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Old 01-17-2012, 10:00 PM   #51 (permalink)
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That would be the obvious 1st issue I have with that chart.

But to be fair ... I do not know the context of the graph.
Turned out that book was one I already had.

After looking again at the Heat Transfer in Engines Section ... where this graph came from ... it actually does explain the greater than 100% ... or at least tries to:

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Figure 10-1 Distribution of energy in a typical SI engine as a function of engine speed. Friction losses, which are generally on the order of 10%, add to the other heat losses and make the total energy distribution greater than 100%
So the % on the left side of the graph is only representing the heat energy content of the fuel ... not all the energy content of the fuel... Which I think they should have included the term 'heat' on the left side to prevent that kind of confusion.

Any chemical energy content of the fuel that does not produce heat during combustion is excluded from the 100% on the left side of the graph ... meaning light energy is not part of that 100% ... pressure changes ( not caused by heat changes ) are also not included in that 100% on the left ... etc... etc.

I think it is a bad approach ... but they are actually not claiming greater than 100% fuel energy content ... they are counting effects of the complete energy content of the fuel in the graph body , but are excluding non-heat energy of the fuel from the % on the left.

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Old 01-18-2012, 06:17 AM   #52 (permalink)
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Hi IamIan,
The chart is for wide open throttle. It appears almost identically here: Internal Combustion Engine Fundamentals by John Heywood but without the troubling "friction" line and with a km/hr scale. Chapter 12 in Heywood has a good discussion of heat balance, but it only discusses WOT conditions. The fomulae can be used for throttled situations. Another book I like is Engine Testing Theory and Practice by Anthony Martyr, Michael Alexander Plint

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The Second thing I noticed is that ... that graph looks to contradict what mort was expecting ...
The situation I was referring to was an engine that was throttled, but producing substantial power, 36 hp in fact.

For WOT, using the graph as an example. The maximum efficiency occurs at a slightly higher rpm than shown. As rpm decreases, particularly below the rpm where the engine has the best breathing, power output falls at a faster rate than the rpm partly because the engine isn't getting the same amount of air-fuel charge as at the higher rpm. There is less charge in the cylinder, so the pre-combustion pressure & temperature are lower. That lowers thermal efficiency. The combustion gases expand into the full cylinder, lowering the exhaust temperature well below the full power temp. At the same time, as rpm drops the amount of time that the high cylinder temperatures of combustion has to dump into the cooling system increases - so the coolant takes away more heat and the exhaust less. Just as the graph shows.

Additionally, if you consider an engine at idle, throttled of course, the bhp is zero so all the heat in the fuel must come out in the coolant and exhaust (plus radiation...) At idle the coolant heat is infinity times the shaft power.

What I expect to happen in a throttled engine, well above idle, is that since the air-fuel charge is reduced, perhaps 1/4 of WOT, the pre-combustion pressure & temperature will also be reduced. Again the gasses expand into the full cylinder and the exhaust temperature is lowered. But the time that the coolant is exposed to the hot cylinders is the same as for WOT at the same rpm. However, the average temperature is reduced by the smaller charge and by having a full expansion stroke. Heat flow is proportional to temperature difference. So I would expect heat put into the coolant to be much less than 1/4 of coolant heat at WOT at that rpm.
So here is a graph I made up of what I think a constant rpm vs throttle heat balance would look like.

As I said before the worst case allowance calls for a radiator capable of dumping 100% of full shaft power. But at reduced power, particularly throttled down at cruise, the radiator only needs to handle a fraction of that. At wide open throttle as rpm drops the shaft power falls. And while the fraction of heat going into the cooling system rises it never gets larger than 100% of maximum output shaft power.
Also see the original WOT graph distorted to show energy balance at lower rpm compared to maximum power. The heat going into the cooling system is at a maximum at full power.
-mort
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Old 01-18-2012, 08:24 PM   #53 (permalink)
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One thing to note about electrolysis is that when you re-combine the H2 and O2 molecules, you get exactly the same amount of energy back that was put in, to crack the water molecules in the first place.

