Intercooling? Or simply cooling the charge air?

[Tom @ Fast Forward]

I am well past 15PSI on 91 octane. If you can tell me how to make water burn, I would switch from fuel. ;)


Well, with the thermal efficiency of the MP62 dropping to ~ 50-55% at the boost I am running, the alternative would be to inject a few gallons of water for every gallon of gas :rofl:

[FormerDatsun510Man]

Well, theoretically at least :biggrin:, the evaporation of water would drop the temps a lot further than the evaporation of fuel. You can check out the calculations on the spreadsheet I attached a few posts back. Also, one could run a mix of water/methanol. Methanol burns :). Just would probably be a PITA to keep it topped off, especially if you go on a trip. One thought with WI though would be to have it only come on at very high rpm and high boost... basically that window you would be in only when racing. That way, you wouldn't use any during "normal" driving. Maybe PC Pro could have an output that would have that sort of functionality? Basically an output that would be a multiplication of boost X rpm = water spray amount, and also that it would only apply the function in a small window of the operating window of the engine... say for example, above 5000rpm and 8psi.

Sorry for the thread drift upon thread drift upon thread drift :biggrin:

Bill

[Tom @ Fast Forward]

There is always give and take in this world. That same thing happens with intercooling. Is an A/A IC more efficient that evaporative cooling? Yes, under certain conditions. Curt, at "Mike's Place" has a WRX. He loves it but can't drive it here in the summer months. It has an A/A IC but pings bad enough to possibly destroy the engine in our Arizona heat. It is stock and I think it is only running about 6PSI. I am running over 15 PSI and can do so on the same 91 octane. DO I think I could add a 5th injector and cure his problem? Yes, I do. Does that make evaporative fuel cooling more efficient? Nope, just different.

Is water injection more efficient at cooling? Yes, it's thermal efficiency is much higher. However, the real question to ask is "Is fuel evaporation properly done efficient enough?" and the answer to that is Yes. Take into account the goal you have set. Is it to make more horsepower or simply to make it safe? If it is to make more HP on less boost, then an A/A IC is way more efficient than WI or FI evaporative cooling. The A/A will make more HP on less boost. However, if the goal is to make a simple light weight system that makes an acceptable amount of HP, then evaporative cooling is the answer. Water will probably do a better job than fuel and make a couple more HP even but, when you have 12 gallons of fuel already available and, if it runs out, you don't destroy the engine, I'll take fuel. WI fans spend a lot of time trying to devise home brew methods of making the engine safe when the water runs dry. With fuel, it is automatic. When the supply runs out, the engine stops. :)

I think we agree that the colder the air the more HP developed simply because there is more oxygem molecules per cubic foot in the air and it is oxygen that burns, not fuel. The fuel is there to burn the oxygen. My kits are not designed for high HP. Never were. Even the Hotside kits I intend to sell are still aimed at the same HP level. I have said repeatedly that the high HP market is not my market. I'll take the 85% share of the market and FM/Bell can have the 220+ market. The really large market is the 160/170 WHP market and that is why I have introduced the new Coldside kits at that level. I digress. :)

Back on track. The first question one has to ask when designing a kit is what HP am I looking to produce and at what price and how many do I think I can sell. For the Miata, I can venture to bet that over 90% of the market is under 200WHP daily driver.

My Coldside always produced more WHP than the same hotside with the same pulley ratio. Now you have the new outlet plenum and a big A/A IC and the R4 intake manifold and you ar making more WHP with the same pulley ratio than my kit. That's a good thing but I can still deliver my 200WHP kit for $3695 and run it on 91 octane if AZ heat on a stock engine. The only real difference is $$$$$. Yes, your setup is more efficient but at what cost? That is the point I keep trying to make. When people on mnet pop in and defend WI as more efficient, I don't argue the point, I just try to say that yes it is BUT FI is efficient enough to meet the needs. Is evaoprative cooling really intercoooling? I think that is a matter of semantics. Give me a real definition of the word and let's see if evaporative cooling fits. :)

It is interesting how many people on mnet will say that E-Cool won't work and they are the same people running WI. LOL They just aren't smart enough to know how much efficiency is enough efficiency to make the system work. SOmewhere I saw a video of WI on a manifold that had a glass window. Water was running down the walls everywhere. The juxt of the video was "see how efficient it is". I guess the WI people think that if a little is good, a lot MUST be better. It's not.

Your turn. ;)

[FormerDatsun510Man]

Why offer a Hotside if it makes no more power than the Coldside? I understand the case of the 1.6, since you would have to make a completely new casting for that head, but what about the 1.8? The cost of the two kits is nearly the same, and actually the new starter Coldside is even less. Additionally, the Coldside has better driveability and lower weight.

[Tom @ Fast Forward]

Because people like to take hotsides and 'enhance' them. :)

[FormerDatsun510Man]

Ok, that enhancement is exactly what I am talking about :). Let's start off with analyzing the net air charge temps for a Hotside running a reasonable rwhp. The analysis would compare E-Cool, WI and an FMIC. The system in the attached spreadsheet example is a '99-'00 Hotside running a 115 pulley. For that setup, the estimate comes out to 215rwhp and it would be running about 10psi peak boost. For the calculations I used inlet air temps of 100deg F with no humidity (a cool Arizona day :biggrin:). There is a seperate tab for each system modeled. The results I am getting with the numbers I plugged in:

Temp Rise from supercharger: 157 deg F
E-Cool temp drop: 4 deg F
E-Cool net temp rise: 153 deg F
WI temp drop: 72 deg F
WI net temp rise: 85 deg F
FMIC temp drop: 126 deg F
FMIC net temp rise: 31 deg F

Other notes are that for the E-Cool I assumed the main injectors running to the duty that calculates to a 13:1 a/f ratio and then turning the E-Cool up to the duty cycle it needs to obtain 12:1 a/f ratio. This figured out to maxing out the main injectors and about 19% duty cycle on the 480cc injector. If the main injectors run a lower duty cycle the E-Cool could run higher... but you mentioned E-Cool was only running 20% max duty in another post. For the WI, so that I wouldn't corrupt the charge air temp drop with E-Cool at the same time I eliminated E-Cool and went with 310cc injectors and it turns out 86% duty on them would yield a 12:1 a/f ratio. I selected the WI size and duty to yield a 15% water to fuel ratio (typical target). For the FMIC, same fuel setup as for WI except I used an 80% efficient intercooler. This is typical efficiency for a properly sized A/A IC and corresponds with what TDR claims for his. It also corresponds with sort of temps I would see on my setup when I had a post IC temp gauge installed.

As you can see the WI drops the air charge temps over 10x compared to E-Cool and the FMIC drops air temps nearly twice as much as WI. Theoretically of course :biggrin:. Now, part of the purpose of this spreadsheet is I would like to get suggestions further refining/improving it so as to be more accurate.

Bill

[Steve in VC]

Enhancements....

I believe the M62 is rated to 16K RPM, up from Tom's 11K RPM, with a pretty much linear increase in power. Add an intercooler to control air temperature, and you can get maybe 300HP?

You can spin the coldside at 16KRPM, but where to put the intercooler?

[FormerDatsun510Man]

Yes, you can try different pulley sizes if you want in that spreadsheet. It does a really simple calculation for rwhp. It is basically blower_airflow * fudgefactor - 26 (approx. difference between crank hp and dynojet rwhp). I think it is reasonably accurate in the range of pulleys from 100mm up to 150mm. I have another spreadsheet that actually looks at heat along with blower pump losses to give a better rwhp approximation. On intercooled Hotsides I found the 150/65 pulley ratio was the one that would yield maximum rwhp with the 7000rpm rev limiter. Probably would be in the 270-280rwhp range. There is a guy who dynoed a Hydra controlled IC'd Hotside that dynoed 262rwhp with the 140/65 pulley.

