Euler’s First Law or the Linear Momentum Principle, or...

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ECH 141 Problem Set #4 1. Euler’s First Law (or the Linear Momentum Principle, or Conservation of Linear Momentum) is sometimes written in words as The rate of change of momentum of a fluid particle = The sum of the forces acting on the particle . In a static fluid, the rate of change of momentum is zero, and the forces acting on a fluid particle are generally the gravitational force and pressure, as described in lecture. We showed therefore that, in a static fluid, p − ρ g = 0 is the equation that determines the pressure distribution. a) Consider a case where the static fluid is in a vehicle that undergoing a constant acceleration a . The rate of change of momentum of a fluid particle with mass m is then ma , or for an infinitesimally small particle with volume dV, it would be ρ a dV. Derive the equation for the pressure distribution in a fluid undergoing constant acceleration a . b) Now consider a cylindrical glass of water with radius R on a horizontal turntable, rotating about its axis of symmetry (the z-axis, where radial position r=0). In the absence of rotation, the height of water in the glass is H. As you learned in physics, each position in an object undergoing pure rotation is accelerating inward, with centripetal acceleration a a = −ω 2 re r , where r is the distance from the axis of rotation in a cylindrical coordinate system, ω is the rotational velocity, and e r is a unit vector pointing away from the z-axis. Use this expression for centripetal acceleration with your result from part “a” to derive an expression for the height h(r) of the air-water interface in a rotating, cylindrical glass of water. Solution
ρ a V m dV = ρ g V m dV p V m dV . Since the material control volume is arbitrary, this integral balance can only be true if at every point the equation p = −ρ a + ρ g is satisfied. b) The vector equation from part (a) can be separated into components. For the cylindrical glass of water, it makes sense to use a cylindrical coordinate system. Then writing out the equation yields e r r + e z z ( ) p = ρω 2 re r − ρ ge z . Dotting both sides of the equation with e Z gives p z = −ρ g and doing the same thing in the radial direction gives p r = ρω 2 r . The difference between this case and one where the fluid is not accelerating is that the top surface is not flat, but rather has some height h(r). At this height, the pressure equals atmospheric pressure. Integrating the two equations, and noting that the derivatives are partial derivatives so that constants of integration can be functions of the variable not being integrated, we find p r,z ( ) =

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