MythBusters Water Heater Rocket

[Aero]

I went a little further, using molecular weight of 18.02 and a steam table. These are the pressures, corresponding temperatures and the calculated Isp values. As you can see, higher values of ISP could occur. Just not when using a household water heater as the pressure vessel.

Code: Select all

deg F 	psig	  Isp
417.35	300	83.36
444.62	400	99.35
467.04	500	113.84
486.24	600	127.24
503.13	700	139.80
518.27	800	151.69
532.02	900	163.01
544.65	1000	173.85
556.35	1100	184.29
567.26	1200	194.36
577.49	1300	204.11
587.14	1400	213.58

With a pressure of 335 psi, altitude and speed at burn-out are 7.04 miles and 745 mph. It takes 1100 psi to achieve altitude and speed at burn-out of 30.06 miles and 1541.33 mph. This is accelerating at 1 gee, so the coast upward against gravity should double the altitude after burnout. Maybe water heater rockets would make a better terrorist weapon than space craft.

[D Tibbets]

Areo, a little further down the page of your link, estimated ISP close to what I initially was calculated (~ 90 sec. ) except he used the Kelvin scale (which near 500 degrees F would give a similar number (within 50 degrees or so). Using the Kelvin scale for the temperature makes the most sinse (has an absolute zero, instead of an artificial 0 starting point), unless an appropiate correction is incorperated into the other scales.



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Quote:
Originally Posted by Sky Captain
Looks like it`s been done - a steam powered model rocket

What is the ISP of that?

"Mmm, a rocket whose exhaust is steam? I'd say 400 seconds; http://en.wikipedia.org/wiki/Vulcain_2. Just kidding, of course. Actually, the exhaust of a cryogenic LOX/LH rocket cannot really be defined as "steam". Isp depends on a few factors, but we can expect it to be pretty low, around 90s, with a chamber pressure of 20 atm, T~500K and a correct expansion to external pressure. The trouble is that pressure will decrease pretty soon... "

Dan Tibbets

[KitemanSA]

With a pressure of 335 psi, altitude and speed at burn-out are 7.04 miles and 745 mph.

Now calculate the energy content of WATER at that temperature and determine how long the 335 psi steam c n be maintained at that pressure to fit into your rocket equation. Sorry, your Isp is transitory at best, negligible unless you have an outside source of energy to maintain the pressure.

Happy Calcing!

[Aero]

With a pressure of 335 psi, altitude and speed at burn-out are 7.04 miles and 745 mph.

Now calculate the energy content of WATER at that temperature and determine how long the 335 psi steam c n be maintained at that pressure to fit into your rocket equation. Sorry, your Isp is transitory at best, negligible unless you have an outside source of energy to maintain the pressure.

Happy Calcing!

KitemanSA - I'm not quite sure I follow you. Remember that we are not talking about a cold water bottle rocket using compressed air to eject cold water. In the bottle rocket the air pressure drops as the water is ejected. Fill your bottle half full of water, put 100 psi of air pressure (for example) on top of it, then launch. As the water is expelled, the air cavity grows to twice its original volume and at "burn-out" the pressure drops to half its original value. That's because there is no energy source to replace the air pressure.

Super heated water is the source of energy that replaces the steam pressure in the hot water rocket. Forty gallons of water at 335 psi and about 430 deg F contains over 30 million BTUs of thermal energy. After ejecting half the water, there remains over 15 million BTUs of thermal energy within the 20 gallons of water remaining in the tank. The only reason the pressure would drop is if the temperature were to drop, but the temperature will drop only marginally if at all. As the last 2 gallon are exiting the nozzle, the tank still contains 1.5 million BTUs of heat energy and the steam pressure remains very high. Tail off will occur but it happens after almost all the hot water is ejected. Remember that the thrust from the hot water rocket results from the water flashing into steam within the rocket nozzle, not significantly from the tank pressure ejecting the water as in a bottle rocket. That's why the nozzle design is so critical.

[Aero]

@ Dan -
You are right that we need to use consistent units in the calculations and I'm not so good at that. You may remember that Lockheed Martin messed up the units on a high visibility space mission a few years back so I'm not alone.

Isp values for Steam rocket motors can't be compared to LH2-LOX rocket motors even though the exhaust products have the same constituents. In steam rockets, the exhaust product is H2O, while liquid hydrogen-liquid oxygen rockets burn at such high temperatures that the oxygen and hydrogen are disassociated in the nozzle as I recall. For some reason that gives a lower molecular weight of the exhaust product. I don't recall why that is so, but I do know that they burn them fuel rich which adds more of the low molecular weight hydrogen to the exhaust, upping the Isp. You can take that with a grain of salt because if the H and O are dissociated then they are not burned it seems to me, so the energy has not been released so its not working right. Oh well, maybe someone who knows will make a post.

[KitemanSA]

KitemanSA - I'm not quite sure I follow you. Remember that we are not talking about a cold water bottle rocket using compressed air to eject cold water. In the bottle rocket the air pressure drops as the water is ejected. Fill your bottle half full of water, put 100 psi of air pressure (for example) on top of it, then launch. As the water is expelled, the air cavity grows to twice its original volume and at "burn-out" the pressure drops to half its original value. That's because there is no energy source to replace the air pressure.

