BTU Q's about burning water vs. burning gassoline

1. Sep 24, 2004

MonstersFromTheId

In another thread (under Dr. Kaku's EXPLORATIONS program page, the thread about the idea of a "Space Elevator"), the conversation turned to the prospect of using something like a "Space Elevator" as the basis for shifting from a petroleum based economic model to essentially a water based economy (or hydrogen based if you prefer), and we need some numbers for a few "back of an envelope" calculations.

Thus the following questions:

1) If you take a gallon of water, and separate it out into h & o2 by way of electrolosis, how many cubic feet of hydrogen and oxygen do you get (at room temp and sea level psi)?

2) If you now take that h & o2 (that you got from one gallon of water), and you just burn it (turning it back into water), how many BTU do you get out of it?

3) How does that compare to gassoline? If you burn a gallon of gassoline, how many BTU do you get out of it?

4) If, instead of immediately burning the resulting h & o2 from a gallon of water, you instead cool it back down into liquid h and liquid o2 (for more compact transport as an energy supply), does the sum of the resulting amount of liquid h and liquid o2 still add up to one gallon?

5) How cold does h have to be before it liquifies at, say, 14.7psia? (I.e. would simply leaving h gas in a container in a shadowed area in orbit where temps can drop to several hundred degrees below zero be enough to liquify the gas?)

6) Ditto for the o2. How cold does o2 have to be before it liquifies at 14.7psia?

7) How much energy is required to separate a gallon of water into hydrogen and oxygen in the first place?

2. Sep 25, 2004

chem_tr

Hello,

In all circumstances, hydrogen would give a better yield compared with gasoline, as burning of the latter produces some unwanted side products like tar, etc.

But please keep in mind that electrolysis require high energy to produce hydrogen and oxygen, so the overall energy gain seem to be somewhat higher in gasoline, I presume.

Regards, chem_tr

3. Sep 25, 2004

chem_tr

As you know, with electrolysis, $$H_2O \rightarrow H_2+1/2~O_2$$ reaction occurs. One US-gallon of water is 3.785 liters, I found like that. As $$d=1 g/mL$$ for water, 3.785 liters of water makes 3.785 kilograms, equal to $$\frac{3.785}{18}=0.2103~kilomols=210.3~mols$$. If we assume that these gases behave ideally, we'll get that $$V=5052.7$$ liters of hydrogen, which is equal to $$1334,9$$ US gallons, if I did the calculations right. As 1 US gallon is $$0.134$$ cubic foot, you may find the rest from there. Note that half a mole of oxygen is produced, and divide the resulting volume by 2 to learn the volume of oxygen.
Gaseous H2O has $$\Delta H^{0}_{f}=-241.818~kJ.mol^{-1}$$. So 210.3 moles of gaseous water gives $$50,854.3~kJ$$ of energy. I have no idea what BTU is, but I think you can do the rest.
Well, this is a hard question; you'll find the formation enthalpies of all of the constituents present in gasoline, and calculate the overall energy gain according to the amounts expressed as moles. If you give us the numbers, maybe we'll be able to calculate. But you may also give the result obtained with a calorimeter bomb, this is far better than calculating, of course.
I don't think so, since "shrinkage constants" would be different as with every single element. Also keep in mind that every gas has a different supercritical point; that is, you have to supply a surrounding temperature lower than this point, otherwise, you cannot liquefy any gas however high pressure you apply (as high as one million atmospheres won't work at all).
As I said in the last answer, you will know it only when you find the supercritical constant of hydrogen. This is the temperature of liquefying of a gas with applied pressure. Note that in orbit there is vacuum, not pressure; I am doubtful about liquefying, indeed. You may have enough temperature to start liquefying, but there is no pressure.
I think it is the reverse of formation enthalpy; we found that a gallon of water is 210.3 moles, and already calculated the energy given out. You should give this energy at least to cause something.

Regards, chem_tr

Last edited: Sep 25, 2004
4. Sep 27, 2004

MonstersFromTheId

Tx! chem_tr. HUGE help!

First;
with re to electrolysis taking quite a bit of energy to drive the process - prolly wouldn't matter TOO much in this particular case.
The idea here is to use water as essentially a clock spring.
You haul the "clock spring" (water) up into orbit on a "Space Elevator" and wind the "clock spring" there (by separating it out into h & o2), where you have access to the essentially open acreage needed to do it on a massive scale (in the form of many fairly large orbital solar collection/electrolysis facilities) without having to cover half of a state the size of Texas with solar collectors to obtain the required energy.
Once the "clock spring" is wound (i.e. the water separated out into h & o2), you send the now wound "clock spring" (the resultant h & o2) back down the "Space Elevator" where it can be used to power things (i.e. unwind), by burning it, thereby turning it back into water, and the cycle repeats.
The only reason you'd want to liquefy the h & o2 is to get it into a compact and more efficiently transportable volume for the trip back down the Space Elevator, and from there to wherever it's going to be used as a power source.

With re to no pressure in orbit, so gaseous h & 02 won't liquefy - not a problem. You wouldn't simply allow the gaseous h & o2 to vent out into the raw vacuum of space.
The idea would be you'd pump the resulting h & o2 into a set of tanks *and as the h & o2 began to build up inside their respective tanks the pressure in the tanks would begin to increase*.
Of course as the pressure increased so would the temp of the gasses, but if the tanks are kept within shadow (where temps are several hundred degrees below zero), and the building heat is allowed to radiate away (through fins on the tanks, or by passing the gasses through a series of pipes that are also in shadow to form a condenser), the gasses may liquefy without having to build and power massive heat pumps to dump the heat energy the gasses need to lose in order to liquefy.
I.e. you make the most of the natural heat sink you've got right there at hand in the way of the very low temps present anywhere in orbit that you don't have impinging sunlight.

As to "BTU";
BTU stands for "British Thermal Units", a unit of measure for energy commonly used by "HVAC" engineers in their calculations. (HVAC stands for "Heating, Ventilating, and Air Conditioning").
And you're quite right, I'm absolutely certain it will be very easy to find the conversion factor between Kilojoules and BTU in any good encyclopedia.

And as to the rest...
HUGE thanks chem_tr! This gives me a very nice starting point.