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Old 10-24-2005, 06:53 AM
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How the MCS Engine Makes Power

In stock form, the supercharger displaces air from the throttle body faster than that air can be ingested by the engine. This produces boost, which increases horsepower. This also increases heat (both due to the heat created by the simple obeyance of the ideal gas law, but also due to the THERMAL efficiency of the supercharger). Roots-type superchargers like the Eaton have pretty crappy thermal efficiency, somewhere on the order of 55%. That means that we are heating up the air A LOT by beating it with the rotors. Luckily, we have an intercooler that removes about 70% of the increase in air temp and transfers it to the atmosphere (thanks, atmosphere!). So, the engine is making lots of power, but wait, the supercharger requires power from the engine in order to spin. This removes some useable power (let's use 20 hp), so instead of having, let's say 188 hp, we only have 168 hp.

Here's how the MCS engine makes power when the supercharger is spun faster:

The supercharger spins faster, and displaces air from the throttle body even faster. This increases boost, but also increases heat. In fact, it increases heat even more than it increases boost. Luckily, we have a net gain since we still have the intercooler, which is still removing about 70% of the increase in air temp and transfers it to the atmosphere (thanks again, atmosphere!). Since the supercharger is spinning even faster, it requires even more power to drive. This removes some more useable power (let's use 30 hp since the power usage goes up exponentially with speed), so instead of having, let's say 210 hp, we only have 180 hp.

Here's how the MCS engine makes power when the supercharger is spun at the stock speed, but a turbo is placed before the throttle body:
Air enters the turbocharger, where power from the exhaust gas (mostly heat, but also pressure) is used to spin the turbine, which is attached to the compressor. Water-cooled turbochargers are water-cooled to preserve bearing life by preventing lubricating oil from overheating, the reason for choosing water-cooling for the turbo has nothing to do with cooling the charge air. Turbochargers have very good thermal efficiency, about 75% is a good rule of thumb. So, the air coming out of the turbocharger is heated due to being compressed (remember the ideal gas law) as well as being heated from being beaten by the compressor blades. That extra heat produced by the turbocharger to make 10 psi is much lower than what the stock supercharger uses to make 10 psi due to the difference in thermal efficiency. Now, you have 10 psi of heated (not too hot, remember) air entering the supercharger, which is spinning at stock speed.

The supercharger could care less what the pressure of the air is coming into it. Its job is to take a handful of stuff from one side of the fence and dump it on the other side of the fence. It only cares how fast it is spinning and what the DIFFERENCE in pressure is between the inlet and outlet. Increase either of those and the supercharger will use more power and make more heat. So, anyway, the supercharger displaces air from the turbocharger faster than that air can be ingested by the engine. This produces boost, which increases horsepower. Because half the compression work is done by the more efficient turbocharger, there is less heating done by the system than if the supercharger had been trying to make that boost all by itself. Also, because the turbocharger requires less horsepower to drive (remember, the main thing that spins the turbo is the heat in the exhaust gas, which was going to be wasted anyway) the system requires less horsepower to drive than if the supercharger had been trying to make that boost all by itself. Let's say the turbo+super engine makes 275 hp and we used 20 of that to drive the supercharger and 5 of it to drive the turbocharger.

Originally Posted by Samurai Will
Mostly that makes sense. But assuming the turbo is mechanically 75% heat efficient, how do you account for heat soaking from the 1000+ degree exhaust gas used to drive the turbocharger? I'm sure that that will bring the temperature up. Or are you accounting for that in the 75% efficiency? If so, how? Is the heat being used, or the exhaust gasses? A water cooled turbo may not be cooling the charge, but it would make sense that being water cooled would account for increasing the thermal efficiency, no? i.e. if it were not water cooled would the intake air heat more passing through the turbo?
I have never seen any data to suggest the reason for water cooling is to cool the intake charge. There may be some slight effect, but I doubt it given the small quantity of water relative to the huge quanity of air. Here's a page that describes the ideas and testing behind water-cooled turbos:

