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**DONOTDELETE**
02-15-2002, 05:58 AM
First I'd like to say that I'm a newb with tuning, so forgive my ignorance.

This is an Idea I've had, based on looking at the e-ram sc. What if you used a better electric motor? The e-ram gives 1psi and uses 50amps of power. I am looking at a blower from Grainger that uses 5amps @ 12volts. Item # 3HV20. They claim it produces 12.2psi max boost. It measures 5x3 inches. Here is the page http://www.grainger.com/Grainger/productdetail.jsp?xi=xi&ItemId=1611764151 .

My question is would it be adaptable to be used for an electic S/C? Its only $153.50. Using it with an IC would seem to be cheaper than a turbo setup.

**DONOTDELETE**
02-15-2002, 06:03 AM
That blower runs only 32.0 CFM. I don't know the exact number, but I recall someone saying that the 240's intake flow, at WOT, is in the hundred's of CFM. 500 or 600, or something like that...

Anyone have exact (or estimated) figures?

In any case, to accomplish anything, you need a high-psi/ high CFM blower...
-B

[ 02-15-2002, 06:05 AM: Message edited by: Senna ]

ADAM HUTCHINSON
02-15-2002, 07:51 AM
we need over 200cfm....

guys..if it was feasable they would have done it already http://www.freshalloy.com/

**DONOTDELETE**
02-15-2002, 08:06 AM
Well I run a BIG 16g turbo on my Eclipse and that flows 560cfm. That is enough air for around 425-450hp You do the math....

ADAM HUTCHINSON
02-15-2002, 08:33 AM
so setting up an electric motor with some huge fan or custom blade contruction with a huge current draw, multiple batteries and alternators to power it... is easier?

remember these devices were all used in miltary planes..and they thought of everything..including using NOS

http://www.freshalloy.com/

[ 02-15-2002, 08:34 AM: Message edited by: ADAM H ]

**DONOTDELETE**
02-15-2002, 08:44 AM
I didn't mean for this thread to become a s/c vs tubo rant.

I was looking at fans that used less than the 50amps that the E-Ram uses, not something that would require an extra battery.

Does anyone know the real no B.S. CFM for the stock KA24DE intake?

AKADriver
02-15-2002, 08:54 AM
There's just no such thing as free energy.

You need a bigger blower than that... about ten times larger. That one's just about the size they use for the E-ram, I believe. So, just tossing out numbers, you're going to need a motor that draws about, say, 1000A+... just to throw out something in the right order of magnitude. Yeah, it's like *that*.

You're not going to be able to power that with a standard charging system. Time to add another alternator... or ten.

Now, you have the added weight of a heavy-duty alternator, plus the rather large electric motor, plus the amount of power lost to the new alternators, and you've accomplished... absolutely nothing that couldn't be done with a single pulley.

Remember how much power it takes to turn a supercharger... that has to be made up somewhere. If you're losing 20hp to the supercharger, that means you'd need a 20hp electric motor to spin it. 20hp is about 15kW... with a typical electric motor efficiency of 85%, your DC motor is going to suck down 18kW from the charging system. In a 12V electrical system, that means 1500A!

If you want a more efficient source of pressurized air, get a turbo. They might be annoying to set up and control, but for the most part the only energy they use to turn is waste exhaust heat.

ADAM HUTCHINSON
02-15-2002, 09:06 AM
yes you need over 200cfm with a forced induction set up...more power more cfm required....

if you want to be silly you could use a 125cc 2 stroke motor to power the supercharger..thus taking away its linear response as well...instead of messing with batteries http://www.freshalloy.com/

SR20VET
02-15-2002, 09:23 AM
Getting off topic but hows this for silly....Honda CRX with 4 Suzuki GSX-R 750 engines w/ the turbo kits (200bhp at the wheel on a GSX-R). Each engine running a seperate wheel. that would be an 800bhp CRX that could rev over 12,000 rpm if you could get it to run right. *imagines hearing 4 turbo'd GSX-R 750s passing him and looking over only to see a CRX* http://www.freshalloy.com/

S14GTR
02-15-2002, 09:24 AM
to create more power...you would need around 350 ~ 400 CFM

Or put 15 X 32 CFM...to get 480 CFM Wow!!!!

