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.)

Respuesta :

Answer:

F = 8552.7N

Explanation:

We need first our values, that are,

[tex]V_{jet} = 40m/s\\V_{Plate} = 10m/s \\D = 11cm[/tex]

We start to calculate the relative velocity, that is,

[tex]V_r = V_{jet}-V_{plate}\\V_r = (40)-(10)\\V_r = 30m/s[/tex]

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

[tex]\dot{m}_r = \rho A V_r[/tex]

[tex]\dot{m}_r = (1000)(30) \frac{\pi (0.11)^2}{4}[/tex]

[tex]\dot{m}_r = 285.09kg/s[/tex]

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

[tex]\sum\vec{F} = \sum_{out} \beta\dot{m}\vec{V} - \sum_{in} \beta\dot{m} \vec{V}\\[/tex]

[tex]F = \dot{m}V_r[/tex]

[tex]F = (285.09Kg/s)(30)[/tex]

[tex]F = 8552.7N[/tex]