = d2/4A = d2/4 V = 4Q/(d2)V = 4Q/(d2)

Kinetic Energy Correction Term:

Kinetic Energy Correction Term:

is a function of the velocity distribution in the pipe.

For a uniform velocity distribution ____ For laminar flow ______ For turbulent flow _____________

Often neglected in calculations because it is so close to 1

is a function of the velocity distribution in the pipe.

For a uniform velocity distribution ____ For laminar flow ______ For turbulent flow _____________

Often neglected in calculations because it is so close to 1

is 1 is 1

is 2 is 2

1.01 < < 1.101.01 < < 1.10

3

3

1

cs

VdA

A Va

æ ö= ç ÷è øò

3

3

1

cs

VdA

A Va

æ ö= ç ÷è øò



datumdatum

2 m2 m2 m2 m4 m4 m

50 L/s50 L/s50 L/s50 L/shhPP = 10 m = 10 mhhPP = 10 m = 10 m

VV = 1.6 m/s = 1.6 m/s = 1.05= 1.05hhLL = 1.44 m = 1.44 m

VV = 1.6 m/s = 1.6 m/s = 1.05= 1.05hhLL = 1.44 m = 1.44 m

m) (1.44

)m/s (9.812

m/s) (1.6(1.05)m) (2.4m) (10)N/m (9810 2

23

2p

m) (1.44

)m/s (9.812

m/s) (1.6(1.05)m) (2.4m) (10)N/m (9810 2

23

2p

2.4 m2.4 m2.4 m2.4 m

= 59.1 kPa= 59.1 kPa

2

2out out

P out out L

p Vh z h

ga

g= + + +

2

2out out

P out out L

p Vh z h

ga

g= + + +

2

2out

out P out out L

Vp h z h

gg aæ ö

= - - -ç ÷è ø

2

2out

out P out out L

Vp h z h

gg aæ ö

= - - -ç ÷è ø

pipe burstpipe burst

Example: Energy Equation(Hydraulic Grade Line - HGL)


We would like to know if there are any places in the pipeline where the pressure is too high (_________) or too low (water might boil - cavitation).

Plot the pressure as piezometric head (height water would rise to in a manometer)

How?

We would like to know if there are any places in the pipeline where the pressure is too high (_________) or too low (water might boil - cavitation).

Plot the pressure as piezometric head (height water would rise to in a manometer)

How?



datum

2 m2 m4 m

50 L/s50 L/sHP = 10 mHP = 10 m

2.4 m2.4 m

p

Pressure head (w.r.t. reference pressure)Pressure head (w.r.t. reference pressure)

EGL (or TEL) and HGLEGL (or TEL) and HGL

g

Vz

p2

EGL2

g

Vz

p2

EGL2

zp

HGL zp

HGL

velocityheadvelocityhead

Elevation head (w.r.t. datum)Elevation head (w.r.t. datum)

Piezometric headPiezometric head

Energy Grade LineEnergy Grade Line Hydraulic Grade LineHydraulic Grade Line

pumppump

coincidentcoincident

reference pressurereference pressure

EGL (or TEL) and HGLEGL (or TEL) and HGL

The energy grade line may never be horizontal or slope upward (in direction of flow) unless energy is added (______)

The decrease in total energy represents the head loss or energy dissipation per unit weight

EGL and HGL are ____________and lie at the free surface for water at rest (reservoir)

Whenever the HGL falls below the point in the system for which it is plotted, the local pressures are lower than the __________________

The energy grade line may never be horizontal or slope upward (in direction of flow) unless energy is added (______)

The decrease in total energy represents the head loss or energy dissipation per unit weight

EGL and HGL are ____________and lie at the free surface for water at rest (reservoir)

Whenever the HGL falls below the point in the system for which it is plotted, the local pressures are lower than the __________________

zz

Example HGL and EGL

z = 0

pump

energy grade line

hydraulic grade line

velocity head

pressure head

elevation

datum

2g

V2

2g

V2

pp

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

Bernoulli vs. Control Volume Conservation of Energy


Free jetpipe

Find the velocity and flow. How would you solve these two problems?Find the velocity and flow. How would you solve these two problems?



