Heat transfer, myth, theory, speculation...
03-29-2008, 02:48 PM
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Heat transfer, myth, theory, speculation...
Something has bothered me for a long time as to the practical implications of water injection.
It's generally accepted, and apparently true, that water injection reduces an engine's tendency towards knock by autoignition and detonation by reducing the temperature of the intake charge. What bugs me is where said temperature reduction takes place.
It seems as though much, if not most of what I read on the subject tends to address temperature differentials within the intake tract, up to and including the intake plenum. Much is spoken of injector placement, whether 'tis best that the injector be placed as near as possible to the intake valves thus to lessen the potential for the water to condense and pool, or that the injector should be placed as far upstream as possible (even prior to the compressor), that it should be allowed to linger in the air for as long as possible, affording it the maximum opportunity to absorb heat. To this end, there is much ado with the placement of air temperature sensors, measuring the temperature of the air at points within the charge piping and even inside the intake plenum. People will say things like "with the MAT sensor placed at the back of the intake manifold, I registered a drop of XX degrees with the water injection on."
I'm pretty sure that these folks are missing the point. Not just that, but also gay. With sheep. Underage sheep.
We all agree that the purpose of spraying water into the system is to absorb heat. But how does heat actually get transferred? Without getting too far into scary science and Greek letters (and assuming heat capacity to be a constant) the transfer of heat from one object into another (ΔQ) is a function of the ratio of the temperatures of the two objects. The labcoat-wearing types call this ΔT (delta-T). That's the last Greek letter, I promise. Clear as mud? Perhaps Dr. Science can spell it out for us.
Thank you, Doctor.
Now, turbochargers heat air. There's really no getting around this. You compress a gas, and its temperature is going to go up. But how hot? Well, if we assume 75° inlet temp, a pressure ratio of 2, and a compressor efficiency of 75%, you're looking at about 230°F. Yeah, it's hot.
But the turbocharger isn't the only compressor in the system. It's not even the biggest. Not by a long shot. The engine is.
Yup. The engine has a compression ratio, remember? Let's say it's 9:1. And to keep it simple, VE is exactly 100%. Each cylinder draws in a volume of air during the intake cycle, then compresses it. This compression process produces heat. Specifically, a whole shitload of heat. Diesel engines, with their 20:1 compression ratios, produce enough heat to cause autoignition of the fuel when it is introduced. But every time I've tried to figure out how much heat a gasoline engine produces during the compression cycle by applying the Combined Gas Law, I keep coming up with ludicrously high numbers, like higher than the melting point of aluminum. I'm clearly doing something wrong.
But some degree of searching did finally turn up a handy reference. It's a book published in 1919 entitled "Diesel Engine Design" which contains a graph that maps the temperature of the air in the combustion chamber throughout the whole compression cycle, prior to the introduction of the fuel.
We'll look at the line n=1.35 for the moment. Now obviously this is for a diesel engine, but the concept is the same. On a 9:1 gasoline engine, assuming atmospheric intake, the theoretical cylinder pressure at the end of the cycle is about 132 PSI. (14.7 * 9). Go to that spot on the chart, and find listed a final temperature of about 750°F. Now they started with a 212°F intake temp, so knock 137° off of that and you've now got vaporized gasoline sitting at 613°F. Wow.
Now, add a turbo into the mix. Let's say you're running 1 atmosphere of boost. We can use the same chart- if we still assume 100% VE, we'll just double the final pressure to 264 PSI. And because the turbo heated up the air, let's just go ahead and keep that 212°F intake number. We're now in the neighborhood of 975°F.
Can you see where I'm going here? It might not be where you think.
Say I'm thinking about that intake temp being 212°. Let's throw an intercooler at it, shall we? A big one in fact- one with zero restriction yet 100% efficiency. No, it's not a powercard, it's just the monster Bar-n-Plate unit that Hustler took out of a Freightliner truck and installed in his Miata. No matter what the temperature of the air going into it, it comes out at exactly SRT: 75°F. So 975° - (212°-75°) = 838°F. Yawn.
