fluid in motion Bernoulli equation from Newtonβs second law
Chapter. 3 Elementary Fluid Dynamics: Bernoulli Equation
Transcript of Chapter. 3 Elementary Fluid Dynamics: Bernoulli Equation
Chapter. 3 Elementary Fluid Dynamics:Bernoulli Equation
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Reference:1. Fluid Mechanics, Frank MWhite, 6th Edition, 2008, McGraw Hill2. Munson et al., Fundamentals of Fluid Mechanics, 8th Edition, 2006, John Wiley &Sons, Inc
1
3-2
Contents
Chap. 1 Introduction
Chap. 2 Fluid Statics
Chap. 3 Elementary Fluid Dynamics : the Bernoulli equation
- Newton's 2nd law; along a streamline or normal to a streamline
- Static, Stagnation, Dynamic, and Total Pressure
- Examples of Bernoulli equation
- Energy and Hydraulic Grade Lines
- Restrictions on use of the Bernoulli Equation
Chap. 4 Fluid Kinetics
Chap 5 Finite Control Volume Analysis
Chap. 6 Differential Analysis of Fluid Flow
Acceleration of Fluid Particle in a Curvilinear Motion
β’ To derive the acceleration vector in tangential and normal components, define the motion of a particle as shown in the figure.
β’ ππ‘ and ππ‘β²are tangential unit vectors for the particle
path at P and Pβ. When drawn with respect to the same origin, β ππ‘ = ππ‘
β²β ππ‘ and βπ is the angle between them.
3-3
The vector β ππ‘/βπ has a magnitude of unity & a direction toward the O'
βππβ²
βππ‘ = 2 sin(βπ
2) lim
βπβ0
βππ‘βπ
= limβπβ0
sin(βπ/2)
βπ/2=1
β΄ limβπβ0
β ππ‘βπ
= ππ =π ππ‘ππ
β’ With the velocity vector expressed as π£ = π£ ππ‘the particle acceleration may be written as
Acceleration of a particle along a streamline
where
After substituting,
where π is the radius of curvature of the motion.
β’ Along a streamline,
3-4
π =π π£
ππ‘=ππ£
ππ‘ ππ‘ + π£
π ππ‘ππ‘
=ππ£
ππ‘ ππ‘ + π£
π ππ‘ππ
ππ
ππ
ππ
ππ‘
π ππ‘ππ
= ππ; πππ = ππ ;ππ
ππ‘= π£
π = ππ‘ ππ‘ + ππ ππ =ππ£
ππ‘ ππ‘ +
π£2
π ππ
ππ‘ =ππ£
ππ‘=ππ£
ππ‘+ π£ β π»π£
=ππ£
ππ‘+ π£
ππ£
ππ
β π£ β π»π£ = π£ ππ‘ βππ£
ππ ππ‘ +
ππ£
ππ ππ = π£
ππ£
ππ
π =ππ£
ππ‘+ π£
ππ£
ππ ππ‘ +
π£2
π ππ
β’ For a steady state, π = π£ππ£
ππ ππ‘ +
π£2
π ππ
The Bernoulli Equation (BE)
3-5
π
π
πΏπ§
Free body diagram Assumptions: Steady state, Inviscid, incompressible flow
πΏπΉπ ,π = πΏm ππ
1) Stream-wise direction along a streamline
= ππΏβ π£ππ£
ππ
πΏπΉπ ,π = πΏππ + πΏπΉππ
πΏππ = βπΏπ sin π = βπΎπΏβ sin π
πΏπΉππ = π β πΏππ πΏnπΏπ¦ β π + πΏππ πΏnπΏπ¦
= β2πΏππ πΏnπΏπ¦ = βππ
ππ πΏsπΏnπΏπ¦
= βππ
ππ πΏβ
πΏπΉπ ,π = β πΎ sin π +ππ
ππ πΏβ
The Bernoulli Equation (BE)
3-6
π
π
πΏπ§
Free body diagram Assumptions: Steady state, Inviscid, incompressible flow
1) Stream-wise direction along a streamline
πππ = ππ£ππ£
ππ = β πΎ sin π +
ππ
ππ
β sin π =ππ§
ππ
ππ£ππ£
ππ = βπΎ
ππ§
ππ βππ
ππ
λΉμμΆμ±
π ππ£2/2
ππ = β
π
ππ π + πΎπ§
β΄π
ππ
ππ£2
2+ π + πππ§ = 0
Integrate against ππ along a streamline
The Bernoulli Equation (BE)
