Monroe L. Weber-Shirk School of Civil and

Environmental Engineering

Finite Control Volume Analysis

CEE 331June 8, 2006

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 = __

mass1mass

cv equation

ˆsy s

c v c s

D Md V d A

D t tρ ρ∂

= + ⋅∂ ∫ ∫ V n

ˆc s c v

d A d Vt

ρ ρ∂⋅ = −

∂∫ ∫V n

ˆsy s

c v c s

D Bb d V b d A

D t tρ ρ∂

= + ⋅∂ ∫ ∫ V n

But DMsys/Dt = 0!

Continuity Equation

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

Conservation of Mass

1 2

1 1 1 2 2 2ˆ ˆ 0c s c s

d A d Aρ ρ⋅ + ⋅ =∫ ∫V n V n

12

V1A1

If mass in cvis constant

Unit vector is ______ to surface and pointed ____ of cv

ˆc s c v

d A d Vt

ρ ρ∂⋅ = −

∂∫ ∫V n

n̂ normal

outˆ

c s

d Aρ ⋅ =∫ V n m±

ˆcs

d AV

A

⋅=

∫ V n

VAρ± =

n̂

We assumed uniform ___ on the control surface

ρ

is the spatially averaged velocity normal to the csV

[M/T]

Continuity Equation for Constant Density and Uniform Velocity

1 2

1 1 1 2 2 2ˆ ˆc s c s

d A d Aρ ρ⋅ + ⋅ =∫ ∫V n V n 0 Density is constant across cs

1 21 1 2 2 0V A V Aρ ρ− + = Density is the same at cs1 and cs2

1 21 2V A V A Q= = [L3/T]

1 1 2 2V A V A Q= = Simple version of the continuity equation for conditions of constant density. It is understood that the velocities are either ________ or _______ ________.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

dtdhAQ res

out −=Example

Constant density

ˆc s c v

d A d Vt

ρ ρ∂⋅ = −

∂∫ ∫V n

ˆc s

Vd At

∂⋅ = −

∂∫ V n

o u t ind VQ Qd t

− = − Velocity of the reservoir surface

Linear Momentum Equation

m=B V mm

=Vb

cv equationˆsy s

c v c s

D Bb d V b d A

D t tρ ρ∂

= + ⋅∂ ∫ ∫ V n

0F ≠∑

momentum momentum/unit mass

ˆcv cs

D m d V d AD t t

ρ ρ∂= + ⋅

∂ ∫ ∫V V V V n

Vectors!

ˆc s

D m d AD t

ρ= ⋅∫V V V n Steady state

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

Linear Momentum Equation

( ) ( )1 1 1 1 2 2 2 2D m V A V A

D tρ ρ= − +

V V V

( ) ( ) 111111 VVM QAV ρρ −=−=

( ) ( ) 222222 VVM QAV ρρ ==

Assumptions

Vectors!!!

Uniform densityUniform velocity

V ⊥ ASteady

ˆcs

D m d AD t

ρ= ⋅∫V V V n

1 2

1 1 1 1 2 2 2 2ˆ ˆcs c s

D m d A d AD t

ρ ρ= ⋅ + ⋅∫ ∫V V V n V V n

V fluid velocity relative to cv

Steady Control Volume Form of Newton’s Second Law

( )1 2

D mD t

= = +∑ VF M M

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

21 MMF +=∑GravityShear at the solid surfacesPressure at the solid surfacesPressure on the flow surfaces

1 2 w a ll w a llp p p τ= + + + +∑ F F F F FW

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 unknownsOften the problem is to calculate the total force exerted by the fluid on the solid surfacesThe magnitude and direction of the force determines

size of _____________needed to keeppipe in placeforce on the vane of a pump or turbine...

1 2p p ss= + + +∑ F F F FW wallwallFF pss τ+=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

WM1

M2

Fssy

Fssx

( )1 1Qρ= −M V( )2 2Qρ=M V

Forces by solid surfaces on fluid

Example: Reducing Elbow

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 ____________ ____________

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 ↑

−ρg*volume=-1961 N ↑

?

?←

Example: What is p2?

