A work of pure genius! - Brilliant "Revetec" Engine

[revetec]

The math here is incorrect. The equation I think you tried to use is Boyle's law which is actually:
pV = k
Where (p) is pressure, (V) is volume, and (k) is a constant (not temperature). So in the case shown the volume would increase thus decreasing the pressure to maintain the constant k.

I was actually refering to Adiabatic cooling but I kind of posted two theories that related to it but not exact. My apologies :-)

If the given parameter for peak pressure is applied to a cylinder with a 10:1 static compression ratio then the exhaust temperature would be roughly 920degC. In order to achieve 700degC you would need a static compression ratio over 19.5:1. If you look at the pressure parameters you gave (peak: 50bar gauge, minimum: 5bar gauge) that only requires a static compression ratio just over 5:1 to achieve at which point the exhaust temperature would be almost 1200degC. At 10:1CR you would only measure 0.6bar gauge for exhaust pressure and at the 19.5:1CR required for the given exhaust temperature of 700degC you would only measure 0.07bar gauge for exhaust pressure.

I'm positive you have never actually tested a real engine. With a sensor in the cylinder head, the combustion temperature is around or just below 2300k or 2023degC. This 10:1 engine's exhaust port temp is around 600-700degC. The exhaust valves open around 50-60deg BBDC where there is still residual cylinder pressure of around 5bar, which you can see in a cylinder pressure graph using a piezo sensor in the chamber. This is not theory, it is actual tests on real engines. With the exhaust manifold heat wrap insulated, the collector temp is around 450deg and then decreases down the pipe.

Firstly turbo-chargers don't increase manifold temperature much by restriction, rather the turbo allows the engine to burn more fuel per cycle, this and the increased exhaust temperature when the exhaust valve opens contribute to manifold heating. Of course the exhaust is cooler after the turbo it was expanded in the turbine section of the charger and heat was extracted and converted to power which was subsequently used to spin the compressor and compress the intake air.

It was expanded in the turbo and heat was extracted and converted to power? The turbo was turned by gas flow and it requires to make a restriction to generate power to produce the intake pressure. Once the gas flows through the turbo, Adiabatic cooling applies to some extent.

Please read the article about BMW's turbosteamer!

Nice article, funny they say they can recoup 80% of exhaust heat and it makes a 15% improvement. So that's 18% exhaust heat losses?
I'm more interested in what they state as the second energy supply... "Most of the remaining residual heat is absorbed by the cooling circuit of the engine, which acts as the second energy supply for the Turbosteamer." This is what I'm intrigued with. It doesn't say how the secondary system works. I would like a complete overview of the system functions. Can you find it for me?

[revetec]

In the BMW Turbo steamer, I wish I could find out the manifold pressure and temp. Maybe they have found with very slight restiction that they increase exhaust temps which can be then recouped to provide higher overall efficiency. I'm very interested in this. Love to get my hands on it and test it.

[hightower99]

I was actually refering to Adiabatic cooling but I kind of posted two theories that related to it but not exact. My apologies :-)

Well what you posted wasn't even close to adiabatic cooling, which is the formula I posted (after Boyle's law)

I'm positive you have never actually tested a real engine. With a sensor in the cylinder head, the combustion temperature is around or just below 2300k or 2023degC. This 10:1 engine's exhaust port temp is around 600-700degC. The exhaust valves open around 50-60deg BBDC where there is still residual cylinder pressure of around 5bar, which you can see in a cylinder pressure graph using a piezo sensor in the chamber. This is not theory, it is actual tests on real engines. With the exhaust manifold heat wrap insulated, the collector temp is around 450deg and then decreases down the pipe.

Actually I have been involved in real engine testing with all these parameters measured and another couple dozen parameters on top. I go to a university that has the equipment to do this sort of thing. My point with doing the quick calculations was that your example was not overly valid as you are assuming too much and therefore further points are invalid. For example you say that during real testing you see a residual pressure of 5bar gauge just as the exhaust valve opens and then state that this means there must be 5bar gauge in the exhaust manifold. Well that is plain wrong because the exhaust flowed through a poppet valve incurring a large pressure drop. Especially as there shouldn't be much residual pressure in the manifold when the exhaust valve opens. This is what drops the temperature to the measured 600-700degC range seen.

I am still trying to fathom your point. You seem to think that because the temperature and pressure drops in the exhaust system that it didn't contain as much energy as conventional theory says it started with? Conventional theory doesn't state that the exhaust maintains all that lost energy all the way to the tailpipe... Thats kinda why turbos tend to be placed as close as possible to the engine...

It was expanded in the turbo and heat was extracted and converted to power?

Yes that is how it works... You mentioned adiabatic cooling... Read up on turbo theory please...

The turbo was turned by gas flow and it requires to make a restriction to generate power to produce the intake pressure. Once the gas flows through the turbo, Adiabatic cooling applies to some extent.

In order for gas to flow there must be a pressure differential, this is created in the turbine section by expanding the exhaust gas as it flows through the turbine. As the gas is forced to expand exactly the same thing happens as when the gas was expanded in the cylinder, namely energy is extracted from the gas. Turbines are not like windmills which are dependant on the Bernoulli principle. If you blew cold air at 5bar gauge through a turbocharger it would not have enough power to produce boost on the intake air side.

Nice article, funny they say they can recoup 80% of exhaust heat and it makes a 15% improvement. So that's 18% exhaust heat losses?

