Question

A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of +18.0 m/s, while the exiting water stream has a velocity of -18.0 m/s. The mass of water per second that strikes the blade is 39.0 kg/s. Find the magnitude of the average force exerted on the water by the blade.

Answer 1

Answer:

1404 N

Explanation:

speed of incident water, u = 18 m/s

speed of exiting water, v = - 18 m/s

mass per unit time, m / t = 39 kg/s

According to the second law of Newton's

force = rate of change of momentum

F = m (v - u) / t

F = 39 x ( -18 - 18)

F = - 1404 N

Thus, the force exerted on the water by the blade is 1404 N.

Answer 2

Answer:

1404 N .

Explanation:

Force = rate of change of momentum

d/dt ( mv₁ - mv₂ )

If v₂ = - v₁

rate of change of momentum

= d/dt ( mv₁+mv₁ )

=2x v₁ dm / dt ( Here velocity of water  v₁ throughout is constant )

Force = 2 xv₁ dm / dt

Given ,

v₁ = 18 m/s

dm / dt = rate of flow of mass of water

= 39 kg / s

Putting the values in the equation above

Force = 2xv₁ dm / dt

= 2x18 x 39

= 1404 N .