bernouli's equation
TRANSCRIPT
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Bernoulli equation
Daniel Bernoulli (1710-1782)
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Bernoulli: preamble
Want to discuss the properties of a moving uid.Will do this initially under the simplest possibleconditions, leading to Bernoullis equation. Thefollowing restrictions apply.
Flow is inviscid, there are no viscous drag forces Heat conduction is not possible for an inviscid
ow
The uid is incompressible . The ow is steady (velocity pattern constant). The paths traveled by small sections of the uid
are well dened.
Will be implicitly using the Euler equations of motion (discussed later)
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Coordinates and streamlines
Each piece of uid has velocity v . Steady ow, nothing changes with time at given
location. All particles passing through (1) end
up at (2) with velocity v
The trajectories followed by the particles arecalled streamlines.
Describe motion is terms of distance traveledalong streamline, s .
Velocity given by |v | = dsdt . Normal to velocityis n .
Stream-line can bend, R is radius of curvature.
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Coordinates and streamlines
Using body xed coordinates. If the particles changespeed along stream-line, or if stream-line bends, thenaccelerations must be present.The tangential acceleration
a s =dvdt
=vs
dsdt
= vvs
The normal acceleration
an =v2
R
The radius of curvature R changes along thestreamline.
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Streamline coordinates
It is convenient to use a coordinate system dened in
terms of the ow streamlines. The coordinate alongthe streamline is s and the coordinate normal to thestreamline is n . The unit vectors for the streamlinecoordinates are s and n .
The direction of s will be chosen to be in the samedirection as the velocity. So v = vs .
s
n^
s^
V
s = 0
s = s 1
s = s 2n = n 2
n = n 1
n = 0
Streamlines
y
x
The ow plane is covered with an orthogonal curvednet of coordinate lines and v = v(s, n ) s ands = s (s, n ) for steady ow.
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Forces on streamlines
Any particle travelling along the streamline will besubjected to a number of forces.
The relevant Forces for Bernoullis equation are
gravity and pressure.
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Streamlines F = ma
Will resolve forces in directions parallel s andperpendicular n to particles motions. y is out of
page, z is down, x is horizontal.
F s = m a s = m vvs
= V vvs
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Streamlines F = ma
Resolve forces
W s = W sin = V sin
W s would be zero for horizontal motion.
The pressure changes with height. Let p be pressurein middle of uid slab. Let p + ps be pressure infront of slab and p ps be pressure behind slab.From Taylors series
ps =ps
s2
The net pressure forceF ps = ( p ps )ny ( p + ps )ny
= 2ps ny = ps
sny = ps
V
Net forceF s = W s + F ps
= sin ps
V
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Bernoulli equation
Equate two expressions for F s
F s = sin ps
V = V vvs
sin ps
= vvs
The change in uid particle speed along a streamlineis accomplished by a combination of pressure andgravity forces.
Now use sin =dz
ds
And v dvds =12
dv 2
ds
And dp = ps ds + pn dn
Along streamline dn = 0
dzds
dpds
=12
dv2
ds
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Bernoulli, compact expression
Now making the assumption that density is constant,Bernoullis equation is obtained
dzds
+dpds
+12
dv2
ds= 0
dds
z + p + 12
v2 = 0
z + p +12
v2 = Constant
The constant density assumption (incompressibleow) is good for liquids (sometimes gases at lowspeed). Bernoullis equation presented in 1738monograph Hydrodynamics by Daniel Bernoulli.
If one has compressible uid
dp + 12v2 + gz = Constantand knowledge of how varies with p .
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Forces normal to streamline
The acceleration normal to the streamline is an = v2
R
where R is the local radius of curvature of thestreamline.
F n =(m)v2
R=
V v2
R
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Forces normal to streamline
A change in stream direction occurs from pressureand/or and gravity forces. Resolve forces
W n = W cos = V cos
W n would be zero for vertical motion.
The pressure changes with height. Let p be pressurein middle of uid slab, p + pn is pressure at top of slab and p pn be pressure at bottom of slab.From Taylors series
pn =pn
n2
The net pressure force, F pn
F pn = ( p
pn )s y
( p + pn )s y
= 2pn n y = pn
n s y = pn
V
Need to combine pressure and weight forces to getnet Force
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Forces normal to streamline
Combine weight and pressure forces
F n = W n + F pn
= cos pn
V =V v2
R
Pressure and weight forces imbalance produces thecurvature. For gas ows it is common to use
pn
= v2
R
The pressures increaseswith distance away fromthe center of curvature( pn is negative sincev2 /R is positive).