Not sure what you find funny? 80% of the heat lost from the engine is recovered and used to power a steam circuit. If we used your values then total heat lost via exhaust from the engine is 37% and the turbosteamer reclaims 29.6% of the total fuel energy (80% of 37%). Steam engines are like any other heat engine in that they can't be 100% efficient so only a portion of the recovered heat is converted to power. In the example BMW gave where the engine gains 15% improvement then the steam circuit must be just over 50% efficient. That means that the heat loss has been reduced from 37% to 22%. Certainly no problem with conventional theory here, unless you thought that there was only 7% heat lost

I'm more interested in what they state as the second energy supply... "Most of the remaining residual heat is absorbed by the cooling circuit of the engine, which acts as the second energy supply for the Turbosteamer." This is what I'm intrigued with. It doesn't say how the secondary system works. I would like a complete overview of the system functions. Can you find it for me?

Here is another article with all the images I have been able to find of the turbosteamer system: Here.
As you can see it gets waste heat from the exhaust as well as from the cooling circuit.

In the BMW Turbo steamer, I wish I could find out the manifold pressure and temp. Maybe they have found with very slight restiction that they increase exhaust temps which can be then recouped to provide higher overall efficiency. I'm very interested in this. Love to get my hands on it and test it.

Well the images in the new article show manifold temperature of 800degC and the performance of the original engine is the same so if any extra restriction is present it is insignificant.

There are still many of my questions that remain unanswered! Can I assume from the lack of response, that I was correct on the relevant points?

[hightower99]

BTW Brad, I saw that you wrote a small section in the Technology Overview section of your website about the turbosteamer concept. It states:

Current development by BMW of their 'Turbosteamer' have claimed that they recover 80% of the exhaust heat which equates to a 15% gain. This means they have proved that 18% of the thermal energy is lost through the exhaust, not over 30%.

Unfortunately that would only be true if the entire turbosteamer system was running at 100% thermal efficiency. This is simply not possible as I have pointed out in my previous post. Maybe it would be prudent to remove this erroneous statement from your website before too many people read it and think you don't have the faintest idea how heat engines work

You are welcome by the way

Maybe I should take a close look at the rest of your website and help you out by making sure all statements are factual and correct...

PS: I was just looking through the Orbital testing report and I noticed that it made a few mentions of a so-called "world wide mapping point" before the actual results are shown then doesn't make mention of the fact that the engine only achieved 490g/kW/h (16.7% thermal efficiency)... Interesting. It looks like other conventional engines are achieving between 21% and 27% thermal efficiency at the world wide mapping point. Very interesting indeed...

[revetec]

Well what you posted wasn't even close to adiabatic cooling, which is the formula I posted (after Boyle's law)

At the time I wasn't aware of your knowledge and was showing the volume temperature connection.

Actually I have been involved in real engine testing with all these parameters measured and another couple dozen parameters on top. I go to a university that has the equipment to do this sort of thing. My point with doing the quick calculations was that your example was not overly valid as you are assuming too much and therefore further points are invalid. For example you say that during real testing you see a residual pressure of 5bar gauge just as the exhaust valve opens and then state that this means there must be 5bar gauge in the exhaust manifold. Well that is plain wrong because the exhaust flowed through a poppet valve incurring a large pressure drop. Especially as there shouldn't be much residual pressure in the manifold when the exhaust valve opens. This is what drops the temperature to the measured 600-700degC range seen.

Haven't you seen an engine run with the exhaust manifold off. There is still residule burning (flames) where the gas is still heating and expanding.

I am still trying to fathom your point. You seem to think that because the temperature and pressure drops in the exhaust system that it didn't contain as much energy as conventional theory says it started with? Conventional theory doesn't state that the exhaust maintains all that lost energy all the way to the tailpipe... Thats kinda why turbos tend to be placed as close as possible to the engine...

That's not what I stated at all. Maybe you should read what I said again.

In order for gas to flow there must be a pressure differential, this is created in the turbine section by expanding the exhaust gas as it flows through the turbine. As the gas is forced to expand exactly the same thing happens as when the gas was expanded in the cylinder, namely energy is extracted from the gas. Turbines are not like windmills which are dependant on the Bernoulli principle. If you blew cold air at 5bar gauge through a turbocharger it would not have enough power to produce boost on the intake air side.

Keep in mind the relationship between heat, volume, and pressure when we talk about gasses. High heat, high pressure, and low volume are all high energy states, low heat, low pressure, and large volumes are low energy states.

The exhaust pulse exits the cylinder at high temperature and high pressure. It gets merged with other exhaust pulses, and enters the turbine inlet - a very small space. At this point, we have very high pressure and very high heat, so our gas has a very high energy level.

As it passes through the diffuser and into the turbine housing, it moves from a small space into a large one. Accordingly, it expands, cools, slows down, and dumps all that energy - into the turbine that we've so cleverly positioned in tho housing so that as the gas expands, it pushes against the turbine blades, causing it to rotate. We've just recovered some energy from the heat of the exhaust, that otherwise would have been lost.

We are agreeing on many points, and the arguement is a bit silly as we are agreeing on the same thing from different views. Mine pressure and yours, temperature, but they go hand in hand.

Not sure what you find funny? 80% of the heat lost from the engine is recovered and used to power a steam circuit. If we used your values then total heat lost via exhaust from the engine is 37% and the turbosteamer reclaims 29.6% of the total fuel energy (80% of 37%). Steam engines are like any other heat engine in that they can't be 100% efficient so only a portion of the recovered heat is converted to power. In the example BMW gave where the engine gains 15% improvement then the steam circuit must be just over 50% efficient. That means that the heat loss has been reduced from 37% to 22%. Certainly no problem with conventional theory here, unless you thought that there was only 7% heat lost

Here's a quote for you... "Turbosteamer converts more than 80 percent of the heat energy in the exhaust into usable power, says Raymond Freymann, head of BMW's advanced research and development subsidiary."

Head of BMW's research and development states they convert 80% of the heat energy in the exhaust into usable power. He does not say they use 80% of the heat to produce 15% power gain. So I gather the system has 20% losses due to further exhaust heat not recouped and system inefficiencies.

