Bernoulli

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Applications of Integral Conservation Equations Bernoulli Equation is one of the most important equations in Fluid Mechanics and finds many applications. One such is the measurement of flow by introducing a restriction within the flow. The restriction may take the form of an orifice plate or a converging-diverging nozzle. The required formula will be first derived for an orifice plate and will be extended to other devices. Flow through a Sharp-edged Orifice Consider an orifice plate placed in a pipe flow as shown in Fig 1. We assume that the thickness of the plate is small in comparison to the pipe diameter. Let the orifice be sharp edged. The effect of a rounded plate is a matter of detail and will not be considered here.

Figure 1: Flow through an Orifice Body A fully developed flow prevails at the upstream of the orifice. The presence of the orifice makes the flow accelerate through it thus increasing the velocity. It may appear that the flow fills the orifice completely and expands downstream of it. But this is not true. As the flow expands downstream it cannot fill the entire diameter of the pipe at once. It requires a distance before it does. A recirculating flow develops immediate downstream of the nozzle. As a consequence the smallest diameter of flow is not equal to the orifice diameter, but smaller than it. The position of the smallest diameter occurs downstream of the orifice. We can deduce the flow rate through the pipe by measuring the pressure difference upstream of the nozzle and at the orifice. We make a few assumptions about the flow as follows, 1. The flow is steady 2. The flow is incompressible 3. There is no friction or losses 4. The velocities at sections (1) and (2) are uniform, i.e, they do not vary in a radial direction. 5. The pipe is horizontal, i.e., z is the same at (1) and (2), i.e., z1 = z2. This assumption can be relaxed easily. It is possible to have a fluid flowing through an inclined pipe. Then the gz term does not cancel out from LHS and RHS of the Bernoulli Equation. The equations we consider are (a) Continuity and (b) Bernoulli equations. a) Continuity Equation.

1 A 1 V 1=2 A 2 V 2

(1)

P1 V 21 P2 V 22  z 1=  z2   2 2

(2)

b) Bernoulli Equation

The Bernoulli Equation gives,

P1−P2=

 2 V 2−V 21   2

(3)

Noting from Continuity Equation that 2 1

2 2

V =V we have

2

     A2 A1

2

2

V2 A P1−P2= 1− 2 2 A1

(4)

Solving for V2 we have,

V 2=

Consequently, the mass flow rate becomes,

        2 P1−P2 

A  1− 2 A1

A  1− 2 A1

A2



(5)

2 P 1−P2 

m= ˙ A2

m= ˙

2

2

 

A 1− 2 A1

2

2 P 1−P2 

(6)

The above equation gives the mass flow rate through the pipe in terms of the pressure drop and the areas. The equation gives only a theoretical value. In order to obtain a more realistic value one need to substitute the actual area at the minimum cross section or the Vena Contracta. This is not easy to measure. In addition losses may not be negligible as we have assumed. Extent of losses is a function of the Reynolds number. In practice, a Coefficient of Discharge is defined such that,

m= ˙

C A2



Further if we define a ratio of diameters

2

 

A 1− 2 A1

(7)

 such that =

m= ˙

2 P 1−P2 

C A2

1−4

D2 D1

2  P1−P2 

(8)



Sometimes the ratio 1/ 1− 4 is referred to as Velocity of Approach Factor. Again it is usual to combine this and the Discharge Coefficient to define a Flow Coefficient given by

K=

C

1=4

(9)

Consequently the mass flow rate is given by,

m=K ˙ A2 2  P1−P2 

(10)

Thus the mass flow rate for a pipe can be calculated with the knowledge of pressure drop, the orifice diameter and the coefficient K. Extensive data exists in handbooks on the coefficient K. Pressure drop is usually measured by using a manometer as shown in Fig. 1. Now the pressure drop is obtained as h, the height of a liquid column (which may be mercury). Accordingly the alternate form of Eqn.10 is

m=K ˙ A2 2  g h

(11)

Flow Through a Nozzle

Figure 2 : Flow through a Nozzle Though geometrically different from an orifice plate, a flow nozzle is conceptually similar to it. Fig 2. shows a flow nozzle which is just a converging nozzle placed in a pipe. The flow mechanism is similar to that for the orifice. Now the area A2 is the throat area of the nozzle. Flow through a Venturi Tube

Figure 3: Venturi Tube A Venturi Tube is a converging-diverging nozzle (Fig.3.) placed in a pipe. The principle of this was demonstrated by Giovanni Battista Venturi(1746-1822) in 1797. It was only later in 1887 that it was employed for flow measurement by Herschel. The mass flow rate is again given by Eqn.11.

