Finite Control Volume Analysis
Application of Reynolds Transport Theorem
CEE 331
March 10, 2016
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Moving from a System to a
Control Volume
Mass
Linear Momentum
Moment of Momentum
Energy
Putting it all together!
Conservation of Mass
B = Total amount of ____
mass in the system
1
b = ____
mass per unit mass = __
DBsys
Dt
DM sys
Dt



t

t
  bdV
ˆ dA cv equation
   bV  n
cv
  dV
cs
ˆ dA But DMsys/Dt = 0!
  V  n
cv

cs  V  nˆ dA   t
cs
  dV
Continuity Equation
cv
mass leaving - mass entering = - rate of increase of mass in cv
Conservation of Mass
If mass in cv

  V  nˆ dA   t   dV is constant 1
cs
òr
n̂
cv
1
V1 ×n
ˆ 1dA +
cs1
òr
2
V2 ×n
ˆ 2 dA = 0 A1
V1
cs2
  V  nˆ dA ± r VA = ± m [M/T]
cs
V 
2
 V  nˆ dA
cs
A
Unit vector n̂ is ______
normal
to surface and pointed
____
out of cv
 on
We assumed uniform ___
the control surface
V is the spatially averaged
velocity normal to the cs
Continuity Equation for Constant
Density and Uniform Velocity
  V  nˆ dA   
1
cs1
1
1
2
ˆ 2 dA  0 Density is constant across cs
V2  n
cs2
 1V 1 A1   2 V 2 A2  0
V 1 A1  V 2 A2  Q
V1 A1  V2 A2  Q
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
________
uniform or _______
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
h
dropping?

ˆ
cs  V  ndA   t cv  dV
V
Constant density
cs V  nˆ dA   t
dV
Qout  Qin  
Velocity of the reservoir surface
dt
dh
Q
Ares dh

Qout  
dt
Ares
dt
Example
Linear Momentum Equation
DBsys
Dt


t
B  mV
DmV


Dt
t
  bdV
cv
ˆ dA cv equation
   bV  n
cs
momentum
  VdV
cv
F  0
b
mV
momentum/unit mass
m
ˆ dA
  V V  n
Vectors!
cs
DmV
Steady state
ˆ dA
  V V  n
Dt
cs
This is the “ma” side of the F = ma equation!
Linear Momentum Equation
DmV
ˆ dA
  V V  n
Dt
cs
DmV
ˆ 1dA   V2  2 V2  n
ˆ 2 dA
  V1 1V1  n
Dt
cs1
cs2
DmV
   1V1 A1  V1    2V2 A2  V2
Dt
M1   1V1 A1 V1   Q V1
M 2    2V2 A2 V2   Q V2
Vectors!!!
Assumptions
Uniform density
Uniform velocity
V  A
Steady
V fluid velocity
relative to cv
Steady Control Volume Form of
Newton’s Second Law
F 
D  mV 
Dt
 M1  M 2
What are the forces acting on
the fluid in the control volume?
Gravity
Shear at the solid surfaces
Pressure at the solid surfaces
Pressure on the flow surfaces
F  W  F
p1
F  M
1
 M2
 Fp2  Fpwall  F wall
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
determines
size of _____________needed
to keep
thrust blocks
pipe in place
force on the vane of a pump or turbine...
F  W  F
p1
 Fp2  Fss
Fss  Fpwall  F wall
=force applied by solid surfaces
Linear Momentum Equation
F  W  F
p1
 Fp2  Fss
Fp2
ma  M1  M 2
M2
M1  M 2  W  Fp1  Fp2  Fss
Fssx
Forces by solid surfaces on fluid
The momentum vectors
have the same direction
as the velocity vectors
M1
Fp1
W
Fssy
M1     Q  V1
M 2    Q  V2
Example: Reducing Elbow
2
Reducing elbow in vertical plane with water flow of
300 L/s. The volume of water in the elbow is 200 L.
1m
Energy loss is negligible.
1
Calculate the force of the elbow on the fluid.
g*volume=-1961 N ↑
W = ________________________
M1  M 2  W  Fp  Fp  Fss
section 1
section 2
z
D
50 cm
30 cm
0.196 m2
A ____________
____________
0.071 m2
1.53 m/s ↑
4.23 m/s →
V ____________
____________
x
?
p
150 kPa
____________
Direction of V vectors
-459 N ↑
1270 N →
M ____________
____________
?←
29,400 N ↑
Fp ____________
____________
1
2
Example: What is p2?
V12
p2
V22
 z1 