As they say in physics "there's no free lunch".

And that is the problem with on-board electrolysis machines. You can't get more energy out of the water than you put into it. So you're not gaining anything. And since the machines you're using to crack the water molecules, and then re-combine the Hydrogen with the Oxygen, are less than 100% efficient, there's actually a net loss of energy.

To get a net gain of energy (for the car, anyway), you need to electrolyze the water off-board (for example, at home), and then put the gas in the car. But this will still be a net loss of energy (in total). It's just that you're shifting the energy loss to someplace other than the car. While your gasoline bill might go down a bit, your electric bill will go up.
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Old 01-18-2012, 08:56 PM   #54 (permalink)
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If the paper is to be believed, the addition of 1.5 lb/hr of hydrogen improved efficiency from 21.95% to 22.78% (bhp/fuel hp)
680 grams of H2 has a lot of energy, with a specific energy of 142 MJ/kg (gas and diesel are around 47 and 45 MJ/kg respectively - lower when ethanol or biodiesel is added)

Essentially, what the study suggests adding in H2, is the energy equivalent of 4.5 lbs or about 2kg/h of fuel .
That's more than half the fuel my car uses ...

If it makes more NOx because of the higher combustion temperatures, it's not a solution, but a problem.
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Old 01-18-2012, 10:22 PM   #55 (permalink)
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Quote:
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It appears almost identically here: Internal Combustion Engine Fundamentals by John Heywood
I have that one as well ... and another good book.


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Another book I like is Engine Testing Theory and Practice by Anthony Martyr, Michael Alexander Plint
Agree ...
I have the 3rd edition in my collection as well... another good one.

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Additionally, if you consider an engine at idle, throttled of course, the bhp is zero so all the heat in the fuel must come out in the coolant and exhaust (plus radiation...) At idle the coolant heat is infinity times the shaft power.
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What I expect to happen in a throttled engine, well above idle,

So I would expect heat put into the coolant to be much less than 1/4 of coolant heat at WOT at that rpm.
For the Coolant system to go from infinity% of Shaft power down to 25% of shaft power ... it transition between them in terms of % of shaft power ... it might not be a perfectly smooth transition along those points but ... the 175% you found incredible therefore must exist between those points ... less than infinity% and more than 25% ... so does 250% and 70% ... etc.... etc...

So we see an offset between the RPM you would expect this condition to happen at and the reported RPM ... which leads to a question of what would cause that kind of offset in that context ... I suspect part of it ( not sure what % ) of that is due to the lean burn operation ... many engines running at lean burn have significantly different heat production than those not running at lean burn and lean burn also significantly reduces the produced shaft power per engine cycle ... it changes the ratio between heat and shaft power as you going into lean burn.

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So here is a graph I made up of what I think a constant rpm vs throttle heat balance would look like.
Even if it worked exactly as you depict ... and that chart were 100% perfect match to a real test engine ... we can find on that chart of yours points where the coolant system exceeds 100% of the shaft power... lower and lower RPMs until as you pointed out yourself the gap between coolant system going to infinity% and shaft going to zero... as long as that relationship exists it is only a question of where does it exceed 100% ... where does it reach 175% , etc.

- - - - - - - - -

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One thing to note about electrolysis is that when you re-combine the H2 and O2 molecules, you get exactly the same amount of energy back that was put in, to crack the water molecules in the first place.

As they say in physics "there's no free lunch".

And that is the problem with on-board electrolysis machines. You can't get more energy out of the water than you put into it. So you're not gaining anything. And since the machines you're using to crack the water molecules, and then re-combine the Hydrogen with the Oxygen, are less than 100% efficient, there's actually a net loss of energy.

