The Impulse-Momentum Principle

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1 Chapter 6 /60 The Impulse-Momentum Principle F F

2 Chapter 6 The Impulse-Momentum Principle /60 Contents 6.0 Introduction 6. The Linear Impulse-Momentum Equation 6. Pipe Flow Applications 6.3 Open Channel Flow Applications 6.4 The Angular Impulse-Momentum Principle Objectives -Develop impulse - momentum equation, the third of three basic equations of fluid mechanics, added to continuity and work-energy principles -Develop linear and angular momentum (moment of momentum) equations

3 6.0 Introduction 3/60 Three basic tools for the solution of fluid flow problems Continuity principle Work-energy principle (Bernoulli equation) Impulse - momentum equation (Momentum equation) Impulse - momentum equation ~ derived from Newton's nd law in vector form F = ma Multiply by dt Σ F dt = madt = d mv c d F = ( mvc) dt ( ) ( )

4 6.0 Introduction 4/60 where v = c mv c = m v c = velocity of the center of mass of the system of mass sys = m linear momentum dm sys vdm ( F) dt = impulse in time dt

5 6.0 Introduction 5/60 - Define the fluid system to include all the fluid in a specified control volume whereas the Euler equations was developed for a small fluid system - Restrict the analysis to steady flow - This equation will apply equally well to real fluids as well as ideal fluids even though shear stress is not explicitly included. - Develop linear and angular momentum (moment of momentum) equations - Linear momentum equation: calculate magnitude and direction of resultant forces - Angular momentum equation: calculate line of action of the resultant forces, rotating fluid machinery (pump, turbine)

6 6. The Linear Impulse Momentum Equation 6/60

7 6. The Linear Impulse Momentum Equation 7/60 Use the same control volume previously employed for conservation of mass and work-energy. For the individual fluid system in the control volume, d d F = ma = mv vdvol (a) dt = dt ρ Sum them all F d vdvol d vdvol ext = = dt dt sys sys ( ρ ) ( ρ ) E = i dm = i ρ dvol system Use Reynolds Transport Theorem for steady flow to evaluate RHS de d = iρ dvol = ( ) + system iρdvol c.. v iρv da dt dt t c.. s Steady flow system

8 6. The Linear Impulse Momentum Equation 8/60 d dt de ( ρvdv ) = = iρv da = ρv ( v da) + ρv ( v da) dt sys cs.. csout.. csin.. i = v for momentum/mass (b) where E = momentum of fluid system in the control volume i = v = momentum per unit mass Because the streamlines are straight and parallel at Sections and, velocity is uniform over the cross sections. The cross-sectional area is normal to the velocity vector over the entire cross section. Thus, integration of terms in Eq. (b) are written as

9 6. The Linear Impulse Momentum Equation 9/60 v ( ρv da) ρv = v n da = ρv vda = ρ V Q c. s. out c. s. out c. s. out v Q v ( ρv da) = c s in ρv v n da = ρv Q c. s. in.. v Flux in through Section Flux out through Section By Continuity eq: RHS of (b) Qρ = Qρ = Qρ = Qρ V V ( ) (c) Substitute (c) into (a) ( ) F = Q ρ V V (6.)

10 6. The Linear Impulse Momentum Equation 0/60 In -D flow, ( ) F = Qρ V V x x x ( ) F = Qρ V V z z z (6.a) (6.b) General form in case momentum enters and leaves the control volume at more than one location: F = Qρv Qρv ( ) ( ) out in (6.3) - The external forces include both normal (pressure) and tangential (shear) forces on the fluid in the control volume, as well as the weight of the fluid inside the control volume at a given time.

11 6. The Linear Impulse Momentum Equation /60 Advantages of impulse-momentum principle ~ Only flow conditions at inlets and exits of the control volume are needed for successful application. ~ Detailed flow processes within the control volume need not be known to apply the principle.

12 6. Pipe Flow Applications /60 Forces exerted by a flowing fluid on a pipe bend, enlargement, or contraction in a pipeline may be computed by an application of the impulse-momentum principle. Case : The reducing pipe bend Knowns: flowrate, Q ; pressures, ; velocities, Find: force exerted by the bend on the fluid, F = equal & opposite of the force exerted by the fluid on the bend) p, p V, V

13 6. Pipe Flow Applications 3/60

14 6. Pipe Flow Applications 4/60 Pressures: For streamlines essentially straight and parallel at Section and, the forces F, and F result from hydrostatic pressure distributions. F F = pa = pa and p p If mean pressure and are large, and the pipe areas are small, then instead of the center of pressure., and assumed to act at the centerline of the pipe h c h p [Cf] Resultant force F = γ ha (.): c p c = γ h c