 z2 
1
2g
2
2g
p1

V12 V22 
p2  p1    z1  z2 


2
g
2
g


2
2


1.53
m/s

4.23
m/s




3
3

p2  150 x 10 Pa    9810 N/m  0  1 m 
2
2  9.8 m/s 




P2 = 132 kPa
Fp2 = 9400 N
Example: Reducing Elbow
Horizontal Forces
M1  M 2  W  Fp1  Fp2  Fss
2
Fp2
Fss  M1  M 2  W  Fp1  Fp2
Fssx  M 1x  M 2x  Wx  Fp1x  Fp2 x
Fss  M 2  Fp
x
x
M2
1
2x
Fssx  1270 N    -9400 N 
Fssx  10.7kN
Force of pipe on fluid
Fluid is pushing the pipe to the ______
left
z
x
Example: Reducing Elbow
Vertical Forces
Fssz  M 1z  M 2z  Wz  Fp1z  Fp2 z
Fssz  M 1z  Wz  Fp1z
Fssz  459N   1,961N    29,400N 
2
1
Fp1
Fssz  27.9kN 28 kN acting downward on fluid
up
Pipe wants to move _________
W
M1
z
x
Example: 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 cm
2.5 cm
Fire nozzle: Solution
Identify what you need to know
P2, V1, V2, Q, M1, M2, Fss
Determine what equations you will use
Bernoulli, continuity, momentum
8 cm
1000 kPa
2.5 cm
Find the Velocities
V12 p2
V22
 z1 

 z2 

2g 
2g
p1
V12 V22


 2g 2g
p1
continuity→
V1 D12  V2 D22
4
p1
2
2
 D2 
2
V 

V
1

D
 1
2
2
2
1
V
V


 2g 2g
4

 D2  
V
1  

p1  

2   D1  


2
2
V2 
2 p1
  D 4 
 1   2  
  D1  
Fire nozzle: Solution
section 1 section 2
D
0.08
0.025 m
A
P
V
Fp
M
Fssx
Q
0.00503
1000000
4.39
5027
-96.8
-4132
22.1
8 cm
1000 kPa
2.5 cm
0.00049 m2
Which direction does the
nozzle want to go? ______
0 Pa
44.94 m/s
Is Fssx the force that the
firefighters need to brace
N
against? ____
__________
NO! Moments!
991.2 N
N force applied by nozzle on water
L/s
Fssx  M 1x  M 2x  Wx  Fp1x  Fp2 x
Example: Momentum with
Complex Geometry
Find Q2, Q3 and force on the
wedge in a horizontal plane.
Q1  10 L/s
V1  20 m/s
q2
cs2
y
Fy  0
q 1  10
q 2  130
  1000 kg / m
q 3  50
x
cs1
3
q1
Q2, Q3, V2, V3, Fx
Unknown: ________________
cs3
q3
5 Unknowns: Need 5 Equations
Identify the 5 equations!
Unknowns: Q2, Q3, V2, V3, Fx
Continuity Q1  Q2  Q3
Bernoulli (2x)
p1
V12 p2
V22
 z1 

