booger

iTrader: (6)

Sentry and I have argued about the amount of loss by a SC [ Vortech , ATI -not twin screw ] . He tends to think the loss isnt as severe as I do . I would like to know what the shops and tuners on My350Z think about the subject .

Also , I would like to know your opinion on doubling the N/A whp with a SC . From what Ive read [ Corky Bell ] it takes 14.7psi of boost to double the N/A whp . And that is if all other factors are included . Fuel , air flow in and out of the motor , and compression . Do you think it can be done on...say...10 psi ????
THANKS !

sentry65

iTrader: (11)

I have read that it takes 14.7 psi to double power, but I'd think that it takes 2x the CFM to double power too. PSI is just a pressure reading, it has no bearing on the volume of air actually going through the engine.

And besides, someone like Dubai's car making 451whp https://my350z.com/forum/forced-induction/255270-dynoed-my-car-451whp-9-5psi-bonus-video.html
on only 9.5 psi of boost seems WAY far off from 14.7 psi, yet that's fairly close to 2x the stock whp...however I bet Dubai has a pretty open exhaust which will have an effect the psi number.
If he was on a totally stock car, maybe he would be close to 14.7 psi

booger

iTrader: (6)

The 14.7 psi doubling the whp is for a centrifugal SC only , not a turbo . And thats if all other things are free flowing also .

sentry65

iTrader: (11)

why does it matter if it's a SC or turbo if the cars were bone stock with stock exhaust etc? PSI is PSI at that point since nothing else has changed

once you change the exhaust, cams, or intake manifold etc, you're basically changing the engine. You're changing how efficient it is at different rpm ranges. When you combine that with a lot of psi of a centrifugal SC, it's totally possible to double the power of a bone stock engine

booger

iTrader: (6)

ummmm Stock motor= higher compression , you'll make a little more power if compression is higher , given everything else is the same ..... stock exhuast = less air flow and less ppower being made at the same psi level

sentry65

iTrader: (11)

we can't have a comparison if the setups are different - the compression would obviously have to be the same - as well as timing and fuel octane etc.

even so, different people are in different locations, dynos, ambient temps, etc - lots of variables.

booger

iTrader: (6)

Looked up the artical again and you're right . Its Turbo's and N/A that are figured together...My bad !

Now lets see if we can get some other input on the subject from shop owners and tuners . We know you have always tried to lesson the difference between SCer's and turbo's .

Resolute

iTrader: (2)

Psi is just a marker in determining potential power gains, it is the actual air mass that you would be interested in to determine doubling the power made by an engine. The same psi values will give different air mass values from kit to kit because adiabatic efficiencies will vary between the compresors. 10 psi on a roots blower will yield a vary different air mass from 10 psi on a Garrett GT series turbo, with the advantage being in the turbo's favor.
As far as parasitic losses go from a SC, it varies on the SC set-up and the friction drive losses. To start, the friction loss on the SC pulley to the compressor wheel on a centriugal SC will run anywhere from 5-3 percent. The absolute best SC in terms of drive loss is the Rotrex, whose planetary gear set-up has an amazing 98.8% drive efficiency. The second issue is the loss of power from the engine to rotate the SC pulley. This is minimal in most cases. If you consider the rotational dynamics of the pulley and the angular acceleration at max rpm, you can calculate fairly easily the loss of drive torque to overcome the pullie's moment of inertia and accelerate it to the given rpm. As an example, accounting for belt friction and corresponding load on the crankshaft, a SC pulley with 3in diameter and a belt width of one inch that is 12 inches long, would only lose about 3.4 hp in first gear- in a perfect world. In second gear, only 1.2 hp, and only .8 in third. This is of course to say that we are running an engine to turn the SC compressor but not actually using the SC to force air into the engine, otherwise there would obvioiusly be no loss of hp. This also does not take into account pumping losses from air pressure and the corresponding friction on the compressor wheel. It is meant to give an idea as to how little power there is to turn the SC compressor wheel. I would venture more power is lost to turn the power steering pump due to the resistance occuring in a hydraulic pump compared to an air compressor. That's my best educated guess. Hope it helps.
Will
Edit: had my numbers backwards, and the calcualtions are assuming a SC pulley that weighs only .5lb (figured it would be aluminum) and does not take into account the final drive, just the fist through third MT gears.

booger

iTrader: (6)

Well . If Im getting it right . Your trying to say , to turn a cetrifugal SC it takes less hp than turning a AC compressor ? Maybe with no IC , IC pipes , or a motor to build up preasure [ boost ] . Because with every psi , the SC is getting harder and harder to turn .