Point to point along streamline Control surface to control surface

No frictional losses Has a term for frictional losses

Based on average velocity

Requires kinetic energy correction factor

Includes shaft work

ph

vg

ph

vg

11

12

22

22

2 2

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

Power and EfficienciesPower and Efficiencies

Electrical power

Shaft power

Impeller power

Fluid power

Electrical power

Shaft power

Impeller power

Fluid power

electricP electricP

waterP waterP

shaftP shaftP

impellerP impellerP

IEIE

TshaftTshaft

TlossTloss

QHpQHp

Motor lossesMotor losses

bearing lossesbearing losses

pump lossespump losses

Example: Hydroplant Example: Hydroplant

Q = 5 mQ = 5 m 33/s/s

2100 kW2100 kW

180 rpm180 rpm

116 kN·m116 kN·m

50 m

Water power = ?Water power = ?total losses = ?total losses = ?efficiency of turbine = ?efficiency of turbine = ?efficiency of generator = ?efficiency of generator = ?

solutionsolution

Energy Equation ReviewEnergy Equation Review

Control Volume equation Simplifications

steady constant density hydrostatic pressure distribution across control

surface (cs normal to streamlines) Direction of flow matters (in vs. out) We don’t know how to predict head loss

Control Volume equation Simplifications

steady constant density hydrostatic pressure distribution across control

surface (cs normal to streamlines) Direction of flow matters (in vs. out) We don’t know how to predict head loss

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

Conservation of Energy, Momentum, and Mass

Conservation of Energy, Momentum, and Mass

Most problems in fluids require the use of more than one conservation law to obtain a solution

Often a simplifying assumption is required to obtain a solution neglect energy losses (_______) over a short

distance with no flow expansion neglect shear forces on the solid surface over a

short distance

Most problems in fluids require the use of more than one conservation law to obtain a solution

Often a simplifying assumption is required to obtain a solution neglect energy losses (_______) over a short

distance with no flow expansion neglect shear forces on the solid surface over a

short distance

to heatto heat

greatergreater

Head Loss: Minor Losses

Head (or energy) loss due to:outlets, inlets, bends, elbows, valves, pipe size changes

Losses due to expansions are ________ than losses due to contractions

Losses can be minimized by gradual transitions

Losses are expressed in the formwhere K is the loss coefficient

Head (or energy) loss due to:outlets, inlets, bends, elbows, valves, pipe size changes

Losses due to expansions are ________ than losses due to contractions

Losses can be minimized by gradual transitions

Losses are expressed in the formwhere K is the loss coefficient

2

2L L

Vh K

g=

2

2L L

Vh K

g=

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

zin = zoutzin = zout

What is pin - pout?What is pin - pout?

Head Loss due to Sudden Expansion:Conservation of Energy

1 2

2 2

2in out out in

L

p p V Vh

gg- -

= +2 2

2in out out in

L

p p V Vh

gg- -

= +

2 2

2in out in out

L

p p V Vh

gg- -

= +2 2

2in out in out

L

p p V Vh

gg- -

= +

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

Apply in direction of flowApply in direction of flow

Neglect surface shearNeglect surface shear

Divide by (Aout )Divide by (Aout )

Head Loss due to Sudden Expansion:Conservation of Momentum

Pressure is applied over all of section 1.Momentum is transferred over area corresponding to upstream pipe diameter.Vin is velocity upstream.

