Bernoulli’s Equation - جامعة نزوى · Applications of the Bernoulli Theorem 4/5 For a...
Transcript of Bernoulli’s Equation - جامعة نزوى · Applications of the Bernoulli Theorem 4/5 For a...
Chapter 4
• Bernoulli’s Equation
(Mechanical Energy Balance)
• Flow Measurements
1
Mechanical Energy Balance From the energy equation, dividing by mass and assuming
incompressible flow, then rearranging
m
Q)uu(
m
W
)zz(g)2
V
2
V()
PP(
net12
netshaft
12
21
2212
Let’s define the
:friction heating per unit mass
m
Qu
F
2
Mechanical Energy Balance
F/g (sometimes symbolized as hL) is the friction or head
loss
Fm
W
)zz(g)2
V
2
V()
PP(
netshaft
12
21
2212
This is the Mechanical Energy Balance or Bernoulli’s
equation
By dividing above equation with g we obtain the “head form” of
the Mechanical Energy Balance:
g
F
mg
W
)zz()g2
V
g2
V()
g
P
g
P(
netshaft
12
21
2212
3
Simplifications Assumptions:
– Steady, incompressible, frictionless (inviscide) flow
– No heat transfer or change in internal energy, no shaft work
– Single input-output
Bernoulli’s equation may reduced to:
In terms of “heads” (units of length)
0)()22
()( 12
2
1
2
212 zzgVVPP
0)zz()g2
V
g2
V()
g
P
g
P( 12
2
1
2
212
4
5
• May be obtained directly from Euler’s equation:
dp/ρg + dz + dV2/2g = 0
• Integrating for incompressible flow:
p/ρg + z + V2/2g = constant
• Multiplying through by ρg:
i.e.
• (Steady, inviscid, incompressible assumptions)
Bernoulli’s Equation
21constant
2p gz V
2 2
1 1 1 2 2 2
1 1
2 2p gz V p gz V
6
Bernoulli’s Equation (Cont.)
Often interpreted as an energy conservation
equation, this may be represented as:
If incompressible and if elevation (z) terms are ignored:
Static pressure Elevation pressure Dynamic pressure
Total Pressure (p0)
21constant
2p gz V
2 2
1 1 2 2
1 1
2 2p V p V
7
(Cont.) Bernoulli’s Equation
May be represented in terms of pressure head terms instead:
In this case the dimensions of all the terms are of length units (metres in SI).
Static head Dynamic head
Elevation head
Total head
2
constant2
p Vz
g g
8
Interpretation of Bernoulli’s Equation
9
10
• Applicable to large number of practical engineering problems though with several limitations.
– Simple form of equation only valid for incompressible fluids.
– No account taken of any mechanical devices in system which add/remove energy (e.g. pumps, turbines, etc.).
– No account taken of heat transfer into or out of fluid.
– No account taken of energy lost due to friction.
-Total energy of any stream line is constant.
-Link between the flow of the velocity and Pressure.
-If the Velocity increase, Pressure decrease.
Bernoulli’s Equation - Restrictions
11
Bernoulli’s Equation 1Example
• Glycerine (SG = 1.26) flows in a processing plant pipe at a rate of 700 L/s. At a point where the pipe diameter is 60 cm, the pressure is 300 kN/m2.
• Find the pressure at a second point, 1 m lower than the first point, with a pipe diameter of 30 cm.
• Neglect any system head losses.
Z2=0 Z1=1
Bernoulli’s Equation Solution • Using Q1 = A V1 (Continuity)
then 0.7 = ( x 0.62 / 4) V1,
V1 = 2.476 m/s
• Using Continuity: Q1=Q2
• then V2 = V1 (A1 / A2)
V2 = 9.9 m/s
• Using Bernoulli:
p1 + ρgz1 + ½ρV12 = p2 + ρgz2 + ½ρV2
2
p2/(ρg) = 300x103/(1260 x 9.81) + (2.4762 – 9.92)/(2 x 9.81) + 1
p2 = 254.5 kN/m2 13
14
15
16
17
18
19
20
Applications of the Bernoulli Theorem 1/5
•Flow and the streamline under consideration are shown in Fig. 5.4. Using the Bernoulli equation, we can form a relation between point (1) and point (2) as
p1 + ½V12 + gz1 = p2 + ½V2
2 + g z2
21
Applications of the Bernoulli Theorem 2/5
At point (1), the pressure is atmospheric (p1 = po), or the gage
pressure is zero, and the fluid is almost at rest (V1 = 0).