To get a net gain of energy (for the car, anyway), you need to electrolyze the water off-board (for example, at home), and then put the gas in the car. But this will still be a net loss of energy (in total). It's just that you're shifting the energy loss to someplace other than the car. While your gasoline bill might go down a bit, your electric bill will go up.
While I agree with a lot of what you wrote ... I will make one point of clarification ... there are endothermic conditions of electrolysis ... where a small % of the energy chemically stored in the H2 & O2 came from heat absorbed from the area ... this heat energy does not have to be 'paid for' by the electricity in the electrolysis ... so yes the energy came out that went in , but some of that 'go in' energy doesn't have to come from the electricity the user is paying for ... unfortunately it is too small of a % to compensate for ( as you already wrote ) all the other steps that are less than 100% efficient... and the final result is still getting out less than one put in... and even if the other steps could somehow be improved enough to get close ... it would just move the question to what one was using as a heat source for that 'free to me' energy input... it can't be something you ever have to pay for or it is not longer 'free' and that greatly limits the already very limited endothermic potential that is still to small to offset the other less than 100% efficient steps.
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Old 01-18-2012, 11:02 PM   #56 (permalink)
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While I agree with a lot of what you wrote ... I will make one point of clarification ... there are endothermic conditions of electrolysis ... where a small % of the energy chemically stored in the H2 & O2 came from heat absorbed from the area ... this heat energy does not have to be 'paid for' by the electricity in the electrolysis ... so yes the energy came out that went in , but some of that 'go in' energy doesn't have to come from the electricity the user is paying for ... unfortunately it is too small of a % to compensate for ( as you already wrote ) all the other steps that are less than 100% efficient... and the final result is still getting out less than one put in... and even if the other steps could somehow be improved enough to get close ... it would just move the question to what one was using as a heat source for that 'free to me' energy input... it can't be something you ever have to pay for or it is not longer 'free' and that greatly limits the already very limited endothermic potential that is still to small to offset the other less than 100% efficient steps.
I am in total agreement with this. I just didn't feel like getting very technical about it. It looks like you took up my slack, though.

Also of note is that if you inject water into the engine directly (i.e. without electrolysis), you're still putting some heat energy into the engine, but without the losses associated with the electrolysis machine. And if you use waste heat from the engine to pre-heat the water, you might get a little more efficiency out of the engine. The big question is whether or not it'll be worth the effort.

Last edited by Blacktree; 01-18-2012 at 11:15 PM..
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Old 01-19-2012, 02:26 AM   #57 (permalink)
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All very good points. A question though. Is it possible that hydrogen injected in a sufficient quantity could serve as a catalyst and thereby contribute to the combustion efficiency? And if so is it possible that it could add more than enough power to compensate for say 20 amps or .24 kw or so required to produce it? Just playing devils advocate.
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Old 01-19-2012, 12:53 PM   #58 (permalink)
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All very good points. A question though. Is it possible that hydrogen injected in a sufficient quantity could serve as a catalyst and thereby contribute to the combustion efficiency? And if so is it possible that it could add more than enough power to compensate for say 20 amps or .24 kw or so required to produce it? Just playing devils advocate.
No. A catalyst facilitates a chemical reaction. Hydrogen cannot act as a catalyst in ANY WAY. It is burned and gives energy. If you read that NASA study, you will realize massive amounts of H2 were burned (relative to what you might electrolyze), and gains of 3% output were realized. There is no catalytic reaction going on, just the extra energy contained in the hydrogen.
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Old 01-19-2012, 01:44 PM   #59 (permalink)
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H not a catalyst, but it can stabilize combustion when trying to ignite and burn ultra-lean mixtures.
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Old 01-19-2012, 02:12 PM   #60 (permalink)
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Seems logical. A couple more questions if you don't mind. Since the timing of the ignition event is so critical in a combustion engine, is it possible that the addition of hydrogen accelerates the process and improves efficiency of the gasoline burn in that way? Possibly similar to the use of higher octane fuel (as I understand it contains less energy than lower octane fuel) in a high compression or boosted engine allowing one to advance the ignition timing and thereby increasing power output?


Last edited by ngrimm; 01-19-2012 at 02:14 PM.. Reason: I typed this before I saw Franks reply
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