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Chapter 6: Momentum Analysis of Flow Systems

Chapter 6: Momentum Analysis of Flow Systems. Siti Kamariah Md Sa’at Pusat Pengajian Kejuruteraan Bioproses September, 2009. NEWTON’S LAWS. Newton’s laws are relations between motions of bodies and the forces acting on them.

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Chapter 6: Momentum Analysis of Flow Systems

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  1. Chapter 6: Momentum Analysis of Flow Systems Siti Kamariah Md Sa’at Pusat Pengajian Kejuruteraan Bioproses September, 2009

  2. NEWTON’S LAWS • Newton’s laws are relations between motions of bodies and the forces acting on them. • First law: a body at rest remains at rest, and a body in motion remains in motion at the same velocity in a straight path when the net force acting on it is zero. • Second law: the acceleration of a body is proportional to the net force acting on it and is inversely proportional to its mass. • Third law: when a body exerts a force on a second body, the second body exerts an equal and opposite force on the first.

  3. NEWTON’S LAWS AND CONSERVATIONOF MOMENTUM • For a rigid body of mass m, Newton’s second law is expressed as • Therefore, Newton’s second law can also be stated as the rate of change of the momentum of a body is equal to the net force acting on the body.

  4. NEWTON’S LAWS AND CONSERVATIONOF MOMENTUM • The product of the mass and the velocity of a body is called the linear momentum. • Newton’s second law  the linear momentum equation in fluid mechanics • The momentum of a system is conserved when it remains constant  the conservation of momentum principle. • Momentum is a vector. Its direction is the direction of velocity. = Momentum

  5. EXAMPLE 1-GRADUAL ACCELERATION OF FLUID IN PIPELINE Q: Water flows through a pipeline 60m long at velocity 1.8m/s when the pressure difference between the inlet and outlet ends is 25 kN/m. What increase of pressure difference is required to accelerate the water in pipe at rate 0.02 m/s2? • Solution: Lets A= cross-sectional of the pipe l = length of pipe ρ= mass density of water a = acceleration of water δp = increase in pressure at inlet required to produce acceleration

  6. EXAMPLE 1- GRADUAL ACCELERATION OF FLUID IN PIPELINE • Not is steady flow state, consider a control mass comprising the whole of the water in the pipe. By Newton’s Second Law: 1. Force due to δp,F = cross-sectional area x δp = A δp 2. Mass of water in pipe,m = Density x Volume = ρ x Al 3. A δp = ρAl a δp = ρl a = 103 x 60 x 0.02 N/m2 = 1.2 kN/m2

  7. LINEAR MOMENTUM EQUATION • Newton’s second law for a system of mass m subjected to a force F is expressed as • During steady flow, the amount of momentum within the control volume remains constant. • The net force acting on the control volume during steady flow is equal to the difference between the rates of outgoing and incoming momentum flows.

  8. LINEAR MOMENTUM EQUATION • Consider a stream tube and assume steady non-uniform flow A2 v2 ρ2 A1 v1 ρ1 v1

  9. LINEAR MOMENTUM EQUATION In time δt a volume of the fluid moves from the inlet at a distance v1δt, so volume entering the stream tube = area x distance = A1 x v1δt The mass entering, mass entering stream tube = volume x density = ρ1A1v1δt And momentum momentum entering stream tube = mass velocity = ρ1A1v1δt v1 Similarly, at the exit, we get the expression: momentum leaving stream tube = ρ2A2v2δt v2

  10. LINEAR MOMENTUM EQUATION By Newton 2nd law Force = rate of change of momentum F = (ρ2A2v2δt v2 - ρ1A1v1δt v1) δt We know from continuity that Q= A1v1 = A2v2 And if we have fluid of constant density, ρ1 = ρ2 = ρ, then F = Qρ (v2-v1)

  11. LINEAR MOMENTUM EQUATION An alternative derivation From conservation of mass mass into face 1 = mass out of face 2 we can write rate of change of mass = ˙ = dm/dt = ρ1A1v1 = ρ2A2v2 The rate at which momentum enters face 1 is ρ1A1v1 v1 = ˙ v1 The rate at which momentum leaves face 2 is ρ2A2v2 v2 = ˙ v2 Thus the rate at which momentum changes across the stream tube is ρ2A2v2 v2 - ρ1A1v1 v1 = ˙ v2 - ˙ v1 Force = rate of change of momentum F = ˙ (v2-v1) m m m m m m