See, the thing about the starting temperature is that it doesn't factor exponentially into the equation or anything, it's just a simple add-subtract variable. If the compression ratio is 1000:1, adding 50° to the starting temp will add 50° to the final temp. Simple.
Well, I am leaving out one humongous variable. We're not just compressing and heating dry air- there's gasoline present as well. ****.
I'm going to admit right here and now that I suck at thermodynamics. Latent heat and phase change can suck on my *****. In fact, I'm not even going to attempt to do all of the maths for what follows, because I'd be wrong. I welcome any physicists or mechanical engineers in the audience to participate in that process. What I will do is lay out some general thoughts.
What we have so far is a fixed space (the combustion chamber, at TDC) containing a mass of air. Since we're in boost at PR=2, and because this is of course a 1.6 engine, we're going to say that there are appx. 0.96 grams of air in play (assuming 1.2g/L at atmo) and with a specific heat capacity of 1.0 kJ/kg.°C, we're looking at roughly 0.503 kJ of energy in that chamber, assuming all the variables above for the non-intercooled example are met. I think.
The heat of vaporization of gasoline is about 304 kJ/kg. Yes, I'm going to totally ignore the energy required to raise it from whatever temperature it happens to be at running down the fuel rail to the vaporization point, and do the phase-change portion only. I'm lazy. And fortunately, the energy required to raise the temperature of a liquid by 50° or 60°C is peanuts compared to the amount of energy actually required to vaporize it.
Say we're running a 12:1 mixture. That 0.96 grams of air is going to have .08 grams of gasoline in it. So 0.00008 kg of gasoline equates to .024 kJ worth of heat absorption.
Water, by way of comparison, has a heat of vaporization of 2,260 kJ / kg. Whoa! I've heard a lot of numbers thrown around as to the proper ratio of water/fuel in a water injection system. 15-20% seems to be common. Let's take the 20% example. With .08g of gasoline, that would give us .016 grams (or 0.000016 kg) of water in the chamber. So assuming that on its way into the chamber the water did absolutely nothing other than come right up to its boiling point, the process of it flashing to steam is going to suck .036 kJ of energy out of the chamber- one and a half times again what the gasoline did.
Now that's what I'm talkin' about. I maintain that the real point to water injection is not to cool the air prior to it's entering the chamber, but during the compression cycle itself. Because that's where the real heat is.
Discuss.
It's generally accepted, and apparently true, that water injection reduces an engine's tendency towards knock by autoignition and detonation by reducing the temperature of the intake charge. What bugs me is where said temperature reduction takes place.
It seems as though much, if not most of what I read on the subject tends to address temperature differentials within the intake tract, up to and including the intake plenum. Much is spoken of injector placement, whether 'tis best that the injector be placed as near as possible to the intake valves thus to lessen the potential for the water to condense and pool, or that the injector should be placed as far upstream as possible (even prior to the compressor), that it should be allowed to linger in the air for as long as possible, affording it the maximum opportunity to absorb heat. To this end, there is much ado with the placement of air temperature sensors, measuring the temperature of the air at points within the charge piping and even inside the intake plenum. People will say things like "with the MAT sensor placed at the back of the intake manifold, I registered a drop of XX degrees with the water injection on."
I'm pretty sure that these folks are missing the point. Not just that, but also gay. With sheep. Underage sheep.
We all agree that the purpose of spraying water into the system is to absorb heat. But how does heat actually get transferred? Without getting too far into scary science and Greek letters (and assuming heat capacity to be a constant) the transfer of heat from one object into another (ΔQ) is a function of the ratio of the temperatures of the two objects. The labcoat-wearing types call this ΔT (delta-T). That's the last Greek letter, I promise. Clear as mud? Perhaps Dr. Science can spell it out for us.
Originally Posted by Dr. Science
Oh Hai!
