# Propulsion enigmas

1. Aug 19, 2004

### Clausius2

I have some doubts waken up by one of the last threads in the Classical physics forum. Before anything, please open the attached file below. It is opened with word.

Please assume subsonic flow in every picture.

Cases a) and b): P<Pa. To where does it move?. You can see I have re-scaled figure b). This is because I want to emphasize the pòssible force exerted by the inlet air jet on the opposite wall. I mean, the jet is not completely dissipated in the inner atmosphere like in case a).

Cases b) and c). They are put across with the same intention as in a) and b). Do you think the exhausting structure in the welding with the vessel would resist the case b)?.

Cases d) and e). P>Pa. Both of them are subsonic rockets. If this sounds wrong imagine a torpedo inside water. They are at rest initially. At instant t=0, both of them init the boosting. But the torpedo b) has an static platform just behind him, where outflow jet is going to shock against. Which of them would you say have the most initial thrust?

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2. Aug 19, 2004

### enigma

Staff Emeritus
Case a-b)

If you integrate the pressure over the surface, you'll see that the only place where there is a non-uniformity is in the hole to the outside.

Pressure only acts perpendicular to the wall. For case b, the sections which are angled inward will have their $\int{PdA}$ broken up into +-y components which will cancel on the other side, and +-x components which will be cancelled by the left wall except for the small hole.

For cases b-c and d-e, the problems cannot be solved so simply. The exhaust gases in both cases will hit and deflect in a non-predictable manner. For c-d), any flow which is diverted -y will contribute thrust in the +y direction. For e-f, the f rocket will probably have marginally more thrust from the flow bouncing back and transferring more momentum to the rocket, but in practice the material wouldn't be able to stand up to the temperatures which is why it isn't done that way.

3. Aug 19, 2004

### Clausius2

I really think you are enlarging the list of factors working in these problems. I feel in some way you are thinking in recirculation zones in case b) and non-symmetry in cases c) and d). My initial idea was simpler. The jet colliding with the opposite wall in case b) is crashing frontally, and I suppose for simplicity there is no recirculations going outwards the vessel. All gas flow does not returns.

Enigma, I have just an idea. I want to comment it with you. If you feel like, write the integral momentum equation for the case a). Neglect viscous, gravitational and unsteady terms. Convective fluid inertia must be balanced with pressure forces. Assume the pressure, just in the outlet of the nozzle (its inner part viewing it out of the vessel), is Pa. That is, discharge pressure is roughly the total pressure. It does not matter. Thus, I work out force exerted for the fluid on the vessel:

F=(rho*U*U*+Pa)*A in x+ direction.

This means if we see it in a conservative manner, and without restricting anything, the vessel always moves towards +x direction. However, I have integrated U over the hole as an uniform (without recirculation) profile. Perhaps in case b) this integration would have negative values and colaborate with a x- direction force, counteracting the suction's one.

4. Aug 19, 2004

### enigma

Staff Emeritus
Not quite. You forgot to take into account the pressure of the fluid at the interface.

The thrust of a rocket engine (which this basically is) is:

$$T=\rho \times U^2 \times A + (P - Pa) \times A[/itex] I'm sorry, I'm not following what you mean here. 5. Aug 20, 2004 ### Clausius2 I have a mistake: [tex]\int{\rho u \overline{v} \cdot \overline{dA}}=-\oint{P\overline{dA}}$$

where first integral is extended over the hole, with dA pointing towards outside the container. (I have chosen a control volume that surrounds the container). The second integral is extended to this control volume I've mentioned. The surface dA of the hole is just positioned at its inner (i.e. narrower) part.

Right hand side of above equation yields:
$$=-\oint{P\overline{dA}}=-(P_{s}A+P_{a}A+\int{PdA})\overline{e_{x}}$$

where Ps is the static pressure just in the narrower part of the nozzle. Are you agree?. Last term is the force exterted on the container by the fluid.

We could fix Ps=P as discharge pressure.

What I was trying to tell you, is that the velocity flux you have used in your rocket equation is

$$\int{ \rho u \overline{v} \cdot \overline{dA}}$$

taking implicitly $$\overline{v}=-u\overline{e_{x}}$$ is an uniform profile always pointing towards inside. Under this hypothesis, the force exterted on the container is always to the right:

$$\overline{F}=(\rho u^2 A + (P_{a}-P)A) \overline{e_{x}}$$

The narrowing of the container in case b) does not make any effect on integral laws. But I felt I understood in your first thread, I think that narrowing would case a flow recirculating in the nozzle, due to the nearby opposite wall. Thus, V would not be an uniform profile. The flux of momentum could not be always positive.

I mean, I am searching a possible explanation for an eventual displacement of case b) container to the right due to jet colliding the frontal wall.

Last edited: Aug 20, 2004
6. Aug 20, 2004

### Clausius2

The first equation is wrong, but the web does not allow me to change it.