To accurately model a pulley size change in the spreadsheet you need to adjust the boost and also the blower efficiency (thermal efficiency). On Hotsides, it is fairly accurate to put the peak boost psi as (rwhp - 110)/10. In other words, every 10rwhp increase over stock is about 1psi of boost. For thermal efficiency note the table on the right with the efficiency related to rwhp for the MP62.



Bill

[Tom @ Fast Forward]

Bill,

Not to be critical of your math but it sucks. 7 degrees? What part of this http://www.FastForwardSuperchargers.com/Brr.wmv don't you believe? We had the whole manifold dropping a couple degrees on the dyno as measured with an infrared thermometer on the outside of the manifold.

When you did the fuel calculations did you remember to use 62PSI instead of 42PSI this time? You forgot that before and it does have an effect. :)

[FormerDatsun510Man]

I appreciate constructive criticism, so show me where my math sucks :). On the spreadsheet you can see I used 62psi fuel rail pressure. If I have the duty cycles off, put in the correct values. The temp drop is based on the latent heat of evaporation, which I have given towards the bottom of the spreadsheet for water and gasoline. Also have the specific heat of air given. The latent heat of evaporation values I researched on to find them. Let me know if you think I have the incorrect values used.

Note, I am not saying I don't believe the brrrr video. Rather, I am trying to look at the numbers based on physics to explain it. So far, this model doesn't explain it admittedly. So that in my mind is either because:

1) I have something wrong with the temp drop calculation? If so, I'd appreciate being shown where it is wrong. I have checked the spreadsheet over myself and as far as I understand it is correct.

2) There is more to the model than simply the latent heat of evaporation? If so, what is it?

[Tom @ Fast Forward]

I'll leave the math to you. I'll do the empirical (real world) data gathering. There has been some criticising of the BRR movie in the past as the thermocouple tip was in the fuel spray. It was considered that the TC was cooling but not the air. Not sure how that can happen?? However, I just went out and pulled the thermocouple back so the tip was hidden inside the hole in the IM wall. It will see the air but not the direct fuel. I made the settings for the 5th injector so that it was adjustable from 0-50% DC based on pressure from 0-19PSI. With E-Cool turned off I was getting serious ping and the main fuel was at about 13.2:1 I stopped at about 12PSI and 4500 RPM as the ping was serious and I only go so far, even for science. ;) IAT sensor in the end of the Air Filter back at the firewall directly below the BTB read 103F. IM temps went to ~225F. With the E-Cool turned to 25% x 12/19 = 15%DC at the same conditions, the IAT was still 103F and the IM was ~150F and A/F was ~11.8:1. I will absolutely guarantee that the TC saw nary a drop of direct fuel as it is totally shielded by the wall so it was only looking at the IM air. I thought that perhaps it wouldn't get a good reading stuffed back up in that hole but if it saw the temp rise to 225, one has to take the 150F as legit too. So you keep doing your math and I will live with reality.

I wrote a piece on mnet one day about the cooling power of a drop of water. :) In the mid 70s I helped a small company called Beldon Wire and Cable bring on line wide band cable for cable TV. Up to that point, all you could get on cable if I remember correctly was channels 2-13 (VHF). Channels 14 and up were UHF and could not make it down the wire. One Sunday morning about 2:00 AM we got the president of BWC out of his New York bed to tell him that we successfully made 100 feet of wide band cable. They had been trying for a year and I solved the problem the day after I got there. The product is a center wire surrounded by a foam shell wrapped in braided wire with a PVC overcoat. The key to success is the uniformity of the diameter of the foam and the density of the foam. That sets up the dielectric parameters and effects the effective capacitance and inductance. If that varies foot by foot, it causes reflections on the cable and the frequency it will carry will be limited. The foam was being extruded onto the center wire with an Entwistle extruder that had a barrel about 12 feet long with 12 zones of temperature control. Each zone had a band heater wrapped around the barrel and a 1/4 copper tube wrapped around that for water cooling. The barrel was about 12" diameter steel with about a 5" hole through the center for the compressive screw. I am going from memory about the barrel dimensions but I am pretty close. The thermocouple well was drilled through the outer steel barrel down to the inner liner (1/4" thick inconel, I believe). So much for the background.

The problem they had was that the temperature would get to 450F, the water cooling would pulse a very small orifice solenoid for one half line cycle (1/120 second). The water it allowed out was two very small drops. The controller then had a delay of a couple seconds before it would allow the solenoid to cycle again to give the thermocouple to react to the change. Why? After all, how much cooling could two small drops of water do to all that mass of steel? Nothing would happen, nothing would happen, nothing would happen. Then, all of a sudden, the temperature would drop 10-12 degrees. I know why, do you? :)

[FormerDatsun510Man]

Well, if in this thread we were going to compare these three different methods of cooling the air charge, shouldn't we show the math? You said the air temps dropped 75 deg F from E-Cool. Based on the latent heat of evaporation for gasoline (not my math... basic physics) that quantity of fuel would drop the air charge by about 4 deg F. What I would like to figure out is what accounts for the additional 71 deg F temp drop that is observed? Is there more to cooling the air than the latent heat of evaporation?

[Tom @ Fast Forward]

Well, I think what I wrote above is a clue and this drawing of the barrel section at BWC is the second clue. The third clue is "What is the boiling point temperature for gasoline. At STP, it is 212F for water. In the IM the water is vaporized but never turned to steam as the temps are always below 212F.

http://www.FastForwardSuperchargers.com/Water-cooling.htm

[FormerDatsun510Man]

Well, I admit, you lost me here :biggrin: ... I never was very good at solving riddles. But if the science behind E-Cool is proprietery I understand, I don't want to step on any toes.

[Tom @ Fast Forward]

Here's the next part to consider:

While it takes about 1 Btu to change the temperature of a pound of liquid water by 1 °F., it takes 144 Btu to freeze one pound of water (latent heat of fusion) and about 1000 Btu to convert one pound of water to steam (latent heat of evaporation).

You keep working with the numbers to raise one pound of water one degree F. Water is 1000 times more efficient when converted to steam. Basically, it takes 1 BTU to raise 1 pound of water from 211F to 212F but 1000 BTUs to raise it from 212F to 213F. Now think about the similar effects of fuel as it boils at a lot lower temperature than water.

In the BWC example, the water delivery system was closed. That is, the return water was returned to the tank below wtaer level. What happened and why 2 drops of water would create so much cooling was that the water would hit the copper tube as water, be raised to 212F and remove a reasonable amount of heat and then instantly turn to steam and remove 1000 times as many BTUs. As soon as it turned to steam and the outlet was plugged by the water in the tank, the pressure in the tube raised rapidly and provided enough pressure to convert the steam back to water. When it turned back to water, the pressure went away as fast as it came and again the water turned to steam. Again taking away lots of BTUs. As soon as it turned to stems and made pressure, it turned back to water and theprocess continued several time until the water finally got out of the tube. The solution was simple. I had them cut off the las couple of inches of pipe in the water tank so that the drain water would enter the tank above water level and we ran the system again. Voila, the same couple drops of water had a very small effect on the temperature of the steel barrl and we were able to control the temperature within 1 degree F and the world had cable TV with hundreds of channels.

The moral of this story is that Thermodynaimcs is not as simple as you would like it to be. Sometimes you have to look beyond the text book. It is like watching a magic act and trying to determine how the magician did it. The first thing you have to remember when trying to figure out how he cut that woman in half is that he really didn't cut her in half. ;)

I think your math is too simple and is missing the point that it really does happen and you need to figure out how, not try to defend that it can't happen. It could be as simple as the water can't turn to steam and the fuel can. maybe eben more than once?

