# How to determine the force exerted by a fluid stream on a plate [closed]

This is related to Problem 1 in Chapter 2 of "Rocket Propulsion Elements, 9th Edition":

1. A jet of fluid hits a stationary flat plate in the manner shown.

(a) If there is 50 kg of fluid flowing per minute at an absolute velocity of 200 m/sec, what will be the force on the plate?

(b) What will this force be when the plate moves in the direction of flow at u = 50 km/h? Explain the methodology.

I went and applied $$F = \frac{\mathrm{d}p}{\mathrm{d}t}$$ naïvely. So $$F = \frac{(\text{initial momentum}) - (\text{final momentum})}{\mathrm{d}t}=\frac{m\cdot c-m\cdot 0}{\mathrm{d}t}=\dot m\cdot c$$

With $$\dot m = 0.833\,\mathrm{kg}.\mathrm{s}^{-1}$$ and $$c=200\,\mathrm{m}.\mathrm{s}^{-1}$$, we get for (a) $$F=167\,\mathrm{N}$$.

Now for (b), the mass flow rate has not changed, neither did the velocity of the stream. So the initial momentum of the fluid has not changed. What changed changed is the final momentum: $$F = \frac{(\text{initial momentum}) - (\text{final momentum})}{\mathrm{d}t}=\frac{m\cdot c-m\cdot u}{\mathrm{d}t}=\dot m\cdot (c-u)=155\,\mathrm{N}$$

I see this as a stream of water coming from a point infinitely far away, hitting a plate that is infinitely large. We ignore all matter interactions so I see it as the water simply stacking up on the plate, transfering its kinetic energy into thermal energy upon hitting the plate, and then staying on it. In (b), the water retains some momentum, given that its speed doesn't go to 0.

The co-author of this book, however, doesn't think this way:

I don't understand how stating that the flow is incompressible makes his calculation correct. It seems, to me, that he places himself both on the plate referential, moving at $$14\,\mathrm{m}.\mathrm{s}^{-1}$$, and on the absolute referential used in (a).

My assumption is that his reasoning would hold true only if the fluid jet is channeled through a fixed-section pipe in (a), and the same pipe in (b). This, with the incompressible flow assumption, would decrease the mass flow rate.

Can someone clarify this for me please?

EDIT because the question is now closed : I now feel stupid for not seeing this earlier. The mass flow rate of the fluid has not changed. However, as stated in the solution, the mass flow rate on the plate has changed. So for every time $$\mathrm dt$$, there is a unit of mass $$\mathrm dm$$ leaving the "source" of the fluid, and a mass $$\mathrm dm'= \frac{\mathrm dm \cdot (c-u)}{c}$$ going on the plate. There is an ever growing mass of fluid staying in suspension between the source of the fluid and the plate.

So the co-author of the book is indeed correct in his calculations. The incompressible flow assumption allows us to state that if the mass flow rate of the fluid leaving the source is : $$\dot m=\rho A c$$ then given that the fluid has a constant density and the jet has a constant cross-section (incompressible), the mass flow rate of the fluid entering the plate's referential is : $$\dot m'=\rho A \cdot (c-u)$$ and the mass flow rate of the fluid staying in suspension is : $$\dot m_{air}= \dot m - \dot m'=\rho A u$$ When computing the force that is exerted on the plate, the mass flow rate we have to consider is $$\dot m'$$, because the mass flow rate of the fluid staying in suspension will never enter the plate's referential. So the force exerted by the fluid flow on the plate is : $$F = \dot m' \cdot (c-u)=\rho A \cdot (c-u)^2=144.24\,\mathrm N$$

• Are you familiar with how doing a "macroscopic momentum balance" is taught in a fluid dynamics course? – Chet Miller Jun 6 at 12:09