Ch5 Energy in Steady Flow
Transcript of Ch5 Energy in Steady Flow
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Ch 5 Energy in Steady Flow
http://e/%ED%8C%91%E7%AF%8B%EE%81%A6%D1%A7/%CD%A2%CE%84/LECTURE%20NOTE/Clips/V4_5.mov -
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Outline
The fundamental equations of fluid dynamics
The continuity equation (principle ofconservation of mass)
The energy equation (Bernoulli equation)
Application of the energy equation
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Derive the Bernoulli (energy) equation
Use of the Bernoulli Equation
Introduce the momentum equation for a fluid
Demonstrate how the momentum equation andprinciple of conservation of momentum is used to
predict forces induced by flowing fluids
Objectives
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5.1 Energies of a Flowing Fluid
kinetic energy
22
2
1
2
1
, VmVKEenergyKinetic ==
g
V
g
VmV
Weight
KE
2
22
212
21
=
=
=
=== === gWweight volumemmass
Kinetic energy per unit weight
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5.1 Energies of a Flowing Fluid (Con)
Potential energy
Due to gravity
hgh
gh
W
EP
Weight
EP=
==
...
ghmghEP == .
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5.1 Energies of a Flowing Fluid (Con)
Pressure head
Fluids have energy in the form of pressure.
Weight
Energyessure
lengthr
p
hhpFrom
Pr
===
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5.2 Equation for Steady Motion of An Ideal
Fluid Along A Streamline
gdAdzds
dzgdAdsagdAdsadG === coscos
Pressure forces:
upstream end +pdA
downstream end -(p+dp)dA
Gravity force:
dt
dudAdsgdAdzdAdpppdA =+ )(
(Streamline dir)
maF=
Deriviation of Bernoull i equation
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)2
(2u
ds
d
ds
duu
dt
ds
ds
du
dt
du===
dt
dudAdsgdAdzdAdpppdA =+ )(
For steady flow ,velocity (u) only varies
with distance(s)
Substituting for du/dt
0)2
(2
=++g
u
g
pz
ds
d
C
g
u
g
pz =++
2
2
Bernouillis equation
(Along the streamline)
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Assumption ()
)(0 fluidInviscid=
)(. fluidibleIncompressconst=
Along the streamline (1-D)
Steady flow
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Bernoulli Equation
=g
p
=+ zg
p
The Bernoulli Equation is a
statement of the conservation
of ____________________Mechanical Energy
Pressure head
z =Elevation head
Velocity head
Piezometric head
Total head
Energy Head Line
Hydraulic Grade Line
Cg
uzg
p =++2
2
=g
u2
2
=++g
uzg
p2
2
Bernoullis eqn. is a useful relationship between p, V and z
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5.3 Energy line (EL) and
Hydraulic Grade line(HGL)
It is often convenient toplot mechanical energy
graphically using heights.
Hydraulic Grade Line
Energy Line (or totalenergy)
PH G L z
g= +
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Fig Energy Grade Line (EGL) and Hydraulic
Grade Line (HGL) for an one-dimensional flow
Energy Grade Line
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)2
(2
g
uz
g
p ++
Energy Grade Line (EGL)= Total head line
Hydraulic grade line (HGL) = Piezometric head line
(Piezometer)
+ z
p
()
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For the two cross- sections
g
u
g
pz
g
u
g
pz
22
2
222
2
111 ++=++
Eliminate the constant in the Bernoulli equation?
Apply at two points along a streamline.
Bernoulli equation does not include
Mechanical energy to thermal energy
Heat transfer
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Solving Steps with Bernoulli Equation
Three selections of steps for solving problems
and one solving-method .
1.Selecting the datum plane.
2.Selecting computation cross sections: It should be
the cross-section in uniform or gradually varied flowswith variables already known as many as possible.
3.Selecting the computed point : For the tube flow, it is
usually on the tube axis; while for the open channelflow, it is usually on the free surface.
4.Listing the energy equation and solving the problem.
Bernoulli Equation: Simple Case
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Bernoulli Equation: Simple Case
(V = 0)
Reservoir (V = 0)Put one point on the
surface, one pointanywhere else
g
uz
g
p=++
2
2
z
Elevation datum
Pressure datum
1
2
Same as we found using statics
We didnt cross any streamlinesso this analysis is okay!
