Whipple application on Duramax - Pressure Ratio

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jfreemanak

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I believe the answers I need will require me to call Whipple, but I thought I would put it out to the community knowledge and experience.

I am looking at using a Duramax in a Hot Rod, with a Whipple as the forced induction method. So, as an example, lets say I want to use the 3.3L Whipple. According to Whipple's website, the max continous RPM of the 3.3L is 18000. So, if I decide to have an RPM limit of the Duramax of 5000, then (18000/5000)=3.6 ratio (for every revolution of the crankshaft, the Whipple has 3.6 revolutions).

Let's assume ideal gas behavior. The crankshaft must turn 4 times for the 6.6L displacement of the engine. So, 3.6 x 4 x 3.3L = 47.52L of air is displaced by the whipple. Using Boyles law, P1V1=P2V2, (1 bar)x(47.52L)=(y bar)x(6.6L), solving for y gives 7.2bar, or 103.68psia. Is this even possible with the whipple?

The website states that the max psi is 30. Is it able to provide more pressure, with the loss of VE and AE? Or does it just leak by the rotors?

Turbo guru's out there, any experience to share? TURBOLVR? Thanks.

Jared
 
Wow, I just realized a flaw in my previous thread. Only two revolutions of the crank (hence, 4 stroke) is needed to displace 6.6L. So, the volume of air that the whipple would displace for every 6.6L of engine displacement would be 23.76L. Following the same calculations as before, should have 51.84psia or 37.44psig. That should be possible with the 3.3L Whipple.

Using the 4L Whipple, the calculations come out to 48.4psig. Does anyone have experience with with a Whipple at these pressures? Thanks.

Jared
 
Boost is a byproduct of resistance to flow. If you have a great cam and awesome heads, then you will have less boost for more power. I dont know how to factor that in.
 
Volumetric efficiency of the engine will help you determine boost.
That will also change as the intake air becomes dense under boost.
So I think the equation might need to look like this P1V1=P2V2, (1 bar)x(47.52L)=(y bar)x(6.6L)x(VE)
Where VE may be around .85 at low boost and 1.02 at 30 psig for example.
To get the actual numbers use a scan tool and read maf, iat and rpm.
 
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to get the correct VE you will have to add and additional swept volume for every 14.7 psi.
 
Thank you for the information cfd. COMP461, I am not sure what you mean by "additional swept volume for every 14.7psi". Would you be able to explain more? Thanks.
 
The power that it takes to drive those blowers at 40psi will be about 300hp. If you're willing to live with that for throttle response, it should work.
 
I doubt the power used would be that high, that said, I also wouldnt recommend driving it that high simply for the IATs created.
Def need a cooler above 25psi.
I drive mine fairly lightly and it makes some heat. I ran it without turbos too, and although it did fine and had lots of off idle grunt, it also had limited top-end capability compared to how it runs in conjunction with turbos.
 
A really nice solution would be a custom intake with a water to air sitting under a blower.
To do this the cp3 has to be moved out of the valley.
 
Steve Morris had a centrifugal blower (F2) on a blower dyno, and it took 353hp to make 32psi, so 300hp is at the low end of the spectrum.
 
Just remember they are 2 different animals. The centrifugals need rpm and have a stepped gear to turn 3-4x pulley speed. the screw types are direct drive.
I have been searching for a HP per psi for them but no dice yet. I DO know that even @ 25psi, there is no belt slip from an 8 rib dayco serp. belt. No cog drive needed!
 
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I ran across one, let me see if I can find it. I think it was in a PDF. I do remember that it was worse than the centrifugals.
 
Screw type superchargers are suppose to be the most efficient in terms of power requirements at least when compared to "roots" blowers. At least they have quite a bit of internal compression. I believe it goes screw > centri > roots.
Just plumb 2 together for some compound action...
 
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JQmile said:
Steve Morris had a centrifugal blower (F2) on a blower dyno, and it took 353hp to make 32psi, so 300hp is at the low end of the spectrum.
Click to expand...

F2 flows around 30% more than a 3.3L whipple. So, depends how much the charger was flowing during that dyno test. to compare it to the whipple in terms of power consumption.
 
