Impact of JetsIntroduction: If the fluid is moving with high velocity, then it is called jet. Generally Nozzle is used to increase the kinetic energy of fluid. The study of forces resulting from the impact of fluid jets and when fluids are diverted round pipe bends involves the application of newtons second law in the form of F = m.a. The forces are determined by calculating the change of momentum of the flowing fluids. In nature these forces manifest themselves in the form of wind forces, and the impact forces of the sea on the harbour walls. The operation of hydro-kinetic machines such as turbines depends on forces developed through changing the momentum of flowing fluids. When the jet moves and strikes the obstruction like flat plate, vane in its path. Then the jet will exert the force on the obstruction, it is known as Impact of Jet. The Impulse Momentum Principle is used to calculate the hydrodynamic force of jet on the vane. This principle is derived from the Newton’s II Law of Motion. Newton’s II Law of Motion: Newtons Second Law can be stated as: The force acting on a body in a fixed direction is equal to rate of increase of momentum of the body in that direction. Force and momentum are vector quantities so the direction is important. A fluid is essentially a collection of particles and the net force, in a fixed direction, on a defined quantity of fluid equals the total rate of momentum of that fluid quantity in that direction “The rate of change of momentum of a moving body is directly proportional to the magnitude of the applied force and takes place in the direction of the applied force”. F mV mU t F ma Where a – acceleration F = k ma If m = 1 and a = 1 then F = 1 k = 1 i.e. F = ma SI unit of force: Newton (N) Momentum: V.SREEDHAR MECHANICAL ENGG., VARDHAMAN COLLEGE OF ENGG. , SHAMSHABAD , HYDERABAD. From the Newton’s IInd law F = ma F = m V U t F x t = mV – mU Impulse = Final momentum – Initial momentum Where F x t is called Impulse.. . Applications of Impulse – Momentum Principle: The followings are the applications of Impulse Momentum Principle. Eg: Kick given to a foot ball. To calculate resultant force on pipe bend 3. Impulse – Momentum Principle: This principle is derived from the Newton’s IInd law . . VARDHAMAN COLLEGE OF ENGG. Impulse of a force = Force x Time interval SI unit: N-sec. 1. Force Of Jet Impinging Normally On A Fixed Plate Consider a jet of water impinging normally on a fixed plate as shown in fig-1. * Impulse of a force is given by the change in momentum caused by the force on the body. V. HYDERABAD. SHAMSHABAD . The momentum of a moving body is given by the product of mass and velocity of the moving body. Impulse of a force is given by the product of magnitude of force and its time of action.SREEDHAR MECHANICAL ENGG.The capacity of a moving body to impart motion to other bodies is called momentum. To calculate the hydrodynamic force on stationary and moving vanes. To calculate the propulsive force on propulsive ships. 2. Momentum = Mass x Velocity =mxV SI Unit: kg-m/s Impulsive Force and Impulse of Force: A force acting over a short interval of time on a body is called impulsive force. . Force Of Jet Impinging On An Inclined Fixed Plate Consider a jet of water impinging normally on a fixed plate as shown in fig-2. F= waV/g X (V-0) F= waV2/g KN. VARDHAMAN COLLEGE OF ENGG. F= mass of water flowing/sec. SHAMSHABAD . V.. is reduced to zero after the impact (as the plate is fixed). X change of velocity. . Therefore force exerted by the jet on the plane. V = Velocity of jet a = Cross sectional area of the jet w = Specific weight of water Mass of water flowing/s = waV/g kg We know that the velocity of jet. in its original direction.Let.SREEDHAR MECHANICAL ENGG. HYDERABAD. and the force exerted by the jet in the direction of flow. VARDHAMAN COLLEGE OF ENGG. force exerted by the jet in a direction normal to flow. Force exerted by the jet in a direction normal to the plate.Let. .SREEDHAR