Notes on Momentum Equation.pdf - Chapter 2 Momentum...

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Chapter 2 Momentum Equations 2.1 Introduction When a fluid flows over/on a solid body, it exert force on it. For example, force exerted on a solid surface by a jet of fluid impinging on it, aerodynamic forces ( lift and drag) on aircraft wing, the force on a pipe-bend by fluid flowing within it, etc. These forces are hydrodynamic forces (due to a moving fluid) and are associated with a change in momentum. The magnitude of the hydrodynamic forces on the body due to a moving fluid is determined by Newton’s second law of motion ” the net force acting on a body in any direction is equal to the rate of increase of momentum of the body in that direction”.i.e (equation 2.1) F = ma (2.1) The law usually need to be expressed in a form particularly suited to steady flow of a fluid. 2.2 Linear Momentum Equations for Steady Flows Consider a steady flow which is non-uniform flowing in a control volume (stream tube) as shown in figure 2.1. Where A is cross sectional area, u is velocity and ρ is density and subscript 1 and 2 represent conditions at entry and exit respectively. In a short interval δt , a volume of the fluid moves from the inlet a distance uδt . The mass of fluid entering the control volume in time δt is given by equation 2.2 Mass entering control volume = volume × density = ρ 1 A 1 u 1 δt (2.2) 16
Figure 2.1: Non-uniform linear flow stream tube Hence momentum entering the control volume is given by equation 2.3 Momentum entering control volume = mass × velocity = ρ 1 A 1 u 1 u 1 δt (2.3) Similarly equation 2.4 expresses the momentum leaving the stream tube Momentum leaving control volume = mass × velocity = ρ 2 A 2 u 2 u 2 δt (2.4) The force exerted by the fluid is calculated using Newton’s 2 nd law, equation 2.5 Force, F = Rate of change of momentum = ρ 2 A 2 u 2 u 2 δt - ρ 1 A 1 u 1 u 1 δt δt (2.5) For a steady flow, continuity requires that Q = A 1 u 1 = A 2 u 2 (2.6) and for a constant density, ρ 1 = ρ 2 = ρ , then equation 2.5 reduces to equation 2.7 F = ( u 2 - u 1 ) = ˙ m ( u 2 - u 1 ) (2.7) 2.3 Angular Momentum Equations for Steady Flows Consider a steady flow which is non-uniform flowing in a control volume (stream tube) as shown in figure 2.2. The inlet velocity vector, u 1 makes an angle θ 1 with the x-axis, while at the outlet velocity u 2 make an angle θ 2 to the x-axis. Therefore, the forces are resolved in the direction of the co-ordinate axes as; 17
Figure 2.2: Non-uniform anhular flow stream tube Force in x direction, F x is given by equation 2.8 F x = Rate of change of momentum in x-direction = ˙ m ( u 2 x - u 1 x ) = ( u 2 cosθ 2 - u 1 cosθ 1 ) (2.8) Similarly, equation 2.9 gives the force in y direction as F y = Rate of change of momentum in y-direction = ˙ m ( u 2 y - u 1 y ) = ( u 2 sinθ 2 - u 1 sinθ 1 ) (2.9) The resultant force, F resultant is given by equation 2.10 (figure 2.3) F resultant = F 2 x + F 2 y (2.10) and the angle, φ at which this force act is given as φ = Tan - 1 ( F y F x ) (2.11) Note : The force exerted by the fluid on the solid body touching the control volume is opposite to F . So the reaction force, R is given as R = - F R . F is the total force

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