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