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Forced Induction Turbo, Supercharger, Methanol, Nitrous |
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#29 | ||
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I gained 100 HP through a combination of lower backpressure AND better compressor efficiency Also, someone else mentioned that if you make the turbos the same, then of course they'll do the same amount of work... Your comparison only works if you have a turbo that has the vortech compressor wheel in it, and you compare it to a vortech supercharger on the same dyno/same conditions. Then you would see the difference between them, but we don't need to. We can already see that turbos are making more power at even lower boost levels. |
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#30 | |
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#31 | |
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The classic, superchargers make more torque argument really does not apply here since the turbo makes fantastic power everywhere. It's kinda funny that the NA stock motor and the supercharged motors feel laggy while the turbocharged FA20 makes great power from idle to redline. |
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#32 | |
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![]() You can pump relevant information in their faces till they're blue but the bunch of armchair critics who've read a few books would certainly be able to tell you otherwise with obscurities to boot. I prefer to let the customer's decide what's best for themselves, but certainly, cutting down on the noise is a different story... Cheers! Bob @ Drift-Office, LLC |
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Still waiting on how this is 'misinformed' since the dyno with sensors is how we quantify the tangibles like, HP/ TQ / AFR / BOOST as units of measure. That is unless, you're saying we've all been doing it wrong all this time?
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#35 | |
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#36 |
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Let's say that, for the sake of argument, the two main forms of FI are used to get towards a goal of maximum power on 93 octane gas on an identical FA20 with a free-flowing exhaust system, upgraded fuel pump, and upgraded fuel injectors. And lets say that, give or take 10~15ft/lbs, that is about 260 ft/lbs of torque. Because, of course, torque is what is being measured to produce hp figures.
Here is the possible issue - say the two systems are similarly efficient and are forcing in roughly the same amount of air into the pistons. Theoretically, they are all making the same power at that point give or take the compressor charge temperatures. What modifies that value further, though, is the amount of parasitic drag. Very unscientifically (since there is not a lot of specific figures): Turbocharged system may reach about 255ft/lbs - loses some power due to the backpressure at the turbo. S/C systems may reach between 210-240ft/lbs - loses some power due to parasitic loss at the pulley. Now, that power lost to a pulley doesn't mean that the s/c engine is running at less power, it means that it's pushing the same amount of air and fuel into the engine, producing the same amount of power (or more) than the turbo, but that the pulley system is then eating up that specific amount of produced power. |
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#37 | |
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Claiming all the differences present in the dyno is purely from the parasitic drag needed to spin the supercharger is what is misinformed. I'm not arguing that one will make more power than the other, just that there is a lot more at play than just the power needed to spin the s/c.
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Even if what you say is true, then how does one explain when boost levels meet at 6500rpm? The Vortech car in this example has a full exhaust, so backpressure should be as low as it can possibly be - giving it a theoretical advantage over the turbocharger's turbine and housing. I have been corrected in that Vortech has indeed published their compressor's map. The AVO kit in question uses a 38-ish lb/min turbocharger (with a smaller hot side) and from the looks from the Vortech map, the supercharger has a 40 lb/min compressor at peak efficiency. So again, advantage supercharger. So yea, the chart does show, as well as it can, the difference between a supercharger and turbocharger's output to be mainly –*mainly –*drag induced. How else can you explain the difference in torque output at >6500 when boost levels become identical? budi |
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#39 |
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Because there are still more variables at play. Just because the s/c flows more at peak efficiency doesn't tell you how close to peak efficiency is it running on this engine at 6500rpm? More flow doesn't mean more efficient in every situation.
It's the same reason two different turbos at the same amount of boost won't make the same power on the same car. The only way to measure the power difference caused by the parasitic drag on the s/c is to run identical compressor wheels and housings and regulate the boost pressure so they match up the same across the RPM range. You can work out theoretically how much hp the s/c needs to spin as well, but that's not measurable.
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#40 |
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Or you could use the compressor work equation and factor in the isentropic efficiency (or adiabatic eff. in case of twin screw SC) to figure out power/torque consumption of an SC at a given boost level...
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It is well known that moving air at pressure requires power. The superchaeger takes out of the crank. The turbo takes it from blowdown pressure. Some pumping hp and some temperature (as seen with the 300 degree temperature drop we have logged on a 12 psi kit across the turbine).
Last edited by Toma; 07-11-2013 at 02:18 AM. |
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#42 |
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If you guys math any harder I will fertilize my keyboard
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