15 6. Pipe Flow Applications 5/60 Body forces: = total weight of fluid, W Force exerted by the bend on the fluid, = resultant of the pressure distribution over the entire interior of the bend between Sections and. ~ distribution is unknown in detail ~ resultant force can be predicted by impulse-momentum eq. F

16 6. Pipe Flow Applications 6/60 Now apply impulse-momentum equation, Eq. (6.) (i) x-direction: F = pa pa α F x cos ( x x ) = ( cos ) Qρ V V Qρ V α V x (a) (b) Combining the two equations to develop an expression for F x pa pacos α F = Qρ( V cos α V) x F = pa pacos α + Qρ( V V cos α) (6.4a) x

17 6. Pipe Flow Applications 7/60 (ii) z-direction F sin z = W pa α + F ( ) ( sin 0 z z = ) Qρ V V Qρ V α W pasinα + F= QρVsinα z z F= W+ pasinα + QρVsinα z (6.4b) [IP 6.] p.93 Water flow through vertical reducing pipe bend 300 l/s of water flow through the vertical reducing pipe bend. Calculate the force exerted by the fluid on the bend if the volume of the bend is m 3.

18 6. Pipe Flow Applications 8/60 F x F z

19 6. Pipe Flow Applications 9/60 Given: 3 3 Q = 300 l s = 0.3 m s; Vol. of bend = m A π π = (0.3) = 0.07 m ; A = (0.) = 0.03 m p = 70 kpa = 70 0 N m Now, we apply three equations to solve this problem. ) Continuity Eq. Q = AV = AV 0.3 V = = 4.4 m/s V = = 9.55 m/s 0.03 (6.5)

20 6. Pipe Flow Applications 0/60 ) Bernoulli Eq. between and p V p V γ g γ g + + z z (4.4) p (9.55) = ,800 (9.8) 9,800 (9.8) p = 8.8 kpa 3) Momentum Eq. Apply Eqs. 6.4a and 6.4b F = pa pacos α + Qρ( V V cos α) x F= W+ pasinα + QρVsinα z

21 6. Pipe Flow Applications /60 F = pa = 4,948 N F 3 = pa = = N W =γ (volume) = = 833 N F = 4,948 (590.6) cos0 + ( )( cos0 ) = 7,94 N x F = (590.6)sin0 + ( )(9.55sin0 0) = 3,80 N z F = F + F = 8,83 N x z F θ = = F z tan 5.7 x

22 6. Pipe Flow Applications /60 Case : Abrupt enlargement in a closed passage ~ real fluid flow The impulse-momentum principle can be employed to predict the fall of the energy line (energy loss due to a rise in the internal energy of the fluid caused by viscous dissipation) at an abrupt axisymmetric enlargement in a passage. Consider the control surface ABCD assuming a one-dimensional flow i) Continuity Eq. Q = AV = AV

23 6. Pipe Flow Applications 3/60

24 6. Pipe Flow Applications 4/60 ii) Momentum Eq. Result from hydrostatic pressure distribution over the area For area AB it is an approximation because of the dynamics of eddies in the dead water zone. F = pa pa = Qρ( V V) x VA ( p p) A = ( V V) g γ p p V = ( V V ) γ g iii) Bernoulli Eq. p V p V H γ + g = γ + g + p p V V = + H γ g g (a) (b)

25 6. Pipe Flow Applications 5/60 where H = Borda-Carnot head loss Jean-Charles de Borda (733~799): French mathematician Nicolas Leonard Sadi Carnot (796~83): French Combine (a) and (b) military engineer V( V V) V V = + H g g g V VV V V ( V V ) H = + = g g g g

26 6.3 Open Channel Flow Applications 6/60 Applications impulse-momentum principle for open channel flows: - computation of forces exerted by flowing water on overflow or underflow structures (weirs or gates) - hydraulic jump - wave propagation Case : Flow under the sluice gate Consider a control volume that has uniform flow and straight and parallel streamlines at the entrance and exit

27 6.3 Open Channel Flow Applications 7/60

28 6.3 Open Channel Flow Applications 8/60 Apply first Bernoulli and continuity equations to find values of depths y and y and flowrate per unit width q Then, apply the impulse-momentum equation to find the force the water exerts on the sluice gate F = Qρ( V V ) x Discharge per unit width ( ) F = F F F = Qρ( V V ) = qρ V V x Q q = = W x where discharge per unit width = yv = yv x x

29 6.3 Open Channel Flow Applications 9/60 Assume that the pressure distribution is hydrostatic at Sections and, and replace V with q/y γy γy Fx = q ρ( ) y y (6.6) [Re] Hydrostatic pressure distribution y γ y F = γha c = γ ( y ) = 3 ( y) Ic l p l = c y la = y = c ( y 6 ) cp = y y = y 6 3