Maybe Im not posing the question right .

What would be the parasitic loss at say - 10 or 12 psi with a Centrifugal SC ?

plumpzz

iTrader: (10)

You can't figure out loss without knowing how much of the volume is being compressed to what pressure and at what temperature. I can give you a rough number if i know.

I need to know atleast some of thsi data:
Volume of air being compressed
Volume of compressed air (volume of the centerfuge trim)
Pressure
Temperature
CFM at the RPM you want the loss at.

Leave me some of this data and I think I can look up the volume of air being compressed. Ill assume its adiabatic compression since its in the compressor for such little time, although I know the compressor eventually gets hot. Just to simplify things. Before even doing calculations, my guess is for 10psi at 6600rpm on the VQ is around 30HP of work required. Itll be more at higher altitudes.

booger

iTrader: (6)

That info would be very hard to come up with unless your data logging rpm, boost , air temps , both intake and ambient , and all the other things you listed . Anything else would be guessing .

The biggest reason I think the loss is quite big . Is the belt slipping at the same psi level every time . Meaning that it gets harder to turn as PSI goes up . And the the actual whp numbers put out with a SC and a turbo at the same psi level .
At 400whp at 10 psi on a SC . The tune must be pretty agressive and at 10psi on a turbo we are seeing whp at around 500whp . So you see the numbers pretty much back up what Im saying .

Oleg

iTrader: (1)

Superchargers is teh suxxors!!!!!!!!! Go twin turbo NOOOOBS!!!!!!!!!!!!!!

sentry65

iTrader: (11)

I still don't think you can accurately compare SC boost to turbo boost unless both cars are running the exact same exhaust setup, compression, displacement, timing, octane, etc

also there's the whole thing where adding a big exhaust on a SC makes the boost psi go down, while on a turbo it'll stay right where the wastegate or boost controller says for it to - hence more air volume going through the engine and more power, but yeah less parasitic loss comes into play also

Oleg

iTrader: (1)

Yeah, but even the PSI reading is flawed. It's taken at the intake manifold, so what you're measuring is how much air DID NOT get into the motor. Exhausts, intakes, TB's etc help you achive the given airflow at less PSI, so you still make same power at less PSI.
Or am I wrong?

sentry65

iTrader: (11)

the psi does have a little to do with how much speed potential or punch the air might have when going through the engine - faster air in means faster air out and more power, but yeah a pressure reading is just a pressure reading.


stuff like ambient temp, boost pressure, displacement, adiabatic efficiency of the compressor, are easy to find out, but what's tricky is how they vary even from month to month. Intake temp is the one that's harder for most people - we can only guess, and it gets even more tricky with water injection or nitrous etc.


it doesn't help that the stock vortech belt design sucks as much *** as you possibly can. It's perfectly fine for the stock 8 psi, but it's far from a serious belt design.

sentry65

iTrader: (11)

here's a question, does the belt slip not as severely in higher gears where there isn't such a drastic change in accelleration????

I'm wondering if it's actually rpm/power based slippage or more of a result because of lower gears accellerating the belt turning faster and causing the belt to studder on the SC pulley



It slips in the first place because it's trying to spin the SC pulley more because of a smaller pulley, and then there's less surface area to grip.

It's the same thing with tire traction and power. You're still accellerating through 1st gear going through the 20mph mark reaching the same speeds turning the crankshaft at the same rpms, but you're doing it so much quicker that the tire has less time to grip the ground. If you mash the gas pedal you spin the tires, but if you ease the throttle on, you do

A T-trim blower I can only assume it takes more power to turn it vs the normal vortech because the T-trim is a physically bigger blower fan even though it's the same outter housing

more power is made by the explosion in the cylinder chamber being bigger and turning the crankshaft faster, and by making the crank turn faster, the accelleration increases. So if the belt is moving faster, it has less time to grip the pulley.


hypothetically, if we took an imaginary 8 liter, 600whp engine and put the 350Z/G35 vortech on it with the exact same belt setup. Because it's such a bigger engine, if it was making 10 psi on a 3.5 liter 287 hp engine, it's now making around 4.375 psi on an 8 liter engine (8/3.5 = 2.28 10psi/2.28= 4.375psi) The belt would still slip if everything else was the same because it's such a horrible belt design. Actually I think it'd be even more likely to slip as a product of the higher power (hence belt accelleration) being made by the engine itself.

