eRAM electric supercharger
03-08-2004, 06:42 PM
#31
I never notice, anyway
Rennlist Member
Rennlist Member
then i got a Q for ya mark, what's the speed of the fan, i just dont think that something like that would have the power and speed to add compression to the system.
03-08-2004, 06:43 PM
#32
Rennlist Member
Be careful before you make statements you cannot support with some facts.
In fact, there are simple equations showing the pressure rise across most any fan (axial or centrifugal) granted, turbo superchargers and belt driven superchargers do well to compress air, but have a hard time making flow. (and its the combo you need to look at) axial flow fans can work as well, but need mutiple stages to generate the same types of pressure and flow. axial flow fans do well with flow, but not that great at pressure. However, this is why it is only a 1psi unit. More specifically, it changes the net pressure in your intake by 1psi. we have measured 1/2" Hg vacuum at 120mph in our 928 race car. with an eRAM we were able to not only neutralize it, but slightly pressurize it. 1psi is 5% and 5% on a 200hp car is 10hp. this is a common result with most cars, especially those with an AFM. we even helped a 16psi turbo MR2 gain 10hp peaks, buy changing the ambient pressure before the turbo. as many of you know, total boost equals ambient pressure plus boost pressure. sorry to say, it doensnt do anything to spool up time.
your right, ram is a joke too. 160mph, with a 75% efficient hood scoop, OUT of the boudary layer ( like a dragster or funny car) only produces .36psi!!!!! (that goes down to .08psi at 80mph)
so, the eRAM is like having some good ram pressure at any speed
its specs are pretty amazing for what it is.
750 watts
57amps
over 1000cfm
25,500 rpm
and about 3.5lbs of thrust, to give a 1psi net change in an intake of cars with 4.7 liter and smaller displacement.
Mk
In fact, there are simple equations showing the pressure rise across most any fan (axial or centrifugal) granted, turbo superchargers and belt driven superchargers do well to compress air, but have a hard time making flow. (and its the combo you need to look at) axial flow fans can work as well, but need mutiple stages to generate the same types of pressure and flow. axial flow fans do well with flow, but not that great at pressure. However, this is why it is only a 1psi unit. More specifically, it changes the net pressure in your intake by 1psi. we have measured 1/2" Hg vacuum at 120mph in our 928 race car. with an eRAM we were able to not only neutralize it, but slightly pressurize it. 1psi is 5% and 5% on a 200hp car is 10hp. this is a common result with most cars, especially those with an AFM. we even helped a 16psi turbo MR2 gain 10hp peaks, buy changing the ambient pressure before the turbo. as many of you know, total boost equals ambient pressure plus boost pressure. sorry to say, it doensnt do anything to spool up time.
your right, ram is a joke too. 160mph, with a 75% efficient hood scoop, OUT of the boudary layer ( like a dragster or funny car) only produces .36psi!!!!! (that goes down to .08psi at 80mph)
so, the eRAM is like having some good ram pressure at any speed
its specs are pretty amazing for what it is.
750 watts
57amps
over 1000cfm
25,500 rpm
and about 3.5lbs of thrust, to give a 1psi net change in an intake of cars with 4.7 liter and smaller displacement.
Mk
Originally posted by dime1622
theres also the fact that it can move air but CAN NOT COMPRESS AIR. our turbos could move a helluva lot more air if it wasnt being compressed. all the air would slip back out of this thing.
think of it this way: ram air inductions dont work because air is virtually incompressible until mach 0.5, or about 370 miles per hour. it take a LOT of force to really compress air. this thing isnt near powerful enough to do anything.
theres also the fact that it can move air but CAN NOT COMPRESS AIR. our turbos could move a helluva lot more air if it wasnt being compressed. all the air would slip back out of this thing.
think of it this way: ram air inductions dont work because air is virtually incompressible until mach 0.5, or about 370 miles per hour. it take a LOT of force to really compress air. this thing isnt near powerful enough to do anything.
03-08-2004, 06:48 PM
#36
Rennlist Member
Its not as intuitive as that, but the rate of flow of the fan is around 1000cfm, and its diameter is 3.5" . A good way to look at the pressure of axial compressor, is to look at its stall pressure, and draw a straight line to zero. (meaning , no pressure at 1000cfm, or max flow) however at every point before max pressure, the air is being compressed. at just before stall, you will have max pressure, but the flow may only be 50cfm. a 2.5 liter engine at 6500rpm may only require 300cfm. so, you would be at the fat part of the curve as far as pressure. Now, there are other dynamics going on , as the air stream is already moving due to the engines speed and displacement under WOT. the pitch of the blades are set to take advantage of this. the net change in the intake from 3000rpm to redine will be in the 1psi range. gains will be in the 5% range.
Here is a dyno done with a 4.7L 928 engine. (in front of lots of people too!)
Mk
Here is a dyno done with a 4.7L 928 engine. (in front of lots of people too!)
Mk
Originally posted by dime1622
then i got a Q for ya mark, what's the speed of the fan, i just dont think that something like that would have the power and speed to add compression to the system.
then i got a Q for ya mark, what's the speed of the fan, i just dont think that something like that would have the power and speed to add compression to the system.