2 21 1 2 2

1 21 22 2

p V p Vz zg gγ γ

+ + = + +

2 21 2

2 1 1 2 2 2V Vp p z z

g gγ ⎡ ⎤

= + − + −⎢ ⎥⎣ ⎦

( ) ( ) ( ) ( )( )

2 23 3

2 2

1 .5 3 m /s 4 .2 3 m /s1 5 0 x 1 0 P a 9 8 1 0 N /m 0 1 m

2 9 .8 m /sp

⎡ ⎤−⎢ ⎥= + − +⎢ ⎥⎣ ⎦

Fp2 = 9400 NP2 = 132 kPa

Example: Reducing ElbowHorizontal Forces

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

1

2

1 21 2s s p p= + − − −F M M F FW

Fp2

M2

1 21 2x x x x xs s x p pF M M F F= + − − −W

xxx pss FMF22 −=

z

x

( ) ( )1 2 7 0 9 4 0 0xs sF N - N= −

1 0 .7 k Nxs sF = Force of pipe on fluid

Fluid is pushing the pipe to the ______left

Example: Reducing ElbowVertical Forces

1 21 2z z z z zs s z p pF M M F F= + − − −W

1

2

11z z zs s z pF M F= − −W

( ) ( )4 5 9 N 1, 9 6 1N 2 9 ,4 0 0 Nzs sF = − − − −

2 7 .9 Nzs sF k= −

Fp1M1

W

Pipe wants to move _________up

z

x

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 2p V p Vz z

g gγ γ+ + = + +

2 21 1 2

2 2p V V

g gγ+ = continuity→ 2 2

1 1 2 2V D V D=4

2 222 1

1

DV VD

⎛ ⎞=⎜ ⎟

⎝ ⎠2 2

1 2 1

2 2p V V

g gγ= −

12 4

2

1

2

1

pVDD

ρ

=⎡ ⎤⎛ ⎞

−⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

422 2

11

12

V DpD

ρ⎛ ⎞⎛ ⎞⎜ ⎟= − ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Fire nozzle: Solution

section 1 section 2D 0.08 0.025 mA 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 xs s x p pF M M F F= + − − −W

Moments!

Example: Momentum with Complex Geometry

L/s 101 =Q

cs1

cs3

0=yF

θ1

θ2cs2

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

x

y

θ3

m/s201 =V

θ 3 50= −θ 2 130=

ρ = 1000 kg / m3

θ 1 = 10

Unknown: ________________Q2, Q3, V2, V3, Fx

5 Unknowns: Need 5 Equations

Unknowns: Q2, Q3, V2, V3, FxIdentify the 5 equations!

Q Q Q1 2 3= +Continuity

cs1

cs3

θ1

cs2

x

y

θ3

Bernoulli (2x)

Momentum (in x and y)

θ2

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 zg gγ γ

+ + = + +

Solve for Q2 and Q3

x

y

θ

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= − + +ρ θ ρ θ ρ θQV Q V Q Vsin sin sin

V sinθ = Component of velocity in y direction

Mass conservationQ Q Q1 2 3= +

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

atmospheric pressure

( )1 1Qρ= −M V

Solve for Q2 and Q3

Eliminate Q30 3= − 1 1 1 2 2 2 3 3+ +ρ θ ρ θ ρ θQV Q V Q Vsin sin sin

Q Q Q3 1 2= −0 1 1 2 2 3 3= − + +Q Q Qsin sin sinθ θ θ

Q Q2 11 3

2 3

=− +− +

sin sinsin sin

θ θθ θ

a fa f

Q2 = 6.133 L / s

Q Q2 110 50

130 50=

− + −− + −

sin sinsin sina f a fa f a f

Why is Q2 greater than Q3?

Q3 = 3.867 L / s1 1 2 2 3 3y y ymV m V m V= ++ + -

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 Vρ θ ρ θ ρ θ= − + +

[ ]1 1 1 2 2 3 3cos cos cosssxF V Q Q Qρ θ θ θ= − + +

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

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

⎡ ⎤−⎢ ⎥⎢ ⎥= +⎢ ⎥⎢ ⎥+ −⎣ ⎦

226ssxF N= − Force of wedge on fluid

Vector solution

M M M F1 2 3+ + = ss

200N111 =−= VQρM

122.66N222 == VQρM

77.34N333 == VQρMsLQ /102 =

sLQ /133.62 =

sLQ /867.33 =

Vector Addition

cs1

cs3θ1

θ2cs2

x

y

θ3

1M

2M3M

ssF

M M M F1 2 3+ + = ss

Where is the line of action of Fss?

