Monroe L. Weber-Shirk S chool of Civil and Environmental Engineering Finite Control Volume Analysis CEE 331 May 4, 2015 Application of Reynolds Transport.

Monroe L. Weber-Shirk School of Civil and

Environmental Engineering

Finite Control Volume AnalysisFinite Control Volume Analysis

CEE 331

April 18, 2023

Application of Reynolds Transport Theorem

Moving from a System to a Control Volume

MassLinear MomentumMoment of MomentumEnergyPutting it all together!

Conservation of Mass

B = Total amount of ____ in the systemb = ____ per unit mass = __

ˆsys

cv cs

DMdV dA

Dt t

V n

ˆcs cv

dA dVt

V n

mass1mass

But DMsys/Dt = 0!

cv equation

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

ˆsys

cv cs

DBbdV b dA

Dt t

V n

Continuity Equation

Conservation of Mass

1 2

1 1 1 2 2 2ˆ ˆ 0cs cs

dA dAr r× + × =ò òV n V n

12

V1A1

If mass in cv is constant

Unit vector is ______ to surface and pointed ____ of cv

ˆcs cv

dA dVt

V n

n̂ normal

out

ˆcs

dA V n m±

ˆcs

dA

VA

V n

VAr± =

n̂

We assumed uniform ___ on the control surface

is the spatially averaged velocity normal to the csV

[M/T]

Continuity Equation for Constant Density and Uniform Velocity

1 21 1 2 2 0V A V A

1 21 2V A V A Q

1 2

1 1 1 2 2 2ˆ ˆ 0

cs cs

dA dA V n V n Density is constant across cs

Density is the same at cs1 and cs2

[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

uniform spatially averaged

Example: Conservation of Mass?

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

resAQ

dtdh

h

dt

dhAQ res

out Example

Constant density

ˆcs cv

dA dVt

V n

ˆcs

VdA

t

V n

out in

dVQ Q

dt Velocity of the reservoir surface

Linear Momentum Equation

mB Vmm

Vb

cv equation

momentum momentum/unit mass

Steady state

ˆsys

cv cs

DBbdV b dA

Dt t

V n

ˆcv cs

DmdV dA

Dt t

VV V V n

ˆcs

DmdA

Dt

VV V n

0F

This is the “ma” side of the F = ma equation!

Vectors!

Linear Momentum Equation

1 1 1 1 2 2 2 2

DmV A V A

Dt

VV V

111111 VVM QAV

222222 VVM QAV

Assumptions

Vectors!!!

Uniform densityUniform velocity

V ASteady

ˆcs

DmdA

Dt

VV V n

1 2

1 1 1 1 2 2 2 2ˆ ˆ

cs cs

DmdA dA

Dt

VV V n V V n

V fluid velocity relative to cv

1 2

D m

Dt

VF M M

Steady Control Volume Form of Newton’s Second Law

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

21 MMF

1 2 wall wallp p p F F F F FW

GravityShear at the solid surfacesPressure at the solid surfacesPressure on the flow surfaces

Why no shear on control surfaces? _______________________________No velocity tangent to control surface

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 determinessize of _____________needed to keep

pipe in placeforce on the vane of a pump or turbine...

1 2p p ss F F F FW wallwallFFpss F

=force applied by solid surfaces

thrust blocks

Linear Momentum Equation

1 2p p ss F F F FW

1 2m a M M

1 21 2 p p ss M M F F FW

The momentum vectors have the same direction as the velocity vectors

Fp1

Fp2

W

M1

M2

Fssy

Fssx

1 1QM V

2 2QM V

Forces by solid surfaces on fluid

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

section 1 section 2D 50 cm 30 cmA ____________ ____________V ____________ ____________p 150 kPa ____________M ____________ ____________Fp ____________ ____________

Example: Reducing Elbow

1 21 2 p p ss M M F F FW

1

2

1 m

z

xDirection of V vectors

0.196 m2 0.071 m2

1.53 m/s ↑ 4.23 m/s →

-459 N ↑ 1270 N →29,400 N ↑

g*volume=-1961 N ↑

?