Important Applications of Control Volume Analysis In this section we consider some of the important applications of the control volume analysis. Every analysis may or may not involve each of the equations we have derived- Continuity, Momentum, Bernoulli and Energy. These applications are important from a physical point of view. Measurement of Drag about a Body immersed in a fluid Consider abody such as an aerofoil placed in a flow, which could be a in a wind tunnel. Far from the body the flow is uniform and inviscid. As the flow approaches the body many dramatic changes take place. The flow will start to depart from uniformity. But as the flow negotiates the body viscosity comes into play. Consequently, the velocity on the body surface is zero. The velocity catches up with the freestream speed as we move away from the body. In other words, a boundary layer develops. A boundary layer is not static. It grows as the flow moves downstream. When the flow leaves the body the centreline velocity is not zero anymore. It starts to build up slowly. This is the Wake region. If a velocity profile is measured across the wake by carrying out what is called a Wake Traverse, we see that it resembles that shown in Fig.4. The wake profile thus carries signatures of the viscous effect.

Figure 4: Measurement of Drag about an immersed body If a force balance is conducted in a region surrounding the body/ aerofoil then a force imbalance is evident. This should be related to Drag. Consider the body/aerofoil placed in a wind tunnel. Let us prescribe a control volume ABCD surrounding it. The left and right hand boundaries AB and CD are far from the body. As a result the flow is uniform ( at a speed U ∞ ) on AB. At the right hand boundary CD is the wake with the velocity profile as sketched. We assume that the top and bottom boundaries of the control volume,AD and BC are far away from the body and the vertical component of velocity namely v is zero across them. We make the following assumptions, 1. Steady Flow 2. Incompressible Flow 3. Static Pressure is same everywhere, which is actually a simplifying assumption. This could be relaxed.

Continuity Equation. Since the flow is steady, we have,

∫CS  V . dA=0 i.e.,

∫AB  V . dA∫BC  V .dA∫CD  V . dA∫DA V . dA=0 since v component of velocity along BC and AD is zero, the equation reduces to

∫AB  V . dA∫CD V . dA=0 B

D

∫A u dy−∫C u dy=0 leading to, B

D

∫A u dy=∫C u dy

(12)

Momentum equation On applying the momentum equation to the control volume we have

F sx F bx =∫AB u  V.dA∫BC u V.dA∫CD u  V dA∫DA u V dA i.e.,

B

C

D

A

F sx F bx =∫A u U ∞ dy∫B u  v dx∫C  u u dy∫D u  v dx

(13) (14)

since v = 0 on BC and AD, we have B

D

F sx F bx =∫A u U ∞ dy∫C  u2 dy

(15)

The body force Fbx on the control volume is zero. The surface forces are drag and that due to pressure. Since we have assumed that pressure is uniform, the latter is zero. Further length AB = length CD but opposite direction,, allowing us to combine the integrals on the RHS. Thus we have, D

D=∫C  u u−U ∞  dy In effect the velocities below C and that above D will be uniform and equal to Consequently the above equation could also be written as

(16)

U∞ .

∞

D=∫−∞  uu−U ∞ dy

(17)

A flaw in the above analysis should be apparent to you. Look at Eqn. 12. This cannot be true. The mass flow going through AB at a uniform velocity U ∞ cannot be equal to that across CD where the velocities are smaller than U ∞ . Some mass has to escape through AD and BC. In other words our assumption of v = 0 on AD and BC is faulty. The equation for drag that we have obtained is inaccurate as a consequence. A more acceptable estimate for drag can be obtained by considering the v component of velocity on AD and BC. The other method is to make these boundaries streamlines of flow. Then AB≠CD . This is left as an exercise. Jet Impingement on a surface Consider a jet with a cross section Aj at a speed vj impinging on a solid surface at an angle Fig 5. It is required to calculate the normal force exerted on the surface.

