Transcript Fluid Flow Concepts and Basic Control Volume Equations

Finite Control Volume Analysis
Application of Reynolds Transport Theorem
CEE 331
July 18, 2015
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
=
¶
r bdV + òr bV ×n
ˆ dA cv equation
ò
¶t cv
cs
=
¶
r dV + òr V ×n
ˆ dA But DMsys/Dt = 0!
ò
¶t cv
cs
¶
r V ×n
ˆ dA = r dV
ò
ò
¶t cv
cs
Continuity Equation
mass leaving - mass entering = - rate of increase of mass in cv
Conservation of Mass
If mass in cv
¶
òr V ×nˆ dA = - ¶t òr dV is constant 1
cs
òr
nˆ
cv
1
V1 ×n
ˆ 1dA +
cs1
òr
V2 ×n
ˆ 2 dA = 0 A1
V1
cs2
òr V ×nˆ dA =r VA =
cs
òV ×nˆ dA
V =
2
2
cs
A
m
[M/T]
Unit vector nˆ is ______
normal
to surface and pointed
____
out of cv
r 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
òr
cs1
1
V1 ×n
ˆ 1dA +
òr
2
V2 ×n
ˆ 2 dA = 0
cs2
- r 1V 1 A1 + r 2 V 2 A2 = 0
V 1 A1 = V 2 A2 = Q
V1 A1 = V2 A2 = Q
[L3/T]
Density is constant across cs
Density is the same at cs1 and cs2
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?
¶
r V ×n
ˆ dA = r dV
ò
ò
¶t cv
cs
¶V
V ×n
ˆ dA = Constant density
ò
¶t
cs
dV
Qout - Qin = Velocity of the reservoir surface
dt
dh
Q
Ares dh

Qout  
dt
Ares
dt
Example
Linear Momentum Equation
DBsys
Dt
¶
=
r bdV + òr bV ×n
ˆ dA cv equation
ò
¶t cv
cs
B = mV
momentum
b=
å
F ¹ 0
mV
momentum/unit mass
m
DmV
¶
=
r VdV + òVr V ×n
ˆ dA
ò
Dt
¶t cv
cs
DmV
Steady state
= òVr V ×n
ˆ dA
Dt
cs
This is the “ma” side of the F = ma equation!
Linear Momentum Equation
DmV
= òVr V ×n
ˆ dA
Dt
cs
DmV
= òV1 r 1V1 ×n
ˆ 1dA + òV2 r 2 V2 ×n
ˆ 2 dA
Dt
cs1
cs2
Assumptions
DmV
= - ( r 1V1 A1 ) V1 + ( r 2V2 A2 ) V2
Dt
M1   r1V1 A1 V1   rQ V1
M 2   r 2V2 A2 V2   rQ V2
Vectors!!!
Uniform density
Uniform velocity
V  A
Steady
V fluid velocity
relative to cv
Steady Control Volume Form of
Newton’s Second Law
å
D (mV )
F=
= M1 + M 2
Dt
 What
are the forces acting on
the fluid in the control volume?
Gravity
Shear forces at the walls
Pressure forces at the walls
Pressure forces on the ends
å F = W +F
p1
F  M
1
 M2
+ Fp2 + Fpwall + Ft wall
Why no shear on control surfaces? _______________________________
No velocity gradient normal to 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 pipe in
thrust blocks
place
 force on the vane of a pump or turbine...