[FormerDatsun510Man]

Intereresting, thanks for sharing! So are you saying that E-Cool works by the gasoline going between liquid and gas phases multiple times and this is consuming a lot more energy (heat) than a single phase change? The spreadsheet calculation I did is based on only one phase change from liquid to gas. Now, I did neglect the energy consumed by raising the temp of the liquid because, as you pointed out, the energy required there is many, many times less than that for a phase change.

I hope you don't think I am trying to debunk E-Cool. I am simply trying to understand the Science behind it. I don't believe I am trying to "defend that it can't happen". I am trying to understand how it works.

[Tom @ Fast Forward]

Actually, I don't know WHY it works, I just know that it does and have done several tests to confirm. I can guess and I guess that it is a lot more than simply the BTUs removed by the latent heat of vaporization. When the pundits said it couldn't work, I even measured the temperature on the outside of the manifold. With E-Cool off, the temperature of the manifold increased. With E-Cool on, the temperature went down. Not much but even 1 degree down is way better than 10 degrees up. The manifold alone is 10 pounds of cast aluminum. The only conclusion is that there is some phenomena taking place that simple thermodynamics doesn't see.

As I said, the first thing is to believe that it works. The second thing is to try and figure out why. The only thing that comes to mind is multiple phase changes. If I had told you that I could prove that 2 drops of water could cool a couple pounds of steel by 10-12 degrees, your simple thermodynamics formula would have proved that it couldn't happen. Now that you know that E-Cool works, I will leave it to you to figure out how. I'm just happy as a pig in slop that it does. ;)

And now the rest of the world knows why you and I have two hour long telephone calls. ;) Some day we will have to get you a car battery for your phone and invite James Low into the conversation. My calls with him are just as philosophical and just as long. LOL

[FormerDatsun510Man]

Well, that (multiple phase changes) is very interesting and something I didn't consider. Do you think the E-Cool method would cool enough to allow the Coldside to hit even higher rwhp levels than the current 200-210rwhp I've seen posted here? Or is there a certain point where you have seen WI works better? In other words, do you think perhaps certain "enhancements" :biggrin: could be done to E-Cool that would allow it to work for high hp Coldsides... say up around 240rwhp?

[Tom @ Fast Forward]

Well, until they can figure out how to burn water, I'll stay with gasoline. :) What really made me drop WI was when my son and I were canyon carving two years ago. He had the 99 with WI and I had the 05 with the 5th injector. Both were running 105/65 pulleys. The stock injectors on the 99 were turned up but had enough fuel to make 12:1 and the WI took the ping away. When we broke for lunch, he said "Dad, it seems to be pinging pretty loud. I think something is wrong." Notice I didn't say we slowed down anywhere along the way. :( Good news is that nothing broke. Sure enough, it was out of water. Aqua Fina is about $6 per gallon and gasoline was around $3 per gallon.

WI fans, don't get me wrong. I think WI is efficient but it is like the ACT HD clutch vs the XT. I probably don't need the XT and the HD is probably good enough but I just like the XT better. The RoadsterSport muffler is a great muffler but I like the quiet of the stock muffler and we found zero WHP difference. It is one of those "Which is better, Ford trucks or Chevy trucks?". Once you decide that both can carry your dog in the back equally well and all you haul around is the dog, the discussion is over. ;)

To answer the question. Well, you always need more fuel. I am a great believer that the stock ECU is the best controller in the world for running the Miata off boost. The stock ECU starts to have idle issues and LTFT CEL issues above 310cc main injectors. That means that you need more fuel from someplace. If I were running your setup, I think your four additional injectors are fine for adding fuel but I think they are too close to the intake valves to provide cooling. Three out of four fuel pulses occur while the valves are closed and don't see any real air turbulence until the valve opens on the fourth pulse. You have a big A/A IC and it should provide more than adequate cooling. Only reason I could think of is, in the interest of science, you could add a 5th injector, turn the four extra injectors down and test it. I don't know how high you can go? I do know I am running about 15% on a 48# injector at 12PSI and making well over 15PSI at the top. With 91 octane and 3* retard above 5000 RPM, I have smooth 12:1 fuel and no ping. Now, I am only making ~220WHP and you are at 240WHP with the same pulley ratio and less boost. I forget if you have engine and head work? Mine is a 137,000 mile old and punished stock engine. I just think the Coldside has a WHP limit and it is about 220WHP?

BTW, I checked the DC on the stock injectors and, as it is now, I am at 75% DC on 280cc injectors. Not 90%. I can turn them up to make the A/F so rich the engine will stumble and cough and blow black smoke out the exhaust. So I know there is a lot of headroom left. Don't know why the math doesn't work there either?????

[wineguy]

I'm gonna stick my nose in here, and I may be way outta my league -- but I'll give it the ol' college try. :surrender: This is a really interesting thread -- I've read through it several times trying to wrap my head around it...

I don't think I have any explanations, but perhaps a few of the assumptions made here may be incorrect. First, latent heat of vaporization is the amount of heat energy required to go from a liquid pahse to a gas phase with no change in temperature. Water is about 2260 J/gm, and gasoline is referenced in a range from 334 to 350 J/gm (probably depends on additives, octane, etc.). Multiple phase changes have been mentioned, however it's not perpetual motion --a phase change from liquid to gas requires energy, typically in the form of heat, and subsequently in this case, air is cooled as the liquid goes to a gas phase. However, if condensation occurs (gas to liquid), heat is transfered back into the system, in this case the air, and it is heated more-or-less back up. So, multiple phase changes could occur in an intake manifold, but the net heat removed from the air is only dependent on that overall/single phase change from liquid to vapor.

Now comes the part I'm still chewing on... superchargers & turbochargers are pressurized systems. Pressure raises the boiling point of whatever liquids are used to cool. Think of a pressure cooker for your carrots -- high pressure results in the water boiling well above 212F/100C and therefore cooks your food faster/better. Forced induction with a pressurized manifold, maybe 10 to 14 psi, will result in higher temperatures required to vaporize the liquid. In this case, water boils at 212F at sea level, but would boil (convert from liquid to gas; requiring latent heat energy) at a higher temperature under pressure. Gasoline (one source I found real quick) boils at about 80 to 100F. Pressure again would raise this figure -- someone else may be able to quote the correct temperatures. So, here's the conundrum... if intake air temperatures are, let's say, only 100F or so, does the water actually vaporize and require the 2260 J/gm of heat energy, thereby cooling the air as much, especially under pressure? I kinda think not, but I could be wrong. Gasoline, on the other hand, with perhaps a lower boiling point, could possibly go through the actual phase change under pressure, and at the temperatures seen in engine conditions. -- I will admit, I don't know for sure. (couldn't find the "don't know" smiley...)

OK, so this doesn't explain the differences between Tom's experience and the spreadsheet math, but maybe take another look at the system as a whole...

This only leads me with more questions:

1. Not knowing terribly alot about water injection to reduce knock, is it the cooling effect of vaporization of the water that reduces the chances of ping occuring by lowering the charge temperature, or is it possibly that the water/air/fuel mixture burns somewhat slower, which eliminates the knock? If I had to guess, especially considering the pressure/temperature/physics of the thing, I would think that water doesn't truly vaporize in the intake manifold, but controls the burn in the combustion chamber, thereby solving the knock problem.

2. Gasoline on the other hand may puddle, evaporate, condense, puddle, evaporate, etc. as it passes through the intact tract. Does this evaporating on the interior surface of the intake manifold create the colder manifold that Tom saw? Eventually, I would guess that would help cool the air charge somewhat by having a cold manifold to travel through. I don't remember enough about thermodynamics and heat transfer rates... that was over 30 years ago. I've remember looking down the throat of an open 4-bbl carb with the throttle open to see the gasoline puddling slightly here and there. I suppose by the time it gets to the combustion chamber it's pretty much history anyway...

