CH7Application on Bernoulli Equation

Ministry of Higher Education & Scientific ResearchFoundation of Technical Education Technical College of Basrah

CH7:Application on Bernoulli Equation

Training Package

in

Fluid Mechanics

Modular unit 7

Application on Bernoulli Equation

By

Risala A. MohammedM.Sc. Civil Engineering

Ass. Lect.Environmental & Pollution Engineering Department

2011

Target population


For the students of second class in

Environmental engineering Department in

Technical College

Central Idea


The main goal of this chapter is to know the

application of Bernoulli equation in the many

situations of fluid mechanics.

Instructions


1 -Study over view thoroughly

2 -Identify the goal of this modular unit

3 -Do the Pretest and if you-: * Get 9 or more you do not need to proceed

* Get less than 9 you have to study this modular

4 -After studying the text of this modular unit , do the post test and if you-:

* Get 9 or more , so go on studying modular unit eight * Get less than 9 , go back and study the modular unit seven

Performance Objectives


At the end of this modular unit the student will be able to :-

1. Describe the pitot tube and calculate the velocity of fluid during

flowing through pipes by using it.

2.Describe the Venturi Venturi meter and calculate the discharge of fluid

during flowing through pipes by using it

3.Calculate the discharge from tanks by using orifices

4.Calculate the hydraulic coefficients ( Cv, Cd and Cc)

5.Calculate the time required for empting tank through an orifices.

6.measuring discharge in open channel by using notches and weirs

Pre test


Q1)) ( 5 mark)Explain briefly how the co-efficient of velocity of a jet "issuing through an orifice can be

experimentally determined. Find an expression for head loss in an orifice flow in terms of co-efficient of velocity and jet velocity, The head lost in flow through a 50 mm diameter orifice under a certain head is 160 mm of water and. the velocity of water in the jet is 7 m/s. If the co-efficient of discharge be 0.6 l, determine: (i) Head on the orifice causing flow; (ii) The co-efficient of velocity; iii) The diameter of the jet.

Q2)) ( 5 mark)A Venturi meter having a throat diameter of 150 mm is installed in a horizonta1300-mm-diameter water main. The coefficient of discharge is 0.982. Determine the difference in level of the mercury columns of the differential manometer attached to the Venturi meter if the discharge is 0.142 m3/s.

Not Check your answers in key answer page

Introduction


• The Bernoulli equation can be applied to a great many

situations not just the pipe flow we have been considering

up to now.

• In the following sections we will see some examples of its

application to flow measurement from tanks, within pipes

as well as in open channels.

Pitot Tube


X0

Streamlines passing a non-rotating obstacle

A point in a fluid stream where the velocity is reduced to zero is known as a stagnation point.

Any non-rotating obstacle placed in the stream produces a stagnation point next to its upstream surface.

The velocity at X is zero: X is a stagnation point.

By Bernoulli's equation the quantity p + ½V2 + gz is constant along a streamline for the steady frictionless flow of a fluid of constant density.

If the velocity V at a particular point is brought to zero the pressure there is increased from p to p + ½V2.

For a constant-density fluid the quantity p + ½V2 is therefore known as the stagnation pressure of that streamline while ½V2 – that part of the stagnation pressure due to the motion – is termed the dynamic pressure.

A manometer connected to the point X would record the stagnation pressure, and if the static pressure p were also known ½V2 could be obtained by subtraction, and hence V calculated.

Pitot Tube


Simple Pitot Tube

A right-angled glass tube, large enough for capillary effects to be negligible, has one end (A) facing the flow. When equilibrium is attained the fluid at A is stationary and the pressure in the tube exceeds that of the surrounding stream by ½V2. The liquid is forced up the vertical part of the tube to a height :

h = p/g = ½V2/g = V2/2g

above the surrounding free surface. Measurement of h therefore enables V to be calculated.

ghV 2

Pitot Tube


Two piezometers, one as normal and one as a Pitot tube within the pipe Two piezometers, one as normal and one as a Pitot tube within the pipe can be used in an arrangement shown below to measure velocity of can be used in an arrangement shown below to measure velocity of flow. flow.