http://www.airpowersystems.com.au/wrx/aps_turbo.htm

If it actually lowered charge temps, I'd think that company would want to use that as a selling point. Logically, I doubt there is much in the way of heat transfer between the small surface area that coolant touches and the large surface area that the charge air touches (there is also a very large flow rate of charge air mass through the turbocharger compared to a very small flow rate of coolant mass through the turbocharger). So, it makes sense that the amount of charge air cooling that can be attributed to the coolant is insignificant. But hey, maybe it's not. Sometimes things are non-intuitive. I've been wrong before and will fess up when I am. If you believe that the amount of charge air cooling that can be attributed to the coolant is NOT insignificant, then feel free to test it and post your results.

Originally Posted by fishbulb
So, more rotations in a fixed amount of time would equal more boost, correct?
Yep, you've got it. Now, there are some interesting nuances to supercharger behavior, some due to the fact that there is clearance between the rotors and the case. As the rotors spin, a small amount of air leaks around the rotors and doesn't end up on the opposite side. Although it is a small effect, it actually lessens the faster you spin the supercharger. But, the faster you spin the rotors, the more violent the whole operation becomes, so the air is heated even more. IMHO, if it could fit under the hood, it would be a good idea to use a larger Eaton blower (M62) which turns slower to make the same boost. GM used an M62 in their 3.8 Regal/Grand Prix/etc. They switched to an M90 and were able to remove the intercooler yet achieve the same power levels with less complexity. FYI, in Eaton-speak, the number refers to the volume in cubic inches inside the supercharger so our M45 transports 45 cu.in. per revolution of the supercharger.

Originally Posted by jlm
the most efficent way to compress a gas is to change its volume very slowly (imagine a piston moving slowly) then you have a good shot at seeing the temp change due to the volume change. (Ideal gas law). now imagine that the piston is moving very fast and mixing up the gas; temp goes up even more than the ideal gas law predicts, much of it due to kinetic energy of the moving molecules. Finally, you don't even have a piston, just a fan that whacks molecules into a tighter space by virtue of imparting directed momentum (the Eaton), even more heat.
In Corky Bell's "Supercharged" are examples of efficiency for centrifugal (like a turbocharger), roots, and lysholm types of compressors. Despite having a similar shape and mass (to transfer heat to the air by conduction), the lysholm is able to compress the air more efficiently (say 65% compared to 55%), imparting less heat on the charge in the process. The reason the lysholm is able to do this is that it has internal compression, that is, as the air is transferred from the inlet to the outlet, it is gradually compressed. This is unlike a roots where the air is suddenly and violently released at the outlet. So, despite being otherwise similar, the lysholm produces cooler boost than the roots, all other things being equal. Likewise, a centrifugal supercharger is even more efficient than the lysholm. Because the air isn't positively displaced at all, but rather accelerated smoothly through the snail-shaped housing, it is able to impart even less heat on the charge than the lysholm.

Of course, there are other considerations when choosing the type of supercharger, namely reliability, shape of the powerband, etc. But it is a fact that a centrifugal compressor such as a turbocharger or a centrifugal supercharger, will generate less heat while compressing the air than will a lysholm or a roots, all other things being equal.

Originally Posted by macncheese
The M45 requires up to 40HP to operate at high RPMS because its inefficient. Do you think your MINI puts this 40HP under its pillow at night for the Boost Fairy? Nope, that energy must be dissapated somewhere and it becomes HEAT. The M45 efficiency is 60% at best and efficiency is measured as a change in temp from inlet to outlet. This is because all wasted energy becomes heat. Turbines on the other hand are very efficient, up around 80% on average. Higher efficiency means less temperature delta. No other numbers necessary.
Here's some rough figures from logs I did on 3 cars on the same day, mine, jlm's, and the Helix turbo car, 2nd gear 2500-6500 rpm. IAT's at 6500:

jlm: 58.5*C (137.3*F)
mine: 54.0*C (129.2*F)
turbo: 38.25*C (100.85*F)

That day, I drove all 3 cars around, hammering on them while logging. The highest peak values I saw during the runs 5-10 minutes each were:

jlm: 65.25*C (149.45)
mine: 60.75*C (141.35)
turbo: 49.5*C (121.1)

That's a pretty big difference, especially since the turbo car makes about 75 more whp than my car and about 50 more whp than jlm's and runs a steady 20 psi.