 z2 
1
2g  2
2g
q2
cs2
V1 V2
V1 V3
Momentum (in x and y)
M1  M 2  M3  W  Fp1  Fp2  Fp3  Fss
y
x
cs1
q1
cs3
q3
Solve for Q2 and Q3
M1  M 2  M3  W  Fp1  Fp2  Fp3  Fss
Fssy  0  M1 y  M 2 y  M 3 y
atmospheric pressure
M1     Q  V1
0   QV
1 1 sin q 1  Q2V2 sin q 2  Q3V3 sin q 3
V sin q  Component of velocity in y direction
y
Q1  Q2  Q3
Mass conservation
q
V1  V2  V3
Negligible losses – apply Bernoulli
x
Solve for Q2 and Q3
0   QV
1 1 sin q 1  Q2V2 sin q 2  Q3V3 sin q 3
0  Q1 sin q 1  Q2 sin q 2  Q3 sin q 3
 sin q  sin q f
a
Q Q
a sinq  sinq f
 sinaf
10  sina
50f
Q Q
 sina
130f sina
50f
2
2
1
3
2
3
Eliminate Q3
Q3  Q1  Q2
Q2  6.133 L / s
1
1
Why is Q2 greater than Q3?
+
+
-
mV
1 1 y  m2V2 y  m3V3 y
Q3  3.867 L / s
Solve for Fssx
Fssx  M1x  M 2 x  M 3 x
Fssx    QV
1 1 cos q1   Q2V1 cos q 2   Q3V1 cos q3
Fssx  V1  Q1 cos q1  Q2 cos q 2  Q3 cos q3 
   0.01 m3 /s  cos 10 



Fssx  1000 kg/m3   20 m/s     0.006133 m 3 /s  cos 130  


   0.003867 m3 /s  cos  50  


Fssx  226 N
Force of wedge on fluid
Vector solution
M1  M 2  M 3  Fss
M1   Q1V1  200N
M 2  Q2V2  122.66N
M 3  Q3V3  77.34N
Q2  10 L / s
Q2  6.133L / s
Q3  3.867 L / s
Vector Addition
M1  M 2  M 3  Fss
q2
cs2
Fss
M3
y
M2
x
M1
cs1
q1
Where is the line of action of Fss?
cs3
q3
Moment of Momentum Equation
DBsys
Dt


t
  bdV
cv
B  mr × V
mr × V
b
m
D  mr × V 
Dt
ˆ dA cv equation
   bV  n
cs
Moment of momentum
Moment of momentum/unit mass


t
  r × VdV
cv
T     r × V  V  nˆ  dA
cs
ˆ  dA
    r × V  V  n
cs
Steady state
Application to Turbomachinery
rVt
Vn
ˆ  dA
T     r × V  V  n
   V  nˆ  dA   Q
cs
cs
Vn
Vt
cs1
r2
cs2
r1
Tz   Q 
 r2 × V2    r1 × V1  

Example: Sprinkler
vt
w
cs2 T   Q  r × V    r × V  
z
2
1
1 
 2
q
0.1w 2   Qr2Vt2
10 cm
Vt2  
Q jet
sin θ  w r2
A jet
Total flow is 1 L/s.
Jet diameter is 0.5 cm. 0.1w 2   Qr  4Q / 2 sin θ  w r 
2 
2 
2

d
Friction exerts a torque of


2
0.1 N-m-s2 w2.
2
2
2
0.1w   Qr2 w   Q r2
sin θ  0
2
q = 30º.
d
Find the speed of rotation.
Vt and Vn are defined relative to control surfaces.
Example: Sprinkler
0.1w 2  Qr22w  Q 2 r2
a = 0.1Nms2
2
sin θ  0
2
d
b
w
b 2  4ac
2a
b   Qr22
b = (1000 kg/m3)(0.001 m3/s) (0.1 m) 2 = 0.01 Nms
2
2
c    Q r2
sin θ
2
d
c = -(1000 kg/m3)(0.001 m3/s)2(0.1m)(2sin30)/3.14/(0.005 m)2
c = -1.27 Nm
w = 127/s
What is w if there is no friction? ___________
w = 3.5/s
0
What is Vt if there is no friction ?__________
= 34 rpm
T   Qr2Vt2
Reflections
Energy Equation
DBsys
Dt

DE


Dt
t

t
  bdV
ˆ dA cv equation
   bV  n
cv
  edV
cs
ˆ dA What is DE/Dt for a system?
   eV  n
cv
cs
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.
ˆ dA
Wpr    pV  n
Qnet  Wnet  E2  E1
in
in
Wnet  Wpr  Wshaft
in
cs
DE
ˆ dA
 Qnet  Wshaft   pV  n
Dt
in
cs
dE/dt for our System?
p  h
F  pA
Wpr   FV
Wpr   pVA
Pressure work
DE
ˆ dA
 Qnet  Wshaft   pV  n
Dt
in
cs
Heat transfer
DE
   pV  n
ˆ dA
Dt
cs
DE
 Qnet
Dt
in
Shaft work
DE
 Wshaft
Dt
General Energy Equation
1st Law of Thermo
DE