From the expressions above,From the expressions above,

2112 2

1Vpp

12

2112

2

2

1

hhgV

Vghgh

Figure 4.3 : A Piezometer and a Pitot tube

Venturi meter


The Venturi meter is a device for The Venturi meter is a device for measuring discharge in a pipe. measuring discharge in a pipe.

It consists of a rapidly converging It consists of a rapidly converging section, which increases the velocity section, which increases the velocity of flow and hence reduces the of flow and hence reduces the pressure. pressure.

It then returns to the original It then returns to the original dimensions of the pipe by a gently dimensions of the pipe by a gently diverging ‘diffuser’ section. By diverging ‘diffuser’ section. By measuring the pressure differences measuring the pressure differences the discharge can be calculated. the discharge can be calculated.

This is a particularly accurate method This is a particularly accurate method of flow measurement as energy losses of flow measurement as energy losses are very small.are very small.

Venturi meter


Applying Bernoulli Equation between (1) and (2), and using continuity equation to eliminate V2 will give :

and Qideal = V1A1

To get the actual discharge, taking into consideration of losses due to friction, a coefficient of discharge, Cd, is introduced.

It can be shown that the discharge can also be expressed in terms of manometer reading :

where man = density of manometer fluid

1AA

ZZρg

pp2g

V 2

2

1

2121

1

1

2

2A

1A

2Z

1Z

ρg2

p1

p2g

1A

dC

idealQ

dC

actualQ

1AA

1ρ

ρ2gh

ACQ 2

2

1

man

1dactual

Example(1)


A Venturi meter with an entrance diameter of 0.3 m and a throat diameter of 0.2 m is used to measure the volume of gas flowing through a pipe. The discharge coefficient of the meter is 0.96. Assuming the specific weight of the gas to be constant at 19.62 N/m3, calculate the volume flowing when the pressure difference between the entrance and the throat is measured as 0.06 m on a water U-tube manometer

Solution

V1 = Q/0.0707,V2 = Q/0.0314

For the manometer :

PwPgg gRRzgPgzP )( 2211

423.587)(62.19 1221 zzPP

Example(1)


For the Venturi meter :

2

222

1

211

22z

g

V

g

Pz

g

V

g

P

gg

221221 803.0)(62.19 VzzPP

smQCQ

smQ

smV

V

ideald

ideal

ideal

/816.085.096.0

/85.02

2.0047.27

/047.27

423.587803.0

3

32

2

22

Example (2)


Example(3)


Example (4)


Example (5)


Sharp edge circular orifice


(1)

(2)

h

datum

streamline

Consider a large tank, containing an ideal fluid, having a small sharp-edged circular orifice in one side.

If the head, h, causing flow through the orifice of diameter d is constant (h>10d), Bernoulli equation may be applied between two points, (1) on the surface of the fluid in the tank and (2) in the jet of fluid just outside the orifice. Hence :

lossesg

V

g

Ph

g

V

g

P 0

22

222

211



Now P1 = Patm and as the jet in unconfined, P2 = Patm. If the flow is steady, the surface in the tank remains stationary and V1 0 (z2=0, z1=h) and ignoring losses we get :

or the velocity through the orifice,

This result is known as Toricelli’s equation. Assuming no loses, ideal fluid, V constant across jet at (2), the discharge through the orifice is

hzzg

V 21

22

2

ghV 22

20 VAQ



where A0 is the area of the orifice

For the flow of a real fluid, the velocity is less than that given by eq. 4.7 because of frictional effects and so the actual velocity V2a, is obtained by introducing a modifying coefficient, Cv, the coefficient of velocity:

Velocity,

gh2AQ 0

ghCV va 22

velocityltheoretica

velocityactualCv (typically about 0.97)



As a real fluid cannot turn round a sharp bend, the jet continues to contract for a short distance downstream (about one half of the orifice diameter) and the flow becomes parallel at a point known as the vena contracta (Latin : contracting vein)..

d0

approx. d0/2

Vena contracta

P = Patm



The area of discharge is thus less than the orifice area and a coefficient of contraction, Cc, must be introduced.