So, what I learned in college physics may be rusty, so please correct me if I'm wrong.

Great thread -- keep it going...

[Steve in VC]

As I understand it, vaporization only has to happen before pressure is high enough to pre-detonate. That occurs after the piston has compressed the charge further, the spark plug has fired, and the burning fuel increases the pressure. Even if the water / gasoline evaporates after the plug fires, it still cools the charge.

Start with 100F air.

Boost it to 10psi, and the temp rises to approximately 159F (55% efficient supercharger).

A 10:1 compression ratio raises it to 650F or more (I don't know the compression efficiency).

The spark plug fires, increasing pressure (peaks around 20 degrees after TDC), plus the heat of burning gasoline.

You are fine as long as the temperature from pressure and ignition stays below temperature of pre-detonation.

Tom's e-cool doesn't pre-detonate.

You can cool the charge by throwing a lot of fuel at it, but would fail CARB. Tom passes with a small reduction in WHP.

It works on stock engines, he didn't lower compression ratio.

If he retarded spark too much, he wouldn't reach the WHP he has.

Could you get more power with an intercooler? Of course, in concert with a bigger crank pulley.

Can water injection work better than gasoline injection? It has 9 times the heat of vaporization, and you can add water without running too rich.

What hasn't been discussed. Ethanol has 3 times gasolines heat of vaporization than water, and an octane rating of 116, raising the average octane of the fuel.

But, water and ethanol can run out under boost. Using gasoline to cool has a built in kill switch when it runs out (great "daily driver" solution Tom!)

Once Tom found e-cool, he could create a cold side system with significant power.

Being stuck with CARB, and there is no better system, turbo, or supercharger. I happy Tom spent the time working out the details, and spent the money on CARB.

[wineguy]

Don't misunderstand what was going through my head... :yes: what you said is great -- I guess I limited my comments to pre-intake valve/intake manifold dynamics. Again, just trying to get a handle on all of this. You're very right, at some point in the combustion chamber, temperatures & pressure shoot throught the roof. I assumed, and maybe improperly, that charge cooling occured in the manifold! In stepping back to look at all of this and get the bigger picture, maybe I didn't step far enough. Thanks...

[Tom @ Fast Forward]

I will try and get back to this later and respond in full as best as I can but your foundation is amiss, I believe. You are missing one word in that sentence. It should read "First, latent heat of vaporization is the amount of heat energy required to go from a liquid pahse to a gas phase with no change in temperature of the liquid. Second, I suppose to be technically correct I should have used the term "Latent heat of evaporation". Water will vaporize at any temperature. It has to hit 212F at STP to go through the phase transformation. My point there was that the temperatures are too low and the pressure too high for that to occur for water. The 'suggestion' is that the temperatures are 'possibly' high enough for it to occur for gasoline. That would aid the explanation.

Another point to ponder is 'Dew Point'. Once upon a time, I saw a video of the inside of a manifold with WI. It showed water (not vapor) running down the sides of the manifold like riverlets. What that suggests is that the temperature and pressure drop below the Dew Point for water and the water saturates out. Once it drops out, it is no longer useful for cooling. Gasoline, on the other hand, probably has a much lower dew point and tends to stay vaporized. We did find under testing that you can lower the temperature too low and cause the gasoline to liquify out of the air. When we cooled the air too low, you would have the engine continue to rev after you lifted. We surmised that the fuel had to come from puddling in the manifold. When you lift, the manifold goes to vacuum and vaporizes the fuel. The A/F would go rich and the engine would race for just a second. It was a strange feeling. :)


First, latent heat of vaporization is the amount of heat energy required to go from a liquid pahse to a gas phase with no change in temperature.

[Steve in VC]

Wineguy,

If it came across that I was pinging on you (pun intended), my apologies.

I was noodling your question. Does the increase in pressure increase the evaporation temperature faster than pressure raises air temp. I don't know. But, both Tom's fuel injection, and water injection works - so I was trying to determine why.

I stepped back to determine why pre-detonation occurs, and why liquid cooling fixes it. Thus my previous post.

Steve

[Tom @ Fast Forward]

Actually, it came across as very complimentary. ;) Thanks

Keep in mind that there are two causes of knock. One is ignition related. The sparkplug fires too advanced and causes the peak power to occur before 20* ATDC. And, second, there is a source of ignition that occurs before the sparkplug fires "Pre-ignition". The first case is correctable by adjusting timing with some ignition retard device. The second is caused by a heat source and the only cure is to correct the heat source. Sometimes this is carbon on the piston that gets red hot and doesn't cool fast enough, sometimes it is caused by the inlet air being too hot and rising above the auto-ignition temperature/pressure of the charge, sometimes it is as simple as the wrong spark plug that stays hot and becomes a source of pre-ignition.

What I have done with the evaporative cooling of the fuel from the 5th injector is simply keep it at enough flow to cool the charge sufficiently to avoid pre-ignition. I have not tried to set it to cool the charge looking for ultimate power. As you said, I had to weigh all of the parameters. CARB was at the top of the list. That is also why you see my competition have a hard time with CARB. They are trying to make ultimate power. I am just trying to make a fun daily driver.

[Tink]

Interesting read, not sure I understand it completely.

What, if any, is the effect on E Cool from using alcohol based gasoline fuels, either E 85 or gasoline with alcohol supplements (about 10% I believe)?

thanks

[Tom @ Fast Forward]

I think the effect of E85 is negligable as it is only 15% ethanol. Ethanol has a higher octane rating but stores less energy (ie, your gas mileage goes down so you need to use more of it). So I think it's benefits or lack of benefits for E-Cool are a net zero.

[Satisaii]

Since my degree is in internal combustion and compressible flow, I should be really getting into this...

I pulled out some text books on it and my brain exploded. I have let my education slip away from me.

[Jeff Albertson]

I think the effect of E85 is negligable as it is only 15% ethanol.
\\\"E85 is an alcohol fuel mixture that typically contains a mixture of up to 85% denatured fuel ethanol and gasoline or other hydrocarbon by volume. On an undenatured basis, the ethanol component ranges from 70% to 83%.\\\"

From: http://en.wikipedia.org/wiki/E85

[Markp]

I think the effect of E85 is negligable as it is only 15% ethanol. Ethanol has a higher octane rating but stores less energy (ie, your gas mileage goes down so you need to use more of it). So I think it's benefits or lack of benefits for E-Cool are a net zero.

You really aren't this ignorant about E85, are you?

E85 is 85% Ethanol, and 15% Gasoline. It has an Octane Rating of 105 and it will absorb 2.4 times the amount of heat that gasoline would.

Latent Heat of Vaporization
Pump Gas: 150 BTU/Gal
E85: 359 BTU/Gal

http://www.drivingethanol.org/motorsports/racing_fuel_characteristics.aspx

So if E-Cool is effective with gasoline, It would be VERY effective with E85 Ethanol.

Mark

[Tom @ Fast Forward]

Oops. My bad. I was thinking of the stuff we have around here that is 85% gasoline and 15% Ethanol. Sorry.

No Mark, I'm not ignorant. I just flipped the numbers. I simply made a mistake for which I apologize profusely and promise it won't happen until the next time.

[Steve in VC]

I am well past 15PSI on 91 octane. If you can tell me how to make water burn, I would switch from fuel.

The power industry has been burning water for years. Coal gassification / syngas happens when you mix water, coal and a ton of heat. Carbon strips the oxygen from water, resulting in carbon monoxide and hydrogen, both burn.