Suppose you have a bowl of water which is at SRT (SoCal Room Temperature), exactly 24°C. You place your hand into the bowl. What do you feel? You feel wet. But not terribly cold. Reason being is that the surface of your hand is only at perhaps 30°C or so. Because the difference in temperature between your hand and the bowl of water is fairly small, very little heat transfer takes place.
What if the water was right at 0° instead? Because there is now a larger difference between the temperature of your hand and the temperature of the water, heat is more rapidly drawn out of your hand and into the water- thus making you feel both wet and cold.
Kthxbye!
Suppose you have a bowl of water which is at SRT (SoCal Room Temperature), exactly 24°C. You place your hand into the bowl. What do you feel? You feel wet. But not terribly cold. Reason being is that the surface of your hand is only at perhaps 30°C or so. Because the difference in temperature between your hand and the bowl of water is fairly small, very little heat transfer takes place.
What if the water was right at 0° instead? Because there is now a larger difference between the temperature of your hand and the temperature of the water, heat is more rapidly drawn out of your hand and into the water- thus making you feel both wet and cold.
Kthxbye!
Now, turbochargers heat air. There's really no getting around this. You compress a gas, and its temperature is going to go up. But how hot? Well, if we assume 75° inlet temp, a pressure ratio of 2, and a compressor efficiency of 75%, you're looking at about 230°F. Yeah, it's hot.
But the turbocharger isn't the only compressor in the system. It's not even the biggest. Not by a long shot. The engine is.
Yup. The engine has a compression ratio, remember? Let's say it's 9:1. And to keep it simple, VE is exactly 100%. Each cylinder draws in a volume of air during the intake cycle, then compresses it. This compression process produces heat. Specifically, a whole shitload of heat. Diesel engines, with their 20:1 compression ratios, produce enough heat to cause autoignition of the fuel when it is introduced. But every time I've tried to figure out how much heat a gasoline engine produces during the compression cycle by applying the Combined Gas Law, I keep coming up with ludicrously high numbers, like higher than the melting point of aluminum. I'm clearly doing something wrong.
But some degree of searching did finally turn up a handy reference. It's a book published in 1919 entitled "Diesel Engine Design" which contains a graph that maps the temperature of the air in the combustion chamber throughout the whole compression cycle, prior to the introduction of the fuel.
We'll look at the line n=1.35 for the moment. Now obviously this is for a diesel engine, but the concept is the same. On a 9:1 gasoline engine, assuming atmospheric intake, the theoretical cylinder pressure at the end of the cycle is about 132 PSI. (14.7 * 9). Go to that spot on the chart, and find listed a final temperature of about 750°F. Now they started with a 212°F intake temp, so knock 137° off of that and you've now got vaporized gasoline sitting at 613°F. Wow.
Now, add a turbo into the mix. Let's say you're running 1 atmosphere of boost. We can use the same chart- if we still assume 100% VE, we'll just double the final pressure to 264 PSI. And because the turbo heated up the air, let's just go ahead and keep that 212°F intake number. We're now in the neighborhood of 975°F.
Can you see where I'm going here? It might not be where you think.
Say I'm thinking about that intake temp being 212°. Let's throw an intercooler at it, shall we? A big one in fact- one with zero restriction yet 100% efficiency. No, it's not a powercard, it's just the monster Bar-n-Plate unit that Hustler took out of a Freightliner truck and installed in his Miata. No matter what the temperature of the air going into it, it comes out at exactly SRT: 75°F. So 975° - (212°-75°) = 838°F. Yawn.
See, the thing about the starting temperature is that it doesn't factor exponentially into the equation or anything, it's just a simple add-subtract variable. If the compression ratio is 1000:1, adding 50° to the starting temp will add 50° to the final temp. Simple.
Well, I am leaving out one humongous variable. We're not just compressing and heating dry air- there's gasoline present as well. ****.
