2.25: Advanced Fluid Dynamics Section 1: Continuum viewpoint and the equation of motion Section 2: Static Fluids Section 3: Mass Conservation Section 4: Inviscid flow - differential approach Section 5: Control Volume Theorums Section 6: Navier-Stokes equation and viscous flow Section 7: Similarity and dimensional analysis Section 8: Boundary layers, separation and effect on drag/lift Section 9: Vorticity and circulation Section 10: Potential flows; lift, drag and thrust production Section 11: Surface tension and its effect on flows Section 12: Introduction to turbulence Back to 2.25 Home | Back to Section 5
Problem 5.26: Lawn Sprinkler
 The sketch shows a lawn sprinkler with two horizontal arms of radial length R, at the termination of which are nozzles with exit area A2 and outward normal vectors in a horizontal plane pointing outward at an angle q relative to the tangent of a circumferential line, as shown.The sprinkler is free to rotate, but the bearing on which it is mounted exerts a torque kw in the direction opposing the rotation, w being the angular rate of rotation. A constant volume flow rate Q passes through the sprinkler, the water being incompressible at density r. (a) Derive an expression for the steady-state angular velocity w¥ of sprinkler in terms of the given quantities R, A2, q, Q, r and k. In this steady state, what is the velocity vector of the fluid emerging from the nozzles, as seen by an observer in the non-rotating reference frame? What is the fluid velocity relative to the ground at the nozzle exit planes if the bearing is frictionless (k=0)? Comment.
 (b) Now consider the startup of the rotation. Let the rotating arms of the sprinkler have a total mass m per unit length in the radial direction (kg/m), including the solid parts and the water contained therein. Suppose the sprinkler is turned on at t=0 in a static state. Derive a differential equation for its angular velocity, making whatever approximations you consider appropriate, and obtain a solution for w(t).