Now all you have to do is haul a train load of coal, and a gassification plant behind your Miata. Be careful in the turns ;-).



As I understand it, E85 has less energy per gallon off boost. E85 supports more boost before pre-detonation due to higher octane, more cooling per gallon, and because you add more fuel (~15%), so more gallons of cooling.

[wineguy]

Aw man... you guys are posting faster than I can read, chew & digest this stuff!

There is no doubt in my mind that E-Cool is effective -- non what so ever! But, I'm still trying to get a better handle on all of this in my skull. Seems there are always more questions than answers sometimes...

Steve, no offense taken -- you asked a lot of questions that helped me clear out some cobwebs!

I assumed, that charge cooling more-or-less took place in the manifold. I also assumed that it was pretty much based on the boiling points of whatever liquid was involved. Wrong assumptions on my part, perhaps. We all know that water will cool air at below 212F -- just spray the garden hose in the air on a 100F (dry) day, and evaporative cooling occurs -- just like a swamp cooler. That kind of cooling is still dependent on temperature & pressure, but also relative humidity -- that's why swamp coolers are less effective in Florida on some days than they would be in say, central California.

Air-Air intercoolers are pretty much nothing more than a heat exchanger -- exhanging ambient air temp at the front of the moving car for the heat in the air flowing through the intercooler. Simple as that, they cool the pressurized air.

Water injection, can cool the charge, however, the physics & dynamics of it are still somewhat beyond my grasp. Again, the PV=nRT stuff and common sense (sometimes mistaken for poor assumptions!) indicate that my gut feeling it is not so much cooling, but water or water vapor induced slow burning in the combustion chamber. So, is WI as effective on a hot dry day as it is on a cold wet, foggy day? Or, is it a negligble variable? Tom reminded me of another aspect in his post - dewpoint. That opens the door to relative humidity and saturation of the air charge with water. In the intake manifold, I would venture to guess that at the pressure & temperatures seen there, that the air charge would be saturated plus some. If, all the water injected did turn to steam in the manifold, which absorbs that 2260 J/gm of heat energy, expansion also occurs, so what happens then? Anybody got their steam tables handy?

E-Cool works. Period. Because it's with gasoline, and perhaps an ethanol (now your talking my language!) mix, there appears to be a greater chance of full vaporization in the manifold. Just the right amount, apparantly. CARB approval indicates the exact right amount, and spark plug color/deposits will tell the rest. Again, the only reason that E-Cool can work is because it cools the charge air, and again I'm assuming here, in the manifold. That's the only difference that can occur. Not to rich or not to lean. Just right -- kudos to Tom.

Somebody else, please, do the math...

Keep this up!

[Tom @ Fast Forward]

I have a trailer hitch. I guess I could pull the gassification plant behind. :)


The power industry has been burning water for years. Coal gassification / syngas happens when you mix water, coal and a ton of heat. Carbon strips the oxygen from water, resulting in carbon monoxide and hydrogen, both burn.

Now all you have to do is haul a train load of coal, and a gassification plant behind your Miata. Be careful in the turns ;-).



As I understand it, E85 has less energy per gallon off boost. E85 supports more boost before pre-detonation due to higher octane, more cooling per gallon, and because you add more fuel (~15%), so more gallons of cooling.

[Tom @ Fast Forward]

Wineguy,

Water will form vapor at basically any temperature. Even ice forms vapor (32F is the tripple point of water. Liquid, solid, gas.). You are correct about the efficiency of cooling of water with respect to humidity. The good news is that a supercharger really heats the air a LOT :) and that corrects the humidity problem pretty fast. It is really "Relative humidity" that we mean when we talk about humidity in the air. That means the amount of water in the air relative to what it could hold. 90% RH at 30F is a lot less water than 90%RH at 100F. When we heat that wet air to 200F, it is dry again. Those of you who live in the north east cold country will know that it can be 100% RH outside and 32F but when that air is brought into the house and heated to just 70F, the air is VERY dry and a lot of static electricity. You basically go from 100% to 10% just raising the temperature 40F, in that case. At 10PSI, the Eaton is trying to raise the temperature 150F or thereabout. All we need to do is take about 75 of that back out to avoid pre-ignition.

[wineguy]

Water will form vapor at basically any temperature. Even ice forms vapor (32F is the tripple point of water. Liquid, solid, gas.).

Sublimation. Watch closely as dry ice evaporates -- never really goes through the liquid phase. Water, yes, but only under a limited number of circumstances... I was in Denver recently, and there was not a lot of "slush" on the roads. Seems that after a snowfall, when it melted it would evaporate quickly.

It would be interesting to look at three "intercooled" applications and compare pre-valve intake manifold temperatures... set the experiment up as something like this:

4 Miatas, all the same size engine, 3 with MP62 hotsides, choose 1 cylinder, and monitor pre-valve temp with some sort of sensor)

1 -- CONTROL: Normally aspirated; no intercooling.
2 -- SUPERCHARGED with air/air intercooler (gotta move forward to make it work though!)
3 -- SUPERCHARGED with water injection
4 -- SUPERCHARGED with E-Cool

Rev each one up similarly to conditions that would trigger boost, and see what happens -- the temperatures will tell the story. :yes:

Any guesses as to the outcome? :dots:

[wineguy]

OK, send me back to High School chemistry... :oops:

Got carried away in replying too quickly.
Just for the record, triple point isn't the same as sublimation.

Sorry. :surrender:

Wikipedia is your friend...

[FormerDatsun510Man]

Some thoughts on WI. The boiling point of water as a function of pressure is 212 deg F at 1 atmosphere and 267 deg F at 2 atmosphere (14.7psi boost). Those numbers are both for at sea level. It would be lower at any altitude above this. For say, 10psi, this would mean the water would boil at about 250 deg F. Based on this chart (http://www.magnusonproducts.com/images/mp62g2.jpg), the MP62 RAISES the air temp by about 180-200 deg F. Thus one would only need an ambient temp of around 50-70 deg F to reach 250 deg F intake manifold temps at that boost pressure. In other words, at 50-70 deg F outdoor temps and higher, running 10psi, the intake manifold temps would be above the boiling point of water. However, once the air temps in the manifold drop below this amount by the water injected and going to vapor state, I guess the water wouldn't continue to vaporize. In fact, that tells me that at 250 deg F intake manifold temp would be the minimum temp that WI could lower it at 10psi boost, strictly by going to vapor state. The remaining cooling of the air would be from the specific heat capacity of water. I.e., the non-vaporized water temps going up. This would explain to me how Tom said he observed streams of water in the inside of the intake manifold for a WI'd setup. Given that the boiling point of gasoline is so much lower, I can see how the E-Cool concept could drop the manifold air temps actually further. However, consider that the cylinder temps during combustion go up ... oh 1400 deg F or so :biggrin:. I'm pretty sure the water would have no problem COMPLETELY going to vapor state by this point, as long as so much isn't injected so as to kill the combustion. And thus, by the end of the power stroke the latent heat energy of water->steam has been completely released. The key thing for controlling detonation however, is reducing the PEAK cylinder temps below the autoignition threshold.

Also, a bit off the topic, it's been a while since I could have time to comment in this thread. Tom asked if my engine had headwork, etc. Nope, as I mentioned many times before I have a STOCK rebuild with 9:1 factory pistons from the 1st gen Miata. The one thing left that I think is possibly holding back a bit of rwhp flowwise is that JR header. In fact, I am about 99% sure it is :biggrin:. If only I could get my hands on a Gibb.

Bill

[Tom @ Fast Forward]

Water will vaporize at any temperature. It turns to steam above the boiling point but will vaporize even at 90F. We cool a lot of homes out here with that. :)

However, once the air temps in the manifold drop below this amount by the water injected and going to vapor state, I guess the water wouldn't continue to vaporize.