Pressure is applied over all of section 1.Momentum is transferred over area corresponding to upstream pipe diameter.Vin is velocity upstream.

sspp FFFWMM 2121 sspp FFFWMM 2121

1 2

xx ppxx FFMM2121

xx ppxx FFMM2121

21x in inM V Ar=- 21x in inM V Ar=- 2

2x out outM V Ar= 22x out outM V Ar=

2 2 inout in

in out out

AV V

p p Agg

--

=

2 2 inout in

in out out

AV V

p p Agg

--

=

AA11

AA22

xx

2 2in in out out in out out outV A V A p A p Ar r- + = -2 2in in out out in out out outV A V A p A p Ar r- + = -

Head Loss due to Sudden Expansion

2 22 2

2

outout in

in in outL

VV V

V V Vh

g g

--

= +

2 22 2

2

outout in

in in outL

VV V

V V Vh

g g

--

= +2 22

2out in out in

L

V V V Vh

g- +

=2 22

2out in out in

L

V V V Vh

g- +

=

( )2

2in out

l

V Vh

g

-=( )2

2in out

l

V Vh

g

-=

22

12

in inl

out

V Ah

g A

æ ö= -ç ÷è ø

22

12

in inl

out

V Ah

g A

æ ö= -ç ÷è ø

2

1 inL

out

AK

A

æ ö= -ç ÷è ø

2

1 inL

out

AK

A

æ ö= -ç ÷è ø

in out

out in

A VA V

=in out

out in

A VA V

=

Discharge into a reservoir?_________Discharge into a reservoir?_________

EnergyEnergy

MomentumMomentum

MassMass

KL=1KL=1

2 2

2in out in out

L

p p V Vh

gg- -

= +2 2

2in out in out

L

p p V Vh

gg- -

= +

2 2 inout in

in out out

AV V

p p Agg

--

=

2 2 inout in

in out out

AV V

p p Agg

--

=

Example: Losses due to Sudden Expansion in a Pipe (Teams!)

Example: Losses due to Sudden Expansion in a Pipe (Teams!)

A flow expansion discharges 2.4 L/s directly into the air. Calculate the pressure immediately upstream from the expansion


1 cm1 cm 3 cm3 cm

We can solve this using either the momentum equation We can solve this using either the momentum equation or the energy equation (with the appropriate term for the or the energy equation (with the appropriate term for the energy losses)!energy losses)!

Solution

SummarySummary

Control volumes should be drawn so that the surfaces are either tangent (no flow) or normal (flow) to streamlines.

In order to solve a problem the flow surfaces need to be at locations where all but 1 or 2 of the energy terms are known

When possible choose a frame of reference so the flows are steady

Control volumes should be drawn so that the surfaces are either tangent (no flow) or normal (flow) to streamlines.

In order to solve a problem the flow surfaces need to be at locations where all but 1 or 2 of the energy terms are known

When possible choose a frame of reference so the flows are steady

SummarySummary

Control volume equation: Required to make the switch from Lagrangian to Eulerian

Any conservative property can be evaluated using the control volume equation mass, energy, momentum, concentrations of

species Many problems require the use of several

conservation laws to obtain a solution

Control volume equation: Required to make the switch from Lagrangian to Eulerian

Any conservative property can be evaluated using the control volume equation mass, energy, momentum, concentrations of

species Many problems require the use of several

conservation laws to obtain a solution

end

Example: Conservation of Mass(Team Work)

Example: Conservation of Mass(Team Work)

The flow through the orifice is a function of the depth of water in the reservoir

Find the time for the reservoir level to drop from 10 cm to 5 cm. The reservoir surface is 15 cm x 15 cm. The orifice is 2 mm in diameter and is 2 cm off the bottom of the reservoir. The orifice coefficient is 0.6.

CV with constant or changing mass. Draw CV, label CS, solve using variables starting

with

The flow through the orifice is a function of the depth of water in the reservoir

Find the time for the reservoir level to drop from 10 cm to 5 cm. The reservoir surface is 15 cm x 15 cm. The orifice is 2 mm in diameter and is 2 cm off the bottom of the reservoir. The orifice coefficient is 0.6.

