ch3 bernoulli equation
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Elementary Fluid Dynamics – The Bernoulli Equation
• Consider inviscid, steady, two-dimensional flow in x-z plane
• Define streamlines
• Select coordinate systems based on streamlines
• Define acceleration
• Define forces
• Apply Newton’s second law of motion along and across streamline
Newton’s Second Law: F = ma
• Prior to apply Newton’s second law of motion to fluid particle:
• Consider motion of an inviscid fluid
• Assume that fluid motion is governed by pressure and gravity forces
• Select an appropriate coordinate system. Consider two dimensional motion (x-z plane)
Newton’s Second Law: F = ma
• Prior to apply Newton’s second law of motion to fluid particle:
• Consider motion of an inviscid fluid
• Assume that fluid motion is governed by pressure and gravity forces
• Select an appropriate coordinate system. Consider two dimensional motion (x-z plane)
Coordinate System
• Motion of a fluid particle is described by its velocity vector
Coordinate System
• Motion of a fluid particle is described by its velocity vector
• As the particle moves, it follows a particular path, the shape of which is governed by velocity vector
Coordinate System
• Motion of a fluid particle is described by its velocity vector
• As the particle moves, it follows a particular path, the shape of which is governed by velocity vector
• If flow is steady, each successive particle that passes through given point (1) will follow the same path. For such cases the path is a fixed line in the x-z plane. The entire x-z plane is filled with such paths.
Coordinate System
• For steady flow each particle slides along its path and its velocity vector is everywhere tangent to the path
Coordinate System
• For steady flow each particle slides along its path and its velocity vector is everywhere tangent to the path
• The lines that are tangent to the velocity vectors throughout the flow field are called streamlines.
Streamlines
• For steady flow each particle slides along its path and its velocity vector is everywhere tangent to the path
• The lines that are tangent to the velocity vectors throughout the flow field are called streamlines.
• We will use coordinates based on streamlines
Streamlines
• Particle motion is described in terms of its distance, s = s(t), along streamline, and local radius of curvature
Particle Motion
• Particle motion is described in terms of its distance, s = s(t), along streamline, and local radius of curvature
• Distance s is related to particle’s speed V = ds/dt, and radius of curvature is related to the shape of streamline
Particle Motion
• Acceleration:
Particle Acceleration
d dta V
• Acceleration:
• Components of acceleration in s and n directions:
Particle Acceleration
2
, s n
V Va V a
s
d dta V
• To determine forces consider free-body diagram of small fluid particle
Forces
F = ma along a Streamline
Free-body diagram of a fluid particle
F = ma along a Streamline
Equation of motion along streamline (details)
sin s
p VV a
s s
Free-body diagram of a fluid particle
Change in fluid particle speed is accomplished by combination of pressure gradient and particle weight along streamline
Example 3.1 Consider the inviscid, incompressible, steady flow along the horizontal streamline A–B in front of the sphere of radius a. From a more advanced theory of flow past a sphere, the fluid velocity along this streamline is
Determine the pressure variation along the streamline from point A far in front of the
sphere (xA = – and VA = V0) to point B on the sphere (xB = – a and VB = 0).
3
0 31
aV V
x
Example 3.1 Consider the inviscid, incompressible, steady flow along the horizontal streamline A–B in front of the sphere of radius a. From a more advanced theory of flow past a sphere, the fluid velocity along this streamline is
Determine the pressure variation along the streamline from point A far in front of the
sphere (xA = – and VA = V0) to point B on the sphere (xB = – a and VB = 0).
Solution Streamline is horizontal, then
Acceleration
3
0 31
aV V
x
p VV
s s
3 32
0 3 43 1
V V a aV V V
s x x x
3 2 3 30
4
3 1a V a xp
x x
632
0 2
a xap V
x
Example 3.1 Consider the inviscid, incompressible, steady flow along the horizontal streamline A–B in front of the sphere of radius a. From a more advanced theory of flow past a sphere, the fluid velocity along this streamline is
Determine the pressure variation along the streamline from point A far in front of the
sphere (xA = – and VA = V0) to point B on the sphere (xB = – a and VB = 0).
Solution Pressure gradient
Pressure distribution
3
0 31
aV V
x
20
2B
Vp
Bernoulli Equation
For incompressible fluid equation of motion along streamline reduces to Bernoulli equation (details)
Restricted to:
- inviscid flow
- steady flow
- incompressible flow
- along streamline
21constant along streamline
2p V z
Example 3.2 Consider the flow of air around a bicyclist moving through still air with velocity V0. Determine the difference in the pressure between points (1) and (2).
Example 3.2 Consider the flow of air around a bicyclist moving through still air with velocity V0. Determine the difference in the pressure between points (1) and (2).
