Understanding why oxyhydrogen does not increase mpg and violates laws

In summary: The first reason is that the engine does not run as hot as it could, and so is less efficient. The second reason is that the hydrogen is not burned completely in the engine, and so something like 3-5% of the hydrogen is wasted as heat in the engine.
  • #1
alpuy
4
0
Hi all, i agree with you that those guys that use Oxyhydrogen to increase MPGs in their vehicles must be violating some physics laws but the aim of this post is to understand why.

Lets call:
Eq : Chemical Enegry
Ee : Electrical Energy
Emec : Mechanical Energy:
Ecal : Calorific Energy

I imagine an engine as a device that turns Eq stored in the fuel into Ee in the alternator, Emec for moving the car and and Ecal for heating the engine and thrown to the air in the exhaust.

So Eq = Ee + Emec +Ecal

When the guys use Oxyhydrogen the engine wastes more energy in electricity let's call it Ee1 and the mecanical energy is reduced by Ee1.

So Eq = Ee + Ee1 + Emec - Ee1 + Ecal

Emec2 = Emec - Ee1

Ee2 = Ee + Ee1

Eq = Ee2 + Emec2 + Ecal, Emec2 < Emec and Ee2 > Ee

k*Ee2 is the energy that the Oxyhydrogen has.

Eq + k*Ee2 = Ee2 + Emec3 + Ecal3

Can it happen that by adding the Oxyhydrogen Ecal3 is reduced more than the increase of Emec3?

Which laws is it violating?

Thank me for helping me understand this matter.

Regards
 
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  • #2
Without even looking at what you did - energy conservation tells that you can't get more out than you put in. So every analysis that shows otherwise must be wrong, period.

It is like if you stated "I found a right triangle in which [itex]a^2 + b^2 \neq c^2[/itex]" - I don't have to check your numbers to know they are wrong.

Beware: what you are trying to discuss borders a banned subject, see the forum rules[url="http://"].[/url]
 
  • #3
I don't think that it will be banned because i try to understand something that i don't get.
I say that an increase in mechanical power can be explained by a decrease in heat.

Please don't ban it and tell me where I am wrong i want to learn.
 
  • #4
Engine efficiency depends on how much heat is wasted, no doubt about it. The higher the working temperature, the higher the efficiency - which is why diesel engines are better in converting chemical energy into the mechanical energy.

I remember reading that in HHO (in a reasonably reliable source) systems working temperature is slightly higher than the working temperature in "normal" cars. Definitely not enough to change the efficiency substantially (especially taking into account fact you need some of the energy for hydrogen generation).

However, this is in general not as simple as just juggling numbers and shifting energy from one place to another. Analysis of such systems requires quite advanced thermodynamics (which I don't feel at home with).

This was discussed many times before. You may try to search the forum for HHO.
 
  • #5
alpuy said:
I don't think that it will be banned because i try to understand something that i don't get.
I say that an increase in mechanical power can be explained by a decrease in heat.

Please don't ban it and tell me where I am wrong i want to learn.
First place to start is studying thermodynamics cycles for a heat engine.

The efficiency of a heat engine is ratio of the useful work output to the energy input or heat input. This can also be related to the temperatures of the hot side T(h) and the cold side T(c), which is also related to heat input Qh at the hot side versus the heat output or heat rejected at the cold side Qc. The greater difference in temperature, the more efficient the heat engine will be. The cold side can be water from a lake for a steam power plant, for example, or the air for a car motor. Since we cannot change the lake temperature or the air temperature, we are left with increasing the Hot temperature, but there we are limited by physical constraints of materials to withstand high temperatures.

Carnot, several years ago determined a most efficient cycle called the Carnot cycle, where this cycle is the most efficient of all, obtaining the most work output W from heat input Qh. Most other cycles are less efficient, but can approach the efficiency of the Carnot cycle to various degrees. The Carnot cycle is practically useless, since its power output is so low, but does serve to illustrate the limitations of converting heat to useful work - there is always some heat rejected to the environment. The next best would be the Stirling cycle, which also theoretically can be as efficient as the Carnot cycle, but to obtain useful power output the efficiency does drop in practice.

A gasoline engine uses what is called the Otto cycle. Being NOT a Carnot cycle, it is less efficient, and more heat is rejected to the environment than for the Carnot cycle, and subsequentially less work W is obtained from the heat input Qh.

