Calling all Engineers

[AdrianParticipant]

I wonder if there is a way to do this with 2 chambers: One containing the liquid, from which the water exits as a superheated liquid, and then a plenum where the water vaporizes due to a pressure drop before exiting the rocket nozzle as a gas. If liquid goes out the nozzle, you would lose a lot of the Isp, I would think. Maybe a top-exit for pulling the vapor from above the waterline is the right approach.

[Kyle Parsons]

So if there might be a problem of the water transfering to steam before exiting the nozzle, how can we fix that? should we make the distance between the tank longer so it has more time to change into steam?

[Chris LaPanse]

Chris,
I think what Shredder is getting at, is assuming that all water in the tank has been heated to the gas phase, no liquid in the tank.

If that’s the case, the pressure in the tank will be absolutely absurd, as will the temperature (assuming it started full of liquid). Some rough calculations indicate that you’d be in the supercritical fluid part of the phase diagram, with pressure of >30kpsi and temperature of > 800F.

[Chris LaPanse]

So if there might be a problem of the water transfering to steam before exiting the nozzle, how can we fix that? should we make the distance between the tank longer so it has more time to change into steam?

It’s not a matter of time, it’s a matter of pressure. For the liquid to change to a gas, there has to be a pressure drop, and with a nozzle designed for gases, I’m not sure you’d see enough of a pressure drop for the phase transition until you reached the diverging section of the nozzle, which is too late. Adrian’s idea seems potentially usable though, If you effectively went through a small expansion chamber to vaporize the water, and then went through the nozzle, that could work. Pulling from above the waterline guarantees that it will work, but it also has some other issues with the geometry of the motor.

[Kyle Parsons]

Would the pressure between the tank opening and the nozzle not be enough to allow the water to flash into steam? That original picture, its twelve inch section, I was thinking in that distance the water would turn into steam. And no the tank wont be full of gas at all. The tank will only hold water and the transition will happen outside the tank. So pressures should be close to a solid rocket motor, 500-1000 PSI and temperatures between 212-400 degrees. Concerning the nozzle being above the water line, I can’t think of any design where this would be possible to build, unless we tested the motor upside down pointing up. Which is realistic for testing but not realistic for an actual launch attempt.

Some equations we went over was of course bernoullies, and we also looked at relationships between pressure and volume of water/steam.

[Chris LaPanse]

The benoulli relation will be relatively useless in this case, since the flow in a rocket motor is definitely compressible (and bernoulli basically assumes incompressible flow). The phase diagram of water is useful though.

Oh, and Ken’s suggestion does bring to mind an interesting possibility. If you intentionally filled the tank only partway, and then heated it such that the final state (just prior to release) was something like 600F, 1000-1500 PSI water vapor. That would make the design of the motor substantially simpler, at the expense of propellant density.

Otherwise, with the primarily liquid chamber conditions, the issue I see right now is simply that the propellant entering the converging section of the nozzle absolutely has to be a gas for it to work properly. If it’s a liquid entering the nozzle, then the throat flow will probably not be choked, which ruins your efficiency. I don’t know any good way to calculate this either – perhaps a small scale test?

[Chris LaPanse]

I just ran some numbers with a chamber pressure of 1000PSI, a chamber temperature of 550F, and an exhaust pressure of 25 PSI (slightly underexpanded), here’s what I get (assuming that the stuff entering the nozzle is 100% pure water vapor, with no liquid):

A/A* = 4.45 (nozzle expansion ratio)
Exit temp = -106 fahrenheit (yes, negative)
Exit mach: 3.05
Exit velocity: 3470 ft/s
Specific impulse: 108 seconds

That exit temp is a problem too, since this analysis assumed gas for the entire time. In reality, at 25psi and -106F, the vapor would quickly condense into liquid water.

After running a couple more numbers, I’ve managed to convince myself that you pretty much can’t do this the way you want to. The problem is that in a nozzle, the flow cools too much, so if it was on the vapor dome in the chamber, it will be solidly within the liquid part of the phase diagram in the nozzle itself. Your best bet would probably be to start with superheated steam rather than a saturated liquid-vapor mix of steam and water.

[Chris LaPanse]

Darn. Double post and I can’t find the delete button…

[Kyle Parsons]

So basically that idea would mean you fill the tank partially of water and heat it till its all vapor? Then release it throught the valve and nozzle? Wouldnt that limit the amount of vapor you could produce? If you have a full tank of water and transform that all to steam you could almost double the amount of steam. If you only partially fill the tank you would only get a fraction of steam.

[Kyle Parsons]

What equations are you using for those numbers? I cant even start to understand the dynamics books I looked at.

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