Moment of Momentum Equation

m=B r × V

mm

=r × Vb

( )( )ˆc s

d Aρ= ⋅∫T r × V V n

cv equation

Moment of momentum

ˆsy s

c v c s

D Bb d V b d A

D t tρ ρ∂

= + ⋅∂ ∫ ∫ V n

Moment of momentum/unit mass

( ) ( ) ( )ˆc v c s

D md V d A

D t tρ ρ∂

= + ⋅∂ ∫ ∫

r × Vr × V r × V V n

∑ Steady state

Application to Turbomachinery

r2r1

VnVt

( ) ( )[ ]2 2 1 1zT Qρ= −r × V r × V

cs1 cs2

( ) ( )ˆc s

d Aρ= ⋅∫T r × V V nrVt Vn ( )ˆ

c s

d A Qρ ρ⋅ =∫ V n

Example: Sprinkler

( ) ( )2 2 1 1zT Qρ ⎡ ⎤= −⎣ ⎦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 ω2.θ = 30º.Find the speed of rotation.

vt ω

10 cmθ 2

220 .1 tQ r Vω ρ− =

2 2s inje tt

je t

QV θ r

Aω= − +

22 22

4 / 20 .1 s inQQ r θ rd

ω ρ ωπ

−⎛ ⎞− = +⎜ ⎟⎝ ⎠

2 2 22 2 2

20 .1 s in 0Q r Q r θd

ω ρ ω ρπ

+ − =

Vt and Vn are defined relative to control surfaces.

cs2

Example: Sprinkler

0sin21.0 2222

22 =−+ θ

drQQr

πρωρω

aacbb

242 −±−

=ω

= 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 Nms22b Q rρ=2

2 22 s inc Q r θd

ρπ

= −

c = -1.27 Nm

a = 0.1Nms2

What is ω if there is no friction? ___________ω = 127/s

ω = 3.5/s

Reflections

What is Vt if there is no friction ?__________0

22 tT Q r Vρ=

Energy Equation

cv equationˆsy s

c v c s

D Bb d V b d A

D t tρ ρ∂

= + ⋅∂ ∫ ∫ V n

What is for a system?

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.n e t n e t 2 1in in

Q W E E+ = −

n e t sh a ftin

ˆc s

D E Q W p d AD t

= + − ⋅∫ V nn e t p r sh a ftin

W W W= +

p r ˆc s

W p d A= − ⋅∫ V n

DE/Dtˆcv cs

D E e d V e d AD t t

ρ ρ∂= + ⋅

∂ ∫ ∫ V n

dE/dt for our System?

n e tin

D E QD t

=

s h a ftD E WD t

=

ˆc s

D E p d AD t

= − ⋅∫ V n

hp γ=

pAF =

p rW F V= −

Heat transfer

Shaft work

Pressure work

n et sh a ftin

ˆc s

D E Q W p d AD t

= + − ⋅∫ V n

p rW p V A= −

General Energy Equation

n e t sh a ftin

ˆ ˆc s c v c s

D E Q W p d A e d V e d AD t t

ρ ρ∂= + − ⋅ = + ⋅

∂∫ ∫ ∫V n V n

n e t sh a ftin

ˆc v c s

pQ W e d e d At

ρ ρρ

∂ ⎛ ⎞+ = ∀ + + ⋅⎜ ⎟∂ ⎝ ⎠∫ ∫ V n

2

2Ve g z u= + +

z

cv equation1st Law of Thermo

Potential Kinetic Internal (molecular spacing and forces)