All three of the cars have the stock IC. The 2.5k-6.5k runs were WOT, which means Wide Open Throttle, which means foot to the floor. The Driver Demand logged for each car was 100%. When I am around traffic, I do not go WOT from 2.5k-6.5k. Of course, there are variations from run to run, and they weren't done with scientific rigor, that wasn't the intent. These are just a ROUGH idea of the temps.

Originally Posted by fishbulb
the turbo is a bit different from the Eaton in this regard. it is not exactly linear, as it is not directly "belted" to the crankshaft.

There is a map that looks like a topographical map that shows the turbo efficiency rates for a given turbo combination when based on pressure vs. air flow.

http://www.turbocalculator.com/compr...maps/t3-40.jpg

first, in turbo-land, boost doesn't 'really' matter, it's airflow...a small frame turbo at 15 PSI may push the same amount of air as a large frame turbo will at 10. Depending on what motor it is bolted to, one may be more or less efficient. (although typically, the larger turbo will be more efficient at the cost of boost response, or "lag". of course, there are exceptions to that rules as well...)

so, you need to figure out how much air you need to flow (every engine set up is different, but it is typically based on the size of the motor) across the bottom scale, and then chart across at different boost levels that you plan to run, and then you can see how efficient this particular turbo is for you particular set up.

so, speeds that fall further & further outside of that center area are less & less efficiaent, they could be slower or faster.
Good points. For anyone who may be unsure how to read a compressor map, it's pretty simple. The x-axis is the amount of airflow, typically is cfm or g/s (but I've seen some more arcane units as well). The y-axis is the pressure ratio in absolute terms. So, if you have an engine that makes 14.7 psi of boost, the pressure ratio is: (14.7 + 14.7 ambient)/14.7 ambient = 2.0. Just trace the lines coming from those two values and they will meet up somewhere on the efficiency scale.

The compressor map looks like a topgraphical map, and in some sense it is. There is an efficiency mountain with its highest level at the summit. Think of the more or less straight line on the left side of the mountain as a steep cliff. Don't go there or you'll fall off! (compressor surge). Since automobile engines run at a variety of speeds and pressures, the airflow and pressure ratio values will vary as well. When turbo sizing, you just try to choose a compressor that gives you the best effciency over the range of driving you'll be doing.
 

Last edited by andy@ross-tech.com; 10-26-2005 at 08:55 AM. Reason: Cleaned up off-topic and extraneous info
  #2  
Old 10-26-2005, 09:31 AM
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As you probably have noticed this thread has been pruned of all the Off-topic discussion and the off-topic remarks. Going forward if there is a thread topic please post on-topic. If you can't start a new thread in the appropriate forum. Additional off-topic posts will be deleted without warning. As before any site guideline violations will result in a first strike warning.

Now back to the discussion at hand...How the MCS engine makes power...

Mark
 
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Old 10-26-2005, 09:42 AM
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Originally Posted by andy@ross-tech.com
When turbo sizing, you just try to choose a compressor that gives you the best effciency over the range of driving you'll be doing.
What about airflow and Pressure Ratio?
 
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Old 10-26-2005, 10:44 AM
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So Andy, what are you trying to say? Add a turbo to your engine?
 
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Old 10-26-2005, 11:13 AM
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In regards to the information on the Turbo sizing, there's a few things I'd like to understand better.