Hence the actual discharge is :

(typically about 0.65)

or introducing a coefficient of discharge, Cd, where :

(typically 0.63)

orificeofarea

contractavenaatjetofareaCc

gh2CACQ v0ca

dischargeltheoretica

dischargeactualCd

gh2ACQ 0da

Cd = Cc . Cv

Experimental Determination of Hydraulic Coefficients


Determination of Co efficient of Velocity of Velocity (Cv)

A Fig. shows a tank containing water at a constant level, maintained by a constant supply. Let the water flow out -of the tank through an orifice, fitted in one side of the tank. Let the section C-C represents the point of vena contracta. Consider a particle of water in the jet at P.

Experimental Determination of Hydraulic Coefficients


Let, x = Horizontal distance travelled by the particle in time 't',

y= Vertical distance between C-C and P, V;:= Actual velocity of the jet at vena-contracta, and

H;:= Constant water head.

Then, horizontal distance, x= V x t

Example (6)


An orifice 50mm in diameter is discharging water under a head of l0 meters. If Cd = 0.6 and Cv

= 0.97, find: (i) actual discharge, and ii) Actual velocity of the jet at vena contraeta

Example (7)


A large tank has a sharp edged circular orifice of 930 mm2 area at a depth of 3 m below constant water level. The jet issues horizontally and in a horizontal distance of 2.4m ,

it falls by 0.53 m, the measured discharge is 4.3 liters\sec. Determine coefficients of velocity, contraction and discharge for the orifice.

Example (8)


The head of water over the centre of an orifice of diameter 30 mm is 1.5m. The actual discharge through the orifice is 2.55 liters/sec. Find the co-efficient of discharge.

Example (9)


Nozzles


In a nozzle, the flow contracts gradually to the outlet and hence the area of the jet is equal to the outlet area of the nozzle.

i.e. Cc = 1.0

therefore Cd = Cv

Taking a datum at the nozzle, Torricelli’s equation gives the total energy head in the system as it assumes an ideal fluid and hence no loss of energy, i.e. theoretical head :

dnozzle

contraction within nozzle

g

Vht 2

22

Nozzles


But the actual velocity is :

V2a = Cv V2 and the actual energy in the jet is :

as P2 and z2 are zero

Therefore the actual energy head is :

And the loss of energy head, hf , in the nozzle due to friction is :

g

Vh a

a 2

22

g

VCh v

a 2

22

2

g

VC

g

Vhhh v

atf 22

22

22

)1(2

22

2vC

g

V

tvf hCh )1( 2

theoretical actual

Example (10)


The nozzle in Fig. below throws a stream of water vertically upward so that the power available in the jet at point 2 is 3.50 hp. If the pressure at the base of the nozzle, point 1, is 36.0 psi, find the (a) theoretical height to which the jet will rise, (6) coefficient of velocity, (c) head loss between 1 and 2, and (4) theoretical diameter of the jet at a point 18 ft above point 2.

Time Required For Empting Tank Through An Orifices


Time Required For Empting Tank Through An Orifices


Example (11)


A circular tank of diameter 3 m contains water up to a height of 4m. The tank is provided with an orifice of diameter 0.4 m at the bottom, Find the time taken by water; (i) to fall from 4 m to 2 m, and (ii) for completely emptying the tank

Example (12)


A 1 m diameter circular tank contains water up to a height of 4 m. At the bottom . of tank an orifice of 40 mm is provided. Find the height of water above the orifice after 1.5 minutes. Take co-efficient of discharge for the orifice Cd= 0.6.

