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 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 - 6-
6.0 Introduction Three basic tools for the solution of fluid flow problems Impulse - momentum equation Continuity principle Work-energy principle Impulse - momentum equation ~ derived from Newton's nd law in vector form F ma Multifly by dt F dt madt d( mv c ) F d ( mvc) dt where vc velocity of the center of mass of the system of mass mvc linear momentum 6-
m sys dm v c m sys vdm ( Fdt ) impulse in time dt - 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 - Shear stress is not explicitly included - This equation will apply equally well to real fluids as well as ideal fluids. - 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) 6-3
6. The Linear Impulse Momentum Equation 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 vdv dt dt (a) Sum them all d d F vdv vdv ext dt dt sys sys Use Reynolds Transport Theorem to evaluate RHS i v for momentum/mass d dt de vdv ivda vvda vvda dt sys c.. s c.. s out c.. s in (b) where E = momentum of fluid system in the control volume 6-4
iv momentum per unit mass Ch. 6 The Impulse-Momentum Principle Because the streamlines are straight and parallel at Sections and, velocity is constant over the cross sections. The cross-sectional area is normal to the velocity vector over the entire cross section. In Eq. (b) vv da v v nda vvda V Q c. s. out c. s. out c. s. out v Q vvda csinvv ndavq csin.... v Q Q Q By Continuity eq: R. H. S. of (b) Q V V (c) Substitute (c) into (a) F Q V V (6.) In -D flow, Fx Q Vx Vx (6.a) Fz Q Vz Vz (6.b) 6-5
General form in case momentum enters and leaves the control volume at more than one location: F Qv Qv 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. - 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. 6-6
6. Pipe Flow Applications Forces exerted by a flowing fluid on a pipe bend, enlargement, or contraction in a pipeline may be computed by a application of the impulse-momentum principle. The reducing pipe bend Known: flowrate, Q; pressures,, p p ; velocities, v, v Find: F (equal & opposite of the force exerted by the fluid on the bend) = force exerted by the bend on the fluid 6-7
Pressures: For streamlines essentially straight and parallel at section and, the forces F, and F result from hydrostatic pressure distributions. If mean pressure p and p are large, and the pipe areas are small, then F pa and F p A, and assumed to act at the centerline of the pipe instead of the center of pressure. Body forces: = total weight of fluid, W Force exerted by the bend on the fluid, F = resultant of the pressure distribution over the entire interior of the bend between sections &. ~ distribution is unknown in detail ~ resultant can be predicted by Impulse-momentum Eq. Now apply Impulse-momentum equation, Eq. (6.) (i) x-direction: 6-8
F p A p A F x cos x cos x x Q V V Q V V Combining the two equations to develop an expression for F x F p A p A cos Q( V V cos ) (6.4.a) x (ii) z-direction F sin z W pa F z sin 0 z z Q V V Q V F W p A sin QV sin (6.4.b) z If the bend is relatively sharp, the weight may be negligible, depending on the magnitudes of the pressure and velocities. 6-9
[IP 6.] 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 0.085 m 3. 590.6 N Given: 3 Q 300 l s 0.3 m s ; Vol. of bend 0.085 m 3 A (0.3) 0.07 m 4 ; A (0.) 0.03 m 4 3 p 70 kpa 70 0 N m ) Continuity Eq. Q AV AV (4.5) 0.3 V 4.4 m/s 0.07 6-0
0.3 V 9.55 m/s 0.03 ) Bernoulli Eq. between and p V p V g g z z 3 70 0 (4.4) p (9.55) 0.5 9,800 (9.8) 9,800 (9.8) p 8.8 kpa 3) Momentum Eq. Apply Eqs. 6.4a and 6.4b F p A p A cos Q( V V cos ) x F W p A sin QV sin z F pa 4948 N F 3 pa 8.80 0.03 590.6 N W (volume) 98000.085 833 N 6-
F 4,948 (590.6) cos0 (9980.3)(4.4 9.55cos0 ) 7,94 N x F 833 (590.6)sin0 (9980.3)(9.55sin0 0) 3,80 N z F F F 8,83 N x z F F z tan 5.7 x 6-
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. Energy loss Consider the control surface ABCD assuming a one-dimensional flow i) Continuity Q AV AV Result from hydrostatic pressure distribution over the area ii) Momentum For area AB it is an approximation because of the dynamics of eddies in the dead water zone. F p A p A Q( V V) x 6-3
VA ( p p) A ( V V) g p p V ( V V ) g (a) iii) Bernoulli equation p V p V H g g p p V V H g g (b) where H Borda-Carnot Head loss Combine (a) and (b) V( V V) V V H g g g V VV V V ( V V ) H g g g g 6-4
6.3 Open Channel Flow Applications Applications impulse-momentum principle for Open Channel Flow - Computation of forces exerted by flowing water on overflow or underflow structures (weirs or gates) - Hydraulic jump - Wave propagation [Case ] Sluice gate Shear force is neglected Consider a control volume that has uniform flow and straight and parallel streamlines at the entrance and exit 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 6-5
F Q( V V ) x Discharge per unit width F F F F Q( V V ) q V V x x x x where q Q W discharge per unit width yv yv Assume that the pressure distribution is hydrostatic at sections and, replace V with q/y y y Fx q ( ) (6.6) y y [Re] Hydrostatic pressure distribution y y F h A y c ( ) 3 ( y) Ic l p l c y la y c ( y 6 ) Cp y y y 6 3 6-6