Solution Apply Bernoulli equation between (1) and (2)
Pressure difference
2 21 1 1 2 2 2
1 1
2 2p V z p V z
2 22 1 1 0
1 1
2 2p p V V
F = ma Normal to a Streamline
Equation of motion along the normal direction (details)
Free-body diagram of a fluid particle
Change in the direction of flow of a fluid particle is accomplished by combination of pressure gradient and particle weight normal to streamline
2dz p V
dn n
Example 3.3 Shown in Fig. a, b are two flow fields with circular streamlines. The velocity distributions are
where C1 and C2 are constant. Determine the pressure distributions, p = p(r), for each, given that p = p0 at r = r0.
1
2
for case ( )
for case ( )
V r C r a
CV r b
r
Example 3.3 Shown in Fig. a, b are two flow fields with circular streamlines. The velocity distributions are
where C1 and C2 are constant. Determine the pressure distributions, p = p(r), for each, given that p = p0 at r = r0.
1
2
for case ( )
for case ( )
V r C r a
CV r b
r
Solution Assume steady, inviscid, and incompressible flow with streamlines in horizontal plane (dz/dn = 0)Since streamlines are circles, /n = - /r and = rThen equation of motion along the normal
becomes
2dz p V
dn n
2p V
r r
Example 3.3 Shown in Fig. a, b are two flow fields with circular streamlines. The velocity distributions are
where C1 and C2 are constant. Determine the pressure distributions, p = p(r), for each, given that p = p0 at r = r0.
1
2
for case ( )
for case ( )
V r C r a
CV r b
r
Solution
For case (a)
For case (b)
Comments: (a) – free vortex, (b) forced vortex
2 2 2 21 1 0 0
1 and
2
pC r p C r r p
r
2222 03 2 2
0
1 1 1 and
2
p Cp C p
r r r r
F = ma Normal to a Streamline
For steady, inviscid, incompressible flow (details)
Restricted to:
- inviscid flow
- steady flow
- incompressible flow
- across streamline
Pressure variation across straight streamlines is hydrostatic
2
constant across streamlineV
p dn z
Physical Interpretation
Work done on a particle by all forces acting on the particle is equal to the change of the kinetic energy of the particle
Each term of Bernoulli equation can be interpreted as:
- head (elevation, pressure, velocity)
- form of pressure (static, hydrostatic, dynamic)
21constant along streamline
2p V z
Example 3.4
Pressure variation across straight streamlines is hydrostatic
Static, Stagnation, Dynamic, and Total Pressure
Useful concept associated with the Bernoulli equation deals with the stagnation and dynamic pressures.
As fluid is brought to rest its kinetic energy is converted to a pressure rise
Static, Stagnation, Dynamic, and Total Pressure
Each term in Bernoulli equation can be interpreted as a form of pressure; static, p, hydrostatic, z, and dynamic, V 2/2 ,
21constant along streamline
2p V z
22 1 1
1
2p p V
Point (2) is a stagnation point
Pressure at the stagnation point is greater than static pressure by dynamic pressure
Static, Stagnation, Dynamic, and Total Pressure
• There is a stagnation point on any stationary body that is placed onto a flowing fluid
• Some of the fluid flows “over” and some “under” the object. Dividing line is termed the stagnation streamline and terminates at the stagnation point on the body
• Location of the stagnation point is function of body shape.
Static, Stagnation, Dynamic, and Total Pressure
• If elevation effect are neglected, stagnation pressure, p + V2/2, is the largest pressure obtainable along a given streamline. It represents the conversion of all of the kinetic energy into a pressure rise.
• Sum of the static pressure, hydrostatic pressure, and dynamic pressure is termed the total pressure, pT
• Bernoulli equation is a statement that total pressure remains constant along a streamline.
21constant along streamline
2 Tp V z p
Fluid Velocity Measurement
3 42 p pV
Pitot-static tubes measure fluid velocity by converting velocity into pressure
Pitot-static tube
Typical Pitot-Static Tube Designs
Measurement of Static Pressure
Incorrect and correct design of static pressure taps
Measurement of Static Pressure
Typical pressure distribution along a Pitot-static tube
Fluid Velocity Measurement
Many velocity measurement devices use Pitot-static tube principle
Cross section of a directional-finding Pitot-static tube
o
1 3
2
2
2 1
If 0 and 29.5
2
2
p p p
Vp p
p pV
Examples of Use of Bernoulli Equation
Free Jets
Exit pressure for an incompressible fluid jet is equal to the surrounding pressure
Velocity:
Vertical flow from a tank
2V gh
Free Jets
For horizontal nozzle velocity is not uniform
If d « h centerline velocity can be used as an “average velocity”
Horizontal flow from a tank
Free Jets
If exit is not smooth, diameter of the jet will be less than diameter of the hole.