At 335psi (~2.3MPa) the TOTAL energy content of each gram of water (~950 J/g) is enough to vaporize 1/2 gram of that water. The AVAILABLE content (call it down to 100degrees C; ~430 J/g) is only enough to vaporize about 1/4 of each gram, at which point it is down to ~14psi, not a very high Isp. You need an EXTERNAL source of energy to maintain that Isp with the entire mass load.

[kunkmiester]

Don't forget your gas laws, Aero. Abiatic expansion will take some of that energy.

[Aero]

@Dan- I have spent hours googling for the formula for specific impulse which uses molecular weight. This is the closest I can find. It actually may be the same thing.
http://www.qrg.northwestern.edu/project ... pulse.html
I have a concern with using consistent units in the formula-
Isp = sqrt(Temp x Press/Molecular wt.)
It seems that temperature should be in degrees Rankine, Pressure in psf but what about Molecular wt?
Temp x Press has units of (deg R x lbs)/ft^2 so molecular wt. must have units of What? These units are meaningless to me, even more so when I take the square root. Factoring out the units of g should give velocity so what are the units of molecular weight?
It seems to me that molecular weight does not have common units. It is measured in units of 1/12 the mass of a carbon-12 atom.

[Aero]

@ kunkmiester and KitemanSA
I'm afraid I forgot my gas laws 40 years ago. That's why I look at fully reduced equations for the rocket nozzle instead of first principles. But remember, throughout this thread I have specified an ideal nozzle. This nozzle has a convergent section, a throat and a divergent section. The gas (steam) expansion happens in the divergent section. There is a shock front at the throat which isolates the tank contents from anything happening outside in the divergent section. What that means it that the energy content of each gram of water within the tank will remain relatively constant. Its like when you turn off the cold water supply then drain your hot water heater. The last gallon of water is almost as hot as the first gallon you drained. (Yes, some of the energy in the superheated tank is consumed to maintain tank pressure, but not much.)

At 335psi (~2.3MPa) the TOTAL energy content of each gram of water (~950 J/g) is enough to vaporize 1/2 gram of that water.

It could be that 335 psi is not enough pressure (heat energy) to permit a fully expanded steam exhaust. That is, it may be that there will be water droplets entrained within the steam, for lack of energy to vaporize them. But if you are saying that steam rockets don't work, then I do have to disagree. I think you're saying that the pressure is not high enough for the MythBuster steam rocket to perform ideally. Again, that may be true.
Here is a fun thread to read for those interested in this topic. This person is making a steam rocket powered ice sled.
http://www.pulse-jets.com/phpbb3/viewto ... &sk=t&sd=a
He did an awfully lot of work but I don't think he got the help he asked for on his nozzle design. His exhaust plume looks more like a bottle rocket jet of very hot water than it does an expanded plume of steam. Remember, the SSME generates an ideally expanded exhaust plume of steam. It is invisible.

[KitemanSA]

...But if you are saying that steam rockets don't work, then I do have to disagree.

I am not saying steam rockets don't work, merely that you won't get that kind of performance with a water heater at 335psi.

Steam rockets with external heat sources would work as you stated, ... almost.

[Aero]

I think we can agree to disagree. That is one thing that differentiates a myth from an accepted fact. So the myth is established.

Myth - "A hot water heater properly configured can deliver enough rocket thrust to launch itself into space."

[blaisepascal]

I think we can agree to disagree. That is one thing that differentiates a myth from an accepted fact. So the myth is established.

Myth - "A hot water heater properly configured can deliver enough rocket thrust to launch itself into space."

I think we can agree that a hot water heater, meaning in the sense of a myth a standard common domestic hot water heater "configured" by the sort of bodging and neglect a typical homeowner might do, can't launch itself into space.

Your contention is that a rocket could be built that consists of a pressure vessel capable of holding the same amount of superheated, pressurized water that a malfunctioning DHW tank could, plus appropriate plumbing to transfer the steam generated to a properly size De Laval nozzle to provide thrust, and that rocket, powered by nothing more than the energy content of the superheated pressurized water, could reach space.

You seemed to feel that the water flashing to steam would maintain a constant pressure head, thus allowing the nozzle to operate at a continuous high Isp and high thrust until almost all the water was used as reaction mass. And that most of the latent energy of the superheated water would be converted to thrust.

I disagree. I feel that a significant portion of the energy in the superheated water would go into the water->steam phase change. Water has a relatively high heat of vaporization.

Here's a more down-to-earth example: The last time I used "canned air" to cool down some computer components, the can was full of a compressed liquid propellant, which when I opened the valve flashed to vapor and escaped out the tube blasting the inside of my case with a fast air stream, cooling the overheated components and making the dust-bunnies scurry for cover. The situation in that can is analogous to the steam rocket: a pressure vessel filled with a fluid which was liquid at that temperature solely because it was under high pressure. When the container was opened via the valve, the liquid flashed to vapor and escaped at high pressure with some force.