There are two reasons why the HHO injection does not increase mpg:
1. A violation of conservation of energy - if one inputs a set quantity of energy into a sytem, the energy output has to be the same amount, whether we add up the work output, heat output, or whatever.
2. A violation of thermodynamics, in that no other cycle or combination of cycles can be more efficient than the Carnot cycle using the same Th and Tc,

Statement 2 is where these things such as HHO falter.

Just consider this, and it is well known for anyone familiar to thermodynamics, and it is usually expalined with a two cycles operating back to back, one being a Carnot and the other some unknown cycle with a better efficiency, with both operating between the same Th and Tc. If the unknown cycle was more efficent, then we would have a situation where the unknown engine could drive the Carnot in reverse ( the Carnot is a reversible heat engine ) with the Carnot then providing the Qh into the unknown engine, which then provides a net work output, by elimination of the hot resevoir at Th.

This is a violation of second law of thermodynamics, statement of Clasius ( that it is impossible to construct a heat engine that transfers heat from a cold resevoir to a hot resevoir without work input ) and the Kelvin-Planc statement ( that one cannot construct a cyclic heat engine to do work using only one heat resevoir )

The automobile HHO arguments violate the same principles. We can re-model and see why so, by the following:
( Remember, we have to use some of the work output from the Otto engine for HHO production, so that work is not available to drive the car )
Instead of injecting the HHO into the original engine, we instead add a second engine to the car that accepts only the HHO produced.
The first engine, an Otto engine, is operating between two resevoirs at Th and Tc and thus has a calculable efficiency, accepting heat input Qh, producing work W, and rejecting heat Qc .We will use the work W to drive an generator to prduce HHO for injection into the HHO engine.
Suddenly we can now see that the HHO engine has to be 100% efficient, with NO heat rejection, all heat input from Th from the combustion of HHO being converted into work Whho, for the same efficiency to drive the car as before using only the Otto engine. Since NO heat engine can be of a greater efficiency than one operating as a Carnot cycle, such a situation is impossible. Not to mention that the HHO production has to be 100% efficient also.

The HHO engine subsequentially, due to thermodynamics, will have to reject heat to the enviroment, and adding that to what the Otto engine already rejects to the environment, we have made the whole system less efficient, producing less useful work as a result.

A few reference you may be interested in for a start on thermodynamics.
http://en.wikipedia.org/wiki/Carnot_cycle
http://en.wikipedia.org/wiki/Carnot_efficiency#Efficiency
 
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1. Why does oxyhydrogen not increase mpg?

Oxyhydrogen, also known as HHO, is a mixture of hydrogen and oxygen gases that is often touted as a way to increase fuel efficiency in vehicles. However, scientific studies have shown that adding oxyhydrogen to a vehicle's fuel supply does not actually increase mpg. This is because the energy released from the oxyhydrogen combustion is not enough to offset the energy required to produce and install the system.

2. How does adding oxyhydrogen to a vehicle's fuel supply violate laws?

The use of oxyhydrogen as a fuel additive violates the first and second laws of thermodynamics. The first law states that energy cannot be created or destroyed, only transformed from one form to another. In the case of oxyhydrogen, the energy used to produce the gas is greater than the energy released during combustion, making it an inefficient use of energy. The second law states that the total entropy of a closed system (such as an engine) will always increase over time. Oxyhydrogen combustion produces more water vapor, which increases the entropy of the system and decreases its efficiency.

3. Can oxyhydrogen actually harm my vehicle?

While oxyhydrogen does not increase mpg, it also does not typically harm a vehicle's engine. However, there have been reports of oxygen sensors and catalytic converters being damaged by the water vapor produced during oxyhydrogen combustion. Additionally, the installation of an oxyhydrogen system may void a vehicle's warranty, so it is important to consult with a mechanic or manufacturer before adding one.

4. Are there any other alternatives to increasing mpg?

Yes, there are many other ways to increase a vehicle's fuel efficiency, such as maintaining proper tire pressure, regular tune-ups and oil changes, and using fuel-saving driving techniques. Additionally, advancements in electric and hybrid vehicles offer even greater fuel efficiency without the need for additives like oxyhydrogen.

5. Why do some people still claim that oxyhydrogen works?

Some people may still claim that oxyhydrogen works because of anecdotal evidence or biased studies. It is important to rely on scientific evidence and research when evaluating the effectiveness of fuel additives. Additionally, some individuals may have a financial interest in promoting oxyhydrogen systems, which can lead to misleading claims about their effectiveness.

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