There are still many of my questions that remain unanswered! Can I assume from the lack of response, that I was correct on the relevant points?

No. I actually work for a living, and I have limited time per day to post on here. Funny for you to think I have time to reply to everything, as this is the least important task I do. I do enjoy the mental stimulation, and I never stop learning as the research is beneficial to me. But I don't have the time to engross myself into it these days as I design, problem solve, correspond, travel, modify designs, calculate, run a business, negotiate, budget, promote, market etc... and I am involved in all aspects of the business.

[revetec]

Unfortunately that would only be true if the entire turbosteamer system was running at 100% thermal efficiency.

Read my previous response and quote from BMW.

BTW Brad, I saw that you wrote a small section in the Technology Overview section of your website about the turbosteamer concept. It states:
Unfortunately that would only be true if the entire turbosteamer system was running at 100% thermal efficiency. This is simply not possible as I have pointed out in my previous post. Maybe it would be prudent to remove this erroneous statement from your website before too many people read it and think you don't have the faintest idea how heat engines work

You are welcome by the way

Maybe I should take a close look at the rest of your website and help you out by making sure all statements are factual and correct...

PS: I was just looking through the Orbital testing report and I noticed that it made a few mentions of a so-called "world wide mapping point" before the actual results are shown then doesn't make mention of the fact that the engine only achieved 490g/kW/h (16.7% thermal efficiency)... Interesting. It looks like other conventional engines are achieving between 21% and 27% thermal efficiency at the world wide mapping point. Very interesting indeed...

The thing is that every engine is different and also when the engines are at different load and RPM conditions. You can easily map a cylinder pressure map against the mechanical design of a crankshaft connecting rod. You can see losses in the design on paper. Then graph and calculate our design and compare the two. You must agree that side and down forces on bearings and other engine components do not contribute to engine power. If you graph a crankshaft efficiency you will see that there is losses which are not included in the losses stated everywhere. These losses in the combustion cycle don't show up in a physical pumping loss tests. So what is wrong, and where does these mechanical losses sit in the losses equation?

We are doing better at this test point and others currently with engine mods and tuning. That testing was performed over 2 years ago. And at the time had only spent 2 hours tuning the engine in ignition and fuel maps.

[revetec]

Anyway off to work now. Have a great day :-)

[hightower99]

Haven't you seen an engine run with the exhaust manifold off. There is still residule burning (flames) where the gas is still heating and expanding.

That depends entirely on the engine. Top Fuel dragsters will blow 1meter long flames out of 1 meter long open pipes, a Peugeot 107 won't blow any flames out the exhaust port unless you floor it past 4000RPM and then wildly feather the throttle (ask me how I know that) I find it alittle hard to believe that your engine blows flames out the ports when running at 15.2:1 A/F and opening the exhaust valves at 130deg ATDC. Also testing with the manifold off introduces unburnt oxygen which normally allows excess fuel to maintain a flame when normally none would exist. Regardless there is still a massive pressure drop over the poppet valve.

That's not what I stated at all. Maybe you should read what I said again.

I do read your posts but I am still confused as to what your actual theory is? I realize that you think there is less energy lost in the exhaust but I fail to understand your rationale.

Keep in mind the relationship between heat, volume, and pressure when we talk about gasses. High heat, high pressure, and low volume are all high energy states, low heat, low pressure, and large volumes are low energy states.

Well you almost got that right. You switched the volume relation. Remember that pV = internal energy, the larger the volume, the more internal energy. I think we might have stumbled into your basic misunderstanding!

The exhaust pulse exits the cylinder at high temperature and high pressure. It gets merged with other exhaust pulses, and enters the turbine inlet - a very small space. At this point, we have very high pressure and very high heat, so our gas has a very high energy level.

Remember that there is another energy involved, namely kinetic energy. When the gas is travelling through the manifold it is travelling relatively slowly. As it enters the turbine housing the volume decreases and that energy is converted to kinetic energy (speeding up the gas).

As it passes through the diffuser and into the turbine housing, it moves from a small space into a large one. Accordingly, it expands, cools, slows down, and dumps all that energy - into the turbine that we've so cleverly positioned in tho housing so that as the gas expands, it pushes against the turbine blades, causing it to rotate. We've just recovered some energy from the heat of the exhaust, that otherwise would have been lost.

You got this completely right! Expansion, cooling, and slowing are the three main processes that release energy from the gas and are converted to mechanical motion by the turbine.

Here's a quote for you... "Turbosteamer converts more than 80 percent of the heat energy in the exhaust into usable power, says Raymond Freymann, head of BMW's advanced research and development subsidiary."
Head of BMW's research and development states they convert 80% of the heat energy in the exhaust into usable power. He does not say they use 80% of the heat to produce 15% power gain. So I gather the system has 20% losses due to further exhaust heat not recouped and system inefficiencies.

We can't really infer anything more from Mr. Freymann's quote other than the fact that the turbosteamer captures 80% of the waste heat in the exhaust (he doesn't quantify "useable power"). It is stated in the article that this leads to a 15% increase in efficiency. This means that 20% of the waste heat is still lost.

However you state that this must mean that the total heat lost in the exhaust is only 18% of total fuel input. This is incorrect as it would only be true if the entire turbosteamer system was 100% efficient. In reality they would struggle to get a steam circuit to perform better than 40% which means total heat loss through the exhaust is closer to 47% of total fuel input. Do you understand???

BTW I appreciate that you are a busy man. I do enjoy the discussion and the fact that you have answered on a daily basis is impressive. However I don't mind waiting if it means you write a more complete response.

The thing is that every engine is different and also when the engines are at different load and RPM conditions.

Yeah but I am talking about a very specific point, namely 2000RPM at 200kPa BMEP.