Example (12)


Flow over notches and weirs


• is an opening in the side of a tank or reservoir, which extends above

the surface of the liquid.

• It is usually a device for measuring discharge.

• A weir is a notch on a larger scale – usually found in rivers.

• It may be sharp crested but also may have a substantial width in the

direction of flow – it is used as both a flow measuring device and a

device to raise water levels.

Notch



Weir Assumptions

assume that the velocity of the fluid approaching the weir is small so that kinetic energy can be neglected.

assume that the velocity through any elemental strip depends only on the depth below the free surface.

These are acceptable assumptions for tanks with notches or reservoirs with weirs, but for flows where the velocity approaching the weir is substantial the kinetic energy must be taken into account (e.g. a fast moving river).



A General Weir EquationTo determine an expression for the theoretical flow through a notch we will consider a horizontal strip of width b and depth h below the free surface, as shown in the figure

velocity through the strip

discharge through the strip,

Integrating from the free surface, h = 0, to the weir crest, h = H gives the expression for the total theoretical discharge,

ghV 2

ghhbAVQ 2

dhbhgQH

ltheoretica 0

21

2



Rectangular Weir

For a rectangular weir the width does not change with depth so there is no relationship between b and depth h. We have the equation, b = constant = B.

Substituting this with the general weir equation gives:

dhhgBQH

O

ltheoretica 21

2

23

23

2HgB

To calculate the actual discharge we introduce a coefficient of discharge, Cd, which accounts for losses at the edges of the weir and contractions in the area of flow, giving :

23

23

2HgBCQ dactual

Example (13)


Water flows over a sharp-crested weir 600 mm wide. The measured head (relative to the crest) is 155 mm at a point where the cross-sectional area of the stream is 0.26 m2. Calculate the discharge, assuming that Cd = 0.61.

As first approximation,

H = 155 mm

Cross sectionalArea = 0.26 m2

23

23

2HgBCQ dactual

23

)155.0(/62.196.03

261.0 2 msmm

= 0.0660 m3/s

Example (13)


Velocity of approach =

= 0.254 m/s

H + V12/2g = (0.155 + 0.00328) m = 0.1583 m

Second approximation:

Further refinement of the value could be obtained by a new calculation of V1 (0.0681 m3/s 0.26 m2), a new calculation of H + V1

2/2g and so on. One correction is usually sufficient, however, to give a value of Q acceptable to three significant figures.

2

3

26.0

/0660.0

m

sm

msm

sm

g

V 32

22

1028.3/62.19

)/254.0(

2

smsmx /0681.0/1583.06.062.1961.03

2 332/3

Post test


Post test

Q1)) ( 5 mark)A tank has two identical orifices in one of its vertical sides, The upper orifice is 1.5 below the water surface and the lower one is 3 m below the water surface . Find the point, at which the two jets will intersect, if the co-efficient of velocity is 0.92 for both the orifices

Q2)) ( 5 mark)Oil is flowing upward through a Venturi. meter as shown in . Assume a discharge coefficient of 0.984 What is the rate of flow of the oil?

Key answer


Pre testQ1((

Key answer


Pre testQ2((

Key answer


Post testQ1((

Key answer


Post testQ2((

References


1. Evett, J., B. and Liu, C. 1989 “2500 solved problems in fluid mechanics and hydraulics” Library of Congress Cataloging- in-Publication Data, (Schaum's solved problems series) ISBN 0-07-019783-0

2. Rajput, R.,K. 2000 “ A Text Book of Fluid Mechanics and Hydraulic Machines”. S.Chand & Company LTD.

3. White, F., M. 2000 “ Fluid Mechanics”. McGraw-Hill Series in Mechanical Engineering.

4. Wily, S., 1983 “ Fluid Mechanics”. McGraw-Hill Series in Mechanical Engineering.

Documents

CH7Application on Bernoulli Equation