For straight parallel streamlines (in gases), pn = 0 .No pressure change across streamlines
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Forces normal to streamline
Will consider uid parameters normal to stream line
cos + pn
+ v2
R= 0
pn = dpdn since s is constant.
cos = dzdnand so for incompressible ows
dpdn
+ dzdn
+v2
R= 0
dpdn
+ dzdn
+ v2
R= 0
ddn
( p + z ) +v2
R= 0
p + z +
v2
Rdn = Constant
For a compressible substance, the best reduction is
dp + v2
R+ gz = Constant
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Interpretation for incompressible ows
Along the streamline
z + p +12
v2 = Constant
Across the streamline
p + z + v2
Rdn = Constant
The units of Bernoullis equations are J m 3 . This
is not surprising since both equations arose from anintegration of the equation of motion for the forcealong the s and n directions.
The Bernoulli equation along the stream-line is a
statement of the work energy theorem. As theparticle moves, the pressure and gravitational forcescan do work, resulting in a change in the kineticenergy.
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Dynamic and static pressures
p +12
v2 + gz = constant
Static pressure is the pressure as measured movingwith the uid. (e.g. static with uid). This is the pterm in Bernoullis equation. Imagine moving alongthe uid with a pressure gauge.
Some times the gz term in Bernoullis equation iscalled the hydrostatic pressure. (e.g. it is the changein pressure due to change in elevation.)
Dynamic pressure is a pressure that occurs when
kinetic energy of the owing uid is converted intopressure rise. This is the pressure associated withthe 12 v
2 term in Bernoullis equation.
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Dynamic and static pressuresThe static pressure at 1 can be estimated by theheight of the column.
p1 = h 3 1 + p3
= h 3 1 + h 4 3
= h
(1) (2)
(3)
(4)
h 3-1
h h 4-3
Open
H
V
V 1
= V V 2
= 0
The dynamic pressure at 2 is estimated by
p2 = p1 +12
v21 (v2 = 0)
The additional pressure due to the dynamic pressurewill cause the uid to rise a height of H > h .The point (2) is called a stagnation point.
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The stagnation point
Stagnation point
(a )
Stagnation streamline
Stagnation point
(b )
When uid ows around any stationary body, someof the streamlines pass over and some pass under theobject. But there is always a stagnation point wherethe stagnation streamline terminates. The
stagnation pressure is pstagnation = p +
12
v2
v is velocity at some point on stream-line away fromobstruction.The total pressure, pT
pT = p +12
v2 + z
is sum of static, dynamic and hydrostatic pressures.
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The pitot tube
Knowledge of static and stagnation pressures makes
it possible to determine the uid velocity. Geometryarranged so that elevation differences have littleimpact. The free stream pressure is p .
Pressure measured at
points 3 and 4Stagnation pressure
p2 = p3 = p +12
v2
V
p
(1)
(2)
(4)
(3)
Static pressure is just p1 = p p4 . Combiningequations
p = p4 = p3 12
v2
Rearranging leads to
v = 2( p3 p4 )
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The pitot tube: complications
The main question is de-sign of pitot tubes is whereto place the orice to mea-
sure the static pressure.
V American Blower company
National Physical laboratory (England)
American Society of Heating & Ventilating Engineers
The static pressure doesvary along the length of the tube. More com-
plicated analysis thanBernoulli required here.
Also make mouth of tube smooth.
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The free jet
The free jet result was rst obtained in 1643 by
Evangelista Torricelli.
p1 +12
v21 + z 1 = p2 +12
v22 + z 2
z 1 = 12
v22
p1 = patm = 0 ; gauge pressure
v1 0 ; large surface, so v1
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The free jet
z 1 =12
v22
v2 = 2h
= 2ghOutside nozzle, stream continues to fall and at (5)
v2 = 2ghv5 = 2g(h + H )Result v = 2gh is speed of freely falling bodystarting from rest.For the uid, all the potential energy is converted tokinetic energy when jet leaves tank. (assume noviscous forces).
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Free jet, ne details
Horizontal nozzle, veloc-ity at center line v2 isslightly smaller than v3and slightly larger thanv1 .
For d h , OK to usev2 as average velocity.
Streamlines cannot follow sharp corner exactly.Would take an innite pressure gradient to achievezero radius of curvature(i.e. R = 0 ). Uniform velocity only occurs at a-a line.
