# What's going wrong?

1. Jun 21, 2005

### sid_galt

Applying the momentum equation to the control volume of a convergent nozzle for inviscid, incompressible, low speed flow, the thrust is

$$mV_{entry} - mV_{exit} + (P_{exit} - P_{ambient})A_{exit}$$

where

$$m=A_{entry}*V_{entry}$$mass flow per unit time
$$V_{entry}=$$entry velocity from the front of the nozzle.
$$V_{exit}=$$exit velocity of the nozzle air.
$$P_{exit}=$$static pressure at exit
$$P_{ambient}$$Ambient static pressure
$$A_{exit}=$$exit area

For mass flow 4 mg, entry velocity 1 m/s, entry area 4 mm2, exit area 1mm2, exit velocity 4 m/s, the difference between exit and ambient static pressure is -9.225 Pa.

Thrust comes to 2.775E-6. Small but still there is thrust. Is it possible assuming the ideal conditions mentioned above?

2. Jun 21, 2005

### FredGarvin

I didn't go through your numbers, but .00000278 N? I challenge you to find something that could reliably measure that force. That is zero for all intents and purposes.

Also, your equation for mass flow is missing density. The equation you have is for volumetric flow rate.

Last edited: Jun 21, 2005
3. Jun 21, 2005

### sid_galt

Oops, forgot that. If density is included force is coming out to be a bit higher.

Still I find it strange that thrust could be produced like that. As for measurement, that's difficult and drag is going to be much more than the thrust anyway.

4. Jun 21, 2005

### Q_Goest

Momentum is mass flow rate times velocity for a fluid, so I would guess that you meant the mass flow is 4 mg/s, not 4 mg. It has to be per unit time. Therefore, density can be ignored, it's already accounted for by the momentum flux.

I'm imagining that you're trying to determine the thrust produced by a nozzle on a tank that has some pressure inside and you've put a control volume around the nozzle. If that's correct, then I think the pressure terms you show can be dropped, but not absolutely sure. Since the nozzle is connected physically to the tank, the axial force produced by the wall on the nozzle holding it to the tank would also have to be accounted for if you're trying to account for internal pressure. A static balance of the nozzle with a control volume around the nozzle that cuts through the tank wall has to include wall stress, but I don't think that's what you're after. It seems more like you're after thrust only, so put the control volume around the entire tank. If you do that, the thrust is just the momentum flux out of the nozzle which is mass flow times velocity. By putting the CV around the entire tank, you get rid of the pressure terms and don't have to worry about accounting for how that pressure is reacted by the shell of the tank and nozzle.

If I look at only the momentum flux out of the nozzle, that's just 4 mg/s * 4 m/s = 0.000016 N.

It's been a long time since I had to calculate something like this so I'll have to give it some more thought.

Last edited: Jun 21, 2005
5. Jun 21, 2005

### FredGarvin

That's true, but if one does not know what the mass flow rate is, it has to be calculated in order to get the momentum change.

6. Jun 22, 2005

### Q_Goest

Fred, I'd agree, the original question is missing some information. Either the "mass flow 4 mg" is actually 4 mg/s or we need to calculate mass flow and if that's the case, there's insufficient information. Pambient isn't given either.

SID, I think I know what your control volume looks like now, you put a control surface along the inside of the nozzle, which gives a force balance on that chunk of fluid. But that doesn't give you thrust, it's just the force balance on the fluid inside the nozzle. To determine thrust, draw the control surface around the entire thing, or just part of a control surface across the nozzle outlet which gives you Thrust = Mdot*V as I showed above.

7. Jun 22, 2005

### sid_galt

But wouldn't the ambient pressure reduce the thrust as the air coming out of the nozzle would actually have less static pressure than the static ambient pressure.

Static ambient pressure is 1E5 Pa
Static pressure of nozzle exit flow is (1E5 - 9.225) Pa

Then nozzle is in an airflow of 1 m/s

8. Jun 22, 2005

### Q_Goest

Hey Sid, I had to think about this and looked around the web a bit, but it seems there IS a small contribution due to pressure as you mention. There's a web site here that gives a similar equation to the one you show:
http://www.braeunig.us/space/propuls.htm
and an example here:
http://www.braeunig.us/space/problem.htm#2.1

Note this equation doesn't show the momentum of the fluid entering the nozzle, possibly because it's simply being neglected.