rocket efficiency

Here are results for the original scenarios at different dead mass values

Dead Mass (g) 50

100 150 200 250 Cutoff Vel (m/s) (1Kg, 20m/s) 60.89045 47.95791 40.73764 35.83519 32.18876 Cutoff Vel (m/s) (.5KG, 40 m/s) 95.91581 71.67038 58.65348 50.11052 43.94449 300 29.32674 39.23317

Optimized results using Cd = 0.65 and pressure = 160 psi (Achievable) Water Temp = 30 deg C

1 Liter: Water Mass = 403 grams Dead Mass = 145 grams

2 Liter: Water Mass = 800 grams Dead Mass = 289 grams

Altitude for both is about 185 feet

- not including the air burst after the water is run out About 200 feet in all from direct observation

More is possible if a long copper tube is used in the launcher. Has the same effect as a piston launcher.

Regards

Reply to
Larry Curcio
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Cool! Now how much horsepower does a water rocket generate? ;)

Reply to
Alan Jones

I don't know what the rules are for water rocket contests, but you would want a separate charge of compressed air to activate the piston.

Reply to
Alan Jones

That depends on whether it's analog or digital. ;)

Reply to
RayDunakin

I have this problem. My ISP isn't displaying all the messages that I see in Google, and I cannot apply to someone who lists hs address as snipped-for-privacy@null.com. Please bear with me. I'm trying to reply to the last message by the above poster.

The reason the two scenarios are not equivalent is that, during operation, the unused propellant (in this case water) must be accelerated. Thus, when the water is half expended, the other half is moving at the same speed as the rest of the rocket. That's not a big deal if the propellant mass is small compared to the rocket mass, but in this case, even the unused water at the halfway mark may be several times the dead mass. That's a lot of kinetic energy put into the propellant that you don't get back when you expend it. There is also the potential energy of lifting that propellant.

(Actually, you can get some of the kinetic energy back by retaining some of the the mass and allowing its associated energy to work against drag. If you were very clever, you could expend it after it did this.)

This is not just a problem with water rockets. It's the reason that low Ve solid propellants are used primarily in boosters. Certainly, if you have an upper stage with a given impulse, you want a lightwaeight propellant with a high Ve, rather than a heavy propellant with a low Ve. After all, you have to lift that propellant before you burn it. Well... it dosn't really matter if the propellant is in a segregated stage.

Hope that helps.

-Larry Curcio

Reply to
Larry Curcio

The same formula that applies to all rockets:

HP = (Ve * F) / 375 Ve in mph, F in pounds

Working through the equations gives me

HP = 2.49*At*P^1.5

where Ax is the throat area in square inches and P is the (relative)pressure in atmospheres. So at 100 psi it's about 44 HP per square inch of throat area.

You had to ask. ;-)

+McG+
Reply to
Kenneth C. McGoffin

I am not sure which is more amusing to me, water rocket boosted darts, or water rocket horsepower calculatons.

:)

Jerry

Reply to
Jerry Irvine

Water rockets have very low Isp, so staging and boosted darts are relatively more important. Rocket horsepower is just silly, but both can be amusing. But ya know, the development of the V2 was eased because lightweight turbo pumps for fire engines were available more or less of the shelf, and turbo pumps require horsepower to produce forceful streams of water. Now if you use your PAD to drive a high pressure turbo pump in your water rocket...

Alan

Reply to
Alan Jones

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