You can easily map a cylinder pressure map against the mechanical design of a crankshaft connecting rod. You can see losses in the design on paper. Then graph and calculate our design and compare the two. You must agree that side and down forces on bearings and other engine components do not contribute to engine power. If you graph a crankshaft efficiency you will see that there is losses which are not included in the losses stated everywhere. These losses in the combustion cycle don't show up in a physical pumping loss tests. So what is wrong, and where does these mechanical losses sit in the losses equation?

I already addressed this but being the busy man you are I guess you missed it. No matter I will reiterate here.

Firstly the aspect of the mechanism you are talking about is called Mechanical Advantage. Basically mechanical advantage is the ratio between a force exerted and the resulting torque created. The first and most important thing to understand about mechanical advantage is that it is not directly equatable to thermal heat loss. Second what is important in engines is the integral of the mechanical advantage function from TDC to BDC. This means that an engine with 100% mechanical advantage has the maximum lever arm length from TDC to BDC.

Your question is how does the loss of mechanical advantage translate to thermal loss? The simple answer is that it increases losses in the other parameters.
For example at TDC in a conventional engine the lever arm is zero/non-existent and zero torque is produced by the force generated by the hot, high pressure gas in the combustion chamber. During this time at TDC the gas will lose heat by radiation and conduction but will not produce any power. If the engine had 100% mechanical advantage at TDC it would generate maximum torque at that moment and the heat losses via radiation and conduction would be less.
Again the important thing to remember is that a loss of mechanical advantage is not directly relative to thermal loss and certainly not on a one-to-one basis ie a 1% loss of mechanical advantage does not necessarily lead to a 1% increase in thermal loss. It could be more or less and is contingent on many other variables.

We are doing better at this test point and others currently with engine mods and tuning. That testing was performed over 2 years ago.

Good to hear. You should definitely be making progress as the values I quoted are between 10 and 15 years old and you have quite a ways to go before you rival those results. It would be interesting to see how you engine stacks up against modern engines once you feel it is adequately modified and tuned.

[revetec]

That depends entirely on the engine. Top Fuel dragsters will blow 1meter long flames out of 1 meter long open pipes, a Peugeot 107 won't blow any flames out the exhaust port unless you floor it past 4000RPM and then wildly feather the throttle (ask me how I know that) I find it alittle hard to believe that your engine blows flames out the ports when running at 15.2:1 A/F and opening the exhaust valves at 130deg ATDC. Also testing with the manifold off introduces unburnt oxygen which normally allows excess fuel to maintain a flame when normally none would exist. Regardless there is still a massive pressure drop over the poppet valve.

It all depends on each individual engine, and fuel used doesn't it. When the manifold is on an no extra oxygen in introduced and excess fuel is now thrown down the exhaust pipe would cause high HC to be eperienced in emission tests. True?

I do read your posts but I am still confused as to what your actual theory is? I realize that you think there is less energy lost in the exhaust but I fail to understand your rationale.

Review BMW's statement again

Well you almost got that right. You switched the volume relation. Remember that pV = internal energy, the larger the volume, the more internal energy. I think we might have stumbled into your basic misunderstanding!

Misunderstanding? Just explained differently to your understanding. No difference really. No work can be done without either pressure or volume.

Remember that there is another energy involved, namely kinetic energy. When the gas is travelling through the manifold it is travelling relatively slowly. As it enters the turbine housing the volume decreases and that energy is converted to kinetic energy (speeding up the gas).

Yes there is a kinetic energy created. but also to drive a turbo there has to be resistance. Resistance creates pressure.

You got this completely right! Expansion, cooling, and slowing are the three main processes that release energy from the gas and are converted to mechanical motion by the turbine.

Cool! we agree on one point.

We can't really infer anything more from Mr. Freymann's quote other than the fact that the turbosteamer captures 80% of the waste heat in the exhaust (he doesn't quantify "useable power"). It is stated in the article that this leads to a 15% increase in efficiency. This means that 20% of the waste heat is still lost.

Actually 80% of the heat from exhaust and he hasn't disclosed how much % of the cooling system. I would imagine that the cooling system transfer is used to bring the water close as they can to 100degC then use the exhaust temp to convert to superheated steam. I would try it this way, but I'm not familiar enough with their system. Yes 20% of the waste heat, but 80% equates to 15% increase to the engine system, so I assume 20% equates to under 4%. That does include heat energy from the cooling system as well... Quite interesting.

However you state that this must mean that the total heat lost in the exhaust is only 18% of total fuel input. This is incorrect as it would only be true if the entire turbosteamer system was 100% efficient. In reality they would struggle to get a steam circuit to perform better than 40% which means total heat loss through the exhaust is closer to 47% of total fuel input. Do you understand???

This is not what he claims. He says that they are able to convert 80 percent into usable power. Not - they use 80% of the exhaust heat energy to produce a 15% gain. So I would assume the 20% is heat lost from exhaust and system losses in efficiency. So 80% from exhaust and (what % of from th cooling system)

"Turbosteamer converts more than 80 percent of the heat energy in the exhaust into usable power, says Raymond Freymann, head of BMW's advanced research and development subsidiary.

BTW I appreciate that you are a busy man. I do enjoy the discussion and the fact that you have answered on a daily basis is impressive. However I don't mind waiting if it means you write a more complete response.

Sometimes if I wait it will never happen. I never say I know everything, I don't, but I know that you can increase mechanical efficiency in an engine. I have done it. The rest is working out how, and why. Unfortunetly people cannot even agree on a convention engine losses. Some websites state that 67% losses occur just from heat, while others quote far diffent figures. For this reason I am writing a comprehensive theory paper on the matter and try to rationalise theory and tests by myself and other parties.