å F = W +F
p1
+ Fp2 + Fss
Fss  Fp
wall
 F
wall
=force applied by solid surfaces
Linear Momentum Equation
å F = W +F +F
F  M  M
p1
1
p2
+ Fss
Fp2
M2
2
M1 + M 2 = W + Fp1 + Fp2 + Fss
Fssx
Forces acting on control volume
The momentum vectors
have the same direction
as the velocity vectors
M1
Fp1
W
Fssy
M1 = - ( r Q ) V1
M 2 = ( r 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.
-1961 N ↑
W = _________
section 1
section 2 M1 + M 2 = W + Fp1 + Fp2 + Fss
D
50 cm
30 cm
0.196 m2 _________
A
_________
0.071 m2
z
m/s ↑ _________
4.23 m/s →
V 1.53
_________
?
p
150 kPa
_________
-459 N ↑ 1269
N → Direction of V vectors
M _________
_________
x
?
N ↑ _________
Fp 29,400
_________
Example: What is p2?
p1
V12
p2
V22
+ z1 +
=
+ z2 +
g1
2g g2
2g
é
V12 V22 ù
p2 = p1 + g êz1 - z2 +
ú
2
g
2
g
ë
û
2
2
é
ù
1.53
m/s
4.23
m/s
(
)
(
)
3
3
p2 = (150 x 10 Pa ) + (9810 N/m ) ê0 - 1 m +
2
2 ú
2
9.8
m/s
2
9.8
m/s
(
) (
) úû
ê
ë
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 2 x - Wx - Fp1x - Fp2 x
Fss  M 2  Fp
x
x
M2
1
2x
Fssx = (1269 N ) - (-9400 N )
Fssx = 10.7kN
Force of pipe on fluid
Pipe wants to move to the _________
left
z
x
Example: Reducing Elbow
Vertical Forces
Fss y = M 1y + M 2 y - Wy - Fp1y - Fp2 y
Fss y = M 1y - Wy - Fp1y
Fss y = - 459N - (- 1, 961N ) - (29,400N )
2
W
1
Fp1
Fss  27.9kN 28 kN acting downward on fluid
M1
z
y
up
Pipe wants to move _________
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
Example: Momentum with
Complex Geometry
Find Q2, Q3 and force on the
wedge.
Q1  10 L/s
V1  20 m/s
q2
cs2
y
Fy  0
q 1  10
q 2  130
r  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 +
g1
2g g2
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
atmospheric pressure
Fy  0  M1 y  M 2 y  M 3 y
0   rQV
1 1 sin q 1  rQ2V2 sin q 2  rQ3V3 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   rQV
1 1 sin q 1  rQ2V2 sin q 2  rQ3V3 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
Q3  Q1  Q2
Q2  6.133 L / s
1
1
Why is Q2 greater than Q3?
Q3  3.867 L / s
Solve for Fx
Fx  M1x  M 2 x  M 3 x
Fx   rQV
1 1 cosq 1  rQ2V1 cosq 2  rQ3V1 cosq 3
Fx  rV1 Q1 cosq 1  Q2 cosq 2  Q3 cosq 3
L
O
c
0.01 m / sh
cosaf
10
M
P
F c
1000 kg / m h
c
0.006133 m / sh
cosa
130fP
a20 m / sfM
M
P
c
0.003867 m / sh
cosa
50fP
M
N
Q
3
3
3
x
3
Fx  226 N
Force of wedge on fluid
Vector solution
M1  M 2  M 3  Fss
M1   rQ1V1  200N
M 2  rQ2V2  122.66N
M 3  rQ3V3  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
=
¶
r bdV + òr bV ×n
ˆ dA cv equation
ò
¶t cv
cs
B = mr × V
Moment of momentum
mr × V
b=
Moment of momentum/unit mass
m
D (mr × V )
¶
=
r r × VdV + òr (r × V )(V ×n
ˆ ) dA
ò
Dt
¶t cv
cs
T=
òr (r × V )(V ×nˆ ) dA
cs
Steady state
Application to Turbomachinery
rVt
Vn
T = òr (r × V )(V ×n
ˆ ) dA
Tz =
ò(rV )( r V dA)
t
cs
cs
n
Vn
Vt
r2
cs1
cs2
) (
)
r1
(
ù
Tz = r Q é
1 t1 û
ë r2Vt2 - rV
Example: Sprinkler
vt
w
(
cs2
) (
)
ù
Tz = r Q é
1 t1 û
ë r2Vt2 - rV
q
- 0.1w 2 = r Qr2Vt2
10 cm
Vt2 = -
Q jet
sin θ + w r2
Ajet
Total flow is 1 L/s.
 4Q / 2