?←

Example: What is p2?

2 21 1 2 2

1 21 22 2

p V p Vz z

g g

2 21 2

2 1 1 2 2 2

V Vp p z z

g g

2 2

3 32 2

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

2 9.8 m/sp

P2 = 132 kPa Fp2 = 9400 N

Example: Reducing ElbowHorizontal Forces

1 21 2 p p ss M M F F FW

1

2

1 21 2ss p p F M M F FW

xxx pss FMF22

1270 9400xssF N - N

10.7kNxssF

Fluid is pushing the pipe to the ______left

z

xForce of pipe on fluid

Fp2

M2

1 21 2x x x x xss x p pF M M F F W

Example: Reducing ElbowVertical Forces

1 21 2z z z z zss z p pF M M F F W

11z z zss z pF M F W

459N 1,961N 29,400NzssF

1

2

27.9 NzssF k

Pipe wants to move _________up

z

x

Fp1M1

W

28 kN acting downward on fluid

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.

8 cm8 cm 2.5 cm2.5 cm

Fire nozzle: Solution

Identify what you need to know

Determine what equations you will use

8 cm2.5 cm

1000 kPa

P2, V1, V2, Q, M1, M2, Fss

Bernoulli, continuity, momentum

Find the Velocities

2 21 1 2 2

1 22 2

p V p Vz z

g g

2 21 1 2

2 2

p V V

g g 2 2

1 1 2 2V D V D

2 21 2 1

2 2

p V V

g g

4

2 222 1

1

DV V

D

422 2

11

12

V Dp

D

12 4

2

1

2

1

pV

D

D

continuity→

Fire 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 NM -96.8 991.2 N

Fssx -4132 NQ 22.1 L/s

2.5 cm8 cm

1000 kPa

force applied by nozzle on water

Is Fssx 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 W

Moments!

Example: Momentum with Complex Geometry

L/s 101 Q

cs1

cs3

0yF

1

2cs2

Find Q2, Q3 and force on the wedge in a horizontal plane.

x

y

3

m/s 201 V

3 50 2 130

1000 kg / m3

1 10

Unknown: ________________Q2, Q3, V2, V3, Fx

5 Unknowns: Need 5 Equations

cs1

cs3

1

cs2

x

y

3

Unknowns: Q2, Q3, V2, V3, Fx

Continuity

Bernoulli (2x)

Momentum (in x and y)

Q Q Q1 2 3

2

1 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 g

Identify the 5 equations!

Solve for Q2 and Q3

x

y

1 2 31 2 3 p p p ss M M M F F F FW

1 2 30ssy y y yF M M M

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

V sin Component of velocity in y direction

Mass conservationQ Q Q1 2 3

V V V1 2 3 Negligible losses – apply Bernoulli

atmospheric pressure

1 1QM V

Solve 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

1 1 2 2 3 3y y ymV m V m V

0 1 1 1 2 2 2 3 3 3 QV Q V Q Vsin sin sin Eliminate Q3

+ + -

Solve for Fssx

1 2 3ssx x x xF M M M

1 1 1 2 1 2 3 1 3cos cos cosssxF QV Q V Q V

1 1 1 2 2 3 3cos cos cosssxF V Q Q Q

226ssxF N

3

3 3

3

0.01 m /s cos 10

1000 kg/m 20 m/s 0.006133 m /s cos 130

0.003867 m /s cos 50

ssxF

Force of wedge on fluid

Vector solution

200N111 VQM

122.66N222 VQM

77.34N333 VQM

sLQ /133.62

sLQ /867.33

sLQ /102

M M M F1 2 3 ss

Vector Addition

cs1

cs3

1

2cs2

x

y

3

1M

2M3M

ssF

M M M F1 2 3 ss

Where is the line of action of Fss?

