Question

A 11-cm-diameter horizontal jet of water with a velocity of 40 m/s relative to the ground strikes a flat plate that is moving in the same direction as the jet at a velocity of 10 m/s. The water splatters in all directions in the plane of the plate. Determine the force exerted by the water stream on the plate. Take the momentum-flux correction factor as unity and the density of water as 1000 kg/m3. (Round the final answer to the nearest whole number.)

Answer 1

F = 8552.7N

Step-by-step explanation:

We need first our values, that are,

`V_(jet) = 40m/s\\V_(Plate) = 10m/s \\D = 11cm`

We start to calculate the relative velocity, that is,

`V_r = V_(jet)-V_(plate)\\V_r = (40)-(10)\\V_r = 30m/s`

With the relative velocity we can calculate the mass flow rate, given by,

`\dot{m}_r = \rho A V_r`

`\dot{m}_r = (1000)(30) (\pi (0.11)^2)/(4)`

`\dot{m}_r = 285.09kg/s`

We need to define the Force in the direction of the flow,

`\sum\vec{F} = \sum_(out) \beta\dot{m}\vec{V} - \sum_(in) \beta\dot{m} \vec{V}\\`

`F = \dot{m}V_r`

`F = (285.09Kg/s)(30)`

`F = 8552.7N`