Resolute

iTrader: (2)

No, no, no. Much simpler than that. I am not even trying to figure out just how much hp it takes to make xxx psi of boost with a SC. I was merely trying to show just how little hp is used to turn the SC pulley. Of course, I used a really light weight for the pulley, but I have no idea what the pulley on something like the Vortech kit actually weighs. That was just an arbitrary best guess. Second, all I did was very simply figure how much hp was lost on a system to turn an additional pulley of 3 inch diameter that is on it's own belt attached to the crank. That's not too tough. Then I made the mistake of calculating what it would be at the wheel, but the hp loss changes in each gear. As the engine accelerates faster- the more hp it takes to turn the pulley. I didn't bother with the final drive because I didn't have that number handy, and I realized half way through the math that inertial dyno's don't work like that so it was all kinda useless anyways.
My original thought was just like I stated, that if someone just belted up a SC and nothing more- no actual plumbing or tuning- and dyno'd the thing, what amount of hp loss there might be at the wheel. This would just show the loss of turning the compressor, which is not very much at all. With as light and small a pulley as I guessed- yeah, turning the AC compressor or power steering pulley would use more hp.
But as Sentry65 said, and just about everyone else has said, the real question is in the pumping losses of moving all that air. I'm not even going to try and find some equations for all that. I just figured it would be worthwhile to show that turning the pulley is not a major power draw, not any more than the steel pullies on the accessories any how. The real work, and power loss, would be in the compression- and I haven't any real answers or guesses for that.
I guess if someone wanted to, they could just bolt up a SC and run it on their engine without hooking anything else up like I suggested earlier. Just a SC belted to the engine for no good reason and doing nothing but soaking up power. Thinking about it now... since the compressor is spinning, it would be working to compress the air. There would be no flow restrictions, obviously, but it might give an idea of the parasitic losses involved.
This is a good question, it's got me thinking...
Will

sentry65

iTrader: (11)

Corky Bell's Supercharged: page 203, says that the power required to drive supercharger is a formula that simplifies to:

(boost x airflow in CFM)/229



CFM takes into account the rpm you're at
The way you calculate CFM (volume of air) is:

(engine rpm x engine cid)/3456




engine displacement converted to cid can be done here:
http://www.4lo.com/calc/literstocid.htm

3.5 liter engine = 213.6 cid
4.24 liter engine = 258.7 cid





so at redline:

(6600 x 213.6)/3456= 407.917 CFM

however, that's just for the raw engine by itself. You gotta add in the power that the boost is adding

so to add 10 psi for instance, you take 10/14.7 = .68

you add 68% more to the 407.917 so:

407.917 x 1.68 = 685.4 CFM



back to the (boost x airflow in CFM)/229 formula:

drive power = (10psi x 685.4CFM)/229 = 29.93 crank hp at 6600 rpms

so that's 29.93 crank hp more needed to drive the supercharger, that a turbo won't require or use up to spin it






that's just the power required to drive the supercharger at that rpm
to calculate the total parasitic power factoring in the longer belt and compressor efficiency (which turbos have to factor in too) then you run things through this equation:

drive power/(.97 x compressor efficiency)

.97 factors in an approximate for belt efficiency.

so in this example:

parasitic loss = 29.93/(.97 x .71) = 43.45 crank hp total
it looks like 10 psi is around 71% efficiency at redline

I'm also attaching the vortech S-trim (green) and T-trim (purple) compressor maps so you can chart where you are and see what efficiency you're at at different rpms

Resolute

iTrader: (2)

That cfm equation doesn't take into account volumetric efficiency, so you might want to multiply your answer by .85 or so. Peak Ev occurs at the max torque point, so to figure the air flow at max rpm, you might actually want to multiply your cfm at redline by .75.
The compressor map you posted is rated in lbs of air a minute, not cfm. At sea level and 15 deg C, you'll want to multiply the cfm figure you have by .075 to get the lb/min rate.
Will

sentry65

iTrader: (11)

that compressor map has both cfm along the bottom of the graph - the pink vertical hash marks

I'm just reciting what corky bell wrote in his book. What you're saying then means the drive power and parasitic loss is even less than what corky bell says it should be