Firstly the aspect of the mechanism you are talking about is called Mechanical Advantage. Basically mechanical advantage is the ratio between a force exerted and the resulting torque created. The first and most important thing to understand about mechanical advantage is that it is not directly equatable to thermal heat loss. Second what is important in engines is the integral of the mechanical advantage function from TDC to BDC. This means that an engine with 100% mechanical advantage has the maximum lever arm length from TDC to BDC.

I agree

Your question is how does the loss of mechanical advantage translate to thermal loss? The simple answer is that it increases losses in the other parameters.
For example at TDC in a conventional engine the lever arm is zero/non-existent and zero torque is produced by the force generated by the hot, high pressure gas in the combustion chamber. During this time at TDC the gas will lose heat by radiation and conduction but will not produce any power. If the engine had 100% mechanical advantage at TDC it would generate maximum torque at that moment and the heat losses via radiation and conduction would be less.

Not if the piston has the same velocity at a given RPM. The radiation and conduction would be the same under the same piston movement conditions. This would be controlled by applying more resistance to rotation, procucing more torque under the same conditions of RPM. More torque at the same RPM increases power with fuel consumption being almost a constant.

Again the important thing to remember is that a loss of mechanical advantage is not directly relative to thermal loss and certainly not on a one-to-one basis ie a 1% loss of mechanical advantage does not necessarily lead to a 1% increase in thermal loss. It could be more or less and is contingent on many other variables.

But it does relate to total efficiency by increasing output. I now know you agree that there is a variance in Mechanical Advantage. My arguement has always been, where is the mechanical losses in the losses equation, and if placed in the equation, what has to be removed or has been misstated? If you can graph the Mechanical Advantage of a crankshaft by applying a pressure map to it, can you quantify where it fits into engine losses? I would appreciate your help with this as I would love another persons quantification.

Good to hear. You should definitely be making progress as the values I quoted are between 10 and 15 years old and you have quite a ways to go before you rival those results. It would be interesting to see how you engine stacks up against modern engines once you feel it is adequately modified and tuned.

Atalan Makine are forging ahead with this one. They told me they will have a CNG and Diesel prototype up and running in approximately 8 months. :-)

[hightower99]

When the manifold is on an no extra oxygen in introduced and excess fuel is now thrown down the exhaust pipe would cause high HC to be eperienced in emission tests. True?

Yes, if measured before a catalytic converter which would burn the left over fuel.

Review BMW's statement again

I know that you think that it proves that there is 18% thermal losses through the exhaust but I have proved that this is not true and that it actually implies a much larger loss approaching 50%.

Misunderstanding? Just explained differently to your understanding. No difference really. No work can be done without either pressure or volume.

What? you said that low volume is a high energy state and that high volume is a low energy state. This is plainly incorrect. You got the relation wrong. Low volume = low energy state, high volume = high energy state.

Resistance creates pressure.

Nope. The pressure is already there in the exhaust system, it doesn't need to be created or increased. You really need to let go of your focus on pressure as it seems it is hindering your understanding. Pressure is just one of the relevant variables. The most important variable is heat.

Actually 80% of the heat from exhaust and he hasn't disclosed how much % of the cooling system. I would imagine that the cooling system transfer is used to bring the water close as they can to 100degC then use the exhaust temp to convert to superheated steam. I would try it this way, but I'm not familiar enough with their system. Yes 20% of the waste heat, but 80% equates to 15% increase to the engine system, so I assume 20% equates to under 4%. That does include heat energy from the cooling system as well... Quite interesting.

I'm not quite sure how you continue to fail to see or even acknowledge my point.

Here is a simple question: If you are right and the total thermal loss via exhaust is only 18% How efficient most the turbosteamer system be to achieve the claimed results? <-this is the most important question in my post

He says that they are able to convert 80 percent into usable power. Not - they use 80% of the exhaust heat energy to produce a 15% gain. So I would assume the 20% is heat lost from exhaust and system losses in efficiency. So 80% from exhaust and (what % of from th cooling system)

Again you infer way too much from a simple quote which doesn't have enough information in it.

My question stands: If you are right how efficient must the entire turbosteamer system be to achieve the claimed results???

Sometimes if I wait it will never happen.

I know the feeling

I never say I know everything, I don't, but I know that you can increase mechanical efficiency in an engine. I have done it. The rest is working out how, and why.

Of course you can increase mechanical efficiency. So far you have increased the overall mechanical advantage of the engine, you have yet to prove an actual increase in mechanical efficiency. Remember: Mechanical Advantage =/= Mechanical Efficiency.

Unfortunetly people cannot even agree on a convention engine losses. Some websites state that 67% losses occur just from heat, while others quote far diffent figures. For this reason I am writing a comprehensive theory paper on the matter and try to rationalise theory and tests by myself and other parties.

Well good luck with the paper. In my opinion it will be difficult and largely inconsequential to define a fixed convention for thermal losses. I believe they should always be measured as every engine is different. That and the fact that all losses already have nice equations to explain and are relatively easy to measure.

Not if the piston has the same velocity at a given RPM. The radiation and conduction would be the same under the same piston movement conditions. This would be controlled by applying more resistance to rotation, procucing more torque under the same conditions of RPM. More torque at the same RPM increases power with fuel consumption being almost a constant.

A conventional crank at TDC has an effective velocity of zero. It is effectively standing still and the force exerted on the piston is not doing anything. Heat is only getting out via radiation and conduction. If the piston had 100% mechanical advantage at TDC it would not be standing still (it would no longer be a reciprocating piston). This means that heat is being converted to mechanical motion which lowers the heat and therefore the rate of loss through radiation and conduction is reduced.

But it does relate to total efficiency by increasing output.

Yes and that increase in output is from energy that would normally be lost via radiation, conduction, or friction.

