# Electric Superchargers

1. Oct 27, 2008

### stickers

Hi all, finally thought i'd post something here.

I'm a 2nd year mech engrg undergraduate and I've come across this electric supercharger called "E-RAM".

There are many other versions out there but this seems to be the most 'promising'.
It's basically an electric motor with an axial type fan connected in-line with the intake piping and it is supposed to act like a turbo/supercharger by increasing the pressure and hence the density of the intake charge.

While viewing the video on youtube, there was one clip where the guy setup an experiment to measure the output of the fan.
He was measuring the current drawn by the motor and the 'wind speed' of the fan by placing an anemometer at the end of the short duct.

Now, for the key question:
From what I understand, the factor that increases the power output of an engine is the increase of mass flow into the cylinders right?
That's how turbochargers are rated, by their CFM.

My aim is to try and prove that this electric supercharger cannot provide enough flow to even meet the demands of a standard 1.6L engine, let alone increase the power output.

Now for the calculations, correct me if I have used the wrong formulas or methods.

Ohm's law: Power = V * I = 13.8V * 62amp = 855 watts.
Assuming 100% efficiency for the motor, that means that 855 watts is 'transferred' into the fluid.

Assuming incompressible flow, perfect gas and no frictional or other losses,
From the power in fluids equation, Power = pressure * flow rate. (bounded flow)
Assuming that the E-RAM does provide 6psi (0.4bar) of positive pressure rise (as seen in another video)
so, the flow rate Q = 0.855kW / 0.4 * 101kPa = 0.0211m3/s.

The formula i got for the flow requirements of a 1.6L engine @ 6500rpm, assuming 0.8 efficiency:
Engine Volumetric Flow = (displacement * rpm * volumetric efficiency/ 3456)
= 97cubic inch * 6500 * 0.8 / 3456
= 146 CFM = 0.0689 m3/s

Thus, at WOT at 6500rpm, the electric supercharger cannot provide enough flow to even meet the demands of the engine, and hence acts as more like a restriction.

Is this statement correct?

2. Oct 27, 2008

### Topher925

No.

You can't just make flow rate and head calculations based off the work going into the electric motor thats powering the pump. Matching pumps to other pumps (the engines a pump) is very difficult and to actually determine if the pump will do any good, you will need its performance chart.

Also, superchargers don't increase the performance of the engine by just increasing the mass flow rate necessarily. They increase the power by putting more air into the cylinder so more fuel can be added, thereby increasing the specific power of the engine. This does not however correlate to a better engine efficiency.

3. Oct 27, 2008

### stickers

Ah.. yep I'm aware that matching pumps to systems requires much more work than that,
and involves coming up with pump and system characteristic curves and such, and that pumps are designed to operate within specific range.

What i tried to do was compare the output of the pump vs the demands of an engine at that particular setting, can't it be done this way?
The electric supercharger could have provided adequately until a certain rpm of the engine, but we'll have to plot graphs to find that.
What i'm only interested is at that particular setting of WOT @ 6500rpm, though I've ignored alot of other losses through the piping, and the equations are probably very basic.

Um.. isn't putting more air = higher mass flow into the engine? That sounds quite confusing. =/
or does pressure, density and others come into play as well.

However, interestingly I've also come up with another set of calculations which don't agree with my previous ones:

The windspeed they measured coming out of the fan was 42 (mph or km/h i think? can't be m/s or ft/s)
Lets say 42km/h = 11.67m/s. X-section Area of 3in piping = 0.00456m2, assuming constant diameter throughout,
Then Flow Q = V * A = 0.0532m3/s. which is double the initial flow calculated from power.

so which flow should I be using? or are my values wrong, because I am plucking values from different videos, they don't agree with each other?

4. Oct 28, 2008

### ray b

fans move air BUT DONOT RAISE PRESSURE

a true supercharger raises pressure by compressing air

el-trick supercharges do exist BUT require huge current flows
one uses three starter motors to drive it
and can be used only for a short time [1/4 mile] due to drain on extra batterys
that are needed to power it

so no the E-ram does not work and just blocks flow

5. Oct 28, 2008

### Mech_Engineer

Using a basic Thermodynamic approximation, it is quite possible to estimate the power the compressor would require if you know how much pressure you're going up (boost) and the initial temperature and pressure.

For example, if your engine is taking in air at 300 K and 0.1 MPa (1 atm), air's enthalpy is approximately 460 KJ/Kg, and if you raise the pressure 3 psi in an isentropic compressor, the enthalpy raises to about 477 KJ/Kg.

This is enough information to calculate how much power the compressor would need, given the amount of air the engine is gulping. Hint- it's about an order of magnitude larger than the 800 W power draw of the compressor you've mentioned.

Last edited: Oct 29, 2008
6. Oct 29, 2008

### stickers

mm.. you mean that an axial type fan cannot compress air? or will it compress it to a certain pressure before the blades 'stall' and the air 'backflows' the blades.

Those electric superchargers that you mention should be a centrifugal or a roots type sort of pump?

Anyway, I just realised that the 6psi raise in pressure was not a valid value because it was taken when the fan was on, but the throttle plate was closed and the engine not running. (so it was in effect just compressing the air against that closed throttle plate.)
It might not have made any positive pressure at all when the engine is at WOT, until someone plugs in a boost gauge and runs it to get a reading.

7. Oct 29, 2008

### Mech_Engineer

Based on my calculations, a supercharger putting out 6psi of boost would require about 3100 W to keep up with a 1.6l engine running at 6500 rpm. It's obvious that the e-ram you're talking about doesn't have that kind of power available.

8. Oct 29, 2008

### stickers

Ah, that sounds more like it. This is not including the mechanical and other losses of the supercharger itself?

What equations did you use to derrive this number, and what are the variables affecting it? I suppose it's not the simple thermo-fluid equations we get in 2nd year huh.. =)

9. Oct 29, 2008

### Mech_Engineer

If you model the compressor as an isentropic thermodynamic process, you only need to figure out air flow into the engine (volume flow), density of the incoming air (assume atmospheric sea level), and the enthalpy of the air going into and out of the compressor.

Using these values, you can calculate the mass flow going through the engine (kg/s) and multiply it by the enthalpy change accross the compressor (approximately 30 kJ/kg at 6psi of boost). This is basically a best-case estimate of how much power the compressor would require, without knowing anything about the compressor's efficiency.

10. Oct 29, 2008

### ray b

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