This phenomena, called a vena contracta effect, is a result of the inability of the fluid to turn the sharp 90º corner indicated by dotted lines in the figure
Vena contracta effect for a sharp-edged orifice
Free Jets
Since streamlines in the exit plane are curved, the pressure across them is not constant.
The highest pressure occurs along the centerline at (2), and lowest pressure, p1 = p3 = 0
Vena contracta effect for a sharp-edged orifice
Free Jets
Assumption of uniform velocity with straight streamlines and constant pressure is not valid at the exit plane
It is valid in the plane of vena contracta, sectшon a-a, provided dj « h
Vena contracta effect for a sharp-edged orifice
Free Jets
Vena contracta effect is a function of the geometry of the ounlet.
Contraction coefficient:
Typical flow patterns and contraction coefficients for various round exit configurations
c j hC A A
Confined Flows
In nozzles and pipes of variable diameter velocity changes from one section to another
For such cases continuity equation must be used along with Bernoulli equation
Continuity equation states that mass cannot be created or destroyed
For incompressible fluid (details)
1 1 2 2 1 2 or AV A V Q Q
Example 3.7 A stream of water of diameter d = 0.1 m flows steadily from a tank of diameter D = 1.0 m. Determine the flowrate, Q , needed from the inflow pipe if the water depth remains constant, h = 2.0 m
Example 3.7 A stream of water of diameter d = 0.1 m flows steadily from a tank of diameter D = 1.0 m. Determine the flowrate, Q , needed from the inflow pipe if the water depth remains constant, h = 2.0 m
SolutionAssume steady, inviscid, incompressible flow.Apply Bernoulli equation between points (1) and (2)
With p1 = p2 = 0, z1 = h and z2 = 0
From continuity equation
Exit velocity
and volume flowrate
2 21 1 1 2 2 2
1 1
2 2p V z p V z
2 21 2
1 1
2 2V gh V
2
1 2
dV V
D
2 4
26.26 m/s
1
ghV
d D
31 1 2 2 0.0492 m /sQ AV A V
Example 3.7 A stream of water of diameter d = 0.1 m flows steadily from a tank of diameter D = 1.0 m. Determine the flowrate, Q , needed from the inflow pipe if the water depth remains constant, h = 2.0 m
SolutionIf D » d, then we can assume V1 ≈ 0. Error associated with this assumption:
4
2
40 2
2 1 1
2 1D
gh d DQ V
Q V gh d D
Example 3.8
Example 3.8
Answers: 13 2
3 22
269.0 m/s =7.67 m/s
0.00542 m / s 2963 N/m
pV V
Q p
Comments: V3 is determined strictly by the value of p1
In absence of viscous effect pressure throughout the hose is constant and equals to p2
Decrease in pressure from p1 to p3 accelerate the air and increase its kinetic energy
Pressure change (density change) is within 3%. Hence, incompressibility assumption is reasonable
Example 3.9
Answer:
Comments:For a given flowrate h does not depend on , but pressure difference, p1 – p2, as measured by pressure gage, does
2 2
2 1
2
1
2 1
A AQh
A g SG
Cavitation
Cavitation occurs when the pressure is reduced to the vapor pressure
Cavitation can cause damage to equipment
Pressure variation and cavitation in a variable area pipe
Cavitation
Tip cavitation from a propeller
Example 3.10
Answer:
Comments: Results are independent of diameter and length of the hose (provided viscous effects are not important
Proper design of hose is needed to ensure that it will not collapse due to the large pressure difference (vacuum) between the inside and outsides of the hose
28.2 ftH
Flowrate Measurement
Typical devices for measuring flowrate
Various flow meters are governed by the Bernoulli and continuity equations
We consider “ideal” flow meters – those devoid of viscous, compressibility, and other effects.
The flowrate is a function of the pressure difference across the flow meter
1 2
2 2
2 1
2
1
p pQ A
A A
Example 3.11
Answer:
Comments: These values represent “ideal” results, and these results are independent of flow meter geometry – an orifice, nozzle, or Venturi meter.
Tenfold increase in flowrate requires one-hundredfold increase in pressure difference. This nonlinear relationship can cause difficulties when measuring flowrates over a wide range of values. An alternative is to use two flow meters in parallel
1 21.16 kPa 116 kPap p
Flowrate Measurement. Sluice Gate
Sluice gate geometry
The flowrate under a sluice gate depends on the water depths on either side of the gate
1 22 2
2 1
2
1
g z zQ z b
z z
In the limit of z1»z2
A vena contracta occurs as water flows under a sluice gate
2 12Q z b gz
Flowrate Measurement. Sharp-crested Weir
Rectangular, sharp-crested weir geometry
Flowrate over a weir is a function of the head on the weir
3 21 12 2Q C Hb gH C b g H
Energy Line and Hydraulic Grade Line
Hydraulic grade line and energy line are graphical forms of the Bernoulli equation
Energy line represents the total head available to the fluid
Locus provided by a series of piezometric taps is termed the hydraulic grade line
Representation of the energy line and the hydraulic grade line
Energy Line and Hydraulic Grade Line
If the flow is steady, incompressible, and inviscid, the energy line is horizontal and at the elevation of the liquid in the tank.