Total

Simplify the Energy Equation

Steady

n e tin

2

sh a ft ˆ2c s

p Vq w m g z u d Aρρ

⎛ ⎞⎛ ⎞+ = + + + ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ V n

cgzp=+

ρ

2

2Ve g z u= + +

Assume...Hydrostatic pressure distribution at cs

ŭ is uniform over cs

0n et sh a ftin

ˆc v c s

pQ W e d e d At

ρ ρρ

⎛ ⎞∂+ = ∀ + + ⋅⎜ ⎟⎝ ⎠∂ ∫ ∫ V n

netin

q mshaftw m

But V is often ____________ over control surface!not uniform

Energy Equation: Kinetic Energy

If V tangent to n

kinetic energy correction term

V = point velocityV = average velocity over cs

32

ˆ2 2c s

V V Ad A ρρ α⎛ ⎞⋅ =⎜ ⎟⎝ ⎠∫ V n

3

31

c s

V d AA V

α ⎛ ⎞= ⎜ ⎟⎝ ⎠∫

3

3

2

2

c s

V d A

V A

ρα

ρ

⎛ ⎞⎜ ⎟⎝ ⎠

=∫

α = _________________________

α =___ for uniform velocity1

Energy Equation: steady, one-dimensional, constant density

n e tin

2

sh a ft ˆ2c s

p Vq w m g z u d Aρρ

⎛ ⎞⎛ ⎞+ = + + + ⋅⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

∫ V n

ˆc s

d A mρ ⋅ =∫ V n mass flux rate

n e tin

2 2

sh a ft 2 2o u t o u t in in

o u t o u t in inp V p Vq w m g z u g z u mα αρ ρ

⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞+ = + + + − + + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

n e tin

2 2

sh a ft2 2in in o u t o u t

in in in o u t o u t o u tp V p Vg z u q w g z uα αρ ρ

+ + + + + = + + +

Energy Equation: steady, one-dimensional, constant density

divide by gn e tin

2 2

sh a ft2 2in in o u t o u t

in in in o u t o u t o u tp V p Vg z u q w g z uα αρ ρ

+ + + + + = + + +

n e tin

2 2sh a ft

2 2

o u t inin in o u t o u t

in in o u t o u t

u u qp V w p Vz zg g g g

α αγ γ

− −+ + + = + + +

n e tin

o u t in

L

u u qh

g

− −=

sh a ftP T

w h hg

= −hPLost mechanical energy

mechanical thermal

2 2

2 2in in o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g gα α

γ γ+ + + = + + + +

Thermal Components of the Energy Equation

2

2Ve g z u= + +

v pu c T c T= ≅

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

For incompressible liquids

n e tin

o u t in

L

u u qh

g

− −=

Change in temperature

Heat transferred to fluid( )n e tin

p o u t in

L

c T T qh

g

− −=

Example

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 o u t Lh z h= + L p o u th h z= −

2.4 mcs1

cs2

hL = 10 m - 4 m

2 2

2 2in in o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g gα α

γ γ+ + + = + + + +

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 α head loss

2 2

2 2in in o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g gα α

γ γ+ + + = + + + +

Example: Energy Equation(pressure at pump outlet)

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)/[ π(0.2 m)2] = 1.6 m/s

How do we get the velocity in the pipe?

How do we get the frictional losses?

What about α?

Q = VA A = πd2/4 V = 4Q/( πd2)

Kinetic Energy Correction Term: α

3

31

c s

V d AA V

α ⎛ ⎞= ⎜ ⎟⎝ ⎠∫

α 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

α is 1

α is 2

1.01 < α < 1.10

Example: Energy Equation(pressure at pump outlet)

V = 1.6 m/sα = 1.05hL = 1.44 m

2

2o u t o u t

P o u t o u t Lp Vh z h

gα

γ= + + +

datum

2 m4 m

50 L/shP = 10 m

⎥⎦

⎤⎢⎣

⎡−−−= m) (1.44

)m/s (9.812m/s) (1.6

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

23

2p

2.4 m

= 59.1 kPa

2

2o u t

o u t P o u t o u t LVp h z h

gγ α⎛ ⎞

= − − −⎜ ⎟⎝ ⎠

Example: Energy Equation(Hydraulic Grade Line - HGL)

pipe burst

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

2p V

gα

γ+

2 2

2 2in in o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g gα α

γ γ+ + + = + + + +

HP = 10 m

p = 59 kPa

What is the pressure at the pump intake?