22
11 z
g
pz
g
p+=+
g
pzz
221 =
Hydraulic and Energy Grade Lines
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Mechanical Energy
Conserved
Hydraulic and Energy Grade Lines
(neglecting losses for now)
The 2 cm diameter jet is
5 m lower than the
surface of the reservoir.
What is the flow rate(Q)?
z
p
g
z
2
2
V
g
Elevation datum
Pressure datum? __________________Atmospheric pressure
z
2
2V
g
2
2
p Vz C
gg+ + =
How do we compensate
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How do we compensate
for energy losses?
We add an energy loss term!
On which side?
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Energy Equation in Steady Total Flow
Real fluid steady total flow energy equation ( per unit
weight fluid )
In which,
zspecific elevation energy elevation head
specific pressure energy pressure head,
piezometric weight )
specific kinetic energy velocity head)
specific potential energy piezometric head)
total specific energy total head
gp
g
v
2
2
g
pz
+
g
v
g
pzH
2
2
++=
wh average specific energy loss (head loss).
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Kinetic Energy Correction Factor
We have assumed in the derivation of Bernoulli
equation that the velocity at the end sections (1)and (2) is uniform. But in a practical situation this
may not be the case and the velocity can very
across the cross section. A remedy is to use a
correction factor for the kinetic energy term in the
equation.
= AVdAu33 /
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0 01
2
z1
hw
1
2
z2z
p1
p2
1v12
2g2v22
2g
Piezometric head
line
Total headline
p
v 2
2g
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v21
2
2
1Water surface
v1
1v12
2g
2v22
2g
z1
z2
hwTotal head line
Piezometric head line
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Summary
By integrating F=ma along a streamline we found That energy can be converted between pressure, elevation, and
velocity
That we can understand many simple flows by applying the
Bernoulli equation
However, the Bernoulli equation can not be applied toflows where viscosity is large or where mechanical energy
is converted into thermal energy.
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Question
A diffuse pipe locates horizontally, as is shown in
the figure. If the head loss is neglected, then the
character of the pressures exerting on the cross-section's center is:
A. p1>p2
B. p1=p2
C. p1
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Applications of Bernoulli Equation
Stagnation tube Pitot tube
Free Jets Orifice Venturi Sluice gate Sharp-crested weir
Applicable to contracting
streamlines (accelerating flow).
Pi T b
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Pitot Tubes
Can connect a differential pressure transducer todirectly measure V2/2g Can be used to measure the flow of water in
pipelinesPoint measurement!
Pit t T b
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Pitot Tubes
g
p
g
p
g
ABB
=+
2
2
Total energy at A = Total energy at B
Static pressure tube
0gHpB =)( 0 hHgpA +=
ghpp BAB 2)(2
==
H. De Pitot
(1675-1771)
Th V t i M t
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The Venturi Meter
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Example Venturi Tube
Given: Water 20oC, V1=2 m/s, p1=150 kPa,
D=6 cm, d=3 cm
Find: p2 and p3
Solution: Continuity Eq.
Bernoulli Eq.
2
12
112
2211
==
=
d
D
VA
A
VV
AVAV
D Dd
1
2
3
( )
( )
kPap
Pa
VdDp
VVpp
g
Vz
p
g
Vz
p
120
2]3/61[2
1000000,150
]/1[2
)(2
22
2
24
21
41
22
2112
22
22
21
11
=
+=
+=
+=
++=++
Similarly for 2 3, or 1 3
Pressure drop is fully recovered, since we
assumed no frictional losses
kPap 1503
=
Nozzle: velocity
increases, pressure
decreases
Diffuser: velocity
decreases, pressure
increases
( ) ]/1[
)(24
212
Dd
ppV
=
Knowing the pressure drop 1 2 and
d/D, we can calculate the velocity and
flow rate
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Example
Given: Velocity in outletpipe from reservoir is 6
m/s and h = 15 m. Find: Pressure at A.
Solution: Bernoulli
equation
kPap
g
Vhp
gVp
gh
g
Vz
p
g
Vz
p
A
AA
AA
AA
A
0.129
)81.9
1815(9800)
2(
20
200
22
2
2
22
11
1
=
==
++=++
++=++
Point 1
Point A
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