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djyliett [7]
3 years ago
7

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.)
Engineering
1 answer:
Reika [66]3 years ago
8 0

Answer:

F = 8552.7N

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) \frac{\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

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