Ch 5 Momentum and forces in fluid flow

Transcription

1 Ch 5 Momentum and forces in fluid flow

2 Outline Momentum equation Applications of the momentum equation

3 Objectives After completing this chapter, you should be able to Identify the various kinds of forces and moments acting on a control volume. Use control volume analysis to determine the forces associated with fluid flow. Use control volume analysis to determine the moments caused by fluid flow.

4 6.1 Development of the Momentum Principle Start with a modified-form of Newton s 2 nd law: r Σ F = F r mv r d r ( mv ) dt = sum of forces on fluid system = linear momentum of system r r ΣF dt = d( mv ) Also, Impulse-momentum principle s

5 6.1 Development of the Momentum Principle Momentum entering = ρa δ 1u1 tu 1 Momentum leaving = ρa δ 2u 2 tu 2 Force = rate of change of momentum F = ( 1 ρ A2u 2δ tu2 ρa1 u1δ tu ) δt Fig 6.1 Momentum in a flowing fluid Q= A 1 u 1 = A 2 u 2, F = ρq( u 2 u1)

6 Momentum equation for twoand three-dimensional flow The force in the x-direction Fig 6.2 Two dimensional flow in a streamtube

7 Momentum equation for twoand three-dimensional flow The force in the y-direction Fig 6.2 Two dimensional flow in a streamtube

8 Momentum equation for twoand three-dimensional flow Total force exerted on the fluid r F = Rate of change of momentum through the control volume r r = ρq( u ) 2 u1 For steady flow with one inlet and one outlet, the momentum equation is r F r r = ρq( β u β ) u β momentum correction factor

9 Momentum correction factor True momentum per unit time = β Mass per unit time Mean velocity = β u 2 da/ V 2 A The momentum equation r F r r = ρq( β V β ) V

10 the resultant force F = F x F y the angle which this force acts at is given by

11 What are the forces acting on the fluid in the control volume? the total force, F T, is given by the sum of these forces r F T r = F R r + F B r + F P

12 Step in Analysis with Momentum Equation 1. Draw a control volume: Based on the problem, selecting the stream between two gradually varied flow sections as the control volume; 2. Decide on co-ordinate axis system: Determining the directions of co-ordinate axis, magnitudes and directions of components of all forces and velocities on each axis. 3. Plotting diagram for computation : Analyzing the forces on control volume and plotting the directions of all forces on the control volume. 4. Writing momentum equation and solving it: Substituting components of all forces and velocities on axes into momentum equation and solving it. All the pressures are relative to the relative pressure.

13 Application of the Momentum Equation 1. Force due to the flow of fluid round a pipe bend. 2. Force on a nozzle at the outlet of a pipe. 3. Impact of a jet on a plane surface. 4. Force due to flow round a curved vane.

14 Force due to Jet Striking Surface

15 Force by Flow Round a Pipe-bend

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17 Example 1 Find the horizontal thrust of the water on each meter of width of the sluice gate shown in the Fig., given y 1 =2.2 m, y 2 =0.4 m, and y 3 =0.5 m. Neglect friction. ( 6.4.2)

18 Solution

19 6.5.5

20 Solution The flow rate Momentum equation

21 6.8 A reducing right-angled bend lies in a horizontal plane. Water enters from the west with a velocity of 3 m/s and a pressure of 30 kpa, and it leaves toward the north. The diameter at the entrance is 500 mm and at the exit it is 400 mm. Neglecting any friction loss, find the magnitude and direction of the resultant force on the bend.

22 Solution Energy Momentum

23 Example Water flows through a reducing 180 bend. The bend is shown in plan. Determine the magnitude of the force exerted on the bend in the x-direction. Assume energy losses to be negligible.

24 Solution:

25 Given: Figure Find: Horizontal force required to hold plate in position Solution: Example T=15 o C Q=0.4 m 3 /s B F pa γ + z A 2 V + A 2g = pb γ + z B 2 V + B 2g p A =75 kpa r r V = Vi pa γ V B 2 V = B 2g p = 2 A ρ = 2*75000/ 999 = 12.3m / s F F = ρq( V V ) 1 2 = ρqv = 999*0.4*12.3 = 4.9kN

26 question If the value of force calculated from momentum equation is negative, what does that mean? Does the magnitude of the unknown force has anything or nothing to do with that of the control volume? How to select control volume in the application?

