Internal combustion engine efficiency

[Hybrid_engine]

In the real world of combustion engine design, power output is the most important factor compared to quantity of fuel.

 

[cjl]

In the real world of combustion engine design, power output is the most important factor compared to quantity of fuel.

Power output per quantity of fuel (per time)= BSFC = efficiency. It's the same thing. I really don't understand why you're trying to make the distinction.

 

[cjl]

OK, I'll show you why I don't understand what you're saying:

Are you aware of the ability to transmit more torque to the crankshaft than pressure inside the combustion chamber?

Torque and pressure are different. They use different units. They cannot be directly compared. The statement "transmit more torque to the crankshaft than pressure" means absolutely nothing. I could similarly ask if you were aware of the ability to make your electric motor go faster by transmitting more voltage than RPMs. Yes, crankshaft torque is related to cylinder pressure, but you really can't directly compare the two, and the relation is complex.

Due to leverage at a given point of rotation? Have you studied any research on torque and combustion pressure relative to crankshaft degrees of rotation and cylinder pressure at intervals of all degrees of rotation with a specific fuel in a standard crankshaft combustion engine?

I am quite familiar with the thermodynamics and general principles of combustion engines. I still don't have the faintest clue what you are trying to get at here. Yes, combustion pressure as a function of crankshaft rotation is important (and things like mixture ratios, charge stratification, fuel injection type/location, ignition location, and spark timing are very important to this), but you really aren't saying anything of use here.

 

[Hybrid_engine]

Power output per quantity of fuel (per time)= BSFC = efficiency. It's the same thing. I really don't understand why you're trying to make the distinction.

My point is, if you say an engine is 50% efficient with given quantity of fuel and power output, and a completely different engine design has 4 times the power, torque and horsepower available at the drive shaft with same quantity of fuel, the way efficiency is calculated needs to be changed to real world horsepower and torque numbers for a predetermined quantity of fuel in each fuel category.

 

[Hybrid_engine]

OK, I'll show you why I don't understand what you're saying:
Torque and pressure are different. They use different units. They cannot be directly compared. The statement "transmit more torque to the crankshaft than pressure" means absolutely nothing. I could similarly ask if you were aware of the ability to make your electric motor go faster by transmitting more voltage than RPMs. Yes, crankshaft torque is related to cylinder pressure, but you really can't directly compare the two, and the relation is complex.I am quite familiar with the thermodynamics and general principles of combustion engines. I still don't have the faintest clue what you are trying to get at here. Yes, combustion pressure as a function of crankshaft rotation is important (and things like mixture ratios, charge stratification, fuel injection type/location, ignition location, and spark timing are very important to this), but you really aren't saying anything of use here.

 

[Hybrid_engine]

Real world useful power is very relative to combustion pressure and the mechanism of transfering the pressure to the power shaft. That is the whole point in combustion engines. Transferring pressure into rotational force or direct pressure.

 

[cjl]

My point is, if you say an engine is 50% efficient with given quantity of fuel and power output, and a completely different engine design has 4 times the power, torque and horsepower available at the drive shaft with same quantity of fuel, the way efficiency is calculated needs to be changed to real world horsepower and torque numbers for a predetermined quantity of fuel in each fuel category.

No, if an engine is 50% efficient with a given fuel flow rate, and a different engine has 4 times the power with the same fuel consumption, the second engine is impossible. The first engine is extracting 50% of the available energy from the fuel as useful work (this is the definition of 50% efficiency), so the second engine (in order to make 4 times the power) would need to extract twice as much energy as the fuel had. This is clearly not possible.

*Minor caveat, just to prevent future nitpicking: most ICE thermal efficiencies use the lower heating value of the fuel for their "fuel energy content" number. Technically, if you could condense the water vapor out of the exhaust and extract the latent heat of vaporization with some kind of thermodynamic cycle/engine on the exhaust, you could extract more energy from the fuel than this lower heating value would indicate, which could (if you fiddle with the numbers just right) allow for a maximum theoretical efficiency of >100% if you're using the standard formula for efficiency of shaft work out divided by the lower heating value of the fuel burned. In real engines though, the water vapor in the exhaust is a gas, and therefore the use of the lower heating value is correct.