My arguement has always been, where is the mechanical losses in the losses equation, and if placed in the equation, what has to be removed or has been misstated?

I am trying to tell you that the losses you are looking for are already included in the equation. Nothing has to be removed and nothing has been misstated. The loss of mechanical advantage leads to increased thermal losses via the conventional parameters.

If you can graph the Mechanical Advantage of a crankshaft by applying a pressure map to it, can you quantify where it fits into engine losses?

Firstly you don't need to apply a pressure map to find mechanical advantage. It is found via a few simple geometric equations. Secondly the real difficulty lies in trying to determine how much thermal loss can be attributed to the lack of mechanical advantage. If it were me I would think that I would try to quantify the conventional losses at various points during the cycle and calculating the actual mechanical advantage at those points. Then compare that information to a theoretical engine that achieves 100% relative mechanical advantage at all times at those points and try to quantify the conventional thermal losses. The difference between the two sets of losses will show what fraction of overall thermal losses can be attributed to low mechanical advantage.

Atalan Makine are forging ahead with this one. They told me they will have a CNG and Diesel prototype up and running in approximately 8 months. :-)

Good to hear. It means I win my wager with my mate as these prototypes won't be out until 2 months after the time limit of the wager.

Hopefully some performance information will be released about these prototypes when they come out and they can be compared to modern engines.

[revetec]

I know that you think that it proves that there is 18% thermal losses through the exhaust but I have proved that this is not true and that it actually implies a much larger loss approaching 50%.

I tend to believe a statement from BMW and it backs up to what I believe.

What? you said that low volume is a high energy state and that high volume is a low energy state. This is plainly incorrect. You got the relation wrong. Low volume = low energy state, high volume = high energy state.

I never said that. Just because the volume is high, without pressure or kinetic energy it can be low energy state.

Nope. The pressure is already there in the exhaust system, it doesn't need to be created or increased.

A few posts back you were saying once the valve opens there is no presssure. :-)

I'm not quite sure how you continue to fail to see or even acknowledge my point.

Here is a simple question: If you are right and the total thermal loss via exhaust is only 18% How efficient most the turbosteamer system be to achieve the claimed results? <-this is the most important question in my post

Read BMW's statement correctly.

"Turbosteamer converts more than 80 percent of the heat energy in the exhaust into usable power, says Raymond Freymann, head of BMW's advanced research and development subsidiary."

We don't know how efficient it is, but we do know the losses are part of the 20% not recouped. This is not my opinion, BMW have stated it. Attack them on it!

Of course you can increase mechanical efficiency. So far you have increased the overall mechanical advantage of the engine, you have yet to prove an actual increase in mechanical efficiency. Remember: Mechanical Advantage =/= Mechanical Efficiency.

Yes, We have increased the mechanical advantage which allowed us to achieve a good result. We had poor combustion lasting too long in time, we wasted more fuel out of the exhaust due to poor head design, our frictional losses were higher, but our total efficiency was higher. This proves the mechanical efficiency was higher. To what degree we don't know in tests, but analysing the CAD model we increased it by over 20%. We now are working on bringing the combustion and frictional values to the same or better of a modern conventional engine. After we test the mods, I will post the results.

Well good luck with the paper. In my opinion it will be difficult and largely inconsequential to define a fixed convention for thermal losses. I believe they should always be measured as every engine is different. That and the fact that all losses already have nice equations to explain and are relatively easy to measure.

Thanks, My opening statement says just that. "Every engine performs differently, and I will define the percentages of each loss as a generalisation."

A conventional crank at TDC has an effective velocity of zero. It is effectively standing still and the force exerted on the piston is not doing anything. Heat is only getting out via radiation and conduction. If the piston had 100% mechanical advantage at TDC it would not be standing still (it would no longer be a reciprocating piston). This means that heat is being converted to mechanical motion which lowers the heat and therefore the rate of loss through radiation and conduction is reduced.

But as the temperature decreases, the surface area exposed to the combustion increases. Different bore and stroke ratios make this variable, and also some of this conducted heat is transfered back into the following charge. Very hard to quantify the % of each process.

I am trying to tell you that the losses you are looking for are already included in the equation. Nothing has to be removed and nothing has been misstated. The loss of mechanical advantage leads to increased thermal losses via the conventional parameters.

Most of the frictional losses are also included in thermal losses. Parasitic losses and some of frictional losses are measured in Pumping losses. Some mechanical losses are measured in Pumping losses. Some mechanical losses are not included in thermal losses. There are many things in very grey areas to consider. This is why I'm writing a theory paper. I know I will not be able to quantify everything, and I don't know if it can be done. This is why it is a theory paper.

Firstly you don't need to apply a pressure map to find mechanical advantage. It is found via a few simple geometric equations.

You do need to as you need to know how much force is being applied at each calculated position to get a result.

Secondly the real difficulty lies in trying to determine how much thermal loss can be attributed to the lack of mechanical advantage. If it were me I would think that I would try to quantify the conventional losses at various points during the cycle and calculating the actual mechanical advantage at those points. Then compare that information to a theoretical engine that achieves 100% relative mechanical advantage at all times at those points and try to quantify the conventional thermal losses. The difference between the two sets of losses will show what fraction of overall thermal losses can be attributed to low mechanical advantage.

Do it!

Good to hear. It means I win my wager with my mate as these prototypes won't be out until 2 months after the time limit of the wager.

Good for you.... But we are coming regardless. :-)

Hopefully some performance information will be released about these prototypes when they come out and they can be compared to modern engines.

I will show them to you if you're nice to me. Hahahah

[revetec]

Hightower99: Your standard at the end of every posting "Power, whether measured as HP, PS, or KW is what accelerates cars and gets it up to top speed. Power also determines how far you take a wall when you hit it. Engine torque is an illusion."