3-6
Assumptions: Steady state, Inviscid, incompressible flow
1) Stream-wise direction along a streamline
π
ππ
ππ£2
2+ π + πππ§ = 0
Integrate against ππ along a streamline
π = π» β π π at steady state
π» =π( )
ππ ππ +
π( )
ππππ +
π( )
ππ¦ππ¦
where π π is the integration path
along a streamline
π = π» β π π =π( )
ππ ππ
1
2 π
ππ
ππ£2
2+ π + πππ§ ππ = 0
1
2
πππ£2
2+ π + πππ§ = 0
ππ£2
2+ π + πππ§
1
= ππ£2
2+ π + πππ§
2
This is a sort of energy equationTotal energy is conserved along a streamline when the fluid inviscid and incompressible and at steady state
π£2
2π+
π
ππ+ π§
1
= π£2
2π+
π
ππ+ π§
2
The Bernoulli Equation (BE)
3-6
Assumptions: Steady state, Inviscid, incompressible flow
2) Normal direction across a streamline
πΏπΉπ,π = πΏππ + πΏπΉππ = πΏπ ππ = πΏππ£2
π
πΏππ = βπΏπ cos π = βπΎπΏβ cos π = βπΎπΏβππ§
ππ
πΏπΉππ = π β πΏππ πΏπ πΏπ¦ β π + πΏππ πΏπ πΏπ¦
= β2πΏπππΏsπΏπ¦ = βππ
πππΏsπΏnπΏπ¦
= βππ
πππΏβ
ππΏβπ£2
π = βπΎπΏβ
ππ§
ππβππ
πππΏβ
ππ£2
π = β
π
πππ + πΎπ§
Integrate against ππ to a direction normal to the streamline
π£2
π ππ +
π
π+ ππ§ = ππππ π‘ across the stream line
incompressible
π + πΎπ§ decreases in the π direction, i.e., toward the rotation center π What happens if R ?
3-9
Bernoulli Equation (BE)
β’ BE is a simple and easy to use relation between the following three variables in a moving fluid
β’ pressure
β’ Velocity
β’ Elevation
β’ It can be thought of a limited version of the 1st law of thermodynamics.
β’ It is valid for steady, incompressible, inviscid flows.
β’ Warning: BE is one of the most used, but also one of the most abused equation influid mechanics. So be careful !!!
β’ BE can be seen as a balance of kinetic, potential and pressure energies.
β’ BE in "Head" form:
π +ππ£2
2+ πππ§ = const along a stream line
Press PE
π
ππ+π£2
2π+ π§ =
KE
const along a stream line
pressure head velocity head elevation head
Total head (βπ)
β’ Consider the flow of water from the syringe. The force applied to the plunger will produce a pressure greater than atmospheric pressure at point 1. The water flows from the needle (point 2) with relatively high velocity and rises up to point 3 at the top of its trajectory (Reference: Munsonβs book).
3
π
2
1
3-10
Applications of the BE
β’ Due to the frictional effects (viscous forces) the water will not go up as much as predicted by the BE.
β’ Such energy-loss effects arise especially at the narrowneedle exit and between the water jet and surroundingair stream.
πΉ
Point Energy type
Kineticππ2/2
Potentialπππ§
Pressure (gage)π
1 Small Zero Large
2 Large Small Zero
3 Zero Large Zero
π +ππ2
2+ πππ§ = constant along a streamline
3-11
Static, Dynamic, Stagnation & Total Pressures
Static pressure
Dynamic pressure
Hydrostatic pressure
Stagnation pressure
BE says that total pressure is constant along a streamline.
