John Roberts {JR} wrote on Sun, 11 May 2008 13:39 |
Steve Weiss wrote on Sun, 11 May 2008 10:45 | OK please tell me this is a late April fools joke????
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No, it's a separate the fools from their money joke.
It takes electrical energy to perform electrolysis on the water, the H and O burn, releasing the energy that was put into it, zero sum game.
I will wait for Randy's actual gas mileage report, not what he believes will happen.
JR
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Its actually a little worse than a zero sum game, JR. I just did the thermodynamics out to double-check.
The heat extractable to the surroundings from the decrease in entropy caused by the formation of the H2O from combustion of H2 and O2 will always be lower, in the case of a reversible, isothermal process (ie the best case scenario).
This is a function of the definition of entropy under these conditions: dS=dQ/T. Since, for irreversible processes dS>dQ/T real processes are worse than in the reversible case.
Even if you consider the behavior of the system as reversible, and use an ideal Carnot cycle (ie two isothermal steps linked to two constant entropy steps) The energy from the chemical reaction of combustion of the hydrogen and oxygen (which takes place at the high temperature point of the carnot cycle) has an inherent penalty.
Ach, that's still complicated. One more try:
1. Put a fixed Q into the system by a chemical reaction at some high temperature: dQ=T_high(dS), or after integration at constant T: deltaQ=T_high*deltaS That means you get an amount of entropy S for a given Q.
2. Now you cool the system to a lower temperature under constant entropy conditions, so S stays the same. dQ is now: dQ=T_low(dS) or deltaQ=T_low*deltaS. Since S is fixed, and T is smaller, the extractable Q is always less at T_low that the Q you put in at T_high.
The difference between the input Q at T_high, and the extracted Q at T_low then represents the total work done by engine during its cycle.
It should be clear then that the lower T_low is, the less heat you have to pull from the system to return to the beginning of the Carnot cycle, and the more was converted to (Pressure)*(Volume) work on the surroundings during the isentropic expansion phase. Hence the desire for the largest temperature gradient possible between the Q in (chemical reaction) and Q out (e.g. radiator sending heat into the ambient air). Since T_low is not 0 Kelvin, and dS for a real engine will be greater than dS_ideal for a reversible system, there is always a penalty for the extraction of mechanical work.
In a real system, the input Q is fixed per amount of fuel burned, in this case from the electrolysis of water.
At a minimum you therefore take the Carnot efficiency penalty, and the efficiency penalty of the battery/alternator combination.
The are only two ways the system would seem to come out ahead. First would be with the production of the hydrogen by an external battery, which of course is just passing the buck for input Q.
The second way the system might possibly come out ahead would be if the injection of H2 and O2 dramatically increased the T_high for the car's engine (ie increasing the average internal cylinder temperature). The efficiency of a Carnot engine is related to the difference in (T_high-T_low)/Absolute temperature. I give this a remote, at best possibility of happening, but it is at least conceivable.
BTW, if anyone reads this and feels just totally confused, don't feel bad! I had two semesters of undergraduate thermodynamics/kinetics, and two semesters of graduate thermo/kinetics, and THEN TA'ed for a graduate level thermodynamics class. It didn't start to make any real sense to me until graduate school and TA'ing. J. W. Gibbs was a genius!
http://en.wikipedia.org/wiki/Carnot_cycle