CV with constant or changing mass. Draw CV, label CS, solve using variables starting

with

2ghCAQ o 2ghCAQ o

ˆcs cv

dA dVt

r r¶

× =-¶ò òV n̂

cs cv

dA dVt

r r¶

× =-¶ò òV n

Example Conservation of MassConstant Volume

Example Conservation of MassConstant Volume

hh

0 ororresres AVAV 0 ororresres AVAV

cs1cs1

cs2cs2

dt

dhVres

dt

dhVres

ororor QAV ororor QAV

02 ghCAAdt

dhorres 02 ghCAA

dt

dhorres

ˆcs cv

dA dVt

r r¶

× =-¶ò òV n̂

cs cv

dA dVt

r r¶

× =-¶ò òV n

1 2

1 1 1 2 2 2ˆ ˆ 0cs cs


1 1 1 2 2 2ˆ ˆ 0cs cs


Example Conservation of MassChanging Volume

Example Conservation of MassChanging Volume

hh

or or

cv

V A dVt¶

=-¶ òor or

cv

V A dVt¶

=-¶ ò

cs1cs1

cs2cs2

ororor QAV ororor QAV

02 ghCAAdt

dhorres 02 ghCAA

dt

dhorres

resor or

A dhdVV A

dt dt=- =- res

or or

A dhdVV A

dt dt=- =-

ˆcs cv

dA dVt

r r¶

× =-¶ò òV n̂

cs cv

dA dVt

r r¶

× =-¶ò òV n

Example Conservation of MassExample Conservation of Mass

th

hor

res dth

dh

gCA

A

002

th

hor

res dth

dh

gCA

A

002

thhgCA

A

or

res 2/10

2/122

thhgCA

A

or

res 2/10

2/122

( )

( ) ( ) ( )( ) ( )( )

21/ 2 1/ 2

22

2 0.150.03 0.08

0.0020.6 2 9.8 /

4

mm m t

mm s

p

-- =

æ öç ÷è ø

( )

( ) ( ) ( )( ) ( )( )

21/ 2 1/ 2

22

2 0.150.03 0.08

0.0020.6 2 9.8 /

4

mm m t

mm s

p

-- =

æ öç ÷è ø

st 591 st 591

Pump HeadPump Head

hhpp

2

2in

in

Vg

a2

2in

in

Vg

a

2

2out

out

Vg

a2

2out

out

Vg

a2

2

2

2

in inin in P

out outout out T L

p Vz h

g

p Vz h h

g

ag

ag

+ + + =

+ + + +

2

2

2

2

in inin in P

out outout out T L

p Vz h

g

p Vz h h

g

ag

ag

+ + + =

+ + + +

Example: VenturiExample: Venturi

Example: VenturiExample: Venturi

Find the flow (Q) given the pressure drop between section 1 and Find the flow (Q) given the pressure drop between section 1 and 2 and the diameters of the two sections. Draw an appropriate 2 and the diameters of the two sections. Draw an appropriate control volume. You may assume the head loss is negligible. control volume. You may assume the head loss is negligible. Draw the EGL and the HGL. Draw the EGL and the HGL.

11 22

h h

Example VenturiExample Venturi

2 2

2 2in out out inp p V V

g gg g- = -

2 2

2 2in out out inp p V V

g gg g- = -

42

12

in out out out

in

p p V dg dg g

é ùæ ö- = -ê úç ÷è øê úë û

42

12

in out out out

in

p p V dg dg g

é ùæ ö- = -ê úç ÷è øê úë û

( )4

2 ( )

1in out

out

out in

g p pV

d dg

-=

é ù-ë û( )4

2 ( )

1in out

out

out in

g p pV

d dg

-=

é ù-ë û

( )4

2 ( )

1in out

v out

out in

g p pQ C A

d dg

-=

é ù-ë û( )4

2 ( )