Entrance loss

Exit lossLoss due to shear

EGL (or TEL) and HGL

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

gV

zp2

EGL2

αγ

++= zp+=

γHGL

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?

EGL (or TEL) and HGL

pump

coincident

reference pressure

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 weightEGL 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 __________________

Example HGL and EGL

z

z = 0

pump

energy grade line

hydraulic grade line

velocity head

pressure head

elevation

datum

2gV 2

α

γp

2 2

2 2in in o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g gα α

γ γ+ + + = + + + +

Bernoulli vs. Control Volume Conservation of Energy

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

pipe Free jet

Bernoulli vs. Control Volume Conservation of Energy

2 21 1 2 2

1 22 2p v p vz z

g gγ γ+ + = + +

2 2

2 2in in o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g gα α

γ γ+ + + = + + + +

Control surface to control surfacePoint to point along streamline

Has a term for frictional lossesNo frictional lossesBased on velocity

Requires kinetic energy correction factor

Includes shaft work

Based on velocity averagepoint

Has direction!

Power and Efficiencies

Electrical power

Shaft power

Impeller power

Fluid power

=electricP

=waterP

=shaftP

=impellerP

EI

Tω

Tω

γQHp

Motor losses

bearing losses

pump losses

Prove this!

P = FV

Example: Hydroplant

Powerhouse

River

Reservoir

Penstock

Q = 5 m3/s

2100 kW

180 rpm

116 kN·m

50 m

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

0.8570.893

0.96

2.45 MW

solution

Energy Equation Review

2 2

2 2in in o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g gα α

γ γ+ + + = + + + +

Control Volume equationSimplifications

steadyconstant densityhydrostatic pressure distribution across control surface (streamlines parallel)

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

Conservation of Energy, Momentum, and Mass

Most problems in fluids require the use of more than one conservation law to obtain a solutionOften a simplifying assumption is required to obtain a solution

neglect 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 changesLosses due to expansions are ________ than losses due to contractionsLosses can be minimized by gradual transitionsLosses are expressed in the formwhere KL is the loss coefficient

2

2L LVh K

g=

2 2

2 2in in o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g gα α

γ γ+ + + = + + + +

When V↓, KE → thermal

Head Loss due to Sudden Expansion:Conservation of Energy

in out

2 2

2 2in in o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g gα α

γ γ+ + + = + + + +

Where is p measured?___________________________At centroid of control surface

z

x

2 2

2in o u t o u t in

Lp p V V h

gγ− −

= + zin = zout

Relate Vin and Vout?2 2

2in o u t in o u t

Lp p V Vh

gγ− −

= +Mass

Relate pin and pout? Momentum

Head Loss due to Sudden Expansion:Conservation of Momentum

Apply in direction of flowNeglect surface shear

Divide by (Aout γ)

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 +=+

21 x in inM V Aρ= − 2

2 x o u t o u tM V Aρ=

2 2 ino u t in

in o u t o u t

AV Vp p A

gγ

−−

=

A1

A2

x

2 2in in o u t o u t in o u t o u t o u tV A V A p A p Aρ ρ− + = −

Head Loss due to Sudden Expansion

in o u t

o u t in

A VA V

=MassEnergy

Momentum

2 2

2in o u t in o u t

Lp p V Vh

gγ− −

= +

2 2 ino u t in

in o u t o u t

AV Vp p A

gγ

−−

=

2 22 2

2

o u to u t in

in in o u tL

VV VV V Vh

g g

−−

= +2 2

2

2 222

o u t in o u t inL

V V V Vhg

− +=

( ) 2

2in o u t

lV V

hg

−=

22

12

in inl

o u t

V Ahg A

⎛ ⎞= −⎜ ⎟⎝ ⎠

2

1 inL

o u t

AKA

⎛ ⎞= −⎜ ⎟⎝ ⎠

Discharge into a reservoir?_________KL=1

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 20.0005 / 6.4 /

0.014

m sV m sm

π

= =

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 o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g gα α