 as shown in

Figure 5 : Jet impinging on a surface Let us consider the physics of the process first. As the jet impinges upon the surface, it splits into two parts. These move tangential to the surface. The normal component of the force however does act upon the surface and is to be countered for stability. Prescribe x and y axes parallel and perpendicular to the surface and chose a control volume as shown. At the entry to the control volume we have the momentum in the y-direction equal to

˙ j cos  ∫Aj v  v j cos dA= A j v j v j cos = mu

(18)

At the solid surface velocity normal is zero and as such there is no normal momentum acting. The normal force acting upon the surface is given by Eqn 18. Force on a Pipe Bend Consider a flow through a pipe bend as shown. The flow enters the bend with a speed V1 and leaves it a speed V2, the corresponding areas of cross section being A1 and A2 respectively. The velocities have components u and v in x and y directions. As the flow negotiates the bend it exerts a force upon it. This force is readily calculated by the momentum theorem.

Figure 6: Force on a Pipe Bend The continuity equation yields

 V 1 A 1= A 2 V 2 =m ˙

(19)

Carrying out a force balance in x-direction, we have

P1 A 1−P2 A 2 cos F x =V 2 A 2 V 2 cos −V 1 A 1 V 1 P1 A 1 P2 A 2 cos F x =mV ˙ 2 cos −mV ˙ 1=mV ˙ 2 cos −V 1

(20)

In the y-direction we have,

0−P 2 A2 sinF y = V 2 A2 V 2 sin −0 giving

P2 A 2 sin F y =mV ˙ 2 sin 

(21)

Thus the force components acting on the bend are

F x =P2 A 2 cos −P 1 A1− mV ˙ 2 cos −V 1  F y =−P 2 A2 sinmV ˙ 2 sin  Froude's Propeller Theory Propellers are a mechanism for the propulsion of an aeroplane. In its generic form a propeller is pair of rotating blades mounted on a shaft that houses the engine as well. As the engine operates the propeller turns sucking a large amount of air. As this air passes through the rotating blades, it gets energised, its speed increases. In the process the required Thrust to propel the aircraft is produced.

Figure 7: Froude analysis of a Propeller The analysis we carry out follows William Froude (1810-1879). We consider the propeller as a thin disc

(22)

rotating in air as shown in Fig 7. Let the pressure and velocity far away from the disc i.e., at section (1) be P1 and V1 respectively. The conditions just at (2) which is the front of the disc are P2 and V2. The disc imparts momentum and energy to the incoming air such that the pressure and velocity just behind the disc (3) are V3 and P3 respectively. At (4), far downstream the conditions are V4 and P4. We assume that the air which is influenced by the disc is confined to a slipstream as shown. Since the disc is thin and the area of cross section at (2) and (3) ar equal, we have

V 2=V 3

(23)

The pressures at (1) and (4) are equal to the freestream value. We consider the control volume formed by slipstream and the ends (1) and (4) and write the momentum equation. Continuity Equation An application of the Continuity Equation gives,

 V 1 A 1= V 4 A4 =m˙

(24)

Momentum Equation Considering first the forces the only force that acts upon the control volume is the net force on the disc or the Thrust, F. Pressures being equal at (1) and (4) does not contribute to the surface force. Since the flow takes place in a horizontal direction there is no body force to be considered. Accordingly,

F=V 4  V 4 A 4−V 1  V 1 A1 =mV ˙ 4−V 1

(25)

Noting that V2 = V3, this force F is equal to A(P3 - P2), where A is the area of cross section of the disc. As a consequence we have,

F=A  P3 −P 2= mV ˙ 4−V 1  dividing by A, we have

 P3−P2 = noting that

we have

m ˙ V −V 1  A 4

m ˙ = A 2 A  P3−P2 = A 2 V 4 −V 1

(26)

Bernoulli Equation It is easy to see that there is no addition of work or heat between sections (1) and (2) and also between (3) and (4). It is possible to apply Bernoulli equation between (1) and (2) and also between (3) and (4) but not between (2) and (3).