30 6.3 Open Channel Flow Applications 30/60 For ideal fluid (to a good approximation for a real fluid), the force tangent to the gate is zero. shear stress is neglected. Hence, the resultant force is normal to the gate. F = F cosθ x We don t need to apply the impulse-momentum equation in the z-direction. [Re] The impulse-momentum equation in the z-direction F = Qρ( V V ) z z z F = F W F = Qρ(0 0) z z OB z F = W F OB Non-uniform pressure distribution

31 6.3 Open Channel Flow Applications 3/60 Case : The two-dimensional overflow structure [IP 6.] p.97 Calculate the horizontal component of the resultant force the fluid exerts on the structure Continuity Eq. q= 5V = V (6.7) Bernoulli's equation between () and () 0 + 5m + V V 0 m g = + + g (6.8)

32 6.3 Open Channel Flow Applications 3/60

33 6.3 Open Channel Flow Applications 33/60 Combine two equations V = 3.33 m s V = 8.33 m s 3 q = 5(3.33) = 6.65 m s m Hydrostatic pressure principle 3 ( γ=9.8 kn m ) y (5) F = γha c = γ y= 9.8 =.5 kn m () F = 9.8 = 9.6 kn m

34 6.3 Open Channel Flow Applications 34/60 ρ =,000 kg m Impulse-Momentum Eq. ( ) F =,500 F 9,600 = (, )( ) x F = 9.65 kn m x x 3 [Cf] What is the force if the gate is closed?

35 6.3 Open Channel Flow Applications 35/60

36 6.3 Open Channel Flow Applications 36/60 Bucket roller

37 6.3 Open Channel Flow Applications 37/60 Case : Hydraulic Jump When liquid at high velocity discharges into a zone of lower velocity, a rather abrupt rise (a standing wave) occurs in water surface and is accompanied by violent turbulence, eddying, air entrainment, surface undulation. such as a wave is known as a hydraulic jump induce a large head loss (energy dissipation)

38 6.3 Open Channel Flow Applications 38/60

39 6.3 Open Channel Flow Applications 39/60 Apply impulse-momentum equation to find the relation between the depths for a given flowrate Construct a control volume enclosing the hydraulic jump between Sections and where the streamlines are straight and parallel γ y γ y Fx = F F = = qρ( V V) where q = flowrate per unit width Substitute the continuity relations V q y ; = V q = y

40 6.3 Open Channel Flow Applications 40/60 γ Rearrange (divide by ) Solve for q y q y + = + gy gy y y y 8 8 q V = = + + y gy gy Set Fr = V gy Fr = V gy

41 6.3 Open Channel Flow Applications 4/60 where Fr = Froude number = Inertia Force Gravity Force = V gy Then, we have William Froude (80~879) y y = + + 8Fr Jump Equation (a) Fr = : critical flow y y = + + = 8 y y = No Jump

42 6.3 Open Channel Flow Applications 4/60 (b) (c) Fr > Fr < : super-critical flow y y > y : sub-critical flow y y < y > y hydraulic jump < y physically impossible ( rise of energy line through the jump is impossible) Conclusion: For a hydraulic jump to occur, the upstream conditions must be such that V gy >

43 6.3 Open Channel Flow Applications 43/60 [IP 6.3] p. 99 Water flows in a horizontal open channel. y = 0.6 m q = m s m Find y, and power dissipated in hydraulic jump. [Sol] (i) Continuity q= yv = yv 3.7 V = = 6.7 m s 0.6 V 6.7 Fr = = =.54> hydraulic jump occurs gy 9.8(0.6)

44 6.3 Open Channel Flow Applications 44/60 (ii) Jump Eq. y y = + + 8Fr 0.6 = + + 8(.54) =.88 m V 3.7 = = ms

45 6.3 Open Channel Flow Applications 45/60 (iii) Bernoulli Eq. (Work-Energy Eq.) V V y+ = y + + E g g (6.7) (.97) = E (9.8) (9.8) E = 0.46 m Power = γ Q E = 9,800( 3.7)( 0.46) = 6.7 kw meter of width The hydraulic jump is excellent energy dissipater (used in the spillway).