My experience with that, and similar situations with liquid propane tanks, tells me that the steam rocket won't work: Before I was fully done getting the dust out of the fan blades and heat sinks of my computer (and verifying that the power supply fan would move freely in an air-stream, but not when supposedly powered by the power supply), the can was (a) cold enough for frost to form on it, (b) no longer capable of producing a stream of "air", and (c) not empty yet, as could be determined by the sloshing of the contents. Similarly, it is well known that liquid propane tanks, after long use, also chill to the point of no longer working because the contents are too cold to vaporize.

The only solution is to provide external heat to bring them back up to operating temperature (the recommended way is to close the valves and let the ambient air temperature provide the heat; don't put a cold can of "air" or a cold propane tank on a fire).

I feel that the same will happen with a superheated water->steam rocket: It will work well for a while, but the vaporization of the water to produce the steam will sap energy (and thus temperature) from the water to the point where it simply will no longer vaporize at tank pressure.

I think this is what KitemanSA was getting at when he was saying that the total energy content of the water was sufficient to vaporize half the water.

[kunkmiester]

Nice explanation, blaisepascal. To expand, once you let the canned air heat up a bit, it works fine again. This happens when it absorbs energy from the atmosphere though, which the theoretical rocket wouldn't do fast enough to maintain the pressure.

If you vaporized the entire quantity of water, you might alleviate some of it, but not all. Condensation point is dependent on pressure, etc. You'll probably run out of pressure before steam.

[Aero]

Yes, I have used canned air to blow dust (and coffee) off my keyboard. The container gets very cold. Its not the same thing.

With canned air, or propane, the liquid phase in the can (tank) evaporates, cooling the liquid and carrying the heat of vaporization out the nozzle with the gas. Simple evaporative cooling of the container and contents. A steam rocket built like that wouldn't work for exactly that reason, and as you wrote.

A steam rocket uses the liquid. The pressure in the tank ejects the high temperature liquid water out the nozzle. The heat energy within the ejected water is lost to the tank but so is the water. (More later) The volume of the ejected water is replaced by steam in the tank. A small amount of the water evaporates to maintain pressure - temperature conditions, but this energy is confined within the tank, not blown across my keyboard. Just how much energy does it take to maintain this pressure-temperature conditions? That is the question. I don't know the answer precisely, but I do know that steam rockets work, so it seems obvious to me that the answer is, "Not so much." Its because steam has a much higher volume than water at the same pressure-temperature conditions so only a little water evaporates to replace the ejected water. This does evaporatively cool the water in the tank, but not much evaporation so not much cooling. You could test this with you can of "liquid air." Just hold it upside down and squirt it out the window. The can will empty before it gets cold.

Back to the ejected water and the thrust produced. As in all rocket engines, the thrust force consists of two parts. Part one is the pressure imbalance between the tank and outside ambient as seen through the engine throat. That is Asub_t (Pt - Pa), throat area times the difference in tank pressure and ambient. Note that this is the term that gives thrust to a bottle rocket. The other term is harder. Understand that the flow in a rocket engine is choked in the throat, that is hot water rocket or chemical rocket exhaust flows through the engine throat at the speed of sound. The engine nozzle downstream from the throat allows the gas or water steam to expand converting the heat energy to directed motion with increased velocity. This change in momentum of the exhaust contributes thrust to the rocket engine. Thrust due to pressure and thrust due to momentum change are about equal, each contributing about half of the total thrust.

In the steam rocket, the energy to drive the expansion comes from the hot water flashing to steam as pressure reduces in the expanding nozzle. The steam cools and accelerates supersonically. In fact, some chemical rockets tested expand the exhaust so much that in the vacuum of space, ice forms at the nozzle exit. (Water is a major component of most chemical rocket engine exhausts.) Thus, the major cooling that you are concerned about occurs in the rocket nozzle, not in the tank. This is illustrated quite well by one of the videos in the blog article linked above where the author had his small hot water rocket strapped to a chair. He turned on the engine which thrust and toppled the chair. He instinctively lunged to catch it and caught the steam blast in his face from about an arms length away from the nozzle. He reported that the steam seemed to be about body temperature.

As an aside, check out my analysis of the blog author's rocket. In the blog, the author reported that his later rocket created thrust for about 25 seconds. The first 10 seconds there was good, constant thrust. After 10 seconds, thrust dropped to about half the initial level and maintained that level for another 10 seconds, then tailed off to zero in the final 5 seconds. His tank is a cylinder for pressurized gas. His problem is that by laying the tank horizontally on his test rig, the water level drops below the level of the exit opening of the tank. Now steam from within the tank, mixed with hot water, is ejected into the engine. This steam does in fact carry the heat of evaporation out the nozzle, just like in the can of liquid air. The mass of steam is less than the mass of water so the mass flow rate drops, hence thrust drops at 10 seconds. At 20 seconds the flow through the throat goes subsonic, nozzle expansion breaks down and all that is left is some water boiling at atmospheric pressure in the tank.

[Josh Cryer]

edit double post, sorry