03-08-2004, 06:50 PM
#37
Rennlist Member
Not actively. The website and business is owned and operated by my brother. I just own the patent and help out with dynos, tech questions, in exhange for some racing support.
Mark
Mark
Originally posted by NZ951
Mark, do you sell a product?
Mark, do you sell a product?
03-08-2004, 06:52 PM
#38
Rennlist Member
It could be a restriction (at WOT) on a 951 with more than 250hp at the wheels. in that case, it certainly would not be a modification of choice!
However, if you are making less than 200hp at the rears, it could be a good mod.
MK
However, if you are making less than 200hp at the rears, it could be a good mod.
MK
Originally posted by NZ951
You the man Danno. I wonder if it would be a restriction on a turbo car once the turbo is spooled and the fan does not turn fast enough or acts like a big as air flow restrictor?
You the man Danno. I wonder if it would be a restriction on a turbo car once the turbo is spooled and the fan does not turn fast enough or acts like a big as air flow restrictor?
03-08-2004, 06:58 PM
#40
Rennlist Member
Good reply, but Ill have to make a couple of comments here.
If you are able to raise density of the intake charge (as a 944 turbo does) and meter its flow ( as the amazing AFM can) you will get hp gains in proportions to that increase in density. 14.7 (stoich.) is for complete burn, not best power. best power happens in the 12:5:1 range for most cars.
as long as the AFM can detect a 5% change in density, which it surely can, then the HP gains will be in the 5% range too. think of flow entering your system, as MASS flow. (both fuel and air) increase the mass flow (burn more air and fuel) and you get a proportional increase in HP.
Still amazed about some of the 944 turbo race cars ive seen that put down 360hp at the wheels and still use a stock AFM!!!
sorry for the bandwidth!
Mark
If you are able to raise density of the intake charge (as a 944 turbo does) and meter its flow ( as the amazing AFM can) you will get hp gains in proportions to that increase in density. 14.7 (stoich.) is for complete burn, not best power. best power happens in the 12:5:1 range for most cars.
as long as the AFM can detect a 5% change in density, which it surely can, then the HP gains will be in the 5% range too. think of flow entering your system, as MASS flow. (both fuel and air) increase the mass flow (burn more air and fuel) and you get a proportional increase in HP.
Still amazed about some of the 944 turbo race cars ive seen that put down 360hp at the wheels and still use a stock AFM!!!
sorry for the bandwidth!
Mark
Originally posted by BoostGuy951
Hey Danno, very good point, but I think your logic is lacking one important point.
I think you are ignoring the huge chemical energy inherent in fuel itself. You are correct in your assertion that this is a basic energy conversion, but what I think is incorrect is the assumption that the Energy needed to move "X" amount of air into the engine is equal to the energy that "X" amount of additional air would produce.
I think this equation is incorrect:
Energy to drive supercharger = Energy produced by supercharger
If this was true, then at 100% efficiency, the mechanical drag produced by a common belt driven supercharger would cancel out the energy it provided. We all know this is not the case. The key being the huge potential energy of fuel. The supercharger allows the engine to injest more air, and thereby add more fuel, and the chemical energy of the added fuel is much more than the energy required to drive the supercharger.
Very Basic stuff:
Obviously, we all know that when tuning cars, we try to combust the most amount of fuel possible, in order to release the most chemical energy stored in the fuel. Since combustion is a reaction that also requires oxygen, it only benefits us to inject fuel in the proper ratio (this case 14.7:1). In a perfect world, where there is perfect mixing of air and fuel, going past the stoichiometric ratio does not yield any more power. We are always after ways to get more oxygen into the cylinders, so we can combust more fuel. I am getting off on a tangent here, let me try to tie this back into my point:
Lets assume this little guy can make 1 psi of boost with a 57 amp draw (which, at low rpms, I think is possible). That 1 psi is responsible for "X" amount of air being moved into the cylinders (whatever "X" may be). Since we have "X" more air, we can add (X/14.7) amount more fuel. When you consider the huge chemical energy stored in even that small amount of fuel, I think its obvious that the enery released by (X/14.7) amount of fuel is much greater than the energy of the 57 Amp draw used to move the original "X" amount of air.
I hope I am making sense.
Hey Danno, very good point, but I think your logic is lacking one important point.
I think you are ignoring the huge chemical energy inherent in fuel itself. You are correct in your assertion that this is a basic energy conversion, but what I think is incorrect is the assumption that the Energy needed to move "X" amount of air into the engine is equal to the energy that "X" amount of additional air would produce.
I think this equation is incorrect:
Energy to drive supercharger = Energy produced by supercharger
If this was true, then at 100% efficiency, the mechanical drag produced by a common belt driven supercharger would cancel out the energy it provided. We all know this is not the case. The key being the huge potential energy of fuel. The supercharger allows the engine to injest more air, and thereby add more fuel, and the chemical energy of the added fuel is much more than the energy required to drive the supercharger.