γ γ+ + + = + + + +

2 21 1 2 2

1 22 2p V p Vz z

g gγ γ+ + = + + 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 knownWhen possible choose a frame of reference so the flows are steady

Summary

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

Scoop Problem

stationary water tank

Scoop Problem:Change your Perspective

moving water tank

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 reservoirFind 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

o rQ C A 2 g h=

ˆc s c v

d A d Vt

ρ ρ∂⋅ = −

∂∫ ∫V n

Example Conservation of MassConstant Volume

h

0=+− ororresres AVAV

cs1

cs2

dtdhV res

−=

ororor QAV =

02 =+ ghCAAdtdh

orres

ˆc s c v

d A d Vt

ρ ρ∂⋅ = −

∂∫ ∫V n

1 2

1 1 1 2 2 2ˆ ˆ 0c s c s

d A d Aρ ρ⋅ + ⋅ =∫ ∫V n V n

Example Conservation of MassChanging Volume

hcs1

cs2

ˆc s c v

d A d Vt

ρ ρ∂⋅ = −

∂∫ ∫V n

o r o rc v

V A d Vt

∂= −

∂ ∫

re so r o r

A d hd VV Ad t d t

= − = −

ororor QAV =

02 =+ ghCAAdtdh

orres

Example Conservation of Mass

∫∫ =− th

hor

res dth

dhgCA

A

002

( ) thhgCA

A

or

res =−− 2/1

02/12

2

( )

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

21 / 2 1 / 2

22

2 0 .1 50 .0 3 0 .0 8

0 .0 0 20 .6 2 9 .8 /

4

mm m t

mm s

π−

− =⎛ ⎞⎜ ⎟⎝ ⎠

st 591=

Pump Head

hp

2

2in

inV

gα

2

2o u t

o u tV

gα2

2

2

2

in inin in P

o u t o u to u t o u t T L

p Vz hg

p Vz h hg

αγ

αγ

+ + + =

+ + + +

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 Venturi2 2

2 2in in o u t o u t

in in P o u t o u t T Lp V p Vz h z h h

g gα α

γ γ+ + + = + + + +

VAQ =2 2

2 2in o u t o u t inp p V V

g gγ γ− = −

in in o u t o u tV A V A=

2 2

4 4in o u t

in o u td dV Vπ π

=42

12

in o u t o u t o u t

in

p p V dg dγ γ

⎡ ⎤⎛ ⎞− = −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦ 2 2

in in o u t o u tV d V d=

( ) 4

2 ( )1

in o u to u t

o u t in

g p pVd dγ

−=

⎡ ⎤−⎣ ⎦

2

2o u t

in o u tin

dV Vd

=

( ) 4

2 ( )1

in o u tv o u t

o u t in

g p pQ C Ad dγ

−=

⎡ ⎤−⎣ ⎦

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

( )n e tin

p o u t in

L

c T T qh

g

− −= ( )( )

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅

=∆

KKgJ4184

m 50m/s 9.8 2

T

n e tin

L

p

g h qT

c

+∆ = KT 117.0=∆

Hydropower

pQ HP γ=

( ) ( ) ( )3 39 8 0 6 / 5 m / 5 0 m 2 .4 5w a terP N m s M W= =

2 .1 0 0 0 .8 5 72 .4 5to ta l

M WeM W

= =

( ) 2 1 m in0 .1 1 6 1 8 0 2 .1 8 7m in 6 0

2 .1 8 7 0 .8 9 32 .4 5

2 .1 0 0 0 .9 62 .1 8 7

tu rb in e

tu rb in e

g en e ra to r

r e v ra dP M N m M Wrev s

M WeM W

M WeM W

π⎛ ⎞= =⎝ ⎠

= =

= =

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 VVg

1 22

1 2

γ=

−

p V V V1 22

1 2= −ρc h

p pV V A

Ag

1 222

12 1

2−=

−

γ

( )

3

1 20.0005 / 6.4 /

0.014

m sV m sm

π

= =

2 0.71 /V m s=

( ) ( ) ( )( )( )21 1000 / 0.71 / 6.4 / 0.71 /p kg s m s m s m s= −

1 4p kPa= − Carburetors and water powered vacuums