  12. LINEAR MOMENTUM EQUATION • So, we know these two expression. Either one is known as momentum equation: F = ˙ (v2-v1) m F = Qρ (v2-v1) The momentum equation: This force acts on the fluidin the direction of the flow of the fluid

  13. Momentum-Flux Correction Factor, b Since the velocity across most inlets and outlets is not uniform, the momentum-flux correction factor, b, is used to patch-up the error in the algebraic form equation. Therefore, Momentum flux across an inlet or outlet: Momentum-flux correction factor:

  14. LINEAR MOMENTUM EQUATION-STEADY FLOW The net force acting on the control volume during steady flow is equal to the difference between the rates of outgoing and incoming momentum flows. Therefore, One inlet and one outlet

  15. LINEAR MOMENTUM EQUATION-STEADY FLOW ALONG COORDINATE The previous analysis assumed the inlet and outlet velocities in the same direction i.e. a one dimensional system. What happens when this is not the case? We consider the forces by resolving in the directions of the co-ordinate axes. v2 v1

  16. The force in x-direction Fx = ˙ (v2 cos θ2 – v1 cos θ1) = ˙ (v2x– v1x ) Or Fx = ρQ (v2 cos θ2 – v1 cos θ1) = ρQ (v2x– v1x ) The force in y-direction Fy = ˙ (v2 sin θ2 – v1 sin θ1) = ˙ (v2y– v1y ) Or Fy = ρQ (v2 sin θ2 – v1 sin θ1) = ρQ (v2y– v1y) The resultant foce can be found by combining these components LINEAR MOMENTUM EQUATION-STEADY FLOW ALONG COORDINATE m m m m

  17. SUMMARY Total Force on the fluid = rate of change of momentum through the control volume F =β ˙ (vout - vin) m F = βQρ (vout - vin) • Remember!!! • We are working with vectors so F is in the direction of the velocity

  18. FORCE ACTING ON CONTROL VOLUME • Consist of two forces: 1. Body forces - such as gravity, electric, magnetic force 2. Surface force - such as pressure, viscous and reaction force • Total force acting on control volume is sum of body force and surface force.

  19. FORCE ACTING ON CONTROL VOLUME Force is made up of these component: • FR = force exerted on the fluid by any solid body touching the control volume • FB = Force exerted on the solid body (eg gravity) • FP = Force exerted on the fluid by fluid pressure outside the control volume Total force is given by the sum of these three forces: FT = FR + FB + FP The force exerted by the fluid on the solid body touching the control volume is opposite to FR So, the reaction force, R is given by R = - FR

  20. APPLICATION OF THE MOMENTUM EQUATION FORCE DUE TO THE FLOW AROUND THE PIPE BEND • A converging pipe bend lying in the horizontal plane turning through an angle ofθ. A2 v2 ρ2 A1 v1 ρ1

  21. FORCE DUE TO THE FLOW AROUND THE PIPE BEND • Why do we want to know the forces here? • As the fluid changes direction a force will act on the bend. • This force can be very large in the case of water supply pipes. • The bend must be held in place to prevent breakage at the joints. • We need to know how much force a support (thrust block) must withstand.

  22. FORCE DUE TO THE FLOW AROUND THE PIPE BEND • Step in analysis • Draw a control volume • Decide on coordinate axis system • Calculate the total force • Calculate the pressure force • Calculate the body force • Calculate the resultant force

  23. EXAMPLE 2 –THE FORCE TO HOLD A DEFLECTOR ELBOW Q: A reducing elbow is used to deflect water flow at a rate of 14 kg/s in a horizontal pipe upward 30° while accelerating it.The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 113 cm2 at the inlet and 7 cm2 at the outlet. The elevation difference between the centers of the outlet and the inlet is 30 cm. The weight of the elbow and the water in it is considered to be negligible. Determine (a) the gage pressure at the center of the inlet of the elbow and (b) the anchoring force needed to hold the elbow in place.

