Bernoulli’s Equation - جامعة نزوى · Applications of the Bernoulli Theorem 4/5 For a...

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Chapter 4 Bernoulli’s Equation (Mechanical Energy Balance) Flow Measurements 1

Transcript of Bernoulli’s Equation - جامعة نزوى · Applications of the Bernoulli Theorem 4/5 For a...

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Chapter 4

• Bernoulli’s Equation

(Mechanical Energy Balance)

• Flow Measurements

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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

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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

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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

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• 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

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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

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(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

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Interpretation of Bernoulli’s Equation

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• 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

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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

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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

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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

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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

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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

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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

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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.

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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.

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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

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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

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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.

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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.

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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

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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)

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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

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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

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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)

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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

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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)

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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

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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

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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

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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).

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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

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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.

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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

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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

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• 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)

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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

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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

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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

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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)

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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

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Page 53: Bernoulli’s Equation - جامعة نزوى · 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

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

Page 54: Bernoulli’s Equation - جامعة نزوى · 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

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

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Page 55: Bernoulli’s Equation - جامعة نزوى · 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

Generalized Chart for Cd

Cd

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