Vena Contracta effect.Jet diameter, d j isslightly smaller thanhole diameter dh .
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Free jet, Vena contracta effect
The contraction coefficient, C d = A j /A h is the ratioof the jet area A j , and hole area Ah .
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Flow rates
How much water ows down a channel or through apipe?
The volume ow rate,Q1 is dened as the vol-ume of uid that owspast an imaginary (orreal) interface.
Volume of uid leaving V = v1 tA 1
Rate of volume changeV t
= v1 A1
The volume ow rate Q = v1 A1 Mass of uid leaving m = v1 tA 1
Rate uid leaving m =dmdt
= v1 A1
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Equation of continuity
For a steady state situation, the mass of uid goinginto the tank must be the same as the mass of uidleaving the tank.
Mass of water in = Mass of water out
1 A1 v1 = 2 A2 v2
This is the continuity equation and forincompressible ow
A1 v1 = A2 v2 or Q1 = Q2
The equation of continuity and the Bernoullisequation are used into conjunction to analyze manyow situations.
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Flow rate: Example 1
Given the water velocityat (2) is 8.0 m/s and
the pipe diameter is 0.10m , what are the volumeand mass ow rates?
Q = vA = v(d/ 2)2
= 8 .0 0.0502 = 0 .06283 m3 /s
The mass ow is just Q sodm
dt= 1000
0.06283 = 62 .83 kg/s
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Flow rate: Example 2A stream of water d =0.10 m ows steadilyfrom a tank of diameterD = 1 .0 m as shownin the gure. What ow-
rate is needed from theinlet to maintain a con-stant water volume in theheader tank depth? Thedepth of water at the out-let is 2.0 m .
Can regard outlet as a free jet (note water level at(1) is not going down).
v2 =
2gh = 2
9.8
2.0 = 6 .26 m/s
Q2 = A2 v2 = (0.050)2 6.26 = 0 .0492 m3 /s
= Q1
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Flow rate measurement
One way to measure ow-rate is to place a
constriction in a pipe. The resulting change invelocity (continuity equation), leads to a pressuredifference. The absolute uid velocity can bedetermined from pressure difference between (1) and(2) .
The Orice, Nozzle and Venturi meters analysis hereignores viscous, compressibility and other real-world
effects.
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Flow rate measurement: 2
Want to determine ow rate, need v2
v1 A1 = v2 A2
v1 =A2A1
v2
p2 +12
v22 = p1 +12
v21
12
v22 12
v21 = p1 p2
12v
2
2 12
A22A21 v
2
2 = p1 p2
v22 =
2( p1 p2 ) 1
A 22A 21
So the ow rate is
Q = A2 2( p2 p1 ) 1 A 22A 21The pressure differences give the ow rate. Real
world ows are 1% to 40% smaller.
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Bernoulli and Cavitation
The temperature at which water boils depends onpressure.
T (oC) pvap (kPa)
10 1.23
20 2.34
30 4.24
40 7.34
Q
p
(Absolutepressure)
(1) (2) (3)
Small Q
Moderate Q
Large Q Incipient cavitation
pv
0 x
The process of cavitation involves
Fluid velocity increases
Pressure reduction
If p < p vap , water boils Bubbles collapse when reach high pressure part
of uid
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Bernoulli and Cavitation
Pressure transients exceeding 100 MPa can be
produced. These transients can produce structuraldamage to surfaces.
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Sluice gate ow rate
The height of water in the channel can be used todetermine the ow rate of water out the reservoir.
Q = z2 b 2g(z1 z2 )1
z 22
z21
z2 b
2gz1
(b) is the width of the reservoir.
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Venturi meter problem
Determine owrate as a functionof the diameter of the tube.
0.2 m
Q0.1 m D
Use venturi meter equation
Q = A2 2( p2 p1 ) 1 A 22A 21 A2 = D
2 / 4 m2
p2 p1 = 0.20 = 98000.20 = 1960 Pa = 1000 kgm
3
1 A22 /A 21 = 1 D 2 / 0.102 = 1 100D 2 m2
Q = D 2
4 39201000(1 100D 2 ) = D 2 0.245
(100D 2 1)
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Sharp crested weir
Between (1) and (2) pressure and gravitationalforces cause uid to accelerate from v1 v2 .
p1 = h and p2 0 forces cause uid toaccelerate from v1 v2 .
Assume ow is like free jet. Average velocityacross weir is C 1 2gH , C 1 = constant.