I'm going to admit right here and now that I suck at thermodynamics. Latent heat and phase change can suck on my *****. In fact, I'm not even going to attempt to do all of the maths for what follows, because I'd be wrong. I welcome any physicists or mechanical engineers in the audience to participate in that process. What I will do is lay out some general thoughts.
What we have so far is a fixed space (the combustion chamber, at TDC) containing a mass of air. Since we're in boost at PR=2, and because this is of course a 1.6 engine, we're going to say that there are appx. 0.96 grams of air in play (assuming 1.2g/L at atmo) and with a specific heat capacity of 1.0 kJ/kg.°C, we're looking at roughly 0.503 kJ of energy in that chamber, assuming all the variables above for the non-intercooled example are met. I think.
The heat of vaporization of gasoline is about 304 kJ/kg. Yes, I'm going to totally ignore the energy required to raise it from whatever temperature it happens to be at running down the fuel rail to the vaporization point, and do the phase-change portion only. I'm lazy. And fortunately, the energy required to raise the temperature of a liquid by 50° or 60°C is peanuts compared to the amount of energy actually required to vaporize it.
Say we're running a 12:1 mixture. That 0.96 grams of air is going to have .08 grams of gasoline in it. So 0.00008 kg of gasoline equates to .024 kJ worth of heat absorption.
Water, by way of comparison, has a heat of vaporization of 2,260 kJ / kg. Whoa! I've heard a lot of numbers thrown around as to the proper ratio of water/fuel in a water injection system. 15-20% seems to be common. Let's take the 20% example. With .08g of gasoline, that would give us .016 grams (or 0.000016 kg) of water in the chamber. So assuming that on its way into the chamber the water did absolutely nothing other than come right up to its boiling point, the process of it flashing to steam is going to suck .036 kJ of energy out of the chamber- one and a half times again what the gasoline did.
Now that's what I'm talkin' about. I maintain that the real point to water injection is not to cool the air prior to it's entering the chamber, but during the compression cycle itself. Because that's where the real heat is.
Discuss.
Last edited by Joe Perez; 03-29-2008 at 04:36 PM.
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03-29-2008, 04:54 PM
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Thats not very much energy. I was under the impression that water injection's main affect was chemical in nature. I.e. it slowed the flame front to prevent the detonation behavior, or affected the hydrocarbon-oxygen mixture to prevent pre-ignition. Got any links I can read supporting the heat transfer theory. Googling it won't be much help since there is too much bad info on web about engines and water.
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03-29-2008, 05:09 PM
#6
wow but i do agree with the statements you have made. only colder air is not going to make as much of a difference as when you add water to the equation. does anyone have a temp sensor on their exhaust mani to measure the EGT? i did see it on a acura before and the guy could tell what his EGT was.
also when you look at methanol when it burns it is carbon dioxide and water i would imagine the water would play a very important factor in absorbing the heat.
also when you look at methanol when it burns it is carbon dioxide and water i would imagine the water would play a very important factor in absorbing the heat.
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03-29-2008, 06:25 PM
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id argue with dr science that putting your hand in room temp water WOULD feel cold since it's such a huge thermal reservoir and seeks equilibrium temperature with your hand (and body). it will effectively suck the heat out of you. brr. go take a 70F bath and tell me I'm lyin.
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03-30-2008, 01:16 AM
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But every time I've tried to figure out how much heat a gasoline engine produces during the compression cycle by applying the Combined Gas Law, I keep coming up with ludicrously high numbers, like higher than the melting point of aluminum. I'm clearly doing something wrong.
If you could transport yourself to the surface of the sun somehow for only 0.0000000000000000000000000000001 seconds you would not fry to death. Probably not even singe a hair.
