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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
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