Flow rate is
Q = ( Hb) C 2gH = C 1 b 2gH 3The parameter C 1 is determined empirically.
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The energy line and hydraulic grade line
Consider Bernoulli equation divided by = g
p
+ 12
v2g
+ z = H = Constant on streamline
The dimensions of the equation are in length. Thereis the pressure head, velocity head, and elevation
head. The sum, H is called the total head.The energy line gives the total head available to auid. It can be measured by measuring thestagnation pressure with a pitot tube.The Hydraulic grade line is the line produced from
the pressure and elevation heads. It is measured witha static pitot tube.
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The energy line and hydraulic grade line
The energy line will be horizontal along thestream line as long as Bernoulli assumptions arevalid.
The hydraulic grade line will not be horizontal if the uid velocity changes along the stream line.
If forces are present (this does occur in pipeows), then there will be a loss in energy andthe energy line will not be constant.
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Example: EL and HGL
A scale drawing can be used to depict the pressure inthe tank/pipe system.
The energy line is horizontal The elevation head at (2) is converted into
increased pressure head p2 / and velocity headv22 / (2g) . The HGL decreases.
At (3) , pressure is atmospheric. So the HGL tothe level of the pipe and the elevation head hasbeen converted entirely into a velocity headv23 / (2g) .
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EL and HGLThe EL and HGL can depict whether there ispositive pressure p > p atm or negative pressure p < patm .
The water velocity will be constant in curvedpipe (equation of continuity).
The pressure head will increase or decrease asthe elevation head changes. Useful to know forleaking pipes.
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Limitations on Bernoulli equation
A number of problems can invalidate the use of theBernoulli equation, these are compressibility effects ,rotational effects , unsteady effects .
Compressibility effects
When can compressibility effects impact on gasows? Consider stagnation point
Stagnation pressure is greater than staticpressure by 12 v
2 (dynamic pressure), provided
constant.
will not changes too much as long as dynamicpressure is not too large when compared tostatic pressure.
So ows at low v will be incompressible But dynamic pressure increases as v
2 , socompressibility effects most likely at high speed.
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Compressibility effects
The isothermal model for an ideal gas, p = R T
C = dp + 12v21 + gz1
C = RT dp p + 12v21 + gz1C = RT ln p1 +
1
2v21 + gz1
This can be used to get
v212g
+ z1 +RT g
ln( p1 /p 2 ) =v222g
+ z2
Now write as p1 /p 2 = 1 + ( p1 p2 )/p 2 = 1 + p/p 2and use ln(1 + x) = x for x1 .v212g
+ z1 +RT g
ln(1 + p/p 2 ) =v222g
+ z2
v21
2g+ z1 + RT
g( p/p 2 ) v
22
2g+ z2
This can be reduced to the standard Bernoulliequation. Bernoulli recovered as long as pressuredifferences are not large.
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Compressibility effects: Isentropic ow
This is the situation that applies when there is no
heat transfer or friction during the ow (reasonablefor many gases). This gas law is p = k D where kdepends on specic heat capacities. Introduce theMach number, Ma = v/c (ratio of the ow speed tothe speed of sound). Consideration of the pressureratio between free stream and stagnation pointsleads to
p2 p1 p1
=kMa 21
2incompressible
p2 p1 p1
= 1 +k 1Ma 21
+k
k 1
1 compressible
The compressible andincompressible expres-sions agree to 2% forMa < 0.3 .
Compressible(Eq. 3.25)
Incompressible(Eq. 3.26)
k = 1.4
0.3
0.2
0.1
00 0.2 0.4 0.6 0.8
Ma1
p 2
p
1
______
p 1
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Unsteady effectsImplicit in the discussion was an assumption thatthe uid ows along steady state streamlines, sov = v(s) is a function of position along the stream
and does not contain any explicit time dependence.If v = v(s, t ) then then it would be necessary toinclude this when integrating along the streamline.
p1 +1
2v21 + z 1 = p2 +
1
2v22 + z 2 +
t 2
t 1
p
sds
The additional term does complicate matters andcan only be easily handled under restrictedcircumstances. There are quasi-steady ows wheresome time dependence exists, but Bernoullisequations could be applied as if the ow were steady(e.g. the draining of a tank).
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Rotational effects
Model of wake behind insect
Y. D. Afanasyev, Memorial University of Newfoundland
Bernoulli equation describes motion of uid particlesalong streamline. If particles spin about the
streamline then Bernoulli is no longer valid.Need to characterize irrotational and rotational ows.