A vertical downward jet of water of density carries at and lands on a horizontal…

PHYSICS · Multi-part · Fermi University Physics · Fluids & Continuum Physics · Kinematics, conservation laws and ideal flow · Kinematics, conservation laws and ideal flow

Fluids and Continuum Physics, Ch Conservation of mass, momentum and energy

Problem

A vertical downward jet of water of density ρ=1000 kg m−3\rho=1000\ \mathrm{kg\,m^{-3}} carries Q=3.0 litres s−1Q=3.0\ \mathrm{litres\,s^{-1}} at 12 m s−112\ \mathrm{m\,s^{-1}} and lands on a horizontal conveyor belt, where it is turned and runs off horizontally, still at 12 m s−112\ \mathrm{m\,s^{-1}}. The flow is steady. Take the control volume to be the water in contact with the belt, entered from above by the jet and left horizontally by the sheet; the weight of the water inside it is negligible, and gauge pressure is used, so the surrounding atmosphere contributes nothing. Because the belt moves horizontally, its motion does not enter the vertical balance. Let zz be measured upwards. A student writes: “Steady flow, so nothing accumulates and the sum of the forces equals the net rate at which momentum flows out. Take the normal on the jet's face along the jet, i.e. pointing downwards into the box; then the jet contributes ρ(u⋅n)A uz=ρQ(−12)=−36 N\rho(\mathbf{u}\cdot\mathbf{n})A\,u_z=\rho Q(-12)=-36\ \mathrm{N}. The water that leaves runs horizontally and carries no vertical momentum, so the net outflow of zz-momentum is 0+(−36)=−36 N0+(-36)=-36\ \mathrm{N}. The only vertical force is that of the belt on the water, so the belt pushes the water down with 36 N36\ \mathrm{N}, and by Newton's third law the water pushes the belt upwards with 36 N36\ \mathrm{N}.” ((a)) Identify the step that is wrong and say what the rule for the normal on a control surface actually is. ((b)) Give the correct magnitude and direction of the vertical force the water exerts on the belt.

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