27 If the value of force calculated from momentum equation is negative, what does that mean? Does the magnitude of the unknown force has anything or nothing to do with that of the control volume? How to select control volume in the application? directions are inverse;independency(when there is no gravitation);calculated crosssection and solid wall

28 Find: Force due to pressure on face of gate Solution: Assume: v 1 and v 2 are uniform (so pressure is hydrostatic) Sluice Gate

29 Application of the Energy, Momentum, and Continuity Equations in Combination In general, when solving fluid mechanics problems, one should use all available equations in order to derive as much information as possible about the flow. For example, consistent with the approximation of the energy equation we can also apply the momentum and continuity equations

30 Example Forces on Transitions

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33 Example: Energy Equation (energy loss) An irrigation pump lifts 50 L/s of water from a reservoir and discharges it into a farmer s irrigation channel. The pump supplies a total head of 10 m. How much mechanical energy is lost? What is h L? cs 2 cs m 2 m 4 m datum Why can t I draw the cs at the end of the pipe? p V p V g 2g g 2g 2 2 in in out out + zin + a in + hp = + zout + a out + ht + hl h p = zout + h h L L = h p - zout

34 Example: Energy Equation (pressure at pump outlet) The total pipe length is 50 m and is 20 cm in diameter. The pipe length to the pump is 12 m. What is the pressure in the pipe at the pump outlet? You may assume (for now) that the only losses are frictional losses in the pipeline. 50 L/s h P = 10 m cs 4 m 2.4 m 2 2 m cs 1 datum p V p V g 2g g 2g 2 2 in in out out + zin + a in + hp = + zout + a out + ht + hl We need in the pipe,, and.

35 Example: Energy Equation (pressure at pump outlet) How do we get the velocity in the pipe? How do we get the frictional losses? What about α?

36 Kinetic Energy Correction Term: α α is a function of the velocity distribution in the pipe. For a uniform velocity distribution For laminar flow For turbulent flow Often neglected in calculations because it is so close to 1

37 Example: Energy Equation (pressure at pump outlet) V = 1.6 m/s α = 1.05 h L = 1.44 m 2.4 m 2 m p V h z h g 2 g h P = 10 m 2 out out P = + out + a out + L 4 m 50 L/s datum p (1.6m/s) = (9810N/m ) (10m) (2.4m) (1.05) (1.44m) 2 2(9.81m/s )

38 Example: Energy Equation (Hydraulic Grade Line - HGL) We would like to know if there are any places in the pipeline where the pressure is too high ( ) or too low (water might boil - cavitation). Plot the pressure as piezometric head (height water would rise to in a piezometer) How?

39 Example: Energy Equation (Energy Grade Line - EGL) p = 59 kpa Loss due to shear Entrance loss 2.4 m 2 m H P = 10 m p g + a V 2 2 g 4 m Exit loss 50 L/s datum What is the pressure at the pump intake? p V p V g 2g g 2g 2 2 in in out out + zin + a in + hp = + zout + a out + ht + hl

40 EGL (or TEL) and HGL EGL 2 p V = + z + α p γ 2 g HGL = + z γ What is the difference between EGL defined by Bernoulli and EGL defined here?

41 EGL (or TEL) and HGL The energy grade line may never be horizontal or slope upward (in direction of flow) unless energy is added ( ) The decrease in total energy represents the head loss or energy dissipation per unit weight EGL and HGL are and lie at the free surface for water at rest (reservoir) Whenever the HGL falls below the point in

42 Example HGL and EGL velocity head α V 2 2g pressure head p γ energy grade line hydraulic grade line z = 0 pump z elevation p V p V g 2g g 2g datum 2 2 in in out out + zin + a in + hp = + zout + a out + ht + hl

43 See you next time.