  24. EXAMPLE 2 • Assumption 1.The flow is steady, and the frictional effects are negligible. 2. The weight of the elbow and the water in it is negligible. 3. The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4. The flow is turbulent and fully developed at both the inlet and outlet of the control volume, and we take the momentum-flux correction factor,βto be 1.03. Figure 2

  25. EXAMPLE 2 • Density,ρ = 1000 kg/m3 • The continuity equation for this one-inlet, one-outlet, steady-flow system is =14 kg/s. • Noting that , the inlet and outlet velocities of water are

  26. EXAMPLE 2 Step 1: Draw Control volume as Figure 2. • We take the elbow as the control volume and designate the inlet by 1 and the outlet by 2. Step 2: Decide on coordinate axis system • Take the x- and z-coordinates as shown. Step 3: Calculate the total force FTx = ˙ (v2 cos 30o- v1) β = 14 kg/s x (20cos 30o – 1.24 m/s) (1N/1kg.m/s2)(1.03) = 231.88 N FTz = ˙ v2 sin 30o β = 14 kg/s x 1.24 x 20sin 30 x (1N/1kg.m/s2)(1.03) = 144.2 N m m

  27. EXAMPLE 2 (a) Step 4: Calculate the pressure force • Using Bernoulli equation, pressure can be calculated if we know pressure in inlet/outlet. • Taking the center of the inlet cross section as the reference level (z1= 0) and noting that P2 = Patm, the Bernoulli equation for a streamline going through the center of the elbow is expressed as

  28. EXAMPLE 2 (b) • Calculate the pressure force: Fp = Pressure force at 1 – Pressure force at 2 Fpx = P1A1 cos 0 – P2A2 cosθ = P1A1 – P2A2 cos 30 = P1A1 = 202.2 kN/m2 x 0.0113 m2 = 2.285 kN Fpz = P1A1 sin 0 – P2A2 sin θ = – P2A2 sin 30 = 0 since P2 = Patm Step 5: Calculate the body force • There is no body force in x or z direction. • The body force is acting on y-direction. So not to be considered. • FBx = FBz = 0

  29. EXAMPLE 2 (b) Step 6: Calculate the resultant force • We let the x- and z-components of the anchoring force of the elbow be FRx and FRz, and assume them to be in the positive direction. • Solving for FRx and FRz, and substituting the calculated values, • FRx = FTx – FPx – FBx = 231.88 – 2285 - 0 = -2053.12 N FRZ = FTz –FPz – FBz = 144.2 -0 – 0 = 144.2 N • And the resultant force on the fluid is given by FR =√( FRx2 – FRx2) = √((2053.12)2 – (144.2)2) = 2048 N The force of the bend is the same magnitude but in opposite direction R = -FR = - 2048 N

  30. EXAMPLE 3- WATER JET STRIKING A STATIONARY PLATE Q: Water accelerated by a nozzle to an average speed of 20 m/s and strikes a stationary vertical plate at rate of 10 kg/s with a normal velocity 20 m/s. After the strike, the water stream splatters off in all directions in the plane of the plate. Determine the force needed to prevent the plate from moving horizontally due to the water stream. Figure 3

  31. EXAMPLE 3 Solution: Assumption • The flow of water at nozzle outlet is steady. • The water splatters in directions normal to the approach direction of the water jet. • β = 1 • We want to find the reaction force of the plate, so that the plate stay in the position.

  32. EXAMPLE 3 • Step 1 & 2 : Control volume and Coordinate axis shown in Figure 3 • Step 3: Calculate the total force In the x-direction : FTx = β˙ (v2x – v1x), = β˙ v1x = (1) (10 kg/s) (20 m/s) (1N/1kg.m/s2) = 200 N The system is symetrical, the force in y-direction is not considered. • Step 4: Calculate the pressure force The pressures at both the inlet and the outlets to the control volume are atmospheric. The pressure force is zero. FPx =FPy =0 m m

  33. EXAMPLE 3 • Step 5: Calculate the body force As the control volume is small we can ignore the body force due to gravity. FBx = FBy = 0 • Step 6: Calculate the resultant force FRx = FTx – FPx – FBx = 200 – 0 - 0 = 200 N The force on the plane is the same magnitude but in the opposite direction R = - FRx = -200 N

  34. Force on angle plane If the plane were at an angle the analysis is the same. It is usually most convenient to choose the axis system normal to the plate.

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