Moment of Momentum Equation

mB r × V

m

m

r × Vb

ˆcs

dA T r × V V n

cv equation

Moment of momentum

Moment of momentum/unit mass

Steady state

ˆsys

cv cs

DBbdV b dA

Dt t

V n

ˆcv cs

D mdV dA

Dt t

r × Vr × V r × V V n

Application to Turbomachinery

r2

r1

VnVt

2 2 1 1zT Q r × V r × V

cs1 cs2

ˆcs

dA T r × V V nrVt Vn ˆ

cs

dA Q V n

Example: Sprinkler

2 2 1 1zT Q r × V r × V

Total flow is 1 L/s.Jet diameter is 0.5 cm.Friction exerts a torque of 0.1 N-m-s2 2. = 30º.Find the speed of rotation.

vt

10 cm 2

220.1 tQr V

2 2sinjett

jet

QV θ r

A

22 22

4 / 20.1 sin

QQr θ r

d

2 2 2

2 2 2

20.1 sin 0Qr Q r θ

d

Vt and Vn are defined relative to control surfaces.

cs2

= 34 rpm

c = -(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 Nms

Example: Sprinkler

0sin2

1.0 2222

22 θ

drQQr

aacbb

242

22b Qr2

2 2

2sinc Q r θ

d

c = -1.27 Nm

= 3.5/s

a = 0.1Nms2

Reflections

What is if there is no friction? ___________

What is Vt if there is no friction ?__________

= 127/s

22 tT Qr V

cv equation

Energy Equation

First law of thermodynamics: The heat QH added to a system plus the work W done on the system equals the change in total energy E of the system.net net 2 1in in

Q W E E

net shaftin

ˆcs

DEQ W p dA

Dt V n

net pr shaftin

W W W

prˆ

cs

W p dA V n

What is for a system?DE/Dt

ˆsys

cv cs

DBbdV b dA

Dt t

V n

ˆcv cs

DEedV e dA

Dt t

V n

dE/dt for our System?

netin

DEQ

Dt

shaft

DEW

Dt

ˆcs

DEp dA

Dt V n

hp

pAF

prW FV

Heat transfer

Shaft work

Pressure work

net shaftin

ˆcs

DEQ W p dA

Dt V n

prW pVA

General Energy Equation

net shaftin

ˆ ˆcs cv cs

DEQ W p dA edV e dA

Dt t

V n V n

net shaftin

ˆcv cs

pQ W e d e dA

t

V n

2

2

Ve gz u

z

cv equation1st Law of Thermo

Potential Kinetic Internal (molecular spacing and forces)

Total

Steady

Simplify the Energy Equation

netin

2

shaftˆ

2cs

p Vq w m gz u dA

V n

cgzp

2

2

Ve gz u

Assume...

But V is often ____________ over control surface!

Hydrostatic pressure distribution at cs

ŭ is uniform over cs

not uniform

0

net shaftin

ˆcv cs

pQ W e d e dA

t

V n

netin

q mshaftw m

If V tangent to n

kinetic energy correction term

V = point velocityV = average velocity over cs

Energy Equation: Kinetic Energy

32

ˆ2 2cs

V V AdA

V n

3

3

1

cs

VdA

A V

3

3

2

2

cs

VdA

V A

= _________________________

=___ for uniform velocity1

Energy 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 u

ˆcs

dA m 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 m

mass flux rate

netin

2

shaftˆ

2cs

p Vq w m gz u dA

V n

Energy 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

netin

out in

L

u u qh

g

shaft

P T

wh h

g hP

Lost mechanical energy

divide by g

mechanical thermal

netin

2 2

shaft2 2in in out out

in in in out out out

p V p Vgz u q w gz u

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g g

Thermal Components of the Energy Equation

2

2

Ve gz u

v pu c T c T

Example

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

For incompressible liquids

Change in temperature

Heat transferred to fluid

netin

out in

L

u u qh

g

netin

p out in

L

c T T qh

g

Example: Energy Equation(energy loss)

datum

2 m4 m

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

p out Lh z h L p outh h z

2.4 m

cs1

cs2

hL = 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 g

What is hL?