Total pressure
BA
πβπ₯
π
π¦
ππ΄ +πππ΄
2
2= ππ΅ = ππ π‘ππ
β’ Between A and B along the streamline
β’ Pitot tube is a device used for speed measurement.
β’ It actually measures pressure and converts it into speed using the BE.
β’ It is a simple tube with a 90o bend.
β’ Read about the role of Pitot tube malfunctions on plane crashes.
http://www.planeandpilotmag.com/article/blocked-pitot-tubes/#.WhPvY1WWaUk
3-12
Simple Pitot Tube
Pitot tube on a F1 race car Pitot tube on an aircraft
β πgβ0 +πππ΄
2
2= πg β0 + β1
3-13
Simple Pitot Tube (contβd)
Exercise: Show that the fluid speed at point A is given by
πA = 2πβ1
β’ With a Pitot tube we actually measure the pressure difference between points βAβ and βπβ and convert this difference to a speed difference using the BE.
β’ Fluid flows in an open channel from left to right.
β’ We want to measure the speed at point A.
β’ Fluid fills the Pitot tube and rises inside it to alevel of β1 above the free surface.
β’ The aim of using a Pitot tube is to create a stagnation flow at point βπβ with zero velocity.
A O
β0
β1πππ‘π
ππ΄ +πππ΄
2
2= ππ Why ππ΄ = πgβ0? even in motion
3-14
Use of Pitot Tube with a Static Tube
β’ Static tube will give :
β’ Pitot tube will give:
β’ BE between points βAβ and βoβ will give:
β’ Combining all these, speed at point A is obtained as
πA = 2π(β1 β β2)
β’ Now the flow is inside a closed channel or a pipe.
β’ We need to use an additional tube called thestatic tube (piezometer).
β’ Static tube is used to measure the static pressure at point A.
β1β2
A π
β0
πππ‘π
ππ΄ = πππ‘π + πg(β0 + β2)
ππ = πππ‘π + πg(β0 + β1)
ππ΄ + πππ΄2/2 = ππ + πππ
2/2
3-15
Combined Pitot-Static Tube (Prandtlβs Tube)
β’ The required pressure difference is π0 β ππ΄ = (ππ β π)πβπ
β’ BE between βAβ and βoβ will give
πA = 2(π0 β ππ΄)/π β
β’ Instead of measuring static pressure at point Ausing a piezometer tube, a second tube is usedaround the Pitot tube.
β’ Static pressure holes (point π) of the outer tube are located such that they measure correct upstream static pressure, i.e. ππ = ππ΄ .
β’ Two tubes provide the necessary pressure difference measurement using the mercury in it.
A O
ππ
ππ
βπ
ππ΄ = 2πβππππβ 1
3-16
Errors & Another variation
What error is expected by the imperfect manufacture of tip?
π1 > π π1 < π π1 = πWhat benefit is expected?
3-17
Combined Pitot-Static Tube (contβd)
β’ Pressure variation along a
combined Pitot-static tube is as shown.
β’ As seen, the static holes are located carefully such that they measure the static pressure ahead of the device.
Exercise: Water flows through the pipe contraction shown. For the given 0.2 m difference in the manometer level, determine the flow rate.
Munsonβs bookπ = ?
πStagnation
pressure at tip
Upstream and static hole pressures are equal
Static holes
Stagnation pressure on
stem
Stem
0.1 mπ = ?
0.2 m
0.05 m
3-18
Bernoulli Equation Exercises: Siphon vs Cavitation
Exercise : A tube can be used to discharge water from a reservoir as shown. Determine the speed of the free jet and the minimum absolute pressure of water that occurs at the top of the bend.
3 m
This is known as siphoning. It can be used to drain gasoline from the tank of an automobile. Onceyou establish the initial flow by sucking gasolinefrom the tube, the gasoline will flow by itself.
1 m
Exercise : Cavitation?
3-19
Use of BE: 1) Free Jets - Toricelli Equation
β’ Consider the discharge of a liquid from a large reservoir through an orifice (hole).
π2 +ππ2
2
2= π1 +
ππ12
2+ ππβ
Why π2 = 0?