I suspect you know all this though
I know some Engineers that do engine development work for a living, Diesel and gas. I'll have to pick their brains on this subject next time we go racing, which is in a couple of weeks. We spend hours in the car to and from the race tracks talking about this sort of stuff
They know a lot about ICE processes, ******* encyclopedic knowledge levels between the two of them, and start throwing around terms and acronyms I have not heard of. You know, the kind of stuff you learn about in the real world after graduation, and is only written in deep dark places in the kinds of books they do not carry at Barnes and Noble.I am particularly interested in this right now since I am deciding if I should go with an FMIC or use only WI with fail-safes. WI, all relative complexity aside, sure is a lot more attractive than a FMIC from a thermodynamics standpoint for the reasons you noted above.
Last edited by ZX-Tex; 03-30-2008 at 01:38 AM.
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03-30-2008, 04:51 AM
#9
To a point i agree with you there joe but there is also alot to be said about, the effeceincy of the air in the intake tract as is enters densely into the combustion chamber. This doesn't 100% make you safer but it does make more power with less boost. The theory you describe goes on to help us run more timming etc and make even more power with our wonderfull WI systems. You are pretty spot on though Joe as usual.
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03-30-2008, 10:18 AM
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Q for joe: you're assuming that neither the water nor the fuel is vapor when it enters the intake ports, correct? just atomized (droplets of liquid, but still liquid)...
because the fluid-fluid heat transfer from the intake air to the gas/water is going to be small compared to the heat removed from the hot cylinder to vaporize both once they get to the valves and beyond.
isn't this why the two ice cubes re-freeze to each other in my scotch? the heat to melt them is pulled out of the room-temperature-scotch so fast that its temp drops to near freezing but the temperature of the ice doesn't change since the heat energy is working purely on phase change.
because the fluid-fluid heat transfer from the intake air to the gas/water is going to be small compared to the heat removed from the hot cylinder to vaporize both once they get to the valves and beyond.
isn't this why the two ice cubes re-freeze to each other in my scotch? the heat to melt them is pulled out of the room-temperature-scotch so fast that its temp drops to near freezing but the temperature of the ice doesn't change since the heat energy is working purely on phase change.
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03-30-2008, 10:50 AM
#11
And not a SINGLE paper made reference to, or presented data on, intake air temperatures so much as in-cylinder temperatures or pressure-time plots.
I dont know if you've read this: "End Zone Water Injection as a means of suppressing knock in a Spark-Ignition engine", NACA Wartime Report, Sept. 1944. It talks about direct injection of water into the cylinder on the compression stroke, and optimising the injection angle and the water-fuel ratio. They found that at 60 deg. BTDC and a water-fuel ratio of 0.5, the octane requirement of the engine reduced by 80 octane points!! In other words, 20 octane + WI = 100 octane with everything else kept constant. The implication for us, since we already have 93 octane available, is that we can run more boost with more advance than is otherwise possible.
Its also a known fact that water slows down the flame front and that is the primary reason for it being an excellent knock suppressant. As you said before, the reduction in intake air temperature due to WI is, at best, insignificant, and at worst, irrelevant. Ricardo is developing a turbocharged engine which they call lean-boost DI: the concept is that the direct injection of fuel takes care of the in-cylinder charge temperature at the end of the compression stroke, and they actually use excess air (like 18:1 AFR) at maximum boost to actually act as a knock suppressant.
I am pretty sure that one could very safely say, that the yo-yo guy would have much, much better luck with this 5th injector device if it were fed with a water-alcohol mixture rather than gasoline.
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03-30-2008, 11:50 AM
#12
all the calculations are probably over my head but in order to find the effectiveness of anything to eliminate preignition, you need to find out at what temperature and afr different fuels will ignite, like say at 12:1 air/fuel 93 octane will ingnite at 975 degrees, but at 838 degrees there will be no preignition.....which may be a small temperate delta but but still could be 100% effective
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03-31-2008, 08:52 PM
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its simple thermodynamics... well at least the principles are simple enough... as for applying these principles it gets very intense.
