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How to find the efficiency of a modified engine?

02/25/2008 12:15 PM

In a 4-stroke PETROL engine if an extra supply of Oxygen(pure O2-A constant supply of oxygen will be given in every suction stroke) is given through the carburator then will it increase the efficiency of the engine by increasing the percentage of fuel combusted?

I also wanted to know, how the increased efficiency can be calculated theoretically?

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#1

Re: How to find the efficiency of a modified engine?

02/25/2008 1:22 PM

A modern, computer-controlled, multi-port fuel-injected engine is already matching the fuel charge to the available oxygen in the air charge. Combustion is already at nearly 100% in order to meet government mileage and emission requirements. The charge of pure oxygen would certainly allow a greater fuel charge to be burnt, but that is what a nitrous system is all about - it results in a great increase in power, but no real change in efficiency. The inefficiency in an IC engine is mainly due to the great amount of energy converted to waste heat, rather than mechanical energy (much like the great amount of heat rather than light generated by incandescent bulbs).

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#2
In reply to #1

Re: How to find the efficiency of a modified engine?

02/26/2008 12:14 AM

I use pure oxygen because of a pulmonary problem. There are warnings all over the place about the dangers of using pure oxygen around any petroleum products. I have seen the same warnings on gas welding equipment. Wouldn't the use of pure oxygen around an internal combustion engine also present the same potentially dangerous conditions?

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#5
In reply to #2

Re: How to find the efficiency of a modified engine?

02/26/2008 5:35 AM

You seem to be the only poster with any knowledge and good sense about the mixing of pure oxygen and anything flammable. I rate your post as good.

What you actually get is a very severe explosion and fire and probably nothing left standing for several 100 meters, plus an awful lot of heavy stuff doing crash landings for even further distances.....for several minutes after the bang, some of the bits may have been part of your body up to very recently....

NEVER EVER EVEN DREAM OF DOING THIS AS THERE IS NO ADVANTAGE ANYWAY, the energy cost of the Oxygen is far more than the extra energy gained...losses as always.

If you want to see a way to get more out of an IC engine with less fuel, then look at Wikipedia and find 6 stroke engines. They run cool enough to put your hand on, need no cooling equipment and appear to be very fuel efficient.

The Crower appears to be a particularly good version and he also has a simple website to visit....check this link out by clicking on it:-

Six stroke engine

Crower six stroke

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#8
In reply to #5

Re: How to find the efficiency of a modified engine?

02/26/2008 9:01 AM

The Velozeta six-stroke engine seems a better option than the Cower since it may be possible to just shut off the fuel for one cycle to accomplish its operation. The Cower needs energy-costly distilled water and the entire injection system its injection entails.

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#3

Re: How to find the efficiency of a modified engine?

02/26/2008 12:34 AM

Any free oxygen left in the engine after combustion of the fuel will corrode the engine parts, especially the outlet valves.

I would propose that pure oxygen be used very carefully as it could have severe side effects like explosion with oil or other chemicals in the engine.

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#4

Re: How to find the efficiency of a modified engine?

02/26/2008 1:08 AM

Since O2 makes up about 21% of the air normally used, you would (in principle) be able to burn (100%/21%) 4.76 times more fuel per power stroke.

But I'd guess that adding anything like that amount of extra fuel will reduce the efficiency greatly. As well as (many) other things the pressure-temp during combustion versus the pressure-temp at the end of the power stroke will have been adjusted during the design and testing cycle to maximise the motor's efficiency.

Still it looks like a fun project, good luck. ffeJ

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#11
In reply to #4

Re: How to find the efficiency of a modified engine?

02/26/2008 4:54 PM

I may be a bit sensitive, but why would this be considered off topic? ffeJ

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#12
In reply to #11

Re: How to find the efficiency of a modified engine?

02/27/2008 5:19 AM

I guess its because you (and many others) did not include in your post a warning about the explosive effects of pure oxygen when brought near anything remotely flammable!!

We try to be as FULLY accurate as possible so that none of the original questioners/posters ends up on the Darwin awards page. See link for further details.

You simply forgot!!

Darwin Awards

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Anonymous Poster
#6

Re: How to find the efficiency of a modified engine?