Hydraulic grade line lies a distance of one velocity head below the energy line
At the pipe outlet the pressure head is zero (gage) so the pipe elevation and hydraulic grade line coincide
Energy line and hydraulic grade line for flow from a tank
Energy Line and Hydraulic Grade Line
The distance from pipe to hydraulic grade line indicates the pressure within the pipe
For flow below the hydraulic grade line, the pressure is positive
For flow above the hydraulic grade line, the pressure is negative
Use of the energy line and hydraulic grade line
Restriction on Use of the Bernoulli Equation
Restrictions on use for the Bernoulli equation are imposed by the assumptions used in its derivation.
To avoid incorrect use of Bernoulli equation one must take into account:
- Compressibility effects;
- Unsteady effects;
- Rotational effects;
- Viscosity effects;
- Presence of mechanical devices (pumps, turbines)
““Change of scene, and absence of the necessity for Change of scene, and absence of the necessity for thought, will restore the mental equilibrium”thought, will restore the mental equilibrium”
(Jerom K. Jerom, “Three Men In a Boat”)(Jerom K. Jerom, “Three Men In a Boat”)
END OF CHAPTER
Supplementary slides
F = ma along a Streamline
s s
V VF ma mV V V
s s
Newton’s second law along streamline
F = ma along a Streamline
sin sinsW W V Gravity force
F = ma along a Streamline
ps s s
pF p p n y p p n y V
s
Pressure force
F = ma along a Streamline
Net force sins s ps
pF W F V
s
back
back
Bernoulli Equation
Consider equation
Along streamline
Also
Finally, along streamline value of n is constant (dn = 0) so that
Hence, along streamline p/s = dp/ds . Then equation (a) becomes
Integration at constant density gives Bernoulli equation
sin (a)p V
Vs s
sindz
ds
21
2
d VVV
s ds
p p pdp ds dn ds
s n s
210 (along streamline)
2dp d V dz
F = ma Normal to a Streamline
Newton second law in normal direction2 2
n
mV V VF
F = ma Normal to a Streamline
Gravity force cos cosnW W V
F = ma Normal to a Streamline
Pressure force pn n n
pF p p s y p p s y V
n
F = ma Normal to a Streamline
Net force cosn n pn
pF W F V
n
back
Continuity EquationConsider a fluid flowing through a fixed volume. If the flow is steady, rate at which fluid flows into the volume must equal the rate at which it flows out of the volume (mass is conserved)
Mass flow rate is given by
Volume flow rate
Conservation of mass requires
If density remains constant
m QQ VA
1 1 1 2 2 2AV A V
1 1 2 2AV A V
back
Compressibility Effects
Bernoulli equation can be modified for compressible flows.
For compressible, inviscid, isothermal, steady flows:
Use of above equation is restricted by inviscid flow assumptions, since most isothermal flows are accompanied by viscous effects.
For compressible, isentropic (no friction or heat transfer), steady flow of a perfect gas:
2 21 1 2
1 22
ln2 2
V RT p Vz z
g g p g
2 21 1 2 2
1 21 21 2 1 2
k p V k p Vgz gz
k k
Compressibility EffectsBernoulli equation for compressible flow can be written for pressure ratio as
Where Ma = V/c is the Mach number; c is local speed of sound
122 11
1
11 Ma 1
2
k
kp p k
p
back
Pressure ratio as a function of Mach number for incompressible and compressible (isentropic) flow
A “rule of thumb” is that the flow of a perfect gas may be considered as incompressible provided the Mach number is less than about 0.3
Unsteady Effects
Bernoulli equation can be modified for unsteady flows.
For incompressible, inviscid, unsteady flows:
Use of this equation requires knowledge of variation of V/t along the streamline
2
1
2 21 2
1 1 2 22 2
s
s
V V Vp z ds p z
t
back
Example 3.12
Answers:
Comments: ?
24.83 m /sQ
b
Example 3.13
Answers:
Comments: ?
2 5 22 2tan 2 tan 2
2 2Q AV H C gh C ghH
0
0
5 2
3 2 05 2
2 0
tan 2 2 3
tan 2 2
H
H
Q C g H
Q C g H
Example 3.14
Answers:
Comments: ?
Example 3.16
Answers:
Comments: ?
2g
l
Example 3.17
Answers:
Comments: ?
21
2 518 kPa2
Vp h
Example 3.18
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