For ideal fluid (to a good approximation, for a real fluid), the force tnagent to the gate is zero. shear stress is neglected. Hence, the resultant force is normal to the gate. F F cos 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 Q( V V ) z z z F F W F Q(0 0) z OB z F W F z OB Non-uniform distribution pressure 6-7
[IP 6.] For the two-dimensional overflow structure, calculate the horizontal component of the resultant force the fluid exerts on the structure Lift gate Ideal fluid Continuity Eq. q5v V (4.7) Bernoulli's equation between () and () 0 5m V V 0 m g g (5.7) Combine two equations V 3.33 m s V 8.33 m s 3 q 5(3.33) 6.65 m s m 6-8
Hydrostatic pressure principle 3 9.8 kn m y (5) F hc A y 9.8.5 kn m () F 9.8 9.6 kn m Impulse-Momentum Eq. ( 3 000 kg m ) F,500 F 9,600 (0006.65)(8.33 3.33) x x F 9.65 kn m x [Cf] What is the force if the gate is closed? 6-9
Jamshil submerged weir (Seo, 999) Jamshil submerged weir with gate opened (Q = 00 m 3 /s) 6-0
Bucket roller Jamshil submerged weir Model Test; Q = 00 m 3 /s (Seo, 999) Jamshil submerged weir Model Test (Q = 5,000 m 3 /s) 6-
[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 large head loss (energy dissipation) Head loss due to hydraulic jump turbulence, eddying, air entrainment, surface undulation in the hydraulic jump Neglect shear force Apply impulse-momentum equation to find the relation between the depths for a given flowrate Construct a control volume enclosing the hydraulic jump between two sections and where the streamlines are straight and parallel y y Fx FF q( V V) where q flowrate per unit width 6-
Substitute the continuity relations V q y ; V q y Rearrange (divide by ) q y q y gy gy Solve for y y y 8 8 q V 3 y gy gy (6.7) Set Fr V gy Fr V gy William Froude (80~879) where Fr Froude number Inertia Force Gravity Force V gy 6-3
Then, we have y y 8Fr Jump Equation (a) Fr : critical flow y y 8 y y No jump (b) Fr : super-critical flow y y yy Hydraulic jump (c) Fr : sub-critical flow y y y 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. 6-4
[IP 6.3] p. 99 ; Water flows in a horizontal open channel. y 0.6 m 3 q 3.7 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 Fr V 6.7.54 hydraulic jump occurs gy 9.8(0.6) (ii) Jump Eq. y y 8Fr 0.6 8(.54).88 m 6-5
V 3.7.97 m s.88 (iii) Bernoulli Eq. (Work-Energy Eq.) V V y y E g g (6.7) (.97) 0.6.88 E (9.8) (9.8) E 0.46 m Power QE 98003.70.46 6.7 kw meter of width The hydraulic jump is excellent energy dissipator (used in the spillway). 6-6
Pulsating jump E/E ~ 85% 6-7
[Case 3] 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 For the steady flow, assign the velocity under the wave as a From continuity 6-8
' ay a y dy From impulse-momentum y y dy ' ay a (6.a) Combining these two equations gives a g y dy Letting dy approach zero results in a gy (6.8) The celerity of the samll gravity wave depends only on the depth of flow. 6-9
6.4 The Angular Impulse-Momentum Principle The angular impulse-momentum equation can be developed using moments of the force and momentum vectors V z V V x (x, z ) (x, z ) Fig. 6.8 0 Take a moment of forces and momentum vectors for the small individual fluid system about d d r F ( r mv) ( r dvolv) dt dt Sum this for control volume d rfext ( rv) dvol. dt sys (a) 6-30
Use Reynolds Transport Theorem to evaluate the integral de d ( r v) dvol. iv da dt dt sys C. S. ( r v ) v da ( rv ) vda CSout.. CSin.. irv (b) where E moment of momentum of fluid system irv = 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 dt ( rv) dvol. ( rv) dq ( rv) dq sys C. S. out C. S. in Because velocity is uniform over the flow cross sections d dt sys ( rv) dvol. Q( r V ) Q( r V ) out out in in Q ( rv) out ( rv) in (c) where r position vector from the moment center to the centroid of entering or leaving flow cross section of the control volume 6-3
Substitute (c) into (a) ( rf ) M Q ( rv) ( rv) ext 0 out in (6.3) In -D flow, M Q( rv rv ) (6.4) 0 t t where Vt 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 Q ( zv xv ) ( zv xv ) (6.5) 0 x z x z where x, z coordinates of centroid of the entering cross section x, z coordinates of centroid of the leaving cross section For the fluid that enter or leave the control volume at more than one cross-section, M 0 ( QrVt) out( QrVt) in (6.6) 6-3
[IP 6.6] Compute the location of the resultant force exerted by the water on the pipe bend. 590 N 8,83 N 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 M Q ( z v x v ) ( zv xv ) 0 x x x x x 0, z 0, x 0.6, z.5 For this case, T r(8,83) 0.55(833).5(590cos60 ) 0.6( 590sin 60 ) (0.3998).5( 9.55 cos60 ) 0.6(9.55sin 60 ) r 0.59 m 6-33
[Re] Torque for rotating system d T ( rf) ( rmvc) dt Where T torque Tdt torque impulse rmv c angular momentum (moment of momentum) r radius vector from the origin 0 to the point of application of a force [Re] Vector product (cross product) V FG -Magnitude: V F G sin -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 z plane. 6-34
Homework Assignment # 6 Due: week from today Prob. 6. Prob. 6.6 Prob. 6.4 Prob. 6.6 Prob. 6.30 Prob. 6.34 Prob. 6.36 Prob. 6.40 Prob. 6.55 Prob. 6.60 6-35