1) How do I determine the Corrected Air Flow in lbs/min?

2a) How do I know what RPM the turbo is running at? (I presume I need to know this since the horizonalish lines represent various turbo RPMs)

2b) Wouldn't the RPMs of the turbo vary along with the Corrected Air Flow (CAF) of the engine, since the turbo spools to different RPMs generating differing amounts of boost, which leads to different amounts of CAFs....or am I thinking sideways again?
 
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Old 10-26-2005, 12:33 PM
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Originally Posted by dominicminicoopers


In regards to the information on the Turbo sizing, there's a few things I'd like to understand better.

1) How do I determine the Corrected Air Flow in lbs/min?
I forget this one. Covered in Max Boost, maybe andy can chime in on that one...

Is short, you need to determine how much air your motor consumes/processes at each RPM level....I had a good page of calculations when i figured it out for my 1.8L miata.


Originally Posted by dominicminicoopers
2a) How do I know what RPM the turbo is running at? (I presume I need to know this since the horizonalish lines represent various turbo RPMs)
you will plot out pressure vs. airflow. at your airflow, crossed with a specific boost level, will give you a specific point on the graph. THat will show for that specific boost/CFM combo, where the turbo will sit in it's efficiency graph and about how fast it is spinning (estimate between the different lines shown)

Originally Posted by dominicminicoopers
2b) Wouldn't the RPMs of the turbo vary along with the Corrected Air Flow (CAF) of the engine, since the turbo spools to different RPMs generating differing amounts of boost, which leads to different amounts of CAFs....or am I thinking sideways again?
AFAIK, RPMs of the turbo will be based on exhaust flow, load and backpressure. Can't give you the specific formulas of that, and there maybe another factor or two involved.

hope that helps the thread!

-jac
 
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Old 10-26-2005, 06:41 PM
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Originally Posted by mferguson
As you probably have noticed this thread has been pruned of all the Off-topic discussion and the off-topic remarks. Going forward if there is a thread topic please post on-topic. If you can't start a new thread in the appropriate forum. Additional off-topic posts will be deleted without warning. As before any site guideline violations will result in a first strike warning.

Now back to the discussion at hand...How the MCS engine makes power...

Mark
Which essentially means you deleted any information that isn't from the NAM favorite sons, including a great quantity of scientific explanation that is very relevant to the topic at hand. You keep a selected post, and answer that post only, disregarding any other argument because it does not serve your purpose. That sound about right?
 
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Old 10-26-2005, 07:12 PM
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Would you prefer that I merge the old posts back in...off-topic and sniping comments included? If so, I'm happy to do that but be aware that if those same issues reappear the thread will be closed. Also, to clarify, there are no NAM favorite sons. If people can't interact with each other in a civil manner on the site (ie - within the posting guidelines) they will receive a 1st strike warning. If their behavior continues their access to the site will be removed.

Mark
 
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Old 10-26-2005, 07:21 PM
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I would like moderation to not mean across the board hacking of relevant information while allowing one person to reign supreme in disseminating imformation. If you feel that hacking the thread due to off topic posts is required, then by all means. What does follow the favored son mentality is allowing Andy to leave in selected posts with his first run answers in place, not allowing for the answers that may in fact refute some of what he says. If you are going to hack the thread, then return Andy's post to the original and leave it at that. It is far from fair to allow one single person the opportunity to control what information remains.
 
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Old 10-27-2005, 01:37 AM
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The only person controlling info here is Mark, as is his right.

The only relevant diversion was questioning the contributory portion of water cooled bearings of a turbo on the overall efficiency/charge temp of a turbo (which was appended in the primary post.) No additional information was provided, only conjecture (and quite a bit of sniping.) I didn't see where anyone contested the 75% efficiency numbers, only why and/or how the efficiency is derived.
If the overall contributory factors to turbo efficiency are of great interest to you, you are free to start a thread which attempts to break down the turbo system into it's components and those which add/dissipate heat, and at what relative rates. Fair enough?
 
