How Does a Jet of Liquid Affect a Stationary and Moving Blade?

[John O' Meara]

A jet of liquid of cross-sectional area A and density p moves with speed vj in the positive x-direction and impinges against a perfectly smooth blade B, which deflects the stream at right angles but does not slow it down. (a) If the blade is stationary, prove that the rate of arrival of mass at the blade is dm/dt = pAvj. (b) If the impulse-momentum theorem is applied to a small mass dm, prove that the x-component of the force acting on this mass for the time interval dt is given by Fx = -(vj)dm/dt. (c) Prove that the steady force exerted on the blade in the x-direction is Fx = pA(vj)^2.
If the blade moves to the right with a seed vb ( vb < vj) derive the equations for (d) the rate of arrival of mass at the moving blade ( e) the force Fx on the blade, and (f) the power delivered to the blade.

Assuming that the deflected jet moves in the positive y-direction (a) p = density = mass/volume then Axvj= volume per unit time therefore pAvj =mass flow rate = dm/dt. (b) The jet of liquid exerts a force F on the blade now the x-component of the force acting on this mass over the time interval dt is the negative of this force =-Fx but Fxdt = dm(vj) therefore this force = -(vj)dm/dt. (c)... (d) pA(vj-vb). (e)... (f)...

 

[anathema]

(c) To prove that the steady force exerted on the blade in the x-direction is Fx = pA(vj)^2, we can use the impulse-momentum theorem. This theorem states that the change in momentum of a system is equal to the impulse applied to the system. In this case, the system is the jet of liquid and the blade. The impulse applied to the system is equal to the force exerted on the blade multiplied by the time interval dt. Therefore, we can write the equation as:

Δp = Fxdt

Where Δp is the change in momentum of the system in the x-direction. Since the blade is stationary, the initial momentum in the x-direction is zero. The final momentum in the x-direction is equal to the mass of the jet of liquid (dm) multiplied by the velocity of the jet (vj). Therefore, we can write:

Δp = dm(vj)

Substituting this into the previous equation, we get:

dm(vj) = Fxdt

Rearranging the equation, we get:

Fx = dm(vj)/dt

But we know from part (a) that dm/dt = pAvj, therefore:

Fx = pAvj(vj) = pA(vj)^2

(d) If the blade moves to the right with a speed vb, the rate of arrival of mass at the moving blade can be calculated by taking into account the relative velocity between the jet and the blade. This can be written as:

dm/dt = pA(vj-vb)

(e) The force exerted on the moving blade can be calculated using the same approach as in part (c). The only difference is that the initial momentum in the x-direction is now equal to the mass of the jet (dm) multiplied by the relative velocity between the jet and the blade (vj-vb). Therefore, we can write:

Δp = dm(vj-vb)

Substituting this into the previous equation, we get:

dm(vj-vb) = Fxdt

Rearranging the equation, we get:

Fx = dm(vj-vb)/dt

But we know from part (d) that dm/dt = pA(vj-vb), therefore:

Fx = pA(vj-vb)(vj-vb) = pA(vj^2-vjvb-vjvb+vb^2) = pA

 

[kyle8921]

To derive the equations for (d), (e), and (f), we first need to consider the change in momentum of the jet as it impinges on the moving blade. The initial momentum of the jet is pAvj in the positive x-direction. After the jet is deflected, its momentum changes to pA(vj-vb) in the positive y-direction. This change in momentum is equal to the impulse exerted on the jet by the blade, which can be calculated using the impulse-momentum theorem:

Impulse = Change in momentum = FΔt

Since we are considering a small mass dm, the time interval Δt can be approximated as dt. Therefore, we can rewrite the impulse as:

Impulse = Fxdt

Setting this equal to the change in momentum, we get:

Fxdt = pA(vj-vb) - pAvj

Solving for Fx, we get:

Fx = pA(vj-vb) - pAvj

Simplifying, we get:

Fx = pA(vj)^2 - pAvjvb

This is the force exerted on the blade in the x-direction, as requested in (e). To calculate the rate of arrival of mass at the moving blade, we can use the same reasoning as in part (a). The only difference is that now the velocity of the jet is (vj-vb) instead of vj, so the rate of arrival of mass at the blade is:

dm/dt = pA(vj-vb)

Finally, to calculate the power delivered to the blade, we can use the definition of power as the rate of work done. In this case, the work done by the jet on the blade is equal to the force Fx multiplied by the distance the blade moves, which is vbdt. Therefore, the power delivered to the blade is:

Power = Fx(vbdt) = pA(vj-vb)(vbdt)

Simplifying, we get:

Power = pA(vj-vb)^2

This is the power delivered to the blade, as requested in (f).