β΄ π2= 2πβ Is not a function of π
π2 = π4 = π1
What about (5)? π5 > π2? π5 > π2? If so, how much?
3-20
Use of BE: 1) Free Jets - Toricelli Equation
β’ Consider the discharge of a liquid from a large reservoir through an orifice (hole).
When exit (2) is small enough to neglect the spatial velocity distribution,flow discharge rate from the exit is
π = π΄2π2= π΄2 2πβ However, this is an ideal value that must be corrected in reality
3-21
Vena Contracta and Contraction Coefficient
π΄ππ΄β
π΄π < π΄β
πΆπΆ = π΄π/π΄β
Contraction Coefficient
π = π΄πππ= π΄π 2πβ
= πΆπΆπ΄β 2πβ
π = πΆπ£πΆπΆπ΄β 2πβ
β‘ πΆππ΄β 2πβ
Further correction
πΆπ: discharge coefficient, experimentally determined; πΆπ£: velocity coefficient for viscous effect
3-22
Obstruction Flow Meters
β’ They are used to measure flow ratesthrough pipes. General idea is
β’ to place an obstacle inside the pipe and force the fluid to accelerate and pass through a narrow area.
β’ measure the pressure differencebetween the low-velocity, high-pressure upstream and the high-velocity, low-pressure downstream.
β’ use the BE to relate this pressuredifference to the flow rate in thepipe.
3-23
Venturi Meter
π π2π1π·
1 2
π΄1 =ππ·2
4π΄2 =
ππ2
4
β’ BE on a streamline along the pipe axisfrom section 1 to section 2
π1 +ππ1
2
2= π2 +
ππ22
2
β’ Continuity equation between section 1to 2 for incompressible flow
π = π΄1π1= π΄2π2 β π·2π1 = π2π2
β’ Combine two equations to eliminate π1and obtain π2 as
π2 =2(π1 β π2)
π(1 β π½4)where π½ =
π
π·
β’ Flow rate through the pipe is given by
π = π΄2π2 = π΄22(π1 β π2)
π(1 β π½4)
Venturi Meter (contβd)
β’ This flow rate can be corrected for viscous effects using the discharge coefficient
Experimentally determined and provided by the manufacturer
(see the Slide 3-26)
β’ The same equation can be used for the nozzle flow meter too.
β’ For both the Venturi meter and the nozzle flow meter, the contraction is mostly smooth so that the contraction coefficient was set to 1 (πΆπΆ = 1)
π = πΆππ΄22(π1 β π2)
π(1 β π½4), π½ =
π
π·
3-24
Orifice Meter
β’ Section o has the orifice plate with the hole diameter π.
β’ Section 2 is the vena contracta section. π2 is measured here.
Exercise: Derive the flow rate equation for this orifice meter.
π π2π1π·
1 2o
π΄1 =ππ·2
4π΄O =
ππ2
4
π΄2 = πΆπΆπ΄O
(vena contracta)
β’ For the orifice meter the expansion is abrupt and πΆπΆ < 1, i.e. vena contractaarea is smaller than the orifice area.
3-25
Typical πͺπ Graphs for Obstruction Flow Meters
π½ = 0.20.58
104 105 106 107 108
0.66
0.64
πΆπ orifice0.62
0.60
Dd
V
π π=πππ·
π
π½ =ππ· = 0.7
0.94104 105 106 107 108
π π
1.00
0.98
πΆπnozzle
0.96π· πV
π½ = 0.2
π½ = 0.8
Range of values depending on specific
Venturi geometry
0.94104 105 π π 107 108
1.00
0.98
πΆπVenturi
0.96
3-26
3-27
Obstruction Flow Meter Exercises
Exercise : (Munsonβs book) a) Determine the flow rate through the Venturi meter shown. b) Flow rate is increased until cavitation is first observed. At that instantπ1 is read as 276 kPa gage. At what flow rate will this occur? Vapor pressure of theflowing fluid is 3.6 kPa and the atmospheric pressure is 100 kPa.