MECHANICAL ENGG. HYDERABAD. = Angle at which the plate is inclined with the jet Force exerted by the jet on the plane = F= waV2/g KN. Similarly. Force Of Jet Impinging On A Moving Plate V.. . SHAMSHABAD . Let. = Velocity of the plate. F= mass of water flowing per sec. . V. X change of velocity F= wa(V-u)/g X((V-u)-0) F=wa(V-u)2/g KN Force Of Jet Impinging On A Moving Curved Vane Consider a jet of water entering and leaving a moving curved vane as shown in fig-4. VARDHAMAN COLLEGE OF ENGG.SREEDHAR MECHANICAL ENGG. as a result of the impact of jet A little conversation will show that the relative velocity of the jet with respect to the plate equal to (V-v)m/s. As a result of the impact of the jet. Therefore force exerted by the jet. let the plate move in the direction of the jet as shown in fig-3. .Consider a jet of water imping normally on a plate. HYDERABAD. For analysis purposes. it will be assumed that the plate is fixed and the jet is moving with a velocity of (V-v)m/s. SHAMSHABAD .. at which the jet enters the vane. at which the jet leaves the vane. v2 = Velocity of the vane (AB. SHAMSHABAD . while entering the vane. while leaving the vane. HYDERABAD.. V1 = Velocity of the jet (EG).Let. = Angle with the direction of motion of the vane. FG) = Angle with the direction of motion of the vane. . v1.SREEDHAR MECHANICAL ENGG. V = Velocity of the jet (AC). which Vr makes with the direction of motion of the vane at inlet (known as vane angle at inlet). VARDHAMAN COLLEGE OF ENGG. V. Vr = Relative velocity of the jet and the vane (BC) at entrance (it is the vertical difference between V and v) Vr1 = Relative velocity of the jet and the vane (EF) at exit (it is the vertical difference between v1 and v2) = Angle. . . equal to the direction of motion of the vane (known as velocity of whirl at outlet). It is a component parallel to Vw1 = Horizontal component of V1 (HG. equal to Vf = Vertical component of V (DC. and (ii)Vr=Vr1 We know that the force of jet. Solution: V. VARDHAMAN COLLEGE OF ENGG. F= waV/g(Vw-Vw1) if F= waV/g(Vw+Vw1) if >90 =90 Example . . in the direction of motion of the vane. Vw = Horizontal component of V (AD. SHAMSHABAD . It is a component at right angles to the direction of motion of the vane (known as velocity of flow at inlet). equal to a = Cross sectional area of the jet.= Angle. Vf1 = Vertical component of V1 (EH. Find the force exerted on the plate. It is a component at right angles to the direction of motion of the vane (known as velocity of flow at outlet).. ).Force of jet impinging normally on a fixed plat 1. HYDERABAD.SREEDHAR MECHANICAL ENGG. As the jet of water enters and leaves the vanes tangentially. which Vr1 makes with the direction of motion of the vane at outlet (known as vane angle at outlet). equal to direction of motion of the ). The relations between the inlet and outlet triangles (untill and unless given) are: (i)v=v1. ). F= mass of water flowing per sec. It is a component parallel to the vane (known as velocity of whirl at inlet).Problem A jet of water of 100 mm diameter impinges normally on a fixed plate with a velocity of 30 m/s. therefore shape of the vanes will be such that Vr and Vr1 will be along with tangents to the vanes at inlet and outlet. X change of velocity of whirl F= waV/g(Vw+Vw1) if <90 ). Determine the pressure on the plat. VARDHAMAN COLLEGE OF ENGG. HYDERABAD. Example .05 m V = 26 m/s v = 10 m/s Pressure on the plate when it is fixed The cross sectional area of the jet. Solution Given. V..The cross sectional area of the jet.Force of jet impinging on a moving plate Problem A jet of water of 50 mm diameter. moving with a velocity of 26 m/s is impinging normally on a plate. Pressure on the plate. Pressure on the plate when it is moving The pressure on the plate when it is moving. SHAMSHABAD . . . d = 50 mm = 0.SREEDHAR MECHANICAL ENGG. when (a) it is fixed and (b) it is moving with a velocity of 10 m/s in the direction of the jet. Force exerted by the jet on the plane.