[FormerDatsun510Man]

I guess we are talking about rates of evaporation then. Obviously water will evaporate at room temperature... just takes a while :). At the boiling point, water immediately turns to steam. Without it turning immediately into steam, I would venture to guess it wouldn't have a chance to evaporate in the intake manifold? Considering that say the volume of the Coldside intake manifold is roughly the displacement of the engine, this would mean it would only take to revs of the engine at WOT (roughly) to consume the air in the manifold. At 6000rpm that would give the water .02 sec to evaporate.

Bill

[MS2KSE]

You'll have lost me:1eye: ,SO, with a coldside, the "e-cool" adds the extra fuel to make the 25+ whp over the base kit plus cools the air enough to stop "ping", BUT WI wouldn't be very useful due to the short time in the manifold, ? BUT injecting pre supercharger would help cool the supercharger ?
WI would be helpful on a hotside due to the lengthened contact time with the hot air ?
So would Pre and post Wi be a better set up on a hotside, then just post Supercharger ?

[Steve in VC]

All is well as long as the water evaporates, and cools the charge, before the temperature of pre-detonation.

If you can eliminate pre-detonation by retarding timing, then you have fire to help evaporate the water (in about 500usec - 5 degree before TDC to 20 after TDC).

[Tom @ Fast Forward]

Well, it really depends on the Relative Humidity of the air as mouc as the temperature. Out here in the summer when it is 100F and 6% RH, it doesn't take a lot of water to cool a large space to 70F and the RH only climbs to about 30%. A lot of shops with large bays, like automotive shops, can cool those large shops for pennies a day on a few gallons of water. To give it a further clue, my pool with about 800 square feet of surface can lose about 1" a day in the summer. Lastly, those of who swim out here can attest to the fact that on a 110F day with the water temperature in the pool at 90F, when you step out of the pool, you will literally freeze. :) Small children actually can turn blue. The good news is that it only lasts for a couple minutes until all the water is gone from the surface of your body and then you realize that, indeed, it is 110F.

Now, what makes it work so well regardless of RH inside the manifold on forced induction is that when you heat the air, you drop the RH significantly. If you had 250F air and it was 100%RH (under pressure, for example), you could add water all day long and all you would make is saturated steam. Again, I stress, this cannot be resolved to simple equations. There are just too many parameters to consider. Temperature, pressure, RH and Dew Point for a start but I'm sure I have neglected even more parameters that will have an effect.


I guess we are talking about rates of evaporation then. Obviously water will evaporate at room temperature... just takes a while :). At the boiling point, water immediately turns to steam. Without it turning immediately into steam, I would venture to guess it wouldn't have a chance to evaporate in the intake manifold? Considering that say the volume of the Coldside intake manifold is roughly the displacement of the engine, this would mean it would only take to revs of the engine at WOT (roughly) to consume the air in the manifold. At 6000rpm that would give the water .02 sec to evaporate.

Bill

[Tom @ Fast Forward]

Steve and MS2KSE, if the conditions (temperature, pressure, RH, dewpoint, etc) are right, water evaporates very very fast. I am not in favor of LI (Liquid injection, be it water, fuel, alcohol, etc) pre-blower for two reasons. First, Magnuson says it could cause fracturing of the coating on the rotors from quick cooling of the surface vs. the rotor; and second, you are heating the liquid. It takes a lot less energy to cool the hot air than it does to cool the blower and all we really care about is the air temperature. With liquid cooling, the hotter the air the better as RH is inversely proportional to the temperature.

Kind of the inverse of making ice in the winter. Hot water will make ice quicker than cold water. If you have a really cold day outside, throw some cold water on you car and note how quick ice forms. Then throw some hot water on your car and see which makes ice quicker. The rate of cooling is so fast with the hot water that it will form ice quicker.

For those who don't know, dewpoint is the temperature at which the RH is exactly 100%. It also defines fog. Fog occurs when RH is exactly 100%. One degree colder and you have dew. One degree warmer and you have less than 100% RH. Either way, fog will go away as it requires the temperature to be exacly = to the dewpoint temperature to occur. The more perfect the dewpoint and actual temperature are, the denser the fog.

[FormerDatsun510Man]

Very fascinating stuff! Tom, you mentioned with E-Cool the temp in the intake manifold went from 225 deg F to 150 deg F for a cooling effect of 75 deg F. And that getting the temps in the manifold down to around 150 deg F is cool enough for quenching detonation issues. I.e., going lower than that would not be beneficial for lowering detonation. Suppose we go the other way though... that is raise the intake manifolds temps up... running a bigger pulley :). Suppose intake manifold temps up around 300 deg F... I am guessing that is where a 130 pulley would bring them under full boost. Can E-Cool bring that also down to 150 deg F? I guess what I am asking is the 75 deg F temp drop at the limits of E-Cool or does it have even more room to pull the temps down?

[Tom @ Fast Forward]

LOL. Well "One IS running a 130mm pulley" :) and it was what was used in those tests. It appears that when IM temps excede about 200F thta detonation occurs. I guess I should say that I am running a 130/65 pulley combo, over 15PSI boost (I hit 10PSI at 3000 RPM, 15PSI at around 4500-5000 and my gauge only goes to 15 so 18?) and IM temps of around 150F. I know that if I just run into boost at partial throttle (as seen in the BRR.wmv video) IM temps will excede 300F in a heartbeat. Luckily, it doesn't matter if it 300F+ at low boost. It is like A/F. Yes, 12:1 is a good number but 13:1 is just fine if you are at low boost and not WOT. 13:1 at 5PSI part throttle is different than 13:1 at 5PSI WOT for some reason? All I can think of is the amount and velocity of the air are different. Frankly, below 3500 RPM, I prefer 13:1, even at 10PSI.


Very fascinating stuff! Tom, you mentioned with E-Cool the temp in the intake manifold went from 225 deg F to 150 deg F for a cooling effect of 75 deg F. And that getting the temps in the manifold down to around 150 deg F is cool enough for quenching detonation issues. I.e., going lower than that would not be beneficial for lowering detonation. Suppose we go the other way though... that is raise the intake manifolds temps up... running a bigger pulley :). Suppose intake manifold temps up around 300 deg F... I am guessing that is where a 130 pulley would bring them under full boost. Can E-Cool bring that also down to 150 deg F? I guess what I am asking is the 75 deg F temp drop at the limits of E-Cool or does it have even more room to pull the temps down?

[FormerDatsun510Man]

Ah, I didn't realize those temps you mentioned were with the 130 pulley. Don't mean to cause another thread drift LOL, but you mentioned you are now making 220rwhp with the 130 pulley. Do you have a dyno plot of it? I'm curious how it compares with running a 130 pulley on my Hotside R4. Also, was this done on 91 octane? If so, there might be more power on 93, which what I am using in my setup out here in Looooisiana :biggrin:.

You're going to laugh at this, but I am even considering going DOWN to the 120 pulley which I have setting on my shelf :). Reason being is that the power and specifically, torque, feels almost a little bit too much. It's not that I can't "handle" it :), just that it feels like the stock drivetrain is being a little bit overpowered and make break someday. Also, the other night I almost spun the car out in 2nd gear just doing a roll-on WOT at about 30mph or so. It now has so much torque that the tires break loose even in a straight line at that speed. Funny thing is that I might find I need to detune it a little which would then possibly beg the question of why not just go Coldside.