Why can’t I draw the cs at the end of the pipe?

Example: Energy Equation(pressure at pump outlet)

datum

2 m4 m

50 L/shP = 10 m

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 at the pump outlet? You may assume (for now) that the only losses are frictional losses in the pipeline.

2.4 m

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

cs1

cs2

0

/ 0

/ 0

/ 0

/velocity 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 g

Expect losses to be proportional to length of the pipe

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

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

Example: Energy Equation (pressure at pump outlet)

How do we get the velocity in the pipe?

How do we get the frictional losses?

What about ?

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

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 1

is 2

1.01 < < 1.10

3

3

1

cs

VdA

A V

Example: Energy Equation (pressure at pump outlet)

datum

2 m4 m

50 L/shP = 10 m

V = 1.6 m/s = 1.05hL = 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

2.4 m

= 59.1 kPa

2

2out out

P out out L

p Vh z h

g

2

2out

out P out out L

Vp h z h

g

pipe 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 piezometer)

How?

Example: Energy Equation(Energy Grade Line - EGL)

datum

2 m4 m

50 L/s

2.4 m

2

2

p V

g

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g g

HP = 10 m

p = 59 kPa

What is the pressure at the pump intake?

Entrance loss

Exit loss

Loss due to shear

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

EGL (or TEL) and HGL

g

Vz

p2

EGL2

zp

HGL

velocityhead

Elevation head (w.r.t. datum)

Piezometric head

Energy Grade Line Hydraulic Grade Line

What is the difference between EGL defined by Bernoulli and EGL defined here?

pump

coincident

reference pressure

EGL (or TEL) and HGL

The energy grade line may never 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 __________________

z

Example HGL and EGL

z = 0

pump

energy grade line

hydraulic grade line

velocity head

pressure head

elevation

datum

2g

V2

p

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g g

Bernoulli vs. Control Volume Conservation of Energy

Free jetpipe

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

Bernoulli vs. Control Volume Conservation of Energy

Point to point along streamline Control surface to control surface

No frictional losses Has a term for frictional losses

Based on velocity

Requires kinetic energy correction factor

Includes shaft work

2 21 1 2 2

1 22 2

p v p vz z

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 g

Based on velocity

Has direction!

point average

Power and Efficiencies

Electrical power

Shaft power

Impeller power

Fluid power

electricP

waterP

shaftP

impellerP

EI

T

T

QHp

Motor losses

bearing losses

pump losses

Prove this!

P = FV

Powerhouse

River

Reservoir

Penstock

Example: Hydroplant

Q = 5 m3/s

2100 kW

180 rpm

116 kN·m

50 m

Water power =Overall efficiency = efficiency of turbine =efficiency of generator =

solution

0.8570.893

0.96

2.45 MW

Energy Equation Review

Control Volume equation Simplifications

steady constant densityhydrostatic pressure distribution across control

surface (streamlines parallel)

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 g

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 solutionneglect energy losses (_______) over a short

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

short distance

to heatmechanical

greater

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 KL is the loss coefficient

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 g

When V, KE thermal

zin = zout

Relate Vin and Vout?

Head Loss due to Sudden Expansion:Conservation of Energy

in out

2 2

2in out out in

L

p p V Vh

g

2 2

2in out in out

L

p p V Vh

g

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g g

Relate pin and pout?

Mass

Momentum

Where is p measured?___________________________At centroid of control surface

z

x

Apply in direction of flow

Neglect surface shear

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.

sspp FFFWMM 2121

1 2

xx ppxx FFMM2121

21x in inM V A 2

2x out outM V A

2 2 inout in

in out out

AV V

p p A

g

A1

A2

x

2 2in in out out in out out outV A V A p A p A

Head Loss due to Sudden Expansion

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

2in out

l

V Vh

g

22

12

in inl

out

V Ah

g A

2

1 inL

out

AK

A

in out

out in

A V

A V

Discharge into a reservoir?_________

Energy

Momentum

Mass

KL=1

2 2

2in out in out

L

p p V Vh

g

2 2 inout in

in out out

AV V

p p A

g

2 2

2

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

A flow expansion discharges 0.5 L/s directly into the air. Calculate the pressure immediately upstream from the expansion

1 cm 3 cm

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

Solution

Use the momentum equation…

3

1 2

0.0005 /6.4 /

0.01

4

m sV m s

m

2 0.71 /V m s

Scoop

A scoop attached to a locomotive is used to lift water from a stationary water tank next to the train tracks into a water tank on the train. The scoop pipe is 10 cm in diameter and elevates the water 3 m.