At point (2), the exit pressure is also atmospheric (p2 = po), and
the fluid moves at a velocity V.
By using point (2) as the datum where z2 = 0 and the elevation of point (1) is h, the above relation can be reduced to
p0 + ½(0)2 + gh = p0 + ½V 2 + g (0)
gh = ½V 2
Hence we can formulate the velocity V
to be
V2 = 2gh
22
Applications of the Bernoulli Theorem 3/5
• Notice that we can also obtain the similar relation by using the relation between point (3) and point (4).
• The pressure and the velocity V2 for point (4) is similar to point (2). However, the pressure for point (3) is the hydrostatic pressure, i.e.
• p3 = p0 + g (h - ) and the velocity V3 is also zero due to an assumption of a large tank. Hence, the relation becomes
[ p0 + g (h - )] + ½(0)2 + g = p0 + ½V 2 + g (0)
and thus: V2 = 2gh gh = ½V 2
we can formulate the 5 & 1 , between Hence
velocity V to be
p1 + ½V12 + gz1 = p5 + ½V5
2 +g z5
p0 + ½(0)2 + g(h+H) = p0 + ½V52
V5 = 2g ( h + H )
(h + H) is vertical distance from point 1 to 5
23
Applications of the Bernoulli Theorem 4/5
For a nozzle located at the side wall of the tank, we can also form a similar relation for the Bernoulli equation, i.e.
V1 = 2g(h – d/2), V2 = 2gh, V3 = 2g (h + d/2),
For a nozzle having a small diameter (d < h), then we can conclude that :
V1 V2 V3 = V 2gh
24
Applications of the Bernoulli Theorem 5/5
i.e., the velocity V is only dependent on the depth of the centre of the nozzle from the free surface h.
If the edge of the nozzle is sharp, as illustrated in Fig. 5.5b, flow contraction will be occurred to the flow. This phenomenon is known as vena contracta, which is a result of the inability for the fluid to turn at the sharp corner 90°. This effect causes losses to the flow.
25
Example : Flow From a Tank- Gravity • A stream of water of diameter d = 0.1m flows steadily from a
tank of Diameter D = 1.0m as shown below. Determine the
flowrate, Q, needed from the inflow pipe if the water depth
remains constant, h = 2.0m.
26
Solution 1/3
2
2
221
2
11zV
2
1pzV
2
1p gg
2
2
2
1V
2
1ghV
2
1
For steady, inviscid, incompressible flow, the Bernoulli
equation applied between points (1) and (2) is
(1)
With the assumptions that p1 = p2 = 0, z1 = h, and z2 = 0, Eq.1 becomes
(2)
Although the water level remains constant (h = constant), there is an
average velocity,V1, across section (1) because of the flow from tank.
For the steady incompressible flow, conservation of mass requires
Q1 = Q2, where Q = AV. Thus, A1V1 =A2V2 , or
2
2
1
2 Vd4
VD4
27
Solution 2/3
2
2
1V)
D
d(V
s/m26.6)m1/m1.0(1
)m0.2)(s/m81.9(2
)D/d(1
gh2V
4
2
42
Hence,
(3)
Equation 1 and 3 can be combined to give
Thus,
s/m0492.0)s/m26.6()m1.0(4
VAVAQ 32
2211
28
Solution 3/3
In this example we have not neglected the kinetic energy of the water in
the tank (V1≠0). If the tank diameter is large compared to the jet
diameter (D >> d), Eq. 3 indicates that V1 << V2 and the assumption
that V1≒0 would be reasonable.
The error associated with this assumption can be seen by calculating
the ratio of the flowrate assuming V1 ≠ 0, denoted Q, to that assuming
V1 = 0, denoted Q0. This ratio, written as
4
4
2
2
0 )/(1
1
2
])/(1/[2
Ddgh
Ddgh
V
V
Q
Q
D
is plotted in Fig. E3.7b. With 0 < d/D < 0.4 it follows that
1 < Q/Q0 <~ 1.01, and the Error is assuming V1 = 0 is less than 1%.
Thus, it is often reasonable to assume V1 = 0.
29
30
Example: Flow in a Variable Area Pipe
• Water flows through a pipe reducer as is shown in Figure below.
• The static pressures at (1) and (2) are measured by the inverted U-tube manometer containing oil of specific gravity, SG, less than one.
• Determine the manometer reading, h.