Jet diameter is 0.5 cm.  0.1w 2  rQr 
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  rQr2 w  rQ 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  rQr22w  rQ 2 r2
a = 0.1Nms2
2
sin θ  0
2
d
b
w
b 2  4ac
2a
b = r Qr22
b = (1000 kg/m3)(0.001 m3/s) (0.1 m) 2 = 0.01 Nms
2
2
c = - r Q r2
sin θ
2
pd
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
Reflections
Energy Equation
DBsys
Dt
=
¶
r bdV + òr bV ×n
ˆ dA cv equation
ò
¶t cv
cs
DE
¶
=
r edV + òr eV ×n
ˆ dA What is DE/Dt for a system?
ò
Dt
¶t 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.
Wpr = - òpV ×n
ˆ dA
Qnet + Wnet = E2 - E1
in
in
Wnet = Wpr + Wshaft
in
cs
DE
= Qnet + Wshaft Dt
in
òpV ×nˆ dA
cs
dE/dt for our System?
p  h
F  pA
Wpr = - FV
Pressure work
DE
= Qnet + Wshaft Dt
in
Heat transfer
òpV ×nˆ dA
DE
=Dt
òpV ×dA
cs
cs
DE
= Qnet
Dt
in
Shaft work
DE
= Wshaft
Dt
General Energy Equation
1st Law of Thermo
DE
= Qnet + Wshaft Dt
in
= Qnet + Wshaft
in
Total
cv equation
¶
pV ×n
ˆ dA =
r edV + òr eV ×n
ˆ dA
ò
ò
¶t cv
cs
cs
æp
ö
¶
=
e r d " + òç + e÷ r V ×n
ˆ dA
ò
èr
ø
¶t cv
cs
z
V2
e = gz +
+u
2
Potential
Kinetic
Internal (molecular
spacing and forces)
Simplify the Energy Equation
q net m
wshaft m
0
Steady
æp
ö
¶
Qnet + Wshaft =
e r d " + òç + e÷ r V ×n
ˆ dA
ò
èr
ø
¶t cv
in
cs
V2
e = gz +
+u
2
2
æ
ö
p
V
æ
ö
q
+
w
+ gz +
+ u ÷ r V ×n
ˆ dA
ç
shaft ÷ m = ò
ç
net
è in
ø
èr
ø
2
cs
in
p
r
 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
Term
3
æ
V ö
r V A V = point velocity
r V ×n
ˆ dA = a
ç
÷
ò
V = average velocity over cs
è2 ø
2
cs
2
a =
æ
V3ö
r dA
ç
÷
ò
è2 ø
cs
If V tangent to n
3
rV A
2
energy correction term
a = kinetic
_________________________
1 æ
V3ö
a = òç 3 ÷ dA
A cs èV ø
a =___
1 for uniform velocity
Energy Equation: steady, onedimensional, constant density
æp
ö
V2
æ
ö
q net + wshaft ÷ m = òç + gz +
+ u ÷ r V ×n
ˆ dA
ç
è in
ø
èr
ø
2
cs
òr V ×nˆ dA = m
mass flux rate
cs
2
éæpout
ö æpin
öù
Vout
Vin2
æ
ö
q
+
w
m
=
+
gz
+
a
+
u
+
gz
+
a
+
u
ç
shaft ÷
out
out ÷
in
in ÷úm
êç
ç
è innet
ø
è
ø
è
øû
2
r
2
ë r
2
pin
Vin2
pout
Vout
+ gzin + a in
+ uin + q net + wshaft =
+ gzout + a out
+ uout
r
2
r
2
in
Energy Equation: steady, onedimensional, constant density
2
pin
Vin2
pout
Vout
+ gzin + a in
+ uin + q net + wshaft =
+ gzout + a out
+ uout
r
2
r
2
in
divide by g
2
pin
Vin2 wshaft
pout
Vout
+ zin + a in
+
=
+ zout + a out
+
g
2g
g
g
2g
in
g
thermal
mechanical
wshaft
= hhPP - hT
g
uout - uin - q net
uout - uin - q net
in
g
= hL
Lost mechanical
energy
2
pin
Vin2
pout
Vout
+ zin + a in
+ hP =
+ zout + a out
+ hT + hL
g
2g
g
2g
Thermal Components of the
Energy Equation
V2
e = gz +
+u
2
u = cvT @c pT
uout - uin - q net
in
= hL
g
For incompressible liquids
Water specific heat = 4184 J/(kg*K)
Change in temperature
Heat transferred to fluid
c p (Tout - Tin ) - q net
in
g
= 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
pin
Vin2
pout
Vout
+ zin + a in
+ hP =
+ zout + a out
+ hT + hL
g
2g
g
2g
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
pin
Vin2
pout
Vout
+ zin + a in
+ hP =
+ zout + a out
+ hT + hL
g
2g
g
2g
/ /
/
/
We need _______
a 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 a?
Kinetic Energy Correction Term:
1 æ
V ö
a = òç
dA
÷
A èV ø
a
3
3
cs
a
is a function of the velocity distribution in
the pipe.
is 1
 For a uniform velocity distribution a
____
 For laminar flow ______
a is 2
 For turbulent flow _____________
1.01 < a < 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
a = 1.05
hL = 1.44 m
2.4 m
2m
2
pout
Vout
hP =
+ zout + a out
+ hL
g
2g
50 L/s
hP = 10 m
4m
datum
pout
2
æ
ö
Vout
= g çhP - zout - a out
- hL ÷
è
ø
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 manometer)
 How?
Example: Energy Equation
(Energy Grade Line - EGL)
p = 59 kPa
HP = 10 m
p
V2
+a
g
2g
50 L/s
4m
2.4 m
2m
datum
What is the pressure at the pump intake?
2
pin
Vin2
pout
Vout
+ zin + a in
+ hP =
+ zout + a out
+ hT + hL
g
2g
g
2g
Hydraulic Grade Line
Energy Grade Line
EGL (or TEL) and HGL
EGL 
p