46 6.3 Open Channel Flow Applications 46/60

47 6.3 Open Channel Flow Applications 47/60

48 6.3 Open Channel Flow Applications 48/60 Case 4: Wave Propagation The velocity (celerity) of small gravity waves in a body of water can be calculated by the impulse-momentum equation. Small gravity waves ~ appears as a small localized rise in the liquid surface which propagate at a velocity a ~ extends over the full depth of the flow [Cf] small surface disturbance (ripple) ~ liquid movement is restricted to a region near the surface

49 6.3 Open Channel Flow Applications 49/60

50 6.3 Open Channel Flow Applications 50/60 For the steady flow, assign the velocity under the wave as a From continuity ( ) ' ay = a y + dy From impulse-momentum ( y + dy) γ y γ = ' ( ay) ρ ( a a) Combining these two equations gives ( ) a = g y + dy Letting dy approach zero results in a = gy The celerity of the samll gravity wave depends only on the depth of flow.

51 6.4 The Angular Impulse-Momentum Principle 5/60 The angular impulse-momentum equation can be developed using moments of the force and momentum vectors Take a moment of forces and momentum vectors for the small individual fluid system about 0 d d r F = ( r mv) ( r d Vol v) dt = dt ρ Sum this for control volume d r Fext = ( r v) ρd Vol. dt sys (6.9) (a)

52 6.4 The Angular Impulse-Momentum Principle 5/60 (x, z ) (x, z )

53 6.4 The Angular Impulse-Momentum Principle 53/60 Use Reynolds Transport Theorem to evaluate the integral de d = ( r v) ρd Vol. iρv da dt dt = sys C. S. ( r v ) v da ( = ρ + r v ) ρv da C. S. out C. S. in (6.0) (b) where E = i= r v moment of momentum of fluid system = moment of momentum per unit mass Restrict to control volume where the fluid enters and leaves at sections where the streamlines are straight and parallel and with the velocity normal to the cross-sectional area d ( r v) d Vol. ( r v) dq ( r v) dq dt ρ = sys ρ ρ C. S. out C. S. in (6.)

54 6.4 The Angular Impulse-Momentum Principle 54/60 Because velocity is uniform over the flow cross sections where d ( r v) ρdvol. = Qρ( rout Vout ) Qρ( rin Vin ) dt sys = Qρ ( r V) out ( r V) in r = (6.) position vector from the moment center to the centroid of entering or leaving flow cross section of the control volume Substitute (c) into (a) (c) ( r F ) = M = Qρ ( r V) ( r V) ext 0 out in (6.3)

55 6.4 The Angular Impulse-Momentum Principle 55/60 In -D flow, M = Qρ( rv rv ) (6.4) 0 t t where V t = component of velocity normal to the moment arm r. In rectangular components, assuming V is directed with positive components in both x and z-direction, and with the moment center at the origin of the x-z coordinate system, for clockwise positive moments, [( ) ( )] M = Qρ zv xv zv xv 0 x z x z (6.5) where x, z = x, z = coordinates of centroid of the entering cross section coordinates of centroid of the leaving cross section

56 6.4 The Angular Impulse-Momentum Principle 56/60 For the fluid that enter or leave the control volume at more than one cross-section, M 0 = ( QρrVt ) out ( QρrVt ) in (6.6) [IP 6.6] p. 06 Water flowing on the pipe bend Compute the location of the resultant force exerted by the water on the pipe bend. Assume that center of gravity of the fluid is 0.55 m to the right of section, and the forces F and F act at the centroid of the sections rather than at the center of pressure. Take moments about the center of section

57 6.4 The Angular Impulse-Momentum Principle 57/60

58 6.4 The Angular Impulse-Momentum Principle 58/60 [( ) ( )] M = Qρ z v xv zv xv For this case, 0 x x x x T = r(8,83) (833) +.5(590cos 60 ) 0.6( 590sin 60 ) = ( ).5( 9.55 cos60 ) 0.6(9.55 sin 60 ) r = 0.59 m [Re] Torque for rotating system d T = ( r F) = ( r mvc) dt

59 6.4 The Angular Impulse-Momentum Principle 59/60 Where r = T = T dt = r mv c = torque torque impulse angular momentum (moment of momentum) radius vector from the origin 0 to the point of application of a force [Re] Vector product (cross product) V = F G -Magnitude: V= FGsinφ -Direction: perpendicular to the plane of F and G (right-hand rule) If F, G are in the plane of x and y, then the V is in the plane. z

60 6.4 The Angular Impulse-Momentum Principle 60/60 Homework Assignment # 6 Due: week from today Prob. 6. Prob. 6.6 Prob. 6.4 Prob. 6.6 Prob Prob Prob Prob Prob Prob. 6.60

Chapter 6 The Impulse-Momentum Principle

Chapter 6 The Impulse-Momentum Principle 6. The Linear Impulse-Momentum Equation 6. Pipe Flow Applications 6.3 Open Channel Flow Applications 6.4 The Angular Impulse-Momentum Principle Objectives: - Develop