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Understanding why oxyhydrogen does not increase mpg and violates laws

  1. Sep 21, 2014 #1
    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 lets 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
     
  2. jcsd
  3. Sep 21, 2014 #2

    Borek

    User Avatar

    Staff: Mentor

    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.
     
  4. Sep 21, 2014 #3
    I dont think that it will be banned because i try to understand something that i dont get.
    I say that an increase in mechanical power can be explained by a decrease in heat.

    Please dont ban it and tell me where im wrong i want to learn.
     
  5. Sep 21, 2014 #4

    Borek

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    Staff: Mentor

    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.
     
  6. Sep 21, 2014 #5
    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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