Draw several streamlines in the left half of the stationary water tank (use the scoop as your frame of reference) including the streamlines that attach to the submerged horizontal section of the scoop.

Use the streamlines to help you draw a control volume and clearly label the control surfaces.

How fast must the locomotive be moving (Vscoop) to get a flow of 4 L/s if the frictional losses in the pipe are equal to 1.8 V2/2g where V is the average velocity of the water in the pipe. (Vscoop = 7.7 m/s)

Scoop

Q = 4 L/s

d = 10 cm3 m

stationary water tank

Vscoop

Scoop Problem:‘The Real Scoop’

moving water tank

2 2

2 2in in out out

in in P out out T L

p V p Vz h z h h

g g

2 21 1 2 2

1 22 2

p V p Vz z

g g Bernoulli

Energy

Summary

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

Summary

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

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

speciesMany problems require the use of several

conservation laws to obtain a solution

end

Scoop Problem

stationary water tank

moving water tank

Scoop Problem:Change your Perspective

Scoop Problem:Be an Extremist!

Very short riser tube

Very long riser tube

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

to integration step

orQ CA 2gh

ˆcs cv

dA dVt

V n

Example Conservation of MassConstant Volume

h

0 ororresres AVAV

cs1

cs2

dt

dhVres

ororor QAV

02 ghCAAdt

dhorres

ˆcs cv

dA dVt

V n

1 2

1 1 1 2 2 2ˆ ˆ 0

cs cs

dA dA V n V n

Example Conservation of MassChanging Volume

h

or or

cv

V A dVt

cs1

cs2

ororor QAV

02 ghCAAdt

dhorres

resor or

A dhdVV A

dt dt

ˆcs cv

dA dVt

V n

Example Conservation of Mass

th

hor

res dth

dh

gCA

A

002

thhgCA

A

or

res 2/10

2/122

21/ 2 1/ 2

2

2

2 0.150.03 0.08

0.0020.6 2 9.8 /

4

mm m t

mm s

st 591

Pump Head

hp

2

2in

in

V

g

2

2out

out

V

g2

2

2

2

in inin in P

out outout out T L

p Vz h

g

p Vz h h

g

Example: Venturi

Example: Venturi

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

1 2

h

Example Venturi

2 2

2 2in out out inp p V V

g g

42

12

in out out out

in

p p V d

g d

4

2 ( )

1

in outout

out in

g p pV

d d

4

2 ( )

1

in outv out

out in

g p pQ C A

d d

VAQ

in in out outV A V A

2 2

4 4in out

in out

d dV V

2 2in in out outV d 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 g

Reflections

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 …? When 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?

Temperature Rise over Taughanock Falls

Drop of 50 metersFind the temperature riseIgnore kinetic energy

netin

L

p

gh qT

c

KKg

J4184

m 50m/s 9.8 2

T

KT 117.0

netin

p out in

L

c T T qh

g

Hydropower

pQHP

3 39806 / 5m / 50m 2.45waterP N m s 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

Solution: Losses due to Sudden Expansion in a Pipe

A flow expansion discharges 0.5 L/s directly into the air. Calculate the pressure immediately upstream from the expansion

1 cm

3 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

3

1 2

0.0005 /6.4 /

0.01

4

m sV m s

m

2 0.71 /V m s

2

1 1000 / 0.71 / 6.4 / 0.71 /p kg s m s m s m s

1 4p kPa Carburetors and water powered vacuums