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Old 10-27-2005, 06:23 AM
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Originally Posted by fishbulb
Is short, you need to determine how much air your motor consumes/processes at each RPM level....I had a good page of calculations when i figured it out for my 1.8L miata.
jac, Yes I know I need to determine how much are the motor consumes at each RPM level. I don't know how to calc this. Was hoping someone already did one for the MCS's 1.6L engine. Or could provide not only the formulas, but also the MCS specific data needed to make the calculations. I don't mind chugging the numbers thru the formulas, I just need to have the formulas and the appropriate data they need to make those calcs. Anyone?
 
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Old 10-27-2005, 06:34 AM
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I posted this last year:

Without a MAF, we don't know EXACTLY how much air is going into the engine, but we can get a pretty good estimate from the speed-density calculation that the DME uses.

Airflow in g/s is a good rough indicator of horsepower, crank hp is about 80% of the peak g/s reading. So, a stock 163 hp MCS should flow roughly 130 g/s.

Since we don't have a MAF, the speed-density formula uses rpm, displacement, and volumetric efficiency to get a "baseline" cfm for a given condition. This baseline is then corrected for conditions like temperature and manifold pressure. So:

airflow = cid x rpm x 0.5 * Ev /1728

To convert cfm to g/s, let's say we have "ideal conditions" of 0° C, 0% Relative Humidity and 760mm of mercury air pressure. The ideal air density is 1292.9 g/m^3 and 1 m^3/sec = 2119 cfm so 1 g/s = 1.639 cfm.

A stock MCS should see roughly 214 cfm.

We know the cid is 97.5. We know the stock hp peak is reached at 6,000 rpm. Let's use 90% as a volumetric efficiency. Plugging this into the airflow equation gets us:

airflow = 97.5 x 6000 x 0.5 x 0.9 / 1728 = 162 cfm.

That way off from the rough calculation of 214 cfm. Why? Well, that doesn't take into account boost. A stock MCS has manifold gauge pressure of about 9.5 psi at 6,000 rpm. That's an absolute pressure that's 1.67 times atmospheric. So:

162 cfm x 1.67 = 270 cfm.

That's closer to the guesstimate but now it is much higher. Why?

Well, for one thing, the engine is really making more than 163 hp, but some of that power is getting used up operating the supercharger. According to the Eaton specs, that's about 20 hp at a blower speed of 12,360 rpm (6,000 engine rpm) and 10 psi boost. So, our 163 hp engine is really a 183 hp engine, but some of the power can't be used by us.

270 cfm x 163 / 183 = 241 cfm

Now, we're getting close. Only 27 cfm to go. Well, we didn't yet take into account how much the air is being heated by the supercharger. According to Eaton's specs, that's about 180F of temp increase at a blower speed of 12,360 rpm (6,000 engine rpm) and 10 psi boost. If our Intercooler is 80 % efficient, that knocks that value down to 36 F. In absolute terms, that's 496 Rankine compared to an STP value of 460 Rankine.

241 cfm x 460 / 496 = 223 cfm

That's close enough to 214 cfm for government work.
 
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Old 10-27-2005, 07:12 AM
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Originally Posted by Eric_Rowland
The only person controlling info here is Mark, as is his right.

The only relevant diversion was questioning the contributory portion of water cooled bearings of a turbo on the overall efficiency/charge temp of a turbo (which was appended in the primary post.) No additional information was provided, only conjecture (and quite a bit of sniping.) I didn't see where anyone contested the 75% efficiency numbers, only why and/or how the efficiency is derived.
If the overall contributory factors to turbo efficiency are of great interest to you, you are free to start a thread which attempts to break down the turbo system into it's components and those which add/dissipate heat, and at what relative rates. Fair enough?
I never said anyone else was controlling the information. I simply stated that that is the problem. If he wants to control the info, fine. But he allowed Andy to use one post of mine and refute it in his limited way, completely disregarding my responses to the argument. You are, I hope, seeing how very biased that is. I provided a great deal of scientific reasoning to support what I was trying to say, most of which was completely ignored, and subsequently chopped. If the biased nature of the act escapes you, Mark, and Andy completely then I would be very surprised.
 