02/26/2008 6:25 AM

O2 is an oxidizer and to get it around petroleum products, you're asking for trouble (read BIG BANG when it oxidizes with the product). Nitrous used in engines has a good bit of oxygen but not pure oxygen. It works by the amount of O2 in the nitrous is more to burn but additional fuel has to be added or it will run lean and burn up the engine. Putting extra O2 into the engine will not increase the efficiency of it. The efficiency of it is determined by the amount of fuel put in and it's BTU content vs. the engine output in HP. You're not going to get much better than what came from the factory in a computer controlled car. There are things you can do that may increase it a couple of percent but that's going to be about it.

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#7

Re: How to find the efficiency of a modified engine?

02/26/2008 8:20 AM

You seem to confuse efficiency and power.

Adding any amount of Oxygen will not improve efficiency. It will not even make any noticeable increase in power, unless more fuel is added to the system. There is a certain and FIXED ratio between the amount of fuel and air that will give you the Ideal mixture. This ratio about 1:15, 1 part of fuel to 15 parts of air ( petrol or gas engines),when changed, will cause a lean or reach mixture. If more fuel is added without any more air it will simply not burn and will evaporate in the exhaust pipe, or may even explode there. (Backfire) If more Oxygen, or air for that matter is added the mixture will become lean, and at best will cause engine heat to rise. By adding pure Oxygen, you are also subjecting the engine to a possibility of oil/Oxygen mixture which is highly explosive, and very very dangerous. In addition free O2, is very aggressive, and highly corrosive to almost all metals, so you are looking now at a whole new set of problems.

Summery: you get nothing but problems, without any added value. Better scrap the idea.

Hope this helps.

Wangito.

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#9

Re: How to find the efficiency of a modified engine?

02/26/2008 10:44 AM

It won't increase the percentage of fuel combusted. In a lean-burn engine, all of it is combusted anyway.

Increasing the oxygen concentration above ambient will reduce the amount of hot nitrogen leaving in the exhaust and increasing the fuel supply in concurrence will increase the available engine output power, up to a limit. Excess of oxygen in some combustion processes generates enough concentrated heat to cut metal. And if the engine block is made of metal that's bad news for the longevity of the engine; one is making a bomb. Plus, some parts of the engine may not be able to withstand the extra torque and speed. All in all it's a good recipe for saying "bye-bye engine".

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#10

Re: How to find the efficiency of a modified engine?

02/26/2008 3:36 PM

When you asked about efficiency of an internal combustion engine, we measure several kinds.

First, there is engine efficiency which is the ratio of power produced by the engine (Brake Horsepower) and the power supplied to the engine (heat content of fuel). By comparing fuel consumption to the engine power output, you can find engine efficiency.

Volumetric effciency is the ratio of actual air (oxygen) drawn into the cylinder and the maximum possible amount of air that could enter the cylinder It is how well an engine can breathe on its intake stroke.If volumetric efficiency were 100%, the cylinder would completely fill with air on the intake stroke. Engines are only capable of about 80 to 90 percent volumetric efficiency. Restrictions in the ports and around the valves limit airflow.

Mechanical efficiency compares brake horsepower and indicated horsepower. It is a measurement of mechanical friction. Indicated horsepower equals theoretical power produced by combustion. Brake horsepwer is the actual horsepower at the engine flywhee. Mechanical efficiency of around 70 to 80 percent is normal. This means about 20 to 30 percent of the engine's power is lost to friction (frictional horsepower loss).

Thermal efficiency is heat efficiency found by comparing fuel burned and horsepower output. It indicates how well an engine can use the fuel's heat energy. Thermal efficency measures the amount of heat energy converted into crankshaft rotation.

A gallon of gasoline, for example, has about 19,000 BTUs (British Thermal Units) of heat energy. One horsepower equals about 42.4 BTUs of heat energy per minute. With information you can find find engine thermal efficiency by using the following formula:

Termal Efficiency = Brake HP x 42.2 BTU/minute

19,000 BTU/gallon x gallons/minute

Generally, engine thermal efficiency is around 20 to 30 percent The rest of the heat is absorbed by the metal parts of the engine or blown out the exhaust.

Hope this helps.

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Andy Germany (1); Anonymous Poster (3); electrone (1); ffej (2); gdelvecchio (1); Johan van Niekerk (1); PWSlack (1); The_curious_one (1); wangito (1)

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