Please check my math about this engine efficiency calculation

Click For Summary

Discussion Overview

The discussion revolves around the efficiency of a car engine when hydrogen and oxygen, produced through electrolysis of water, are injected alongside gasoline. Participants explore the calculations involved in determining the energy required for electrolysis compared to the energy released from combustion, focusing on the mathematical steps and assumptions made in the process.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the energy required for electrolysis of water and the energy released from the combustion of hydrogen and oxygen, but expresses uncertainty about the accuracy of their calculations, particularly in step 7.
  • Another participant questions the use of the term "efficiency" in the context of the calculations, noting that efficiency is typically expressed as a percentage and suggesting that the calculations do not reflect this.
  • Concerns are raised about the conservation of energy, with participants discussing the implications of energy input and output in the context of reversible reactions.
  • One participant acknowledges a potential error in their calculations regarding the energy required for electrolysis and revises their figures, questioning if the revised number is sufficiently close to the energy output from combustion.
  • There is a discussion about the impact of temperature on electrolysis and how it may affect the energy calculations.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and the concept of efficiency, with no consensus reached on the accuracy of the mathematical steps or the implications of energy conservation in the context of the reactions discussed.

Contextual Notes

Participants highlight potential limitations in their calculations, such as the assumptions made regarding the energy values used for electrolysis and combustion, as well as the effects of temperature on the electrolysis process.

Le Luc
Messages
10
Reaction score
2
Hello. I wanted to see how much more efficient a car would be if hydrogen and oxygen (created through electrolysis of water) were injected into the engine alongside gasoline. I got lost when I found out that I would need 241.8 kJ/mol of water to electrolyze it (https://chemistry.stackexchange.com...ch-energy-is-needed-to-break-a-water-molecule), but I would get 572 kJ just from the combustion of 2H2+O2 (https://www.quora.com/How-much-energy-is-released-in-hydrogen-combustion-reaction).

So here are the steps I took:
1) 1 gallon water is 210 moles
2) 241.8 kJ to split 1 mole of water
3) 241.8*210 = 50778 kJ
4) 50778 kJ to split 210 moles
5) 2H2(g) + O2(g) --> 2H2(g) + 572kJ
6) 210/2 = 105 (because the formula produces 2 H2O)
7) 105*(6.02214179*10^23)*572 = 3.616898359*10^28 kJ

Sorry if the math is incorrect beyond explanation.
To clarify step 7, I divided the # of moles because there were 2 water molecules produced.

I think step 7 is incorrect.
 
Engineering news on Phys.org
Le Luc said:
Hello. I wanted to see how much more efficient a car would be if hydrogen and oxygen (created through electrolysis of water) were injected into the engine alongside gasoline. I got lost when I found out that I would need 241.8 kJ/mol of water to electrolyze it (https://chemistry.stackexchange.com...ch-energy-is-needed-to-break-a-water-molecule), but I would get 572 kJ just from the combustion of 2H2+O2 (https://www.quora.com/How-much-energy-is-released-in-hydrogen-combustion-reaction).

So here are the steps I took:
1) 1 gallon water is 210 moles
2) 241.8 kJ to split 1 mole of water
3) 241.8*210 = 50778 kJ
4) 50778 kJ to split 210 moles
5) 2H2(g) + O2(g) --> 2H2(g) + 572kJ
6) 210/2 = 105 (because the formula produces 2 H2O)
7) 105*(6.02214179*10^23)*572 = 3.616898359*10^28 kJ

Sorry if the math is incorrect beyond explanation.
To clarify step 7, I divided the # of moles because there were 2 water molecules produced.

I think step 7 is incorrect.
It isn't clear to me what you are trying to do. You used the word "efficiency", but efficiency is expressed in percent and there are no percentages in your math. It almost looks like you are trying to calculate how much energy you get back by burning hydrogen and oxygen, which should be exactly how much you put into split it from water (per conservation of energy)...

...In step 5 you made the oxygen disappear...

But yeah, your big error is in step 7, in which you multiplied by Avagadro's number for no reason that I can see. You already had units of gallons, so there is no conversion needed.

Ultimately all you really did is do a really roundabout calculation of the heat of vaporization of water.
 
russ_watters said:
It isn't clear to me what you are trying to do. You used the word "efficiency", but efficiency is expressed in percent and there are no percentages in your math. It almost looks like you are trying to calculate how much energy you get back by burning hydrogen and oxygen, which should be exactly how much you put into split it from water (per conservation of energy)...

...In step 5 you made the oxygen disappear...

But yeah, your big error is in step 7, in which you multiplied by Avagadro's number for no reason that I can see. You already had units of gallons, so there is no conversion needed.

Ultimately all you really did is do a really roundabout calculation of the heat of vaporization of water.
Thank you very much. Since the energy I get out of combustion needs to be less than the energy I put in through electrolysis, would that mean that I would need to have more than 60060 kJ of input to get 60060 kJ of output, since step 7 without Avagadro's number is 105*572 = 60060 kJ?

In step 3, I got 50778 kJ to split 210 moles of water, so where would the missing 9282 kJ be? EDIT: I think I got the wrong number for energy required for electrolysis/mole. Now I got 285.8*210 = 60018 kJ and I see now that it is much closer to 60060 kJ. Is 60018 close enough to 60060?
 
Last edited:
Le Luc said:
Thank you very much. Since the energy I get out of combustion needs to be less than the energy I put in through electrolysis...
Do you understand that that is impossible? The reactions are reversible - the energy is the same going forwards and backwards. This is demanded by conservation of energy.
...would that mean that I would need to have more than 60060 kJ of input to get 60060 kJ of output, since step 7 without Avagadro's number is 105*572 = 60060 kJ?
Pay attention to the state (liquid, gas) of the products and reactants. If the energy you get back isn't exactly the same as the energy you put in, it is probably because you started with one and ended with the other.
 
russ_watters said:
Do you understand that that is impossible? The reactions are reversible - the energy is the same going forwards and backwards. This is demanded by conservation of energy.

Oh. I said that because of electrical resistance. Sorry for not stating that.

Since temperature affects electrolysis, then my missing 42 kJ was probably from using the wrong temperature for water. Finally. I think I get it now.

Thank you very much.
 
Last edited:
  • Like
Likes   Reactions: jim mcnamara and russ_watters