When anyone makes a statement like this I also tell them to go drive their car and slowly bring the RPM to just below peak power, then floor it. How well does your car accelerate at this point? Then drive you car just below peak torque and do the same. You find it accelerates harder with far less power. No doubt that we need both, and they go hand in hand. We need torque and RPM to produce power from an engine.

[hightower99]

I tend to believe a statement from BMW and it backs up to what I believe.

I have already said this several times: You are misunderstanding the quote! You are reading more information into the quote then exists and therefore your conclusion is invalid. "Useable power" is not defined or quantified in the quote.

A few posts back you were saying once the valve opens there is no presssure. :-)

Yes I said there is little residual pressure in the manifold when the exhaust valve opens. But as you said yourself, there is 5bar gauge residual pressure in the cylinder when the exhaust valve opens. The pressure is already there. Kinda irrelevant though as it is the heat content of the exhaust that is important, not the pressure.

"Turbosteamer converts more than 80 percent of the heat energy in the exhaust into usable power, says Raymond Freymann, head of BMW's advanced research and development subsidiary."

From this quote one can only conclude that 80% of the wasted heat energy in the exhaust is recovered and that it is this energy that is used as the input for the auxiliary power system.

Thats it!

You infer more than is possible with the given information if you read the quote to mean that 80% of total heat loss in the exhaust is equal to 15% total fuel energy input. This is what you are doing and it is quite simply an incorrect interpretation of the quote!

Think about it: The steam circuit makes 14hp. If 14hp is equal to 15% of the total fuel energy then total fuel energy input is only a little over 93hp... The stock engine makes 115hp and with the steam system total output is 129hp. Clearly you think that the whole engine is actually 138% thermally efficient!

Again I have no issues with what BMW says about their system. I take issue with your interpretation of the facts!

We don't know how efficient it is, but we do know the losses are part of the 20% not recouped. This is not my opinion, BMW have stated it.

BMW has not stated this at all... This is only true given your false interpretation of the facts. Even given your incorrect interpretation you should be able to quickly and easily calculate the required efficiency of the system to achieve the given results. The problem is that if you did you would arrive at the fact that your version of the turbosteamer would have to be 100% thermally efficient in order for your interpretation to be correct. This is simply impossible, so you are wrong.

Thanks, My opening statement says just that. "Every engine performs differently, and I will define the percentages of each loss as a generalisation."

The question then becomes: What exactly do you plan on using a set of generalized losses for?

But as the temperature decreases, the surface area exposed to the combustion increases. Different bore and stroke ratios make this variable, and also some of this conducted heat is transfered back into the following charge. Very hard to quantify the % of each process.

Calculating the parameters at instantaneous points is not difficult at all. Calculating them over any spread is then a relatively simple calculus operation away really.

Most of the frictional losses are also included in thermal losses. Parasitic losses and some of frictional losses are measured in Pumping losses. Some mechanical losses are measured in Pumping losses. Some mechanical losses are not included in thermal losses. There are many things in very grey areas to consider. This is why I'm writing a theory paper. I know I will not be able to quantify everything, and I don't know if it can be done. This is why it is a theory paper.

All losses eventually boil down to thermal losses. I am very much looking forward to reading your paper.

You do need to as you need to know how much force is being applied at each calculated position to get a result.

No you don't. It all boils down to lever arm length. The longer the lever arm, the higher the mechanical advantage, regardless of the forces involved. A simple formula that calculates lever arm length by degrees of crank rotation is all you need to integrate to find the overall average mechanical advantage.

If you apply a pressure curve you get a composite torque curve which, when integrated from TDC to BDC, would give you a value more closely related to thermal efficiency than mechanical advantage. This could very well help with calculating the relationship between mechanical advantage and thermal efficiency. Hmmm I will take a look at a few equations...

Do it!

You never know your luck

I will show them to you if you're nice to me. Hahahah

Hopefully you will settle for civil as opposed to overtly "nice"

When anyone makes a statement like this I also tell them to go drive their car and slowly bring the RPM to just below peak power, then floor it. How well does your car accelerate at this point? Then drive you car just below peak torque and do the same. You find it accelerates harder with far less power.

Hahaha yes that is the old test that seems to confuse people who have no understanding of how kinetic energy works. In your given test I assume that both peak power and peak torque acceleration are to be tested in the same gear, correct? If that is true then the car accelerates relatively more quickly around peak torque for the same reason that 0-60 times tend to be far less than 100-160 acceleration times. Namely less power is required to accelerate at low speeds then at high speeds.

Try doing the same test but this time change gears so that you start accelerating from the same speed both from peak power and from peak torque. Peak power acceleration will win this fair test every time

The whole point is that thinking about engine torque is irrelevant for performance if you already know the power curve.

[revetec]

I have already said this several times: You are misunderstanding the quote! You are reading more information into the quote then exists and therefore your conclusion is invalid. "Useable power" is not defined or quantified in the quote.

I am reading the quote literally as it is written. You are the one who is modifying it.

Yes I said there is little residual pressure in the manifold when the exhaust valve opens. But as you said yourself, there is 5bar gauge residual pressure in the cylinder when the exhaust valve opens. The pressure is already there. Kinda irrelevant though as it is the heat content of the exhaust that is important, not the pressure.

Heat alone means nothing as energy contained, if volume and pressure are not included.

From this quote one can only conclude that 80% of the wasted heat energy in the exhaust is recovered and that it is this energy that is used as the input for the auxiliary power system.

That is not what was quoted and is your interpretation.

Thats it!

You infer more than is possible with the given information if you read the quote to mean that 80% of total heat loss in the exhaust is equal to 15% total fuel energy input. This is what you are doing and it is quite simply an incorrect interpretation of the quote!

If the quote is clear, and if BMW was incorrect in their statement you should ask them for clarification.