But draw 15 X 5 Amp = 75 AMP...close to full power from the alternator (80 AMP)

http://www.freshalloy.com/

**DONOTDELETE**
02-15-2002, 09:38 AM
Originally posted by 240 KSX:
to create more power...you would need around 350 ~ 400 CFM

Or put 15 X 32 CFM...to get 480 CFM Wow!!!!

But draw 15 X 5 Amp = 75 AMP...close to full power from the alternator (80 AMP)

http://www.freshalloy.com/ <font size="2" face="Verdana, Arial[/img]I was looking at some Delta fans, but the max operating preasure is pretty low.

I'll prolly just end up with NOS, since i hear it has the highest performance:price ratio.

ADAM HUTCHINSON
02-15-2002, 09:50 AM
battery technology as far as advances is still in the drak ages compared to most of the other advances in all the other areas of electronics...other wise you would see battery powered cars all over the place..they are just to heavy and do not have enough storage capacity...and are affected by temp greatly...and really good ones are very expensive..

problem is a SC would have huge drain..it just takes to much energy to power the damn thing...


Originally posted by Replicant_s14:
</font><blockquote><font size="1" face="Verdana, Arial[/img]quote:</font><hr /><font size="2" face="Verdana, Arial[/img]Originally posted by ADAM H:
so setting up an electric motor with some huge fan or custom blade contruction with a huge current draw, multiple batteries and alternators to power it... is easier?
http://www.freshalloy.com/ <font size="2" face="Verdana, Arial[/img]Allow myself to quote.........myself:) " ...it's cheaper and easier to use a pully."

Battery tech has gotten huge so it's a possibility but a more likely scenario is to have a sort of capacitive discharge type hybrid like the one panasonic is sitting on. Store and release when needed. Controlling it is another thing all together but I'm sure there are some sharp folks who can figure it out (not me, that's for sure).

I still say hydraulic is the way to go. I don't see any advantage power wise but it would be easier to control (than electric)and you would still have options for the physical location of the SC. I think Allied is already on this anyway.

None of the above would be as cheap as just plunking down $3k for an off the shelf turbo.</font><hr /></blockquote><font size="2" face="Verdana, Arial[/img]

**DONOTDELETE**
02-15-2002, 10:03 AM
Something like this is more along the lines of where you might want to start... they're 12v Dc at least... from www.jabsco.com (http://www.jabsco.com)
http://www.jabsco.com/images/hdblo.jpg

"HEAVY DUTY BLOWERS
Our new heavy duty blowers are available in either Flange or Flexmount form. Motor life of 10,000 hours for service in the toughest conditions. Designed for use wherever continuous ventilation is required. Outputs of 7.1m/min (250 cfm)."
-B

[ 02-15-2002, 10:05 AM: Message edited by: Senna ]

redlyne
02-15-2002, 10:23 AM
Not to mention, the electrical drain would slow the engine, both in terms of cooler spark, lower fuel pressure, and increased effort needed to turn the alternator, which, BTW, wouldn't last long at all, because it'd be at 100% capacity all the time...

Sean

**DONOTDELETE**
02-15-2002, 02:28 PM
not too keen on the idea of an electric SC on a mechanical motor. too much work for the engine. if it was to be done, why not use a hybrid car (gasoline/electric). wouldn't that make more sense?

**DONOTDELETE**
02-16-2002, 08:59 PM
In-Line Blower 100mm (4") ducting giving 6.7m/min (240 cfm) using 12Vdc 5 amps. Measures 5.25" by 7".

If it was controlled by a NOS type switch that activated on WOT, then there wouldn't be a constant load on the alternator, even at the low 5 amps.

http://www.jabsco.com/images/lineblo.jpg

'97 S14 SE Turbo
02-16-2002, 11:38 PM
When it's all said and done, a turbocharger is still more efficient since it captures the energy from exhaust gas that typically get's tossed overboard and make it do work instead of a crank driven device (direct and indirect {indirect: ie electrical SC}).