Great discussion guys. So, work is defined as dW=-PdV. Assuming ideal behavior and reversible, the situation with the 3.3L Whipple becomes dW=-(1bar)*(0.02376-0.0066)=1716J. Now, power is defined as a Joule/sec, or Watt and 1Hp=746Watts. At 5000prm, the crank is spinning at 83.3 revs/sec. Plugging everything in comes to 95.84Hp. Now this is an ideal situation with no heat generated (hence reversible), which also means that it is path independent, or that it doesn't matter how the compression took place.

There are significant differences between a positive displacement compressor and a centrifugal compressor, but barely any difference on the compressor side between a turbo and centrifugal supercharger. Therefore the amount of work required to move a certain amount of air into a certain volume should be the same between a turbo and centrifugal supercharger. It is important to remember that turbos are not driven by HEAT, as many people believe. A simple experiment to verify this would be to heat the turbine section of a turbo up with a weed burner (or some type of torch). When it is glowing red, look to see how fast the turbine wheel is spinning. There will be no movement. Turbos operate off of pressure differences.

TURBOLVR - I have avidly followed your adventure with forced induction on diesels. I am just now getting to a point where I am able to start experimenting myself. I remember you posting at one time that whipples don't like to be fed 50psig of boost. Let's say I was to put the 3.3L on the duramax, and feed it 20psig from a large turbo (gated to never produce more than that). In your experience, would the Whipple be able to handle that?

I would like to pose more questions, but must leave. Thanks for all responses.

Jared
 
JQmile said:
The power that it takes to drive those blowers at 40psi will be about 300hp. If you're willing to live with that for throttle response, it should work.
Click to expand...

JQmile said:
Steve Morris had a centrifugal blower (F2) on a blower dyno, and it took 353hp to make 32psi, so 300hp is at the low end of the spectrum.
Click to expand...

jgsturbo said:
Screw type superchargers are suppose to be the most efficient in terms of power requirements at least when compared to "roots" blowers. At least they have quite a bit of internal compression. I believe it goes screw > centri > roots.
Just plumb 2 together for some compound action...
Click to expand...
jgsturbo is correct. it takes almost 275ish hp to run our little 8-71 on our blown alky 565 at 50% over driven it makes only 18psi. it takes almost zero hp to run the newer whipples these days. the old 980r whipples that are the size of a trashcan make HUGE power but are HUGE blowers and weight a ton. we just put a 8l whipple on and its AMAZING how much power you gain just going from a roots to a screw. its WAY more effiecient and makes a TON more power. with the 8-71 we were at 1720hp and with the whipple at 20psi is just at 2k. we also have a procharger setup on a little 383 and its a F1R and its more effiencient than a roots only because its a true compressor as well. a roots in only a air pump. i have seen many many whipples and psi c rotors turn near 60psi.
 
you can always compound a procharger setup. that has been done on a gasser and makes CHIT ton of power and boost.
 
svyus.jpg


and the video of said car.

MOV1FD - YouTube
 
jfreemanak said:
TURBOLVR - I have avidly followed your adventure with forced induction on diesels. I am just now getting to a point where I am able to start experimenting myself. I remember you posting at one time that whipples don't like to be fed 50psig of boost. Let's say I was to put the 3.3L on the duramax, and feed it 20psig from a large turbo (gated to never produce more than that). In your experience, would the Whipple be able to handle that?

Jared
Click to expand...

It could handle it, but asking the super to take-in 2.5 atmospheres and force-out 3 will consume a healthy amount of crank power and produce a lot of heat.
In your case, I would suggest running the blower with the rotor case upside-down so the air enters the back and exits the top. Then pipe that to a W/A IC or directly to the turbo then the A/A IC and into the engine. This will offer the least parasitic loss from the super, while still offering fast low-end response. Using a large bypass will allow the turbo to flow beyond what the super can when called for. :Cheer:
 
TURBOLVR said:
Just remember they are 2 different animals. The centrifugals need rpm and have a stepped gear to turn 3-4x pulley speed. the screw types are direct drive.
I have been searching for a HP per psi for them but no dice yet. I DO know that even @ 25psi, there is no belt slip from an 8 rib dayco serp. belt. No cog drive needed!
Click to expand...

Why can't we just find a map on them. I usually use 9 rwhp per lb/min of airflow on turbos. In theory wouldn't this be a good rule on any charger?
 
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