Volumetric Efficiency.

[oab_au]

Not wanting to keep any body from there work. I thought we could discuss the affects of forced induction, on the Volumetric Efficiency of the engine here.

On another thread, the VE is quoted as 95% and used as the basis for predicted output. It probable is 95% at some rpm, but it is not a constant figure. The VE is the % of air that the engines cylinder can capture, with the swept vol given as 100%.

This figure is controlled by the mechanics of the engine, the inlet and exhaust tracts, and the cams profile. These components are chosen to suit a particular rpm. So it is at this engine speed that the maximum VE, thus torque is produced. The VE figure for the EG33 reaches 105% at 4000 rpm, 110.8 at 5000 rpm. This is achieved, due to the effect, that resonate action in the inlet and exhaust tracts, and the cam profile, has on cylinder filling.

If a positive displacement pump is fitted to increase the air pressure in the inlet, the VE will increase in proportion to the increase in applied pressure, but the rate of change of the VE will still be dependent on the mechanical components that the engine is fitted with.

The extra pressure will not cause the resonate frequency to occur at a higher rpm, so the VE curve, hence the torque, will still drop off at the same rpm as the original engine.

Harvey.

[GreenMarine]

I love reading your posts Harvey... They actually make me think and use the knowelage that I am gaining in school

Do you use AOL Instant Messenger?? If so, you should drop me a line sometime... I'd love to crunch numbers with ya on the mechanical workings of diffrent components in the car

[Phast SVX]

Originally Posted by oab_au

Not wanting to keep any body from there work. I thought we could discuss the affects of forced induction, on the Volumetric Efficiency of the engine here.

On another thread, the VE is quoted as 95% and used as the basis for predicted output. It probable is 95% at some rpm, but it is not a constant figure. The VE is the % of air that the engines cylinder can capture, with the swept vol given as 100%.

This figure is controlled by the mechanics of the engine, the inlet and exhaust tracts, and the cams profile. These components are chosen to suit a particular rpm. So it is at this engine speed that the maximum VE, thus torque is produced. The VE figure for the EG33 reaches 105% at 4000 rpm, 110.8 at 5000 rpm. This is achieved, due to the effect, that resonate action in the inlet and exhaust tracts, and the cam profile, has on cylinder filling.

If a positive displacement pump is fitted to increase the air pressure in the inlet, the VE will increase in proportion to the increase in applied pressure, but the rate of change of the VE will still be dependent on the mechanical components that the engine is fitted with.

The extra pressure will not cause the resonate frequency to occur at a higher rpm, so the VE curve, hence the torque, will still drop off at the same rpm as the original engine.

Harvey.

I was under the influence that HP will continue to increase in a positive displacement blower. I dont think tq was really ever the issue.

[oab_au]

Originally Posted by Phast SVX

I was under the influence that HP will continue to increase in a positive displacement blower. I dont think tq was really ever the issue.

Your bin spending too much time with Rob.

HP can't keep increasing, if the torque is dropping off. Over the same rpm range, the torque has to increase, for the HP to increase. HP is just an effect of torque over time. Anything you do to the engine to increase power, has to increase the torque, or there will be no HP difference.

Harvey.

[Phast SVX]

then how do you explain this
http://www.whipplesuperchargers.com/...obra_graph.gif
http://www.whipplesuperchargers.com/...gers_Graph.gif

[oab_au]

Originally Posted by Phast SVX

then how do you explain this
http://www.whipplesuperchargers.com/...obra_graph.gif
http://www.whipplesuperchargers.com/...gers_Graph.gif

Whats to explain Phil? The first is a comparison of an Eaton V a Whipple.
The second shows an increase in torque, with a corresponding increase in HP.

Is that what you see?

Harvey.

[Trevor]

Originally Posted by oab_au

Your bin spending too much time with Rob.

HP can't keep increasing, if the torque is dropping off. Over the same rpm range, the torque has to increase, for the HP to increase. HP is just an effect of torque over time. Anything you do to the engine to increase power, has to increase the torque, or there will be no HP difference.

Harvey.

NO not so, but yes you correctly said, "torque over time". More correctly "torque within a period of time".

Put simply one could say, shortening the time by increasing RPM provides more periods of applied torque within a given length of time and therefore an increase in resultant power.

In order to increase power output the torque does not have to increase and in fact can fall off if offset by increased "effort/energy". This is indicated exactly by riding a bicycle in a lower gear. Less applied torque, but increased RPM can result in an increase in speed from the generation of more power. Not rocket science. (In this instance the gear ratio makes this possible but has nothing to do with "Torque/RPM/Power" as a factor.)

[oab_au]

Originally Posted by Trevor

NO not so, but yes I guess this means, you agree with me? you correctly said, "torque over time". More correctly "torque within a period of time".

Put simply one could say, shortening the time by increasing RPM provides more periods of applied torque within a given length of time and therefore an increase in resultant power. Yes thats the way it normalys goes

In order to increase power output the torque does not have to increase and in fact can fall off if offset by increased "effort/energy"another name for torque?.

