Momentum

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.

NEWTON’S LAWS AND CONSERVATION OF 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.

NEWTON’S LAWS AND CONSERVATION OF 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

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 2. What increase of pressure difference is required to accelerate the water in pipe at rate 0.02 m/s 2 ? 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 A ρ δp a l

EXAMPLE 1- GRADUAL ACCELERATION OF FLUID IN PIPELINE Note 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 = 10 3 x 60 x 0.02 N/m 2 = 1.2 kN/m 2

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.

LINEAR MOMENTUM EQUATION Consider a stream tube and assume steady non-uniform flow A2v2ρ2A2v2ρ2 v1v1 A1v1ρ1A1v1ρ1

LINEAR MOMENTUM EQUATION In time δ t a volume of the fluid moves from the inlet at a distance v 1 δ t, so volume entering the stream tube = area x distance = A 1 x v 1 δ t The mass entering, mass entering stream tube = volume x density = ρ 1 A 1 v 1 δ t And momentum momentum entering stream tube = mass x velocity = ρ 1 A 1 v 1 δ t v 1 Similarly, at the exit, we get the expression: momentum leaving stream tube = ρ 2 A 2 v 2 δ t v 2

LINEAR MOMENTUM EQUATION By Newton 2nd law Force = rate of change of momentum F = (ρ 2 A 2 v 2 δt v 2 - ρ 1 A 1 v 1 δt v 1 ) δt We know from continuity that Q= A 1 v 1 = A 2 v 2 And if we have fluid of constant density, ρ 1 = ρ 2 = ρ, then F = Qρ (v 2 -v 1 )

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 = ρ 1 A 1 v 1 = ρ 2 A 2 v 2 The rate at which momentum enters face 1 is ρ 1 A 1 v 1 v 1 = ˙ v 1 The rate at which momentum leaves face 2 is ρ 2 A 2 v 2 v 2 = ˙ v 2 Thus the rate at which momentum changes across the stream tube is ρ 2 A 2 v 2 v 2 - ρ 1 A 1 v 1 v 1 = ˙ v 2 - ˙ v 1 Force = rate of change of momentum F = ˙ (v 2 -v 1 ) m m m m m m

LINEAR MOMENTUM EQUATION So, we know these two expression. Either one is known as momentum equation: F = ˙ (v 2 -v 1 ) m F = Qρ (v 2 -v 1 ) The momentum equation: This force acts on the fluid in the direction of the flow of the fluid

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

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

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. v1v1 v2v2

LINEAR MOMENTUM EQUATION-STEADY FLOW ALONG COORDINATE The force in x-direction Fx = ˙ (v 2 cos θ 2 – v 1 cos θ 1 ) = ˙ (v 2x – v 1x ) Or Fx = ρQ (v 2 cos θ 2 – v 1 cos θ 1 ) = ρQ (v 2x – v 1x ) The force in y-direction Fy = ˙ (v 2 sin θ 2 – v 1 sin θ 1 ) = ˙ (v 2y – v 1y ) Or Fy = ρQ (v 2 sin θ 2 – v 1 sin θ 1 ) = ρQ (v 2y – v 1y ) The resultant foce can be found by combining these components m m m m

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

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.

FORCE ACTING ON CONTROL VOLUME Force is made up of these component: 1.F R = force exerted on the fluid by any solid body touching the control volume 2.F B = Force exerted on the solid body (eg gravity) 3.F P = Force exerted on the fluid by fluid pressure outside the control volume Total force is given by the sum of these three forces: F T = F R + F B + F P The force exerted by the fluid on the solid body touching the control volume is opposite to F R So, the reaction force, R is given by R = - F R

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 of θ. A1v1ρ1A1v1ρ1 A2v2ρ2A2v2ρ2

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.

FORCE DUE TO THE FLOW AROUND THE PIPE BEND Step in analysis 1.Draw a control volume 2.Decide on coordinate axis system 3.Calculate the total force 4.Calculate the pressure force 5.Calculate the body force 6.Calculate the resultant force

Example (1)

Example (2)

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

EXAMPLE 3 Solution: Assumption 1.The flow of water at nozzle outlet is steady. 2.The water splatters in directions normal to the approach direction of the water jet. 3.β = 1 We want to find the reaction force of the plate, so that the plate stay in the position.

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 ˙ (v 2x – v 1x ), = β ˙ v 1x = (1) (10 kg/s) (20 m/s) (1N/1kg.m/s 2 ) = 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. F Px =F Py =0 m m

EXAMPLE 3 Step 5: Calculate the body force As the control volume is small we can ignore the body force due to gravity. F Bx = F By = 0 Step 6: Calculate the resultant force F Rx = F Tx – F Px – F Bx = 200 – = 200 N The force on the plane is the same magnitude but in the opposite direction R = - F Rx = -200 N

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.