Think about it: The steam circuit makes 14hp. If 14hp is equal to 15% of the total fuel energy then total fuel energy input is only a little over 93hp... The stock engine makes 115hp and with the steam system total output is 129hp. Clearly you think that the whole engine is actually 138% thermally efficient!

Not if other quoted losses are incorrect. Such as friction being quoted as an extra loss which may be included in a pumping loss test and heat loss created by that friction. This is why I'm writing a paper on the inconsistancies of what is preached.

Again I have no issues with what BMW says about their system. I take issue with your interpretation of the facts!

I have made no interpretation other tha reading it literally. You are the one changing the quote. Read it carefully.

The question then becomes: What exactly do you plan on using a set of generalized losses for?

Calculating the parameters at instantaneous points is not difficult at all. Calculating them over any spread is then a relatively simple calculus operation away really.

Doubling up on losses, incorrect interpretation, unrealistic calculations, not taking into consideration of molecule buffers on surfaces against transfering heat.

No you don't. It all boils down to lever arm length. The longer the lever arm, the higher the mechanical advantage, regardless of the forces involved. A simple formula that calculates lever arm length by degrees of crank rotation is all you need to integrate to find the overall average mechanical advantage.

That's fair enough, but in the real world there are varing pressures on that lever arm, which are different under varying load and RPM conditions.

If you apply a pressure curve you get a composite torque curve which, when integrated from TDC to BDC, would give you a value more closely related to thermal efficiency than mechanical advantage. This could very well help with calculating the relationship between mechanical advantage and thermal efficiency. Hmmm I will take a look at a few equations...

This is what I did 14 years ago, do it.

Hahaha yes that is the old test that seems to confuse people who have no understanding of how kinetic energy works. In your given test I assume that both peak power and peak torque acceleration are to be tested in the same gear, correct? If that is true then the car accelerates relatively more quickly around peak torque for the same reason that 0-60 times tend to be far less than 100-160 acceleration times. Namely less power is required to accelerate at low speeds then at high speeds.

My point is that torque and power are related and no less important than the other.

Try doing the same test but this time change gears so that you start accelerating from the same speed both from peak power and from peak torque. Peak power acceleration will win this fair test every time

Changing to a lower gear to match RPM creates more torque. Like I say, they both are as important as each other, but your scenario uses a lot more fuel!

The whole point is that thinking about engine torque is irrelevant for performance if you already know the power curve.

And the power curve is provided by torque and RPM

[hightower99]

I am reading the quote literally as it is written. You are the one who is modifying it.

No you aren't reading it literally and I'm not modifying it at all. Unfortunately I don't know how to prove how a quote should be understood other than testing the interpretation against the facts and relevant laws of physics. I don't know the laws of language that would prove your interpretation wrong directly, but I have already shown that your interpretation doesn't stand up to further analysis. You sole defence so far is that if you are wrong then BMW is, which is a highly arrogant position to take. I'm not sure how to proceed on this matter? Maybe a public poll to see how many see it your way in the hopes that a massive majority against your interpretation would convince you of your fault?

Heat alone means nothing as energy contained, if volume and pressure are not included.

Remember that Heat =/= Temperature. Heat content is everything when it comes to energy. If you know that a system contains 100 Joules of heat energy then it is largely irrelevant what the volume, pressure, or temperature is as you can manipulate those 3 variables however you wish in accordance with gas laws.

That is not what was quoted and is your interpretation.

It is the only literal and logical interpretation that can be drawn from the given quote. Your interpretation is neither literal nor logical as I have shown.

If the quote is clear, and if BMW was incorrect in their statement you should ask them for clarification.

But I don't need clarification from BMW. My interpretation of their quote fits perfectly well with conventional laws of physics and the laws of logic. Your interpretation doesn't. I need clarification from you about your own interpretation. Other than your sole defence that you are reading the quote directly you have made no further clarification as to why your interpretation is correct! Your interpretation leads to a turbosteamer system that must be 100% efficient as well as an overall efficiency of 138% for the whole engine. There must be something wrong with your interpretation, not with BMW's quote.

I have made no interpretation other tha reading it literally. You are the one changing the quote. Read it carefully.

This is your sole defence. It is a highly arrogant position. You should check your interpretation.

That's fair enough, but in the real world there are varing pressures on that lever arm, which are different under varying load and RPM conditions.

So what? The mechanical advantage doesn't change from varying forces, only by lever arm length. The varying forces change the composite torque curve.

This is what I did 14 years ago, do it.

Good for you. You certainly aren't the first and you won't be the last either.

My point is that torque and power are related and no less important than the other.

But once you know the power, why would you need to know the torque? All performance parameters can be calculated from power...

Changing to a lower gear to match RPM creates more torque. Like I say, they both are as important as each other, but your scenario uses a lot more fuel!

The point is to change gears to match speed not RPM. Your scenario took accelerating from two different starting speeds to be a fair comparison. This is blatantly false. If you accelerated from the same speed in both cases then peak power wins every time. There is no relation between torque and acceleration. If I ask you to estimate how quickly a car can accelerate with Xlbs/ft of torque from the engine, you couldn't possibly tell me without calculating power first, which is my point. BTW: My scenario involves lower speeds and the same usage of power as yours so my scenario uses the same of less fuel than your scenario. What exactly is the point of your scenario???

And the power curve is provided by torque and RPM

So what? Once you know the power curve the torque is irrelevant.

BTW: I have been looking at my own equations for composite torque that we have been talking about. Do you happen to have a pressure over crank angle graph that can be made public? The few that I have from testing aren't allowed to be made public unfortunately. I am trying to find a generic one to use in comparison with you. To see if we can figure out the relationship with thermal efficiency. Failing that I will construct a simple one.