Bernoulli Principle

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1. Bernoulli principle Fluid Mechanics - I Assigned by : Dr. Manzoor ul Haq Rajpoot Presented by: Sarwan Ahmed Ursani (18CH101) MEHRAN UNIVERSITY OF ENGINEERING & TECHNOLOGY

2. What will we discuss? Introduction of Bernoulli Principle Limitations on which this principle is based Derivation of Bernoulli Equation Applications of Bernoulli Principle Problem

3. Introduction Named after Daniel Bernoulli , who published this principle in his book Hydrodynamics in 1738. Leonhard Euler derived Bernoulli’s equation in its usual form in 1752 . For a perfect incompressible fluid flowing in a continuous stream, the total energy of particle remains the same, while the particle moves from one point to another point.

4. Introduction (continue) Conservation of energy principle for perfect incompressible fluids ( ideal fluids ) flowing in a pipe. Total Energy/ Total head of a system remains constant. OR  

5. Limitations The velocity of every liquid particle across any cross section of pipe is uniform but in actual practice the velocity is maximum at centre and gradually decreases at the edges due to friction There is no external force on the moving particles except the gravitational force but in actual practice there are other external forces such as pipe friction. There is no loss of energy of liquid particle while flowing but in real life in a turbulent flow some K.E is converted to heat energy and in viscous flow some energy is lost due to shear force. Force due to centrifugal force should be accounted (curved path).

6. Limitations’ simulation

7. Flow of Real fluid in pipe

8. Derivation

9.   We know that The change in potential energy is = mgz 2 –mgz 1 OR         2     V   W/ A. dx   W/ A 1 dx 1 =A 2 dx 2 = A.dx       Total work done (W)    

10. CONTINUE.. The change in Kinetic energy = OR Work done = Change in K.E + Change in P.E From above equation we can say that the total energy per unit weight of an ideal fluid remains constant when it moves from one point to another point. ½ m(v 2 2 -v 1 2 ) K.E = ½ W(v 2 2 -v 1 2 )/g   = W(z 2 -z 1 ) + ½ W(v 2 2 -v 1 2 ) / g   = W{(z 2 -z 1 ) + ½ (v 2 2 -v 1 2 ) / g} P1/   = ½ v 2 2 /g – ½ v 1 2 /g + z 2 -z 1 - P2/    

11. Applications of Bernoulli principle The Bernoulli’s theorem or Bernoulli’s equation is the basic equation which has the widest applications in Hydraulics and Applied Hydraulics. Since this equation is applied for the derivation of many formulae, therefore its clear understanding is very essential. The Bernoulli’s equation has a number of practical applications. 1. Venturi meter 2. Orifice meter 3. Pitot tube.

12. Venturi meter was designed by an Italian engineer Venturi in 1791 . It’s working principle is based on Bernoulli principle and its equation is also derived using Bernoulli equation. It is used to find out the discharge of liquid flowing from the pipe. Venturi meter Q = C.a 1 .a 2 √ (2gh) / √(a 1 2 – a 2 2 )

13. Clemons Herschel developed this instrument in 1886 . This instrument is used to find out flow rate/discharge of pipe with the help of reducing pressure working on the Bernoulli principle Orifice meter Q = C.a 1 .a 2 √ {2(p1-p2)/ ρ } / √(a 1 2 – a 2 2 )

14. This tube was designed by Pitot Henry a French scientist. This instrument is designed to measure the velocity of flow at a required point. It is used to measure the velocity of an aeroplane. Pitot tube V = √ (2gh)

15. problem The diameter of pipe changes from 200 mm at section 5 m above datum to 50 mm at section 3 m above datum. The pressure of water at first section is 500 KPa. If the velocity of flow at the first section is 1 m/s, determine the intensity of pressure at the second section. Data d 1 = 200 mm = 0.2 m z 1 = 5 m P 1 = 500 KPa v 1 = 1 m/s d 2 = 50 mm = 0.05m z 2 = 3 m P 2 = ??

16. CONTINUE.. Solution To determine the pressure intensity at section 2, we can use Bernoulli equation. First we need to calculate velocity at section 2. Using continuity equation Eq (II) To calculate the area at both ends we will use the formula Z 1 + ½ v 1 2 /g + P 1 / = Z 2 + ½ v 2 2 /g + P 2 /   A 1 V 1 = A 2 V 2 V 2 = A 1 V 1 /A 2 A = πd 2 / 4 Eq (I) A 1 = (3.14)(0.2) 2 / 4 A 2 = (3.14)(0.05) 2 / 4

17. CONTINUE.. A 1 = 31.42 x 10 -3 m 2 A 2 = 1.964 x 10 -3 m 2 Now substitute values of A 1 , A 2 and V 1 in Eq (II), we get V 2 = 16 m/s Now put all the known variables in Eq (I) to determine the value of pressure intensity at section 2 5 + ½ (1) 2 /9.81 + 500/ = 3 + ½ (16) 2 /9.81 + P 2 /   5 + ½ (1) 2 /9.81 + 500/ - 3 - ½ (16) 2 /9.81 = P 2 /   40 = P 2 /   P 2 = 40 x 9.81 P 2 = 392.4 KPa V 2 = (31.42 x 10 -3 ) x 1 / (1.964 x 10 -3 )

18. Thank you The floor is open for question.