NOSTALGIC_HERO
02-17-2002, 08:06 AM
your electric supercharger sounds neat, but energy is exchanging too many times, each time you lose a little more, turbos are about as direct as you can get in transferring energy from the motor and using it to compress air.

97KAT
02-17-2002, 09:25 AM
OMG, It is only 5"X3" the inlet must be like 1/4 inch, if this thing were any smaller you'd suck it in. Remember CFM's are to PSI as AMPS are to Voltage.

It is like trying to power you car with 8 1.5v AA sized batteries, you got 12 volts but no amps so noo good.


Originally posted by Tex:
First I'd like to say that I'm a newb with tuning, so forgive my ignorance.

This is an Idea I've had, based on looking at the e-ram sc. What if you used a better electric motor? The e-ram gives 1psi and uses 50amps of power. I am looking at a blower from Grainger that uses 5amps @ 12volts. Item # 3HV20. They claim it produces 12.2psi max boost. It measures 5x3 inches. Here is the page http://www.grainger.com/Grainger/productdetail.jsp?xi=xi&ItemId=1611764151 .

My question is would it be adaptable to be used for an electic S/C? Its only $153.50. Using it with an IC would seem to be cheaper than a turbo setup.<font size="2" face="Verdana, Arial[/img]

[ 02-17-2002, 09:59 AM: Message edited by: 1997 240 sx se ]

**DONOTDELETE**
02-17-2002, 09:25 PM
From the responses I got to the first post, I figured out the fan that I originally found would be too small.
Closer to the bottom of the page, where the most recent posts go, are other higher power fans.
I did not get any PSI info on the last fan I found, see 4 posts up.

Do you have anything constructive, like a formula to figure out the PSI for a fan that generates 240cfm thru a 4" hole?


Originally posted by 1997 240 sx se:
OMG, It is only 5"X3" the inlet must be like 1/4 inch, if this thing were any smaller you'd suck it in. Remember CFM's are to PSI as AMPS are to Voltage.

It is like trying to power you car with 8 1.5v AA sized batteries, you got 12 volts but no amps so noo good.

</font><blockquote><font size="1" face="Verdana, Arial[/img]quote:</font><hr /><font size="2" face="Verdana, Arial[/img]Originally posted by Tex:
I am looking at a blower from Grainger that uses 5amps @ 12volts. Item # 3HV20. They claim it produces 12.2psi max boost. It measures 5x3 inches. Here is the page <font size="2" face="Verdana, Arial[/img]</font><hr /></blockquote><font size="2" face="Verdana, Arial[/img]

97KAT
02-18-2002, 09:40 AM
I am sorry to kid, I was going to try to explain better, but then I ran into this site that has the formulas and not only that but it also discuss the e-ram. ifeel it is well explained here:

web page (http://home.att.net/~t.vago/eram-waste-howto.html)

Is that so? Well, let's use a little something I like to call the Law of Atmospheres. This is something I'm using from college thermodynamics, so it's not something I just pulled out of thin air, either. Basically, it states that:

P1 * V1

--------------------------------------------------------------------------------
T1 = P2 * V2

--------------------------------------------------------------------------------
T2
Where
P1 = Gas Pressure at State 1
V1 = Volume that gas occupies at State 1
T1 = Absolute temperature of gas at State 1
and
P2 = Gas Pressure at State 2
V2 = Volume that gas occupies at State 2
T2 = Absolute temperature of gas at State 2

This law states that a given volume of gas at a given temperature and a given pressure at state 1 will occupy a different volume at a different pressure at the same temperature (This is actually a simplified explanation, but will do fine for my reply).

Now, consider 211 cubic feet at 15.7 psia and at 70 degrees Fahrenheit (530 degrees Rankine for absolute temperature). Now, keeping the temperature the same, and dropping the pressure to 14.7 psia, the volume increases to:

15.7 psia * 211 cf

--------------------------------------------------------------------------------
530 R = 14.7 psia * V2

--------------------------------------------------------------------------------
530 R

530 R * 15.7 psia * 211 cf

--------------------------------------------------------------------------------
530 R = 14.7 psia * V2

15.7 psia * 211 cf = 14.7 psia * V2

15.7 psia * 211 cf

--------------------------------------------------------------------------------
14.7 psia = V2

225.4 cf = V2

So, in other words, to flow 211 cfm at 1 pound of boost (15.7 psia, or 1 psi over atmospheric pressure), you'd need to flow 225.4 cfm at atmospheric pressure.