ˆ dA 
 Qnet  Wshaft   pV  n
Dt
t
in
cs
Qnet  Wshaft
in
cv equation
  edV
cv
ˆ dA
   eV  n
cs

p


e

d



e
V  n
ˆ dA




t cv


cs 
z
V2
e  gz 
u
2
Total
Potential
Kinetic
Internal (molecular
spacing and forces)
Simplify the Energy Equation
q net m
wshaft m
0
Steady
 p


ˆ dA
Qnet  Wshaft 
e d      e   V  n

t cv

in

cs 
V2
e  gz 
u
2
2


p
V


ˆ dA
q

w
m


gz

 u  V  n
shaft 

 net


2
 in


cs 
in
p

 gz  c
Assume...
Hydrostatic pressure distribution at cs
ŭ is uniform over cs
not uniform over control surface!
But V is often ____________
Energy Equation: Kinetic Energy
3
V 
V A V = point velocity
cs  2   V  nˆ dA   2 V = average velocity over cs
2
V 3 
cs  2   dA
 
3
V A
2
If V tangent to n
energy correction term
 = kinetic
_________________________
1 V 3 
    3  dA
A cs  V 
 =___
1 for uniform velocity
Energy Equation: steady, onedimensional, constant density

 q net  wshaft
 in
 p

V2

ˆ dA
 m      gz  2  u   V  n


cs 
  V  nˆ dA  m
mass flux rate
cs

 q net  wshaft
 in
2
 pout
  pin

Vout
Vin2

m


gz



u


gz



u

out
out 
in
in   m


2
2

  

 
2
Vin2
pout
Vout
 gzin   in
 uin  q net  wshaft 
 gzout   out
 uout

2

2
in
pin
Energy Equation: steady, onedimensional, constant density
2
Vin2
pout
Vout
 gzin   in
 uin  q net  wshaft 
 gzout   out
 uout

2

2
in
pin
divide by g
2
Vin2
wshaft
pout
Vout
 zin   in


 zout   out


2g
g

2g
pin
in
g
thermal
mechanical
wshaft
 hhPP  hT
g
uout  uin  q net
uout  uin  q net
in
g
 hL
Lost mechanical
energy
2
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL

2g

2g
pin
Thermal Components of the
Energy Equation
V2
e  gz 
u
2
u  cvT  c pT
uout  uin  q net
 hL
in
g
For incompressible liquids
Water specific heat = 4184 J/(kg*K)
Change in temperature
c p Tout  Tin   q net
in
g
Heat transferred to fluid
 hL
Example
Example: Energy Equation
(energy loss)
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? What is hL?
cs2
4m
2.4 m
2m
cs1
datum
Why can’t I draw the cs at the end of the pipe?
2
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL

2g

2g
pin
hp  zout  hL
hL  hp  zout
hL = 10 m - 4 m
Example: Energy Equation
(pressure at pump outlet)
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.
50 L/s
hP = 10 m
4m
cs2
2.4 m
2m
cs1
0 0
0 datum
0
2
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL

2g

2g
pin
/ /
/
/
We need _______
 and ____
velocity in the pipe, __,
head ____.
loss
Example: Energy Equation
(pressure at pump outlet)
How do we get the velocity in the pipe?
Q = VA
A = d2/4
V = 4Q/( d2)
V = 4(0.05 m3/s)/[ 0.2 m)2] = 1.6 m/s
How do we get the frictional losses?
Expect losses to be proportional to length of the pipe
hl = (6 m)(12 m)/(50 m) = 1.44 m
What about ?
Kinetic Energy Correction Term:
1 V 
  
 dA
A V 

3
3
cs
 is a function of the velocity distribution in
the pipe.
is 1
For a uniform velocity distribution 
____
For laminar flow ______
 is 2
For turbulent flow _____________
1.01 <  < 1.10
Often neglected in calculations because it is so
close to 1
Example: Energy Equation
(pressure at pump outlet)
V = 1.6 m/s
 = 1.05
hL = 1.44 m
2.4 m
2m
hP 
pout

 zout   out
2
Vout
 hL
2g
50 L/s
hP = 10 m
4m
datum
2


Vout
pout    hP  zout   out
 hL 
2g


2


(1.6
m/s)
3
p2  (9810 N/m ) (10 m)  (2.4 m)  (1.05)
 (1.44 m)  = 59.1 kPa
2
2(9.81 m/s )