31
Solution 1/3
2211VAVAQ
With the assumptions of steady, inviscid, incompressible flow, the
Bernoulli equation can be written as
The continuity equation provides a second relationship between V1
and V 2 if we assume the velocity profiles are uniform at those two
locations and the fluid incompressible:
By combining these two equations we obtain
….(1)
2
2
221
2
112
1
2
1zVpzVp gg
])/(1[2
1)( 2
12
2
21221 AAVzzpp g
32
Solution 2/3
h)SG1()zz(pp1221
gg
0)( 2121 phSGhzzp ggggg
2
1
22
2 12
1)1(
A
AVhSG g
This pressure difference is measured by the manometer and can be
determine by using the pressure-depth ideas. Thus,
or
…..(2)
Equations 1 and 2 can be combined to give the desired result as
follows:
( ( ( SG1g2
A/A1A/Qh
2
122
2
or since V2=Q/A2
(Ans)
33
Solution 3/3
The difference in elevation, z1-z2, was not needed
because the change in elevation term in the Bernoulli
equation exactly cancels the elevation term in the
manometer equation.
However, the pressure difference, p1-p2, depends on the
angle θ, because of the elevation, z1-z2, in Eq. 1. Thus,
for a given flowrate, the pressure difference, p1-p2, as
measured by a pressure gage would vary with θ, but the
manometer reading, h, would be independent of θ
34
Example: Siphon and Cavitation
• Water at 60 ℉ is siphoned from a large tank through a constant diameter hose as shown in Figure. Determine the maximum height of the hill, H, over which the water can be siphoned without cavitation occurring. The end of the siphon is 5 ft below the bottom of the tank. Atmospheric pressure is 14.7 psia.
Vapor pressure
Solution 1/5
3
2
332
2
221
2
112
1
2
1
2
1zVpzVpzVp ggg
If the flow is ready, inviscid, and incompressible we can apply
the Bernoulli equation along the streamline from (1) to (2) to (3)
as follows:
With the tank bottom as the datum, we have
z1 = 15 ft,
z2 = H, and
z3 = -5 ft. Also,
V1 = 0 (large tank),
p1 = 0 (open tank) and
p3 = 0 (free jet),
(1)
36
Solution 2/5
From the continuity equation
A2V2 = A3V3, or
Because the hose is constant diameter V2 = V3.
Thus, the speed of the fluid in the hose is determined from Eq. 1 by
taking positions between points 1 & 3, then :
2
2
313 /9.35)]5(15)[/2.32(2)(2
V
sftftsftzzgV
37
Solution 3/5
2
221
2
2
21
2
112
2
1)(
2
1
2
1
Vzz
zVzVpp
g
gg
Use of Eq. 1 between point (1) and (2) then gives the pressure
p2 at the top of the hill as
From Table B.1, the vapor pressure of water at 60℉ is 0.256 psia.
Hence, for incipient (initial) cavitation, the lowest pressure in the
system will be p = 0.256 psia.
(2)
38
Solution 4/5
233
222
)/9.35)(/94.1(2
1)15)(/4.62(
)/.144)(./4.14(2
sftftslugsftHftlb
ftininlbP
Careful consideration of Eq. 2, will show that this lowest pressure at
point (1) (p1 = 0), and the lowest pressure at point (2) alsowe must
use gage pressure. Thus,
p2= 0.256 – 14.7 = -14.4 psi
and, equation 2 gives :
ftH 2.28(Ans) 39
This gives the limit or the maximum height H above which the
cavitation will occur:
2
22122
1)( Vzzp g
Solution 5/5
•For larger value of H, vapor bulbles will form at point (2) and the
siphon action may stop.
•Note that we could have used absolute pressure throughout
(p2 = 0.256 psia and p1 = 14.7 psia) and obtained the same results.
•The lower the elevation of point (3), the larger the flowrate and,
therefore, the smaller the value of h allowed.
We could also have used the Bernoulli equation between (2) and (3),
with V2 = V3, to obtain the same value of H. In this case it would
not have been necessary to determine V2 by use of the Bernoulli
equation between (1) and (3). The above results are independent of the diameter and length of
the hose (provided viscous effects are not important). 40
Applications: Flow Measurement
The Bernoulli equation can be applied to several commonly
occurring situations in which useful relations involving pressures,
velocities and elevations may be obtained.
A very important application in engineering is :
fluid flow measurement
Measurement of velocity : Pitot-static tube
Measurement of flow rate: 1-Venturi meter,
2- Orifice meter &
3- Rotameter
41
Flow Measuring Devices
Pitot Tube
• This is an open-ended tube, bent through 90o with the nose facing the direction of the oncoming flow.