 z a
V2
2g
HGL 
Elevation head (w.r.t.velocity
datum)
head
Pressure head (w.r.t.
reference pressure)
p

z
Piezometric
head
EGL (or TEL) and HGL




The energy grade line may never be horizontal or 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
coincident
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
__________________
reference pressure
Example HGL and EGL
velocity head
a
V2
2g
pressure head
p

energy grade line
hydraulic grade line
z elevation
pump
z=0
2
in
datum
2
pin
V
pout
Vout
+ zin + a in
+ hP =
+ zout + a out
+ hT + hL
g
2g
g
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
 h1 
  h2 

2g 
2g
p1
2
pin
Vin2
pout
Vout
+ zin + a in
+ hP =
+ zout + a out
+ hT + hL
g
2g
g
2g
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
Power and Efficiencies
P = FV
 Electrical
power
Motor losses
Pelectric  IE
 Shaft
power
Pshaft 
 Impeller
bearing losses
Tw
power
pump losses
Pimpeller  Tw
 Fluid
power
Pwater  QHp
Prove this!
Example: Hydroplant
Water power = ?
total losses = ?
efficiency of turbine = ?
efficiency of generator = ?
2100 kW
50 m
116 kN·m
180 rpm
solution
2
pin
Vin2
pout
Vout
+ zin + a in
+ hP =
+ zout + a out
+ hT + hL
g
2g
g
2g
Energy Equation Review
 Control Volume
equation
 Simplifications
 steady
 constant
density
 hydrostatic pressure distribution across control
surface (cs normal to streamlines)
 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
to heat over a short
energy losses (_______)
distance with no flow expansion
 neglect shear forces on the solid surface over a
short distance
 neglect
2
pin
Vin2
pout
Vout
+ zin + a in
+ hP =
+ zout + a out
+ hT + hL
g
2g
g
2g
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 Remember vena contracta
 Losses can be minimized by gradual
transitions
V2
 Losses are expressed in the form hL = K L
2g
where KL is the loss coefficient
Head Loss due to Sudden Expansion:
Conservation of Energy
1
2
2
pin
Vin2
pout
Vout
+ zin + a in
+ hP =
+ zout + a out
+ hT + hL
g
2g
g
2g
2
pin - pout
Vout
- Vin2
=
+ hL
g
2g
2
pin - pout Vin2 - Vout
hL =
+
g
2g
zin = zout
What is pin - pout?
Head Loss due to Sudden Expansion:
Conservation of Momentum
A2
A1
x
2
1
M1  M 2  W  Fp  Fp  Fss Apply in direction of flow
1
2
M 1x  M 2 x  Fp  Fp
1x
M 1x = - r Vin2 Ain
Neglect surface shear
2x
2
M 2 x = r Vout
Aout
2
- r Vin2 Ain + r Vout
Aout = pin Aout - pout Aout
pin - pout
=
g
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
2
pin - pout Vin2 - Vout
+
Energy hL =
g
2g
2
2 Ain
Vout - Vin
Aout
Momentum pin - pout =
g
g
2
Vout
- Vin2
hL =
Vin (
=
Vout
Vin
g
2
2
in
2
out
V -V
+
2g
Ain
Vout
=
Mass A
Vin
out
2
Vout
- 2VinVout + Vin2
hL =
2g
2
2
Ain ö
V æ Ain ö K = æ
1hl =
1L
ç
÷
ç
÷
Aout ø
è
2g è
Aout ø
2
in
Vout )
hl
2g
Discharge into a reservoir?_________
KL=1
Example: Losses due to Sudden
Expansion in a Pipe (Teams!)
 A flow
expansion discharges 2.4 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
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
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
¶