31 mm 19 mm
π2 = 550 kPaπ1 = 735 kPa
Q
πΎ= 9.1 kN/m3
π = π΄2π2 = π΄22(π1 β π2)
π(1 β π½4)
Note that p1 is a gage pressure and vapor pressure is in absolutepressure unit. Recall at which total pressure cavitation begins.
3-28
Obstruction Flow Meters (contβd)
β’ Comparison of obstruction type flow meters
β’ Other types flow meters
β’ Rotameter
β’ Thermal flow measurement
β’ Vortex type flow meter
β’ Ultrasonic flow meter
β’ Coriolis flow measurement
β’ Turbine flow meter
β’ Weirs (for open channels)
( youtube.com/watch?v=2dfIWNYJwZM )
( youtube.com/watch?v=YfQSf2NBGqc )
( youtube.com/watch?v=GmTmDM7jHzA )
( youtube.com/watch?v=Bx2RnrfLkQg )
( youtube.com/watch?v=XIIViaNITIw )
Cost Ease of Installation Pressure Loss
Orifice meter Cheap Difficult High
Nozzle flow meter Medium Difficult Medium
Venturi meter High Difficult Low
Be Careful in Using the Bernoulli Equation
β’ The simplest and the most commonly used BE that we studied in the previous slides may lead to unphysical results for problems similar to the following ones.
Γengelβs book
1
1
1
1
1
2
2
2
2
2
A sudden expansion
A longnarrowtube
Flow through a
valve
Flow through an energy delivering
or extracting device
Flow in highlyviscous regions
A boundary layer
A wake
Flow with excessive heat
transfer
3-29
3-30
Extended Bernoulli Equation (EBE)
β’ It is a modified version of the BE to include effects such as viscous energy-loss effect and shaft works.
β’ Remember the energy conservation equation for a single inlet (1), single exit (2)CV with uniform properties.
π
π+π2
2+ ππ§
1
=π
π+π2
2+ ππ§
2
+π€π +π€π
where π€π is a sharp work out per unit mass, for example, π€π is negative for pump work in, but positive for turbine work out; π€π is an energy loss by friction.
Streamline
1
2
π€π
π€π
β’ "Head form" of the BE
π
ππ+π2
2π+ π§
1
=π
ππ+π2
2π+ π§
2
+ βπ + βπ
Pressurehead
Velocityhead
Elevationhead
Frictionhead
Pump head in < 0, turbine head out > 0
Total head at 2(βπ2)
Pump and Turbine Head (hπ)Pump head βπ is related to the power delivered to the fluid by thepump (π«π ) as follows
β’ Power delivered to the fluid (π«π) is related to the power consumed by the pump (π«pum p ) through thepump efficiency
π =pump
π«π
π«pump
β’ For a turbine, power extracted from the fluid is calculated in a similar way.
π«π = πππβπ
β’ Power produced by the turbine (π«turbine) is smaller than the extracted fluid power
πturbine =π«turbine
π«π
1
2
π
π
Centrifugal pump
3-31
ππ = ππ€π = πππ€π β ππ= πππβπ
3-32
Extended BE Exercises
Exercise : The pump shown below pumps water steadily at a volumetric rate of0.005 m3/s through a constant diameter pipe. At the end of the pipe there is a nozzle with an exit area that is equal to half of the pipe area. Neglecting frictional losses, determine the power that must be supplied to the pump, if it is working with 70 % efficiency.
Pump
25 m
7 m
15 m
9 m
3-33
Extended BE Exercises (contβd)
Exercise : A pump is used to transport water between two large reservoirs. Desired volumetric flow rate through the pipes is 0.016 m3/s. Cross-sectional area of the pipes are 0.004 m2. Total frictional head losses between two reservoirs is estimated to be 2 m. Efficiency of the pump is 75 %. Determine
a) the required pump head.
b) the power delivered to the water by the pump.
c) the power required to drive the pump.