I agree with Joe, however thinking about it in any detail hurts.
as for schooling, most engineering majors require thermodynamics/heat transfer course. I have to take them as a structural engineer
I agree with Joe, however thinking about it in any detail hurts.
as for schooling, most engineering majors require thermodynamics/heat transfer course. I have to take them as a structural engineer
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04-11-2008, 03:17 PM
#18
I hate when Joe tries to debate something, my productivity at work drops to zero as i can't quit reading. Keep it up Joe, lol.
Plenty good points in this discussion, unfortunately I cannot contribute anythign usefull.
Only thing i care about and can comprehend would be nozzle placement? Where do i drill and tap Joe? lol
Keep up the good work, i am very curious to see where we will end up. I personaly always though that water should be injected same as fuel, in the runners or chamber directly.
Pipefather-one thing about your post that has my mind tickling...
If 0.5 water to fuel ratio reduces octane requirement by 80 points at 60*BTDC, what does it do at 10*, 20* etc. reduces octane requirement more or less? Lets say it stays unchanged, would it be more efficient and sane for us to just run water in factory made cars and use less fuel all togther since you no longer need high quality fuels, 20 octane will do. This is where i'm lost. Why not just mix water with gas and inject it together?
One thing i want to conclude based on the combined info from this thread alone is that water does need to get injected just like gasoline does, as close to combustion chamber as possible. Also, if water to gas ratio of 0.5 reduces octane demand so significantly, why not assume that it will do the same with high grade fuel (93), would that mean that you can now run 40-50* of timing advance and be on the safe side? At what point is your charge too cold to burn completely and efficiently? Also, assuming that you run 4 550cc injectors, can you just run 4 275cc nozzles spraying water simultaniously and run much more advance at any given time and also gain spool due to increase exhaust mass?
**** I'll stop now, i am way out of place, I'm just a sleazy salesman.
Thanks for reading and thanks for entertaining my mind.
Plenty good points in this discussion, unfortunately I cannot contribute anythign usefull.
Only thing i care about and can comprehend would be nozzle placement? Where do i drill and tap Joe? lol
Keep up the good work, i am very curious to see where we will end up. I personaly always though that water should be injected same as fuel, in the runners or chamber directly.
Pipefather-one thing about your post that has my mind tickling...
If 0.5 water to fuel ratio reduces octane requirement by 80 points at 60*BTDC, what does it do at 10*, 20* etc. reduces octane requirement more or less? Lets say it stays unchanged, would it be more efficient and sane for us to just run water in factory made cars and use less fuel all togther since you no longer need high quality fuels, 20 octane will do. This is where i'm lost. Why not just mix water with gas and inject it together?
One thing i want to conclude based on the combined info from this thread alone is that water does need to get injected just like gasoline does, as close to combustion chamber as possible. Also, if water to gas ratio of 0.5 reduces octane demand so significantly, why not assume that it will do the same with high grade fuel (93), would that mean that you can now run 40-50* of timing advance and be on the safe side? At what point is your charge too cold to burn completely and efficiently? Also, assuming that you run 4 550cc injectors, can you just run 4 275cc nozzles spraying water simultaniously and run much more advance at any given time and also gain spool due to increase exhaust mass?
**** I'll stop now, i am way out of place, I'm just a sleazy salesman.
Thanks for reading and thanks for entertaining my mind.
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04-11-2008, 08:25 PM
#19
Well that 60 BTDC number was the only one reported in the paper, and frankly, to do something like that you need to come up with something that injects directly into the cylinder. Not a viable option.
Also its tough to answer the questions you raised without access to, at the very least, a dyno, cylinder pressure sensors, and various other instrumentation that costs more than the car anyway. But hey, there are a few brave souls out there willing to push the envelope so all we have to do is wait until someone comes up with the right set of parameters that gives good results.
Also its tough to answer the questions you raised without access to, at the very least, a dyno, cylinder pressure sensors, and various other instrumentation that costs more than the car anyway. But hey, there are a few brave souls out there willing to push the envelope so all we have to do is wait until someone comes up with the right set of parameters that gives good results.
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