Bernoulli application, Lecture notes of Fluid Mechanics

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Ministry of Higher Education & Scientific Research Foundation of Technical Education Technical College of Basrah

Training Package in

Fluid Mechanics

Modular unit 7

Application on Bernoulli Equation

By

Risala A. Mohammed

M.Sc. Civil Engineering Asst. Lect. Environmental & Pollution Engineering Department 2011

1- Over view

1-1 Target population

For the students of second class in

Environmental engineering Department in

Technical College

1-2 Rationale

Lies the importance of studying the applications

on Bernoulli's equation in the use of this equation as

principle in the design of speed and flow rate

measurement devices in the pipes. in addition to its

application for measuring the discharge through the

orifices in the reservoirs.

1-3 Central Idea

The main goal of this chapter is to know the

application of Bernoulli equation in the many

situations of fluid mechanics.

1-4 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

1-5 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 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

2- 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.

Q 2 )) ( 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 m^3 /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

0 X

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 + ½ V2 + 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 + ½ V^2.  For a constant- density fluid the quantity p + ½ V^2 is therefore known as the stagnation pressure of that streamline while ½ V^2 – 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 ½ V^2 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 ½ V^2. The liquid is forced up the vertical part of the tube to a height : h = p/g = ½ V^2 /g = V^2 /2g above the surrounding free surface. Measurement of h therefore enables V to be calculated.

V  2 gh

Pitot Tube

 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 flow.  From the expressions above,

2 2 1 1

p  p   V

 2 1 

2 2 1 1

2

2

1

V g h h

gh gh V

 

    

Figure 4.3 : A Piezometer and a Pitot tube

Venturi meter

 The Venturi meter is a device for measuring discharge in a pipe.  It consists of a rapidly converging section, which increases the velocity of flow and hence reduces the pressure.  It then returns to the original dimensions of the pipe by a gently diverging „diffuser‟ section. By measuring the pressure differences the discharge can be calculated.  This is a particularly accurate method of flow measurement as energy losses are very small.

Venturi meter

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

and Qideal = V 1 A 1

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

A^ A^1

2 gp ρg^ p Z Z V (^2) 2

1

(^1212) 1 ^  

 



^     

1

2 A 2

A 1

ρg^2 Z^1 Z^2 2g p^1 p Qactual CdQideal CdA 1 ^  

   





 

     

AA^1

2gh ρ ρ 1 Q C A 2

2

1

man actual d 1 ^  

 





 

 (^)  

Example(1)

A Venturi meter with an entrance diameter of 0.3 m and a throat diameter of 0. 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/m^3 , 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

V 1 = Q /0.0707, V 2 = Q /0.

For the manometer :

P 1   (^) g gz 1  P 2   gg ( z 2  RP )  wgRP

P 1  P 2  19. 62 ( z 2  z 1 ) 587. 423

Example(1)

For the Venturi meter :

2

2 2 2 1

2 1 1 2 2

z g

V

g

P

z g

V

g

P

g g

    

 

2

P 1  P 2  19. 62 ( z 2  z 1 ) 0. 803 V 2

Q C Q m s

Q m s

V m s

V

d ideal

ideal

ideal

0. 96 0. 85 0. 816 /

0. 85 /

27. 047 /

0. 803 587. 423

3

3

2

2

2 2

   

 











 







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 :

losses g

V g

P h g

V g

P      0  2 2

2 2 2

2 1 1  

Sharp edge circular orifice

 NowP 1 = Patm and as the jet in unconfined,P 2 = Patm. If the flow is steady, the surface in the tank remains stationary andV 1  0 (z 2 =0, z 1 =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

z z h

g

V

 1  2 

2 2

V 2  2 gh

Q  A 0 V 2

Sharp edge circular orifice

whereA 0 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,

Q  A 0 2 gh

V 2 a  Cv 2 gh

theoretical velocity

actual velocity

Cv ^ (typically about 0.97)

Sharp edge circular orifice

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).

d 0

approx. d 0 /2

Vena contracta

P = Patm

Sharp edge circular orifice

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)

areaof orifice

areaof jet at venacontracta Cc 

Q a  Cc A 0 Cv 2 gh

theoreticaldischarge

C actual discharge d 

Q (^) a  Cd A 0 2 gh

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 sectionC-C represents the point of vena contracta. Consider a particle of water in the jet at P.