1in out

v out

out in

g p pQ C A

d dg

-=

é ù-ë û

VAQ VAQ

in in out outV A V A=in in out outV A V A=

2 2

4 4in out

in out

d dV V

p p=

2 2

4 4in out

in out

d dV V

p p=

2 2in in out outV d V d=2 2in in out outV d V d=

2

2out

in outin

dV V

d=

2

2out

in outin

dV V

d=

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g ga a

g g+ + + = + + + +

Fire nozzle: Team WorkFire nozzle: Team Work

Identify what you need to know

Count your unknowns

Determine what equations you will use

Identify what you need to know

Count your unknowns

Determine what equations you will use

8 cm8 cm2.5 cm2.5 cm

1000 kPa1000 kPa

Find the VelocitiesFind the Velocities

2 21 1 2 2

1 22 2p V p V

z zg gg g

+ + = + +2 2

1 1 2 21 22 2

p V p Vz z

g gg g+ + = + +

2 21 1 2

2 2p V V

g gg+ =

2 21 1 2

2 2p V V

g gg+ = 2 2

1 1 2 2V D V D=2 21 1 2 2V D V D=

2 21 2 1

2 2p V V

g gg= -

2 21 2 1

2 2p V V

g gg= -

4

2 222 1

1

DV V

D

æ ö=ç ÷è ø

4

2 222 1

1

DV V

D

æ ö=ç ÷è ø

422 2

11

12

V Dp

Dr

æ öæ ö= -ç ÷ç ÷è øè ø

422 2

11

12

V Dp

Dr

æ öæ ö= -ç ÷ç ÷è øè ø

12 4

2

1

2

1

pV

DD

r

=é ùæ ö-ê úç ÷è øê úë û

12 4

2

1

2

1

pV

DD

r

=é ùæ ö-ê úç ÷è øê úë û

Fire nozzle: SolutionFire nozzle: Solution

section 1 section 2D 0.08 0.025 m

A 0.00503 0.00049 m2

P 1000000 0 PaV 4.39 44.94 m/s

Fp 5027 N

M -96.8 991.2 N

Fssx -4132 N

Q 22.1 L/s

2.5 cm2.5 cm8 cm8 cm

1000 kPa1000 kPa

force applied by nozzle on water

Is this the force that the firefighters need to brace against? _______

Which direction does the nozzle want to go? ______

NO!

1 21 2x x x x xss x p pF M M F F= + - - -W1 21 2x x x x xss x p pF M M F F= + - - -W

Temperature Rise over Taughanock Falls

Temperature Rise over Taughanock Falls

Drop of 50 meters Find the temperature rise

Drop of 50 meters Find the temperature rise

netin

L

p

gh qT

c

+D =

netin

L

p

gh qT

c

+D =

KKg

J4184

m 50m/s 9.8 2

T

KKg

J4184

m 50m/s 9.8 2

TKT 117.0 KT 117.0

( )netin

p out in

L

c T T qh

g

- -=

( )netin

p out in

L

c T T qh

g

- -=

HydropowerHydropower

pQHP g= pQHP g=

( ) ( ) ( )3 39806 / 5m / 50m 2.45waterP N m s MW= =( ) ( ) ( )3 39806 / 5m / 50m 2.45waterP N m s MW= =

2.1000.857

2.45total

MWe

MW= =

2.1000.857

2.45total

MWe

MW= =

( ) 2 1min0.116 180 2.187

min 60

2.1870.893

2.452.100

0.962.187

turbine

turbine

generator

rev radP MNm MW

rev s

MWe

MWMW

eMW

pæ ö= =è ø

= =

= =

( ) 2 1min0.116 180 2.187

min 60

2.1870.893

2.452.100

0.962.187

turbine

turbine

generator

rev radP MNm MW

rev s

MWe

MWMW

eMW

pæ ö= =è ø

= =

= =

Solution: Losses due to Sudden Expansion in a Pipe

Solution: Losses due to Sudden Expansion in a Pipe



1 cm1 cm

3 cm3 cm

AA

VV

1

2

2

1

p V V Vg

1 22

1 2

p V V V1 22

1 2 c h

p pV V

AA

g1 2

22

12 1

2

V

m s

mm s1

3

2

0 0024

0 01

4

30 56 . /

.. /

a fV m s2 3 39 . /

p kg s m s m s m s121000 3 39 30 56 3 39 / . / . / . /a fa f a fa fd i

p kPa1 92

Finite Control Volume Analysis

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Transcript of Finite Control Volume Analysis