[Tom @ Fast Forward]

I don't have a dyno plot at the moment. We were rolling off at a little over 200WHp last time on the dyno. It was headed for 220 but we couldn't figure out why it was dieing above 200. Used to make 212 on a 115mm pulley. Then the Split Second coils melted on cylinders 2&3. Then I installed the new stock coils and wires. Now it runs up to redline like a scalded cat. Just a guess, but I 'think' the missing power is back. It was 212 with the 115 and we did see roughly 170/180 ft-lbs even though the power was falling off above 200. I promise to get back to the dyno when I am not doing someone elses car and get a new graph. At the moment, I am having problems keeping up with sales, getting the 91 to run and getting the parts into inventory for the $2995 Coldside. We did finish the automatic kit and get it shipped.

PS, I am not complaining about the increase in sales. :)

Oh yeah. the 130mm pulley is on to stay. It is way too much fun to take off. And, yes, I am running on 91 octane.

[ROZ]

Man I have been devouring this thread like a madman, trying to get to the end and screaming "sensible heat, sensible heat".:taz:

The difference in the earlier thought processes between Bill and Tom is the difference between physics and engineering, one is why it happens, the other is how. Now I haven't checked the math(s) because I just giggle at your quaint archaic measurement of temperature (;)) but I think you're underestimating the sensible heat that the E-cool takes out.

Yes the specific heat capacity of water is significantly less thanthe latenet heat of vapourisation, but you need to include the temperature difference between the incoming fuel and the manifold/air. With a pressurised system the boiling point will be higher, so the delta between the two is large, say 100F Arizonia ambient and 300F in the boosted manifold, then the fuel has to "absorb" 200F of sensible heat before it makes the jump to vapour and pull sout the large latent heat*. So comparing 1 degree difference in sensible heat with latent heat is misleading, you need to compare a 200 degree difference, and therefore it does have a significant effect and shouldn't be omitted from Bills calcs. This may go some well to explaining the difference between the numbers and reality...

Also, careful with the multi-phase question. First thing is that the air/fuel is going through the manifold at a hell of a lick - not much time for things to oscillate if it were possible - but essentially it will be in equilibreum and the rate of vapourization will level out instantaneously (in our terms), so you will not get a cycle between gas/liquid. The liquid will only vapourise as much as the thermodynamic system can handle.

* Careful with water can vapourise at any temperature Tom. True in any multistate system there will be an equilibreum between solid/liquid/vapour, and as temperature and pressure change, so does the equilibreum point, but this is not the same as the spontaneous change from liquid to gas when it reaches it's boiling point.

Darn I've got a call with the US in 20 minutes and need to prepare, but I think I've got my points over. I was tempted to draw in the PhD Chem Eng "petrolhead" colleague from across the corridor, but I'll wait to see what you guys think, and whether I need to explain myself! :biggrin:

[Tom @ Fast Forward]

E-Cool is NOT an intercooler. It does cool the charge air between the blower and the intake valves but it is NOT an intercooler and should not be mistaken for an intercooler. Real intercoolers can't add fuel. :)

[ROZ]

er yes, I'd go with that. I thought that was what I was saying!

An IC is a heat exchanger. There's no net heat transfer in E-cool, conservation of energy and all that. So it's clearly not an IC - who said it was? But the addition of cooler fuel that gets heated and vapourises does lower the temp. All the energy (other than small heat losses) goes into the jinny.

Ps. Thought you were on vacation?

[Tom @ Fast Forward]

Sorry. That was tongue in cheek humor. The oldtimers will understand. ;)

[ROZ]

phew!
Thought I'd lost the thread there (in more ways than one)

[socal pat]

phew!
Thought I'd lost the thread there (in more ways than one)

Hehehe I don't have an intercooler too.

[Tom @ Fast Forward]

Besides, it is a LOT more fun to have a signature that says I am at 18PSI NON-INTERCOOLED. LOL I don't know any other Miata kits that can say that.

:grouphug: :party:

[Steve in VC]

I got an email from Tom where he gave me the history of E-Cool (after the WI ran out of water one time too many).

Using just the stock injectors to get a 12:1 A/F ratio gave him lots of ping.

He leaned out the main injectors, and added lots of fuel with E-Cool (cooled the intake manifold almost to ambient), but enough fuel was condensing out that the engine revved after he lifted off the throttle.

Through experimentation, he found if 7% to 10% of the fuel came from E-Cool, and the rest from the main injectors, he could cool enough to run 12lbs of boost (actually 18lbs in his case) without pinging, in the cool Arizona summers no less.....

(Tom, I cut down a long email, hopefully this is accurate).


If it was simply fuel cooling air, it shouldn't matter where it's added, but it does.

This isn't magic, there has to be a reason why this works.

I have a concept that isn't jelling, maybe someone else knows. If E-Cool keeps the intake manifold cool enough that more of the fuel from the main injectors enters the cylinders as droplets, instead of vapor...

Does the heat of vaporization change with temperature?

Do droplets lose some heat to the cylinder walls?


Where does the additional cooling come from?

[Tom @ Fast Forward]

Yes. When you go above ~5-6PSI boost and/or 160/170 WHP, you need to provide some cooling of the charge air (A/A, W/A, WI or E-Cool). You could change the main injectors to 550cc and it would still tend to knock even at 10:1 A/F if based solely on the main injectors. It is just way too late to cool the air.

Most people think that the sequence is that the intake valve opens and a spray of fuel is injected and then the intake valve closes and the piston compresses the air, ignites and so on. Wel, close but not close enough. In a stock Miata (94+), the injector on time at WOT is about 10-11 mS to get enough fuel and the intake valve is ONLY open for only 8.5mS at 7000 RPM. Most of you are now saying "Well, that wont work". :) To understand how it does work, let's step back to the 90-93 1.6L Miata. Like all MIata ignition systems with wasted spark, the coil for 1/4 fire every time the cylinders come to TDC. That means one cylinder is on the ignition stroke and the other has the exhaust valve wide open and it's spark is "wasted". On the next stroke 2/3 cylinders do the same. Well, in the 90-93, the fuel was injected in basically the same way. The fuel injectors for 1/4 fired every time the cylinders came to TDC. That means that one cylinder has the intake valve open and the other has the intake valve closed. The one that is closed simply uses the fuel to cool the intake valve which vaporizes the fuel and makes it ready for that intake valve to open on the next stroke along with a little more fresh fuel as the intake valve is pulsed again. So, if 10mS of fuel was needed, it would fire 5mS when the valve was open and 5mS when the valve was closed. My guess is that, if it weren't for emissions, they would have stayed with this as it is very efficient (cooling) but not perfect enough for California. ;)

With the 94+ models, each injector is individually fired but some of the fuel is injected when the intake valve is open and some when closed. At 7000 RPM, that is about 8mS when open and 2mS when closed. One pulse, not two. For those of you who are interested in duty cycle, at 7000 RPM it is about 17mS from intake valve open to intake valve open again. DC would be 10/17 or about 58%. As you can see, they gave up about 30% of the cooling time it had before (2mS during valve closed rather than 5mS). :( Emissions were better but it is not as knock resistant. That time with the fuel present and the valve closed is needed for cooling.

Anyhow, that is my story and I'm sticking to it. ;)

[ROZ]

He leaned out the main injectors, and added lots of fuel with E-Cool (cooled the intake manifold almost to ambient), but enough fuel was condensing out that the engine revved after he lifted off the throttle.

I have a concept that isn't jelling, maybe someone else knows. If E-Cool keeps the intake manifold cool enough that more of the fuel from the main injectors enters the cylinders as droplets, instead of vapor... At the temperatures of an engine during operation you will not get consensation of fuel, even at boost pressures.

Does the heat of vaporization change with temperature? No, the latent heat doesn't change with temperature, but it does change with pressure. "..the higher the pressure then the smaller the amount of latent heat required." [Eastop and McConkey, Applied Thermodynamics]. Temperature and pressure are linked though in a "sealed" system according to a pv chart (pressure/volume)

Do droplets lose some heat to the cylinder walls? You should get droplets at working termperature, but my Phd across the corridor tells me you can get "bore washing" where poor injectors spray a stream into the cylinder rather than a mist. At starting temperatures this can wash the walls of the cyclinder and lead to errosion.