• The fluid is brought to rest at the nose so it measures the fluid’s stagnation or total pressure (po).
42
Stagnation Conditions
• Fluid on streamline OX forced to zero velocity at
X – this is a stagnation point.
• The term stagnation is often used to describe
the total pressure conditions as this corresponds to
when the dynamic pressure term is zero.
• The static pressure at zero-velocity conditions thus
relates to the total or stagnation pressure. 43
Flow Measuring Devices
Pitot-Static Tube
• Comprises pitot tube surrounded by concentric outer tube with a ring of small holes, carefully positioned to give representative static pressure reading.
44
Pitot-Static Tube
P1,V1 Stagnation
Point V2=0
1 2
P1 is a Static pressure: It is measured by a device (static tube) that
causes no velocity change to the flow. This is usually accomplished by
drilling a small hole normal to a wall along which the fluid is flowing.
P2 is a Stagnation pressure: It is the pressure measured by an open-
ended tube facing the flow direction. Such a device is called a Pitot
tube.
P2
45
Pitot-Static tube
Bernoulli equation between 1 and 2: 0
2
)VV()PP( 2
1
2
212
(Recall that position 2 is a stagnation point: V2= 0)
2/1
121
)PP(2V
We can measure pressures P1 and P2 using hydrostatics:
P1=Patm + ρgh1, P2=Patm + ρgh2
or using a Pressure Gauge, see next slide
2
112 V2
1PP Stagnation Pressure is higher
than Static Pressure
46
• Or using Bernoulli’s equation for a constant horizontal datum level head:
p + ½ ρ V2 = po
po – p = ½ ρ V2
= h ρm g
2 mh gV
(6a)
47
Venturi meters, Orifice and Nozzle
• Fluid is accelerated by forcing it to flow through a constriction,
thereby increasing kinetic energy and decreasing pressure energy.
• The flow rate is determined by measuring the pressure difference
between the meter inlet and a point of reduced pressure.
• Desirable characteristics of flow meters:
Reliable, repeatable calibration
Introduction of small energy loss into the system
Inexpensive
Minimum space requirements
Basic principle: Increase in velocity causes a decrease in pressure
48
Various flow meters are
governed by the Bernoulli
and continuity equations.
2211
2
22
2
112
1
2
1
VAVAQ
VpVp
( 2
12
21
2
)A/A(1
pp2AQ
The theoretical flowrate
49
Venturi Meter This device consists of a conical contraction, a short
cylindrical throat and a conical expansion. The fluid is accelerated by
being passed through the converging cone. The velocity at the “throat” is
assumed to be constant and an average velocity is used. The venturi tube
is a reliable flow measuring device that causes little pressure drop. It is
used widely particularly for large liquid and gas flows.
P1 P2
])/(A-[1
)(2C
2
12
212
A
PPV d
Where the discharge
coefficient, Cd =f(Re), can
be found in Figure shown
later
Flow Measuring Devices
50
Flow Measuring Devices
Venturi Meter
Ignoring friction losses and applying Bernoulli’s equation from (1) to (2).
p1 + ½ ρ V12 = p2 + ½ ρ V2
2
Assuming incompressible flow (ρ = constant):
p1 - p2 = ½ ρ (V22 - V1
2) _______ (eq.1)
Applying continuity equation from (1) to (2):
V2 = (A1 / A2) V1 ____________ (eq.2)
51
Flow Measuring Devices
Venturi Meter (Cont.)
Substituting (eq.2) into (eq.1):
p1 - p2 = ½ ρ [(A1/A2)2 V12 – V1
2]
Rearranging:
And
Discharge coefficient (Cd) typically 0.97.
(6b)
(6c)
( 1 2
1 2
1 2
2( )
/ 1
p pV
A A
( 1 2
1 1 1 2
1 2
2( )
/ 1d d
p pQ C AV C A
A A
52
Flow Measuring Devices
Flow Nozzle
• Similar to venturi-meter but no diffuser so cheaper and more compact.
(6d)
( 1 2
1 2
1 2
2( )
/ 1d
p pQ C A
A A
( 1 2
1 2
2
/ 1
md
h gQ C A
A A
53
Flow Measuring Devices
Orifice Plate
Cheaper and takes up less space than a venturi-meter but has larger total pressure drop.
As with others:
Cd typically 0.6 - corrects for non-uniformity & jet contraction.
( 1 2
1 0
2
/ 1
md
h gQ C A
A A
54
Generalized Chart for Cd
Cd
55