Unfortunately, the E-Ram can't even do that. It's best "boost" for a 2.5L V-6 at WOT is just barely enough to budge a boost meter's needle. This pitiful performance is far below the 1 psi needed to substantiate the E-Ram claim to fame.

--------------------------------------------------------------------------------
Quote:
This is what the eRAM does. A better test, would be to figure out the mass the eRAM has to flow in the form of thrust. Through some simple calculations, the eRAM would have to produce 2-2.5 lbs of thrust. (roughly the amount of thrust to levitate the device) Shown on our website, the eRAM can levetate itself proving the 2-3lbs necessary to produce pressure in your intake under WOT.
--------------------------------------------------------------------------------

No, not really. Let's see what the E-Ram actually does, shall we?

Given that the E-Ram weighs 2.5 pounds, and it has 8 fan blades with each blade having 0.75 square inches of area. Since it's the moving fan blades that actually work to compress the air going through the E-Ram to generate air pressure, thus generating thrust, we'll limit the evaluation to just the fan blades. Now to levitate itself, the E-Ram would have to develop 2.5 pounds of thrust to overcome its weight, right? Let's see...

2.5 pounds

--------------------------------------------------------------------------------
8 * 0.75 square inches = 2.5 pounds

--------------------------------------------------------------------------------
6 square inches

= 0.417 pounds/(square inch)
= 0.417 psi

Note that this pressure is able to just barely levitate the E-Ram. This leads to the other part of the questionable claim put forth by Mr. Kibort. If this E-Ram is actually able to flow 400 to 500 cfm of air at atmospheric, then it ought to not just levitate itself, it really should violently try to fly up into the sky. Why? Consider this:

One cubic foot of air weighs about 0.0753 pounds. At 400 cfm, that E-Ram would have to move 30.1 pounds of air every minute, or 0.502 pounds per second. Let's use a pipe with a diameter of 2.7 inches (such as the effective pipe size Mr. Kibort is going to use later on). To flow 400 cubic feet per minute through such a pipe, one would have to pass that air through a cross-sectional area of 0.040 square feet. This gives a velocity of that 400 cubic feet of air per minute, through that 0.040 square foot restriction, of 10060 feet per minute, or 167.7 feet per second.

Using physics momentum and impulse equations, one can see that:

f * t = m * (v2 - v1)
Where
f = average force acted on object
t = time over which average force acted on object
m = mass of object
v1 = initial speed of object
v2 = final speed of object

Now, assuming t = 1 second (since it would take 1 second to make 0.501 pounds of air move, using the above numbers), we get:

f * 1 sec = 0.501 lbm * (167.7 foot/sec - 0 foot/sec) / (32.2 lbm-foot/lbf-sec-sec)

f * 1 sec = 2.6 lbf-sec

f = 2.6 lbf

Now, keep in mind this thrust is only due to conservation of momentum from moving the air mass through the E-Ram. This does not take into account the relatively static mass of air behind the E-Ram, from which the moving mass is able to push against, therefore creating pressure at the exhaust of the E-Ram. Nor does it take into account the partial vacuum developed at the inlet of the E-Ram. We'll assume, for the sake of argument, that the E-Ram is able to affect air pressure within 2 cubic feet of the inlet and exhaust. Allowing for the energy imparted to the moving air of about 600 joules (Anybody who ever took college physics ought to be able to get that figure from the mass and velocity of the air), and considering that this energy is being dissipated in the form of pressure only (no heat losses), we find that the E-Ram is able to generate a 0.7 psi boost. Now, multiply that by the 6 square inches of the fan blades, and you get 4.54 pounds of thrust due to pressure differences.

Hm... 4.54 pounds due to pressure, and 2.6 pounds due to momentum conservation... That's 7.14 pounds of total thrust acting against a 2.5 pound object. That's a thrust-to-weight ratio of 2.85! This thrust would tend to make the E-Ram take off for parts unknown rather than just levitate.