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Old 10-27-2005, 07:22 AM
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Originally Posted by andy@ross-tech.com
I posted this last year:

Without a MAF, we don't know EXACTLY how much air is going into the engine, but we can get a pretty good estimate from the speed-density calculation that the DME uses.

Airflow in g/s is a good rough indicator of horsepower, crank hp is about 80% of the peak g/s reading. So, a stock 163 hp MCS should flow roughly 130 g/s.

Since we don't have a MAF, the speed-density formula uses rpm, displacement, and volumetric efficiency to get a "baseline" cfm for a given condition. This baseline is then corrected for conditions like temperature and manifold pressure. So:

airflow = cid x rpm x 0.5 * Ev /1728

To convert cfm to g/s, let's say we have "ideal conditions" of 0° C, 0% Relative Humidity and 760mm of mercury air pressure. The ideal air density is 1292.9 g/m^3 and 1 m^3/sec = 2119 cfm so 1 g/s = 1.639 cfm.

A stock MCS should see roughly 214 cfm.

We know the cid is 97.5. We know the stock hp peak is reached at 6,000 rpm. Let's use 90% as a volumetric efficiency. Plugging this into the airflow equation gets us:

airflow = 97.5 x 6000 x 0.5 x 0.9 / 1728 = 162 cfm.

That way off from the rough calculation of 214 cfm. Why? Well, that doesn't take into account boost. A stock MCS has manifold gauge pressure of about 9.5 psi at 6,000 rpm. That's an absolute pressure that's 1.67 times atmospheric. So:

162 cfm x 1.67 = 270 cfm.

That's closer to the guesstimate but now it is much higher. Why?

Well, for one thing, the engine is really making more than 163 hp, but some of that power is getting used up operating the supercharger. According to the Eaton specs, that's about 20 hp at a blower speed of 12,360 rpm (6,000 engine rpm) and 10 psi boost. So, our 163 hp engine is really a 183 hp engine, but some of the power can't be used by us.

270 cfm x 163 / 183 = 241 cfm

Now, we're getting close. Only 27 cfm to go. Well, we didn't yet take into account how much the air is being heated by the supercharger. According to Eaton's specs, that's about 180F of temp increase at a blower speed of 12,360 rpm (6,000 engine rpm) and 10 psi boost. If our Intercooler is 80 % efficient, that knocks that value down to 36 F. In absolute terms, that's 496 Rankine compared to an STP value of 460 Rankine.

241 cfm x 460 / 496 = 223 cfm

That's close enough to 214 cfm for government work.
Fabulous calculations, Andy. I wonder, though, would it be more accurate to just nab someone with an AFC's AF ratio from a 'dialing in' dyno run? Those with AFC controllers would be ideal sources of info, because they can tell how much 'dialing fuel back' has affected their curves, right? If you take two [before and after] points, you wouldn't need to know EXACTLY how much fuel was being delivered- you could simply go by amt. of increased fuel divided by difference in AFR]. Figure out the amount of gas being delivered and then multiply by the AF proportion? Just a thought... :smile:
 
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Old 10-27-2005, 07:40 AM
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for you calculator types: if you could calculate the btu's needed to be removed from the turbo airstream to drop it's output temp to an acceptable level (at various boost levels), then you would have the heat load for the water cooling (or any other) scenario and could determine what sort of heat exchanger would be needed.
 
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Old 10-27-2005, 07:43 AM
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Samurai Will had some pretty decent info in that last thread. I was really bummed to see his stuff go.
 