òr V ×nˆ dA = -
cs
r dV
ò
¶t
cv
Example Conservation of Mass
Constant Volume
¶
r V ×n
ˆ dA = r dV
ò
ò
¶t cv
cs
òr
1
V1 ×n
ˆ 1dA +
cs1
òr
2
V2 ×n
ˆ 2 dA = 0
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
¶
r V ×n
ˆ dA = r dV
ò
ò
¶t cv
cs
Vor Aor
Vor Aor
¶
=dV
ò
¶t cv
Ares dh
dV
==dt
dt
Vor Aor  Qor
dh
dt
Ares  CAor
2 gh  0
cs1
h
cs2
Example Conservation of Mass
 Ares
CAor
2g
 Ares
CAor
2g
h

dh
h
h0
t

 dt
0


2 h1 / 2  h01 / 2  t
2
- 2 (0.15m )
2
æ
p (0.002m ) ö
2
2
9.8
m
/
s
(0.6) ç
(
)
÷
4
è
ø
t  591s
1/ 2
1/ 2
0.03
m
0.08
m
(
)
(
)
(
)=t
Pump Head
2
Vout
a out
2g
pin
Vin2
+ zin + a in
+ hP =
g
2g
2
pout
Vout
+ zout + a out
+ hT + hL
g
2g
hp
Vin2
a in
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
pin
Vin2
pout
Vout
+ zin + a in
+ hP =
+ zout + a out
+ hT + hL
g
2g
g
2g
2
pin
pout
Vout
Vin2
=
g
g
2g
2g
pin
p
- out
g
g
Vout =
4
é
æd out ö ù
V
=
ê1 - ç
÷ ú
2g ê
d
è
ø
in
ú
ë
û
2
out
2 g ( pin - pout )
4
é
g ë1 - (d out din ) ù
û
Q = Cv Aout
2 g ( pin - pout )
4
ù
gé
1
d
d
ë ( out in ) û
Q  VA
Vin Ain = Vout Aout
2
p din2
p d out
Vin
= Vout
4
4
2
Vin din2 = Vout d out
Vin = Vout
2
d out
d in2
Fire nozzle: Team Work
Identify what you need to know
Count your unknowns
Determine what equations you will use
8 cm
1000 kPa
2.5 cm
Find the Velocities
p1
V12 p2
V22
+ z1 +
= + z2 +
g
2g g
2g
p1 V12 V22
+
=
g 2g 2g
V1D12 = V2 D22
4
2
2
æD2 ö
V ç ÷ = V12
è D1 ø
2
2
2
1
p1 V
V
=
g 2g 2g
4
æ
æD2 ö ö
V
p1 = r
1- ç ÷ ÷
ç
2 è è D1 ø ø
2
2
V2 =
2 p1
é æD ö4 ù
r ê1 - ç 2 ÷ ú
ê
ë è D1 ø ú
û
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 this the force that the
firefighters need to brace
N
NO!
against? _______
991.2 N
N force applied by nozzle on water
L/s
Fssx = M 1x + M 2 x - Wx - Fp1x - Fp2 x
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 …?
 Why 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
c p (Tout - Tin ) - q
net
in
g
ghL + q net
DT =
= hL
in
cp

9.8 m/s 50 m 
DT 
2


J
 4184



Kg

K


DT  0.117 K
Hydropower
P = g QH p
Pwater = (9806 N / m3 )(5m3 / s ) (50m ) = 2.45MW
etotal
2.100 MW
=
= 0.857
2.45MW
rev 2p rad 1min ö
æ
Pturbine = (0.116 MNm ) 180
= 2.187 MW
è
min rev
60 s ø
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 2.4 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
0.0024m / s
V1 
2  30.56m / s
2
A
V
p1 V2  V1V2
1
0.01m
 2


A
V

g
2
1
4
V2  3.39m / s
p1  r V22  V1V2

a f
c h
p a
1000kg / sfa
d3.39m / sf  a30.56m / sfa3.39m / sfi
2
1
p1  92 kPa
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
Scoop Problem:
‘The Real Scoop’
stationary water tank
stationary water tank