1. Kinematics of Flow

2. Fluid Kinematics  Fluid motion -Types of fluid - Velocity and acceleration - Continuity equation  Potential Flows -Velocity Potential - Stream function - Flow net (Cir and Vorticity) - Source, Sink and Doublet

3. Section I

4. Fluid Kinematics: It is defined as the branch of science which deals with motion of particles without considering the force causing the motion. Types of fluid flow: 1)Steady flow and unsteady flow 2)Uniform and no-uniform flow 3)Laminar and turbulent flow 4)Compressible and in compressible flow 5)Rotational and irrigational flows 6)One, two and three dimensional flow Rate of flow or Discharge Q: m 3 /s or liters/s Q = A V

5. Continuity equation The equation based on the principle of conservation of mass is called continuity equation. Thus the quantity of fluid per second is constant. Consider two cross-section of a pipe Let V 1 = Average velocity Section 1-1 ς 1 = Density at section 1-1 A 1 = Area of pipe at section 1-1 And V 2, ς 2, A 2 are corresponding values at section 2-2 Rate of flow at section 1-1 = ς 1 A 1 V 1 Rate of flow at section 2-2 = ς 2 A 2 V 2 ς 1 A 1 V 1 = ς 2 A 2 V 2 Continuity equation A 1 V 1 = A 2 V 2 Incompressible fluids

6. Ex The diameter of a pipe at the section 1 and 2 are 10 cm and 15 cm respectively. Find the discharge through the pipe if the velocity of water flowing through the pipe at section 1 is 5 m/s. Determine also the velocity at section 2. D 1 =10cmD 2 =15cm V 1 =5 m/s (1) (2) Ans Q = 0.03927 m 3 /s, V 2 = 2.22 m/s

7. Ex A 30 cm diameter pipe, conveying water, branches into two pipes of diameters 20 cm and 15 cm repectively. If the average velocity in the 30 cm diameter pipe is 2.5 m/s, find the discharge in this pipe. Also determine the velocity in 15 cm pipe if the average velocity in 20 cm diameter pipe is 2 m/s. V 1 = 2.5 m/s D 1 = 30 cm V 3 = ? D 3 = 15 cm V 2 = 2 m/s D 2 = 20 cm Ans Q 3 = 0.1139 m 3 /s, V 3 = 6.44 m/s

8. Ex Water flows through a pipe AB 1.2 m diameter at 3 m/s and then passes through a pipe BC 1.5 m dia. At C, the pipe branches. Branch CD is 0.8 m in diameter and carries one-third of the flow in AB. The flow velocity in branch CE is 2.5 m/s. Find the i) Volume rate of flow in AB, ii) The velocity in BC, iii) The velocity in CD and iv) The diameter of CE. Ans Q = 3.393 m 3 /s, V BC = 1.92 m/s, V CD = 2.25 m/s, D CE = 1.07 m

9. Ex A 25 cm diameter pipe carries oil of Sp.gr. 0.9 at a velocity of 3 m/s. At another section the diameter is 20 cm. Find the velocity at this section and also mass rate of flow of oil. Ans: V 3 = 4.68 m/s, Mass rate of flow = 132.23 kg/s

10. Fluid element of length dx, dy and dz (In x, y, z direction) Velocity = u, v and w Mass of fluid entering the face ABCD = ς x Velocity-x x Area ABCD = ς u dy dz Mass of fluid leaving the face EFGH Continuity equation in Three-Dimensions Gain of mass in x-direction = Mass through EFGH- Mass ABCD

11. Similarly the net gain of mass in y and z direction Net gain of mass Since there is no accumulation of mass

12. Velocity function: Acceleration function:

13. Ex The velocity vector in a fluid flow is given. Find the velocity and acceleration of a fluid particle at (2, 1, 3) at time t=1. Ans V, u=32, v=-40, w=2, R=51.26: a, a x =1536, a y =320, a z =2, R=1568.9

14. Section II

15. Velocity Potential Function: it is define as a scalar function of space and time such that its negative derivative with respect to any direction gives the fluid velocity in that direction. It is defined by ø

16. Stream Function: it is define as a scalar function of space and time such that its partial derivation with respect to any direction gives the velocity component at right angles to that direction. It is defined by Ψ

17. Equipotential line: A line along which the velocity potential ø is constant, is called equipotential line. ø = Constant dΨ =0 dø = 0

18. Line of constant Stream Function: Ψ = constant dΨ =0

19. Flow Net: A grid obtained by drawing a series of equipotential lines and stream lines is called a flow net. Relation between stream function and velocity Potential function: and

20. Ex The velocity potential function is given by Calculate the velocity components at the point (4, 5). Ans u = -40 units, v =50 units

21. Ex A Stream function is given by Calculate the velocity components and also magnitude and direction of the resultant velocity at any point. Ans u = 6 units/sec, v =5 units/sec, R =7.81 unit/s, θ=39 48

22. Types of Motion: 1)Linear Translation or Pure Translation: It is define as the movement of a fluid element in such a way that it moves bodily from one position to another position and the two axes ab and cd represented in new position by a’b’ and c’d’.

23. 2) Linear deformation: It is define as the deformation of a fluid element in linear direction when the element moves. 3) Angular deformation or shear deformation: It is define as the average change in angle contained by two adjacent sides.

24. 4) Rotation: it is defined as the movement of a fluid element in such a way that both of it’s axes (H & V) rotate in the same direction.

25. Vortex flow: Vortex flow is defined as the flow of a fluid along a curved path or the flow of a rotating mass of fluid is known a ‘Vortex flow’. 1) Force vortex flow: it is defined as that type of vortex flow, in which some external torque is required to rotate the fluid mass. The fluid mass in this type of flow, rotates at constant angular velocity, ω The tangential velocity of any fluid particle is given by v = ω x r

26. Examples of forced vortex: 1.A Vertical cylinder containing liquid which is rotated about its central axis with a constant angular velocity ω 2.Flow of liquid inside the impeller of a centrifugal pump 3.Flow of water through the runner of turbine.

27. 2) Free vortex flow: When no external torque is required to rotate the fluid mass, that type of flow is called free vortex flow. Example 1)Flow of liquid through a hole provided at the bottom of a container 2)Flow of liquid around a circular bend in a pipe 3)A whirlpool in river 4)Flow of fluid in a centrifugal pump casing Angular momentum = Mass x Velocity = m x v Moment of momentum = Momentum x r = m v r Time rate of change of angular momentum = For free vortex mvr = Constant

28. Ex A fluid is given by V = 8x 3 i-10x 2 yj. Find the shear strain rate and state whether the flow is rotational or irrotational. Ans Shear strain= -10xy, W z =-10xy

29. Ex The velocity components in two-dimensional flow are u=y 3 /3 +2x-x 2 y v = xy 2 -2y-x 3 /3 Show that these components represent a possible case of an irrotational flow. Two-dimensional flow:

30. Prepared, Prepared by, Dr Dhruvesh Patel Prepared, Prepared by, Dr Dhruvesh Patel