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 (_________)
pipe burst 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)
Loss due to shear
HP = 10 m
Entrance loss
2.4 m
2m
2
p
V


2g
4m
p = 59 kPa
Exit loss
50 L/s
datum
What is the pressure at the pump intake?
2
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL

2g

2g
pin
Hydraulic Grade Line
Energy Grade Line
EGL (or TEL) and HGL
EGL 
p

 z 
V2
2g
HGL 
Elevation head (w.r.t.velocity
datum)
head
p

z
Piezometric
head
Pressure head (w.r.t.
reference pressure)
What is the difference between EGL defined by Bernoulli
and EGL defined here?
EGL (or TEL) and HGL
 The energy grade line may never slope upward (in
direction of flow) unless energy is added (______)
pump
 The decrease in total energy represents the head
loss or energy dissipation per unit weight
 EGL and HGL are ____________and
lie at the
coincident
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 __________________
reference pressure
Example HGL and EGL
velocity head 
V2
2g
pressure head
p

energy grade line
hydraulic grade line
z elevation
pump
z=0
pin

 zin   in
2
in
datum
2
V
p
V
 hP  out  zout   out out  hT  hL
2g

2g
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
v12
p2
v22
 z1 

 z2 

2g 
2g
p1
2
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL

2g

2g
pin
Point to point along streamline
Control surface to control surface
No frictional losses
Has a term for frictional losses
Based on point velocity
Based on average velocity
Requires kinetic energy
correction factor
Includes shaft work
Has direction!
Power and Efficiencies
P = FV
Electrical power
Motor losses
Pelectric  EI
Shaft power
Pshaft 
bearing losses
Tw
Impeller power
pump losses
Pimpeller  Tw
Fluid power
Pwater  QHp
Prove this!
Example: Hydroplant
Water power = 2.45 MW
Overall efficiency = 0.857
efficiency of turbine = 0.893
efficiency of generator = 0.96
50 m
Reservoir
Powerhouse
2100 kW
116 kN·m
180 rpm
River
solution
2
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL

2g

2g
pin
Energy Equation Review
 Control Volume equation
 Simplifications
 steady
 constant density
 hydrostatic 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
solution
Often a simplifying assumption is required
to obtain a solution
mechanical
to heat over a short
neglect energy losses (_______)
distance with no flow expansion
neglect shear forces on the solid surface over a
short distance
2
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL

2g

2g
pin
Head Loss: Minor Losses
Head (or energy) loss due to:
outlets, inlets, bends, elbows, valves, pipe
size changes
Losses due to expansions are ________
greater than
losses due to contractions When V, KE  thermal
Losses can be minimized by gradual
transitions
V2
Losses are expressed in the form hL  K L 2 g
where KL is the loss coefficient
Head Loss due to Sudden Expansion:
Conservation of Energy z
x
in
out
At centroid of control surface
Where is p measured?___________________________
2
pin
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL

2g

2g
pin  pout

hL 
2
Vout
 Vin2

 hL
2g
pin  pout

2
Relate Vin and Vout?
Vin2  Vout

2g
Relate pin and pout?
zin = zout
Mass
Momentum
Head Loss due to Sudden Expansion:
Conservation of Momentum
A2
A1
x
2
1
M1  M 2  W  Fp1  Fp2  Fss Apply in direction of flow
M 1x  M 2 x  Fp1 x  Fp2 x
M 1x   Vin2 Ain
Neglect surface shear
2
M 2 x  Vout
Aout
2
 Vin2 Ain  Vout
Aout  pin Aout  pout Aout
pin  pout

2
Vout
 Vin2

g
Ain
Aout
Pressure is applied over all of
section 1.
Momentum is transferred over
area corresponding to upstream
pipe diameter.
Vin is velocity upstream.
Divide by (Aout )
Head Loss due to
Sudden Expansion
Energy hL 
pin  pout