Pump
6 m
Suction pipe
Discharge pipe
Suctionreservoir
Dischargereservoir
3-34
Hydraulic Grade Line vs Energy Line in Head form of BE
βπ‘ππ‘ = π» = π§2 +π2πΎ+π22
2π= π§3 +
π3πΎ+π32
2πEL π§1 +π1πΎ> π§2 +
π2πΎ> π§3 +
π3πΎ
HGL
3-35
Hydraulic Grade Line vs Energy Line in Head form of BE
3-36
Hydraulic Grade Line vs Energy Line in Head form of BE
3-37
Restrictions on Use of BE
1) Compressibility Effect
β’ For a simple compressible gas flow where temperature remains constant along a streamline,the ideal gas law can be used to consider the density variation along the streamline.
ππ
π+1
2π£2 + ππ§ = ππππ π‘ β π π
ππ
π+1
2π£2 + ππ§ = ππππ π‘
The constant of integration is readily obtained if the state of gas flow is known in terms of π§1, π1, and π1 at any location on the streamline. The result is
π12
2π+ π§1 +
π π
πππ
π1π2
=π22
2π+ π§2
Valid for inviscid, isothermal, compressible, stead-state, along the gas streamline
β’ For a more common condition, say, isentropic expansion or contraction process which means there is no friction or heat transfer
π
ππ= πΆ β
ππ
π+1
2π£2 + ππ§ = ππππ π‘ β πΆ1/π
ππ
π1/π+1
2π£2 + ππ§ = ππππ π‘
π
π β 1
π1π1
+π12
2+ ππ§1 =
π
π β 1
π2π2
+π22
2+ ππ§2
Valid for inviscid, isentropic, compressible, stead-state, along the gas streamline
3-38
Restrictions on Use of BE
1) Compressibility Effect
π
π β 1
π1π1
+π12
2+ ππ§1 =
π
π β 1
π2π2
+π22
2+ ππ§2 Difference b/w incompressible BE
21
π1
π π1
ππ1
π1
β’ For a stagnation flow, let us learn the difference between compressible and incompressible.
From Chap. 11 (compressible)
π2 β π1π1
= 1 +π β 1
2ππ1
2
π/(πβ1)
β 1
When ππ1 β 0, consider
β 1 + ν π= 1 + πν + n n β 1ν2
2+ β―
π2 β π1π1
=πππ1
2
21 +
1
4ππ1
2 +2 β π
24ππ1
4 +β―ππ1β0 πππ1
2
2(compressible)
β ππ1=π1π1
=π1
ππ π1π1 +
ππ12
2= π2 β
π2 β π1π1
=π12
2π π1=πππ1
2
2
β’ For an incompressible stagnation flow
3-39
Restrictions on Use of BE
2) Unsteady Effect
π =ππ
ππ‘+ π
ππ
ππ ππ‘ +
π£2
π ππ β π = π
ππ
ππ ππ‘ +
π2
π ππ(unsteady) (steady)
β a change in velocity of a particle resulting from a change in its positionalong a streamline
β‘ a change in velocity of a particle at a fixed position
ππ
ππ‘ππ +
ππ
π+ π
π2
2+ πππ§ = 0
π1 +ππ1
2
2+ πΎπ§1 = π
π1
π2ππ
ππ‘ππ + π2 +
ππ22
2+ πΎπ§2
Valid for unsteady, incompressible, inviscidalong a streamline
Uneasy to calculate this integral, but simplified for irrotational flow (Chap. 6)
3-40
Restrictions on Use of BE
3) Rotational Effect: e.g., for incompressible, uniform flow w/ V
π1 +ππ1
2
2+ πΎπ§1 = π2 +
ππ22
2+ πΎπ§2 = πΆ12
π1 = π2 = π0, π§1 = π§2 = 0, π1 = π2 = V
πΆ12 = π0 +ππ2
2
π3 +ππ3
2
2+ πΎπ§3 = π4 +
ππ42
2+ πΎπ§4 = πΆ34
π3 = π4 = π0 β πΎβ,π§3 = π§4 = β, π3 = π4 = V
πΆ34 = π0 β πΎβ +ππ2
2+ πΎh = πΆ12
For a uniform flow that is "irrotational flow", there is no restriction of "along the streamline"
BE w/ a same constant can be applied for (1-2), (3-4), (2-4), but not for (4-5)