But in answer to your question, you would get a temp drop if it happened...

Where does the additional cooling come from? cooling comes from the fuel being taken from it's ambient temperature up to it's boiling point at the system pressure (=sensible heat/heat capacity = 2.22KJ/Kg.K) and then the absorption of heat by moving the fuel into the vapour phase (latent heat of vapoursation = 293 KJ/Kg).

So
raising gasoline by say 100degC or K = 222KJ/Kg
Vapourising it at that temperature = 293KJ/Kg
There is also the specific heat of raising temperture of the fuel vapour but I can't find that number...

So we're looking at about 400KJ/Kg of heat to be taken out of the incoming air based on the example above. Then you need to look at the secific heat capacity of air (~1.01KJ/Kg.K at constant P) to see what temperature difference this would make.

Using the example above at stoich of 14/1 of air we would cool by 28degC or Kelvin. Of course we need actual ambient and engine temperatures, plus any heat taken out by the vapourised fuel. Bill had all the numbers I believe.

[Tom @ Fast Forward]

ROZ, If it helps, I believe air is about 0.075 pounds per cubic foot and 200HP (170WHP for a Miata) is about 330 SCFM of air or 24.8 pounds per minute. At 12:1 A/F (12 pounds of air per 1 pound of fuel), that would be 2.06pounds of fuel per minute. With 10% of that coming from E-Cool, that would be about 0.206 pounds of fuel/minute from E-Cool.

That's not too far off of the 'rule of thumb' that stoich (14.7:1) is ~0.5 pounds of fuel per HP per hour. 1.67#/Hr for 200HP at 14.7:1. 14.7/12 = 1.225. 1.225 * 1.67 = 2.05. Pretty close to 2.06, huh? :) Isn't math fun?

[ROZ]

ROZ, If it helps, I believe air is about 0.075 pounds per cubic foot and 200HP (170WHP for a Miata) is about 330 SCFM of air or 24.8 pounds per minute. At 12:1 A/F (12 pounds of air per 1 pound of fuel), that would be 2.06pounds of fuel per minute. With 10% of that coming from E-Cool, that would be about 0.206 pounds of fuel/minute from E-Cool.

That's not too far off of the 'rule of thumb' that stoich (14.7:1) is ~0.5 pounds of fuel per HP per hour. 1.67#/Hr for 200HP at 14.7:1. 14.7/12 = 1.225. 1.225 * 1.67 = 2.05. Pretty close to 2.06, huh? :) Isn't math fun?

So the question is, who now wants to work out the actual temp drop...?

[FormerDatsun510Man]

Actually, if you don't round the numbers off before the completed calculation it is 2.04. :devil2:

Somewhat seriously (but not too much), the calculation for fuel flow should be based on crank hp not rwhp. 200rwhp ~ 226 crank hp according to the wisdom at Miata.net. Based on that with .5lb/min you would get 226 * .5 / 60 * 14.7 /12 = 2.31 lb/min. Comparing with your number for 2.06 (for your measured CFM and a/f ratios at 200rwhp), that would mean the estimated BSFC for a Miata at stoich would be 2.06/2.31 * .5 = .45

Actually, I would tend to believe .45, because the rule of thumb I have seen is that .5, however they are referring to full power not cruise. Typically at full power normally aspirated around 13:1 would be ideal. 13/14.7 * .5 = .44 Isn't math fun? :biggrin:

Bill

[maggie]

Thanks all for the lessons in injector timing and everything else. I love learning about our cars and although I am pretty strong in math and the sciences, I am just beginning to put it together in the application of internal combustion and FI.:cool:

[ROZ]

So come on Bill, are you going to step back up to the plate and modify your spreadsheet to include the sensible heat as well??

Just thinking though :ack2: some of my logic above was off. Here I am talking about the engine temperature, but we're interested in the temp of the air that's been compressed - darn.

So we need ambient air, temperature increase due to compression but do we need the throughput through the engine? If the ratio of the mass of air to fuel is the important element, the division of heat absorption by the heat capacity of the air knocks out mass.

ie KJ/Kg divided by KJ/Kg.K = K = temperature change, so if you've got the ratio the flowrate doesn't matter.......?????

This has all given me a headache (and I've got no "real" work done today) :surrender:

Loving this thread though - thanks all.

[Tom @ Fast Forward]

ROZ,

I'm driving and can't run the math right now. You have simplified the thermo-dynamics pretty well. Do me a favor and run an example for me and let me know if E-Cool works. ;)

[ROZ]

(third time lucky, the N95 kept overwriting when I tried to put in smilies - back on PC now:mad:)

Driving AND typing :eek:!!

Once bitten twice shy - we all know that it works; how many units are running around??:tongue: I just got hooked like a fish by the earlier threads, and the Engineer in me just wants to see HOW MUCH it works. Prove it with thermodynamics and the naysayers look even more stupid than they do already.:stupid:

If someone else doesn't beat me to it I'll try and work it all tomorrow. Perhaps I can fool the boss into thinking I'm desining a new heat exchanger network :devil:

[Steve in VC]

If it was simply fuel cooling air, it shouldn't matter where it's added, but it does.

This isn't magic, there has to be a reason why this works.

I have a concept that isn't jelling, maybe someone else knows. If E-Cool keeps the intake manifold cool enough that more of the fuel from the main injectors enters the cylinders as droplets, instead of vapor...

Does the heat of vaporization change with temperature?

Do droplets lose some heat to the cylinder walls?


Where does the additional cooling come from?

I was looking for information on detonation and stumbled into this that seems to provide a possible explanation:

The air fuel mix as it flows into the combustion chamber is not perfectly atomized, that is the fuel vapor droplets consist of larger droplets of fuel molecules surrounded by air. It takes additional energy to further atomize this vapor, that is to break the hydrostatic forces (the surface tension of the fuel droplet). This additional energy can be taken from a hot surface (such as the piston crown, etc.), which then leads to a cooling of the piston. The additional energy can also be imparted via large turbulences and pressure waves, as in a squish band-type motor, which will help to further atomize the fuel. Note that the term "atomize" is actually misleading since the molecules are still left intact, that is the hydrocarbon chains (and oxygen bonds for alcohols) are not broken.


If E-Cool cools the air, more of the fuel from the main injectors remains droplets, requiring more energy than gaseous gasoline to burn.

I recently read a explanation why water has a high specific heat. There is an electrostatic field that must be overcome for the molecules to move with heat. In this case, surface tension provides a bit more energy than just thermodynamics.

Ramon also asks:Also, if the fuel does evaporate quickly and creates additional pressure - thus reducing the amount of fresh charge - then the engine will produce less horsepower, right?

Correct. The horsepower will depend on the volumetric efficiency of the engine which is a function of the pressure difference between ambient air and cylinder pressures. If additional fuel is vaporized inside the combustion chamber the pressure in the cylinder will rise, and, while the valves/ports are still open, reduce the volumetric efficiency, and thus the power output.



Lets make an assumption that uncooled air vaporizes the fuel droplets before 20 degrees after top dead center, and E-Cooled air delays vaporization until after. That reduces the peak pressure, thus peak temperature.

Either, or both of these can mechanisms may be at play. Or it may be nothing at all....

Enjoy,
Steve

[Tom @ Fast Forward]

Well I, for one, still think it's magic. ;)

The scientists also said 'Cold Fusion' couldn't be done. I not only did it but got my Cold Fusion in the Guinness Book of World Records. LOL