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Old 10-27-2005, 08:06 AM
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If you have complaints about moderation or about other threads, please post in the Feedback forum. Please keep this thread (and this forum) on topic.
 
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Old 10-27-2005, 08:07 AM
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Here's another component to the MCS's power-generation scheme: the intercooler.

The MCS uses an air/air intercooler that has two perpendicular flow paths. Inside the core, charge air flows from the supercharger outlet on the passenger side to the intake manifold inlet on the driver's side (reversed for UK-types ). Meanwhile, outside the core, ambient air from the atmosphere travels through the hood scoop and down through the intercooler fins into the engine bay. As the cool, fresh air passes through the intercooler, it exchanges heat with the charge air, making the fresh air hotter and the charge air cooler. In an ideal situation, the ambient air would take on all of the charge air's extra heat and the temp of the charge air leaving the intercooler would be the same as ambient. Unfortunately, intercoolers are not 100% efficient so the charge air is still warm, but nowhere near as hot as when it entered.

The way to measure the efficiency of the intercooler is to record the temperature of the charge air as it enters, the temperature of the charge air as it leaves, and the temperature of the ambient air coming into the hood scoop. By knowing these three values, it's easy to calculate how efficient the intercooler is, compared to the 100% ideal.

With the stock IC, I picked up a pair of inexpensive laboratory temp probes:

http://scientificsonline.com/product.asp?pn=3105400

I recorded the three values of interest, in a variety of conditions, and plugged them into this formula:

Ei = (Tco - Tio)/(Tco-Ta)

where:

Ei = intercooler efficiency
Tco = intercooler inlet temp
Tio = intercooler outlet temp
Ta = ambient temp

Here's what the values look like plotted:


The efficiency of the stock IC is actually pretty good, despite how it looks.
 
  #19  
Old 10-27-2005, 03:45 PM
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Originally Posted by andy@ross-tech.com
If you have complaints about moderation or about other threads, please post in the Feedback forum. Please keep this thread (and this forum) on topic.
It wouldn't exactly be relevant if not connected with the topic. Since it directly relates to on topic discussion, I'll post my reasoning on topic.
 
  #20  
Old 10-27-2005, 03:53 PM
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Originally Posted by andy@ross-tech.com
If you have complaints about moderation or about other threads, please post in the Feedback forum. Please keep this thread (and this forum) on topic.
What exactly is the topic??

Is it:

1. How the MCS engine makes power.
2. How the MCS engine makes power with a turbo added.
3. How Andy thinks the MCS engine should make power.
4. Unproven calculations on how the MCS engine makes power.

I'm still confused as to which it is.

Having said that... Thanks for the first post. Had some good reading in it.
 
  #21  
Old 10-27-2005, 04:33 PM
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Let's all just calm down and enjoy the coming winter weather! Colder air means MORE POWER, right? Let's burn some rubber, boyyyyyeeeeeZ!
 
  #22  
Old 11-01-2005, 06:47 PM
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Shouldn't this thread be called: How [boosted] engines work? This "information" isn't specific to the MINI
 
  #23  
Old 11-02-2005, 06:46 AM
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Right~0n

;O)
Originally Posted by Ryephile
Shouldn't this thread be called: How [boosted] engines work? This "information" isn't specific to the MINI
 
  #24  
Old 11-03-2005, 02:04 PM
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Thanks, guys. Keep the good tech info coming.

Originally Posted by ingsoc
Let's all just calm down and enjoy the coming winter weather! Colder air means MORE POWER, right? Let's burn some rubber, boyyyyyeeeeeZ!
Originally Posted by Ryephile
Shouldn't this thread be called: How [boosted] engines work? This "information" isn't specific to the MINI
Originally Posted by joker
;O)
 
  #25  
Old 11-03-2005, 10:07 PM
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No prob. ....

=0)
Originally Posted by andy@ross-tech.com
Thanks, guys. Keep the good tech info coming.
 


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