Alas, the E-Ram is just barely able to levitate...

--------------------------------------------------------------------------------
Quote:
your comment of restriction is incorrect as the Throttle plate provides all the restriction at part throttle conditions and we have flow benched the eRAM in the "off" condition and found it to flow as much as a 2.7" clean pipe. (the eRAM diameter is 3.5") This is a non issue as the eRAM would never be "off" under WOT, so it doesn't matter.
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Ah, so the air filter and the intake piping don't provide any restriction, either? These are also non-issues, is that correct? In case you've forgotten, those items are also present before the throttle plate. Those items will also hurt intake performance, or else there wouldn't be such a market for performance aftermarket induction tubes and high-flow filters.

How can you say that the E-Ram flows as much when off, as a 2.7" diameter pipe? How long would that pipe be? How much airflow did you use to determine your assertion? You do know that flow ability is related to airflow and pipe length, right?

--------------------------------------------------------------------------------
Quote:
if your eRAM doesn't draw 40-50amps, there is something wrong with it. (ie worn brushes or some other mechanical failure) generally, the eRAM lasts years. ours has been on our race car that participates in SCCA pro racing, NASA, TCC, and SSF motorsports events on a monthly basis and it is still running strong today (51amps measured just last week after 3 years of operation)
--------------------------------------------------------------------------------

No, if my E-Ram did draw 40-50 amps, it would quickly turn itself into a hot pile of molten slag. Passing 40 amps through a resistance of 0.9 ohms (such as the E-Ram's motor coil resistance) would give a power draw of 1360 watts. Your typical high-performance audio amp is only rated at 500 watts. A blow-dryer on high draws about 1500 watts. Add this to the fact that there really is no good way that the E-Ram is able to shed the heat generated by this much wattage (remember that the amp has a butt-ton of cooling fins and a large metal heat sink, and that the blow-dryer is constantly passing air past those spaced-out heater coils that consume the 1500 watts). The E-Ram would very quickly fail due to overheating, and would probably cause a good deal of damage to the electrical system of the vehicle that had this device installed. Thank God, in this case, that the E-Ram only draws about 15 amps.

About your product's claim to fame with racing: So what? If your product doesn't perform (which it doesn't), why should I care what application it's used for?

--------------------------------------------------------------------------------
Quote:
We got over 10 hp on a 3.2 liter porsche engine and thats 7ft lbs more torque from 3000rpm all the way to redline. Terry at Frey Racing did the independent test and he has been dynoing and building Winston cup car engines for years.
--------------------------------------------------------------------------------

Let me get this straight. A close friend of yours did the "independent" test that supposedly proves your E-Ram's performance gains. Oh, yeah, that really sounds objective... Not.

Also, how am I supposed to relate your Porsche engine numbers? Was this E-Ram alone providing your gains, or was it (as I suspect) an intake tube better than stock, that the E-Ram was installed into? Where's your dyno chart "proving" this claim of yours?

--------------------------------------------------------------------------------
Quote:
So, please get your facts straight before bashing a product or a concept.
--------------------------------------------------------------------------------

Okay, I will. By the way, you do have a point. I was actually mistaken about the amount of power it takes to boost 212 cfm of air to 1 psi. Turns out it actually takes about 3.9 HP, and not the 1 HP I claimed earlier. So, let's see...

3.9 HP... That is equivalent to about 2900 watts. At 13 volts DC, that turns out to be 223 amps! Remember, this is merely the power needed to boost 212 cfm of air to 1 psi over atmospheric. I haven't even touched on the power consumed by the E-Ram in the form of mechanical and electrical losses!

I look forward from Mr. Kibort's reply to the above information...



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[QUOTE]Originally posted by Tex:
[QB]From the responses I got to the first post, I figured out the fan that I originally found would be too small.
Closer to the bottom of the page, where the most recent posts go, are other higher power fans.
I did not get any PSI info on the last fan I found, see 4 posts up.

Do you have anything constructive, like a formula to figure out the PSI for a fan that generates 240cfm thru a 4" hole?

[ 02-18-2002, 10:04 AM: Message edited by: 1997 240 sx se ]