Momentum pin  pout

2
Vin2  Vout

2g
2
2 Ain
Vout  Vin
Aout

g
2
2 Vout
V

V
2 out 2 in
2
Vin
Vin2  Vout
hL 

2g
2g
2
Vout
 2VinVout  Vin2
hL 
2g

Ain 
Vin  Vout 

1 

hl 
A
out 

2g
Discharge into a reservoir?_________
KL=1
2
2
in
V
hl 
2g
Ain
Vout
Mass A  V
out
in
2

Ain 
K L  1 

A
out 

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
3
0.0005
m
/s
energy losses)!
V1 
 6.4m / s
2
0.01m 

Use the momentum equation…

4
V2  0.71m / s
Solution
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 cm
3m
Vscoop
stationary water tank
Scoop Problem:
‘The Real Scoop’
2
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL

2g

2g
pin
V12 p2
V22
 z1 

 z2 

2g 
2g
p1
Energy
Bernoulli
moving water tank
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 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 long riser tube
Very short riser tube
Example: Conservation of Mass
(Team Work)
 The flow through the orifice is a function of the depth
of water in the reservoir
Q  CAor 2gh
 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

  V  nˆ dA     dV
cs
t
cv
Example Conservation of Mass
Constant Volume
  V  nˆ dA  
cs

t
  dV
cv
  V  nˆ dA   
1
1
1
cs1
2
ˆ 2 dA  0
V2  n
Vres 
dt
h
cs2
 Vres Ares  Vor Aor  0
dh
cs1
 dh
dt
Ares  CAor
2 gh  0
Vor Aor  Qor
cs2
Example Conservation of Mass
Changing Volume
  V  nˆ dA  
cs
Vor Aor
Vor Aor


t

t
  dV
cv
cv
Ares dh
dV


dt
dt
dt
Ares  CAor
h
 dV
Vor Aor  Qor
dh
cs1
2 gh  0
cs2
Example Conservation of Mass
 Ares
CAor
h
2g
 Ares
CAor
2g

dh
t

h
h0
 dt
0


2 h1 / 2  h01 / 2  t
2  0.15m 
   0.002m 
 0.6  
4

t  591s
2




2
2  9.8m / s 2 

 0.03m 
1/ 2
  0.08m 
1/ 2
t
Pump Head
 out
Vin2
 zin   in
 hP 

2g
pin
pout

 zout   out
2
Vout
 hT  hL
2g
hp
Vin2
 in
2g
2
Vout
2g
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.
Dh
1
2
Example Venturi
2
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL

2g

2g
pin
pin

pin


Vout 

pout

pout

2
Vout
Vin2


2g
2g
4




V
d out
1  

 
2g 
 d in  


2
out
2 g ( pin  pout )
4

 1   d out din  


Q  Cv Aout
2 g ( pin  pout )
4
 1   d out din  


Q  VA
Vin Ain  Vout Aout
Vin
 din2
4
 Vout
2
 d out
4
2
Vin din2  Vout d out
Vin  Vout
2
d out
d in2
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 meters
Find the temperature rise
Ignore kinetic energy
c p Tout  Tin   q net
in
g
DT 
ghL  q net
in
cp
 hL
9.8 m/s 50 m 
DT 
2


J
 4184


Kg  K 


DT  0.117 K
Hydropower
P   QH p
Pwater   9806 N / m3  5m3 / s   50m   2.45MW
2.100 MW
etotal 
 0.857
2.45MW
rev 2 rad 1min 

Pturbine   0.116 MNm   180
  2.187 MW
min rev
60 s 

2.187 MW
eturbine 
 0.893
2.45MW
2.100 MW
egenerator 
 0.96
2.187 MW
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
A
1 cm
V22  V12 1
p1  p2
A2

3 cm
3

g
0.0005m / s
V

 6.4m / s
1
2
2
A1 V2
p1 V2  V1V2
0.01m 




A
V

g
2
1
4
c
V  0.71m / s
h
p  1000kg / s    0.71m / s    6.4m / s  0.71m / s  
p1   V22  V1V2
2
2
1
p1  4kPa
Carburetors and water powered vacuums