1 1 P1  V 21=P2 V 22 2 2 1 1 P3  V 23=P 4  V 24 2 2

(27)

Since V2 = V3and P1 = P4 we have from the above equations

1 2 2  P3−P2 = V 4 −V 1  2

(28)

Eliminating P3 - P2 from Eqns 26.and 28 we have,

V 2=

V 1V 4 2

(29)

If the velocities are referred to the freestream air speed, i.e., V1, we see that the propeller moves at a velocity V1. The work done by the propeller on the air stream or the power output is then,

Power out =F V 1= mV ˙ 4−V 1 V 1

(30)

In addition some kinetic energy is added to the air stream, which goes as a waste. The power input therefore is given by

1 2 Powerinput =mV ˙ 4−V 1 V 1 mV ˙ 4−V 1  2

(31)

From Eqns. 30 and 31 the efficiency of the propeller will be

fr =

The term

V1 1 V 1 V 4 −V 1 2

(32)

fr is called the Froude Efficiency.

Analysis of a Wind Turbine

Figure 8: Analysis of a Wind Turbine A wind turbine (Fig 8) extracts energy from an air stream while a propeller adds energy to the air stream. The

analysis follows the same lines. The wind turbines are smaller in size compared to the propellers. For the wind turbine too we have the result that

V 2=

V 1V 4 2

Power output =F V 1= mV ˙ 4−V 1V 1

(33)

Now the efficiency is given by the Kinetic energy extracted divided by the kinetic energy in the free stream. Thus,

1  A V 2 V 21 −V 24  V 1V 4 V 21−V 24  2 th = = 3 1 2 2V1   A V 1 V 1 2

(34)

The above expression has a maximum when V4/V1 = 1/ The maximum theoretical efficiency is 59.3%. Total Pressure Loss through a Sudden Expansion

Figure 9 : Losses through a Sudden Expansion, Borda-Carnot Equation Consider a sudden expansion placed in a duct (Fig 9 ). The flow does not follow the area changes as suddenly as the geometry does. Any flow will find the sudden area increase difficult to negotiate. In fact a recirculating flow develops as was seen in case of the orifice flow. This gives to losses which are reflected in the total pressure at downstream being reduced. It is possible to calculate this loss from a control volume analysis. Continuity Equation

 A 1 V 1= A 2 V 2

(35)

Momentum Equation The forces upon the control volume sum up to

 F x =P1 A 1−P2 A 2P1  A 2− A 1=P1 A 2−P2 A 2

(36)

Substituting in the momentum equation, we have,

P1 A 2−P2 A 2=−V 21 A1V 22 A 2 i.e. 2

2

P1−P2=V 2 −V 1

  A1 A2

(37)

(38)

Bernoulli Equation We remind ourselves that we cannot connect stations (1) and (2) with the Bernoulli Equation. But we just use

the total pressure relation at (1) and (2). Accordingly,

1 P1  V 21=PO1 2 1 P2  V 22=PO2 2

(39)

 PO1−PO2=P1 −P 2 V 21−V 22  2

(40)

Total pressure loss is hence equal to

Measurement of Airspeed Bernoulli equation readily allows one to determine the flow speed once the static and stagnation pressures are known. Rewriting the equation for Stagnation Pressure we have

V=



2 PO −P 

(41)

It is therefore a matter of measuring the static and stagnation pressures at a given location. Static Pressure is conveniently measured by drilling a hole in the wall or the pipe, called the Pressure Tap (Fig. 10). A manometer or a pressure gauge is connected to the tap. During flow static pressure is communicated to the measuring device. Alternately one could use a Static Pressure probe shown in Fig. 11. This has holes which communicate the pressure to a measuring device. To Manometer or a gauge

To Manometer or a gauge Figure 10: Pressure Tap to measure Static Pressure

Figure 12: Stagnation Tube

Figure 11: Static Pressure Probe

Figure 13: Pitot Tube

Measurement of stagnation pressure requires that the flow be brought to rest. A glass tube or a hypodermic needle aligned with the flow and facing upstream as shown in Fig. 12 will do the job. Alternately, what is called a Pitot Tube shown in Fig.13, with a hole facing upstream of the flow may be employed. The method shown in Fig. 14 suggests itself. But for an accurate determination of flow speed, static and stagnation pressures are to be measured simultaneously . This is made possible by a Pitot-Static tube shown in Fig. 15. This combines the static pressure probe and the pitot tube. The "static holes" and the "stagnation hole" are as near to each other as possible.

Figure 14: Pitot tube used with a static pressure tap

Figure 15: Pitot-static tube

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