Very Basic stuff:
Obviously, we all know that when tuning cars, we try to combust the most amount of fuel possible, in order to release the most chemical energy stored in the fuel. Since combustion is a reaction that also requires oxygen, it only benefits us to inject fuel in the proper ratio (this case 14.7:1). In a perfect world, where there is perfect mixing of air and fuel, going past the stoichiometric ratio does not yield any more power. We are always after ways to get more oxygen into the cylinders, so we can combust more fuel. I am getting off on a tangent here, let me try to tie this back into my point:
Lets assume this little guy can make 1 psi of boost with a 57 amp draw (which, at low rpms, I think is possible). That 1 psi is responsible for "X" amount of air being moved into the cylinders (whatever "X" may be). Since we have "X" more air, we can add (X/14.7) amount more fuel. When you consider the huge chemical energy stored in even that small amount of fuel, I think its obvious that the enery released by (X/14.7) amount of fuel is much greater than the energy of the 57 Amp draw used to move the original "X" amount of air.
I hope I am making sense.
03-08-2004, 07:35 PM
#41
I never notice, anyway
Rennlist Member
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ok well it would run off this equation:
pA/gamma + VA^2/2g + zA = pB/gamma + VB^2/2g + zB + hs - hl
where hs = W/(gamma*Q)
basically, lets assume that the engine is NA with WOT and pressure in the manifold is 0.4 bara, or 5.88 psia. now, this pressure would be existent before the pump, and would be built after the pump. assume that the flowrate at this level is 284 cfm (.087533 ft^3 displacement * 6500/2 times per minute). 284 cfm is 490752 in^3/min. work input from the pump is 750 watts, so 1.007 hp, or 553.85 ft*lb/sec or 6646.2 in*lb/sec or 398772 in*lb/min. we can negate the velocity effects because flowrates are equal and areas are equal, using the continuity equation Q = V * A. zA and zB cancel because they are at the same height. hl will also be taken out assuming no loss.
so, assuming pump efficiency of 100% and frictional losses at 0%, we get:
pA/gamma = pB/gamma + W/(gamma*Q)
so pA = pB + W/Q
so:
(x lb/in^2) = (5.88 lb/in^2) + (398772 in*lb/min)/(490752 in^3/min)
our final answer for pA is 6.69257 psi, or a net change of +0.812 psi.
this is with a perfect system, mind you.
but wait, our motors arent ideal systems, are they? this is a system that requires a much smaller fan running at extremely high speeds without being able to lose the pressure out the back of the blades. this little toy fan isnt designed to hold a pressure. this fan is designed for flow, not pressure. a turbo is designed for pressure, not flow.
so, taking into account the frictional losses, inefficiencies in the motor, and its inability to hold a pressure, you may get a net charge of +0.1 psi, if youre lucky.
i wouldnt put it on if it was free. the weight of it probably discounts any power gains you would get from it.
pA/gamma + VA^2/2g + zA = pB/gamma + VB^2/2g + zB + hs - hl
where hs = W/(gamma*Q)
basically, lets assume that the engine is NA with WOT and pressure in the manifold is 0.4 bara, or 5.88 psia. now, this pressure would be existent before the pump, and would be built after the pump. assume that the flowrate at this level is 284 cfm (.087533 ft^3 displacement * 6500/2 times per minute). 284 cfm is 490752 in^3/min. work input from the pump is 750 watts, so 1.007 hp, or 553.85 ft*lb/sec or 6646.2 in*lb/sec or 398772 in*lb/min. we can negate the velocity effects because flowrates are equal and areas are equal, using the continuity equation Q = V * A. zA and zB cancel because they are at the same height. hl will also be taken out assuming no loss.
so, assuming pump efficiency of 100% and frictional losses at 0%, we get:
pA/gamma = pB/gamma + W/(gamma*Q)
so pA = pB + W/Q
so:
(x lb/in^2) = (5.88 lb/in^2) + (398772 in*lb/min)/(490752 in^3/min)
our final answer for pA is 6.69257 psi, or a net change of +0.812 psi.
this is with a perfect system, mind you.
but wait, our motors arent ideal systems, are they? this is a system that requires a much smaller fan running at extremely high speeds without being able to lose the pressure out the back of the blades. this little toy fan isnt designed to hold a pressure. this fan is designed for flow, not pressure. a turbo is designed for pressure, not flow.
so, taking into account the frictional losses, inefficiencies in the motor, and its inability to hold a pressure, you may get a net charge of +0.1 psi, if youre lucky.
i wouldnt put it on if it was free. the weight of it probably discounts any power gains you would get from it.
03-08-2004, 07:42 PM
#43
I never notice, anyway
Rennlist Member
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btw what i said above means you can throw as many pies as you want at a fat kid but he can only eat so quickly. throw more pies and hell still eat the same. the only thing that will help is literally shoving it down his throat. these engines need pressure, NOT FLOW.
the +0.812 psi is ASSUMING the engine can take the greater flow. which it cant.
the +0.812 psi is ASSUMING the engine can take the greater flow. which it cant.