 

[Hybrid_engine]

Everything you mention about timing, spark, air fuel ratio are what impact internal pressure. The mechanism of transferring the pressure is the engine design. To calculate torque you need to add the positive pressure and subtract the negative pressure transferred to the crank. Actually what you mentioned can be left out of the equation. Only need to measure pressures produced and calculate the leverage factors and mechanical efficiency of the engine. What you mentioned has impact on pressure increase or decrease. Not needed to calculate useable torque or horsepower. With exact same airfuel ratio and fuel and temperature in two different engines you get different power output as we all know. My point is, there exist more efficient transfer of energy, based on mechanical advantage, centrifugal and centripetal force, friction, pressure, all determine the final power output.

 

[Hybrid_engine]

No, if an engine is 50% efficient with a given fuel flow rate, and a different engine has 4 times the power with the same fuel consumption, the second engine is impossible. The first engine is extracting 50% of the available energy from the fuel as useful work (this is the definition of 50% efficiency), so the second engine (in order to make 4 times the power) would need to extract twice as much energy as the fuel had. This is clearly not possible.

*Minor caveat, just to prevent future nitpicking: most ICE thermal efficiencies use the lower heating value of the fuel for their "fuel energy content" number. Technically, if you could condense the water vapor out of the exhaust and extract the latent heat of vaporization with some kind of thermodynamic cycle/engine on the exhaust, you could extract more energy from the fuel than this lower heating value would indicate, which could (if you fiddle with the numbers just right) allow for a maximum theoretical efficiency of >100% if you're using the standard formula for efficiency of shaft work out divided by the lower heating value of the fuel burned. In real engines though, the water vapor in the exhaust is a gas, and therefore the use of the lower heating value is correct.

 

[Hybrid_engine]

You are assuming that the combustion pressure produced can only equal double the power if 100% efficient when compared to an engine you call 50% efficient. When you truly understand efficiency and mechanical output you will realize the engine you call 50% efficient is not delivering 50% of the torque and power potential of the fuel. It makes sense to you only if you look at the energy transfer system in a fixed design. Design an engine with constant offset and leverage factor, quadruple the leverage factor, reduce friction with air or magnetic bearings, eliminate g forces from directional changes, designed to give the optimum compression ratio for any fuel, use centrifugal and centripetal force to the maximum advantage in the given space. Is this impossible?

 

[cjl]

You are assuming that the combustion pressure produced can only equal double the power if 100% efficient when compared to an engine you call 50% efficient. When you truly understand efficiency and mechanical output you will realize the engine you call 50% efficient is not delivering 50% of the torque and power potential of the fuel. It makes sense to you only if you look at the energy transfer system in a fixed design. Design an engine with constant offset and leverage factor, quadruple the leverage factor, reduce friction with air or magnetic bearings, eliminate g forces from directional changes, designed to give the optimum compression ratio for any fuel, use centrifugal and centripetal force to the maximum advantage in the given space. Is this impossible?

Combustion pressure and details of the engine are irrelevant - the efficiency is defined as energy out/energy in. You can look at the whole system as a black box and not care about the details as far as an efficiency calculation is concerned - you just have to know the fuel flow in, and the shaft power out. Acceleration from directional changes isn't inherently lossy either - the amount of energy lost depends on the details of the linkages and the movements.

Also, what the heck do you mean by "leverage factor"? You can increase the leverage by increasing the stroke, but this also increases the piston velocity for a given RPM, which tends to decrease maximum RPM (this can be good for efficiency though, depending on the details of the engine). You really should use more standard terminology, and be clearer with your descriptions (though it sounds like you really don't know much about internal combustion engine design).

 

[russ_watters]

Locked pending moderation.
[edit]
I don't really have anything to add to cjl's excellent responses, so I'll just reiterate for emphasis:

Efficiency is simply the output power (mechanical work) divided by the input power (chemical energy per unit time) and is the same whether expressed as a percentage or brake specific fuel consumption. If a typical engine is 25% efficient and a claimed alternate design gets 8x the power output for the same fuel consumption it must be 25%*8=200% efficient, an obvious violation of conservation of energy.

Hybrid_engine, it appears to me that you are getting bogged down in details about internal combustion engines that you don't understand and allowing your misunderstandings to cause you to lose sight of the simple issue of how efficiency and conservation of energy work. Instead you should be using conservation of energy as a guiding principle to help focus your understanding of these other concepts. And you will need to start from scratch and learn quite a lot of thermodynamics before you can attempt to design your own new type of engine.

This thread will remain locked.