This is indicated exactly by riding a bicycle in a lower gear. Less applied torque, but increased RPM can result in an increase in speed from the generation of more power. Not rocket science. (In this instance the gear ratio makes this possible but has nothing to do with "Torque/RPM/Power" as a factor.)

Yes it has nothing to do with increasing HP without increasing RPM. "Lowering the gearing. Less applied torque, but increased RPM" will not increase the speed or generate more power. Just allows you to remain at the same speed, with reduced torque.

Harvey.

[Trevor]

Originally Posted by oab_au

Yes it has nothing to do with increasing HP without increasing RPM. "Lowering the gearing. Less applied torque, but increased RPM" will not increase the speed or generate more power. Just allows you to remain at the same speed, with reduced torque.

Harvey.

Harvey you are creating a smoke screen by twisting your words during a debate and one which you initiated. You have previously stated here, quote - "anything you do to the engine to increase power, HAS to increase the torque." This statement is incorrect.

Alterations such that effective RPM are increased CAN and often do, increase power, even at the expense of torque; i.e. even if less torque is available there can be more power as a result of alterations.

Regards, Trevor.

[mbtoloczko]

Welcome back Trevor. You picked a fine thread to make an appearance! :-)

Originally Posted by Phast SVX

I was under the influence that HP will continue to increase in a positive displacement blower. I dont think tq was really ever the issue.

Seems to me that this was the crux of the discussion on the SC thread. I am getting the impression that people think that since the Whipple SC is a fixed displacement blower, the amount of air being forced into the motor is increaing at a rate proportional the the rpm of the SC. This isn't the situation though. We are controlling boost pressure and air flow is the dependent variable. As the boost pressure rises, air flow into the motor will increase. The Whipple SC will make peak boost at very low rpm, and then control it at the setpoint with the bypass valve. The flow is whatever the engine can do with that increase in pressure. Its really not much different than raising the atmospheric pressure by 7 psi.

[Phast SVX]

Originally Posted by mbtoloczko

Welcome back Trevor. You picked a fine thread to make an appearance! :-)



Seems to me that this was the crux of the discussion on the SC thread. I am getting the impression that people think that since the Whipple SC is a fixed displacement blower, the amount of air being forced into the motor is increaing at a rate proportional the the rpm of the SC. This isn't the situation though. We are controlling boost pressure and air flow is the dependent variable. As the boost pressure rises, air flow into the motor will increase. The Whipple SC will make peak boost at very low rpm, and then control it at the setpoint with the bypass valve. The flow is whatever the engine can do with that increase in pressure. Its really not much different than raising the atmospheric pressure by 7 psi.

guys im done arguing with this. we never said it would be a perfect line,, we said that it would be a differnt curve, and nothing you are saying is telling me anything but the fact that, mychailo, you are simply making your best guess at what you think will happen. I dont know that you have any direct experience with boosting cars of any kind. And im not saying that some of you dont( i know some of you do!) but there is no doubt going to be a differet curve, and a positive displacement blower, with their high adbiatic efficiency, tend to create symmetrical power curves.
anyways im unsubscribed.
phil

[longassname]

Phil, I can understand your frustration but don't get too po'd. Everyone looks at things starting from a different perspective based on their past understandings. They don't mean to missunderstand. Their past beliefs are just strongly held and make it difficult to accept, believe, or even pay attention to something that is conflict with them.

Originally Posted by Phast SVX

guys im done arguing with this. we never said it would be a perfect line,, we said that it would be a differnt curve, and nothing you are saying is telling me anything but the fact that, mychailo, you are simply making your best guess at what you think will happen. I dont know that you have any direct experience with boosting cars of any kind. And im not saying that some of you dont( i know some of you do!) but there is no doubt going to be a differet curve, and a positive displacement blower, with their high adbiatic efficiency, tend to create symmetrical power curves.
anyways im unsubscribed.
phil

[longassname]

Actually that is exactly incorrect. The whipple supercharger is a fixed displacement blower. It does put out an exact amount of air with each rotation and the only thing that will prevent that is what trevor said..a vacuum on the inlet side. If there is no restriction on the inlet side preventing the blower from getting the air into it's lobes it will put the same amount of air out the discharge side with each rotation and it's output will be a 45 degree angle line. On a side note, that is why all of the dyno plots of engines with the whipple super charger are a 45 degree angle line.

The bypass valve does not control "boost." It does not open when boost gets to the desired level. It works exactly the oposite of that. With a positive displacement blower boost is always at the desired level. At any rpm, be it 1 rpm or 9000 rpms boost will always be the same..even if you unplug all the coil packs and the engine wasn't actually running but you used something else to turn the motor over boost would be the same because the blower puts out exactly the same amount of air with each rotation which is always the same ratio of air to the volume of the engine.

The bypass valve opens when the throttle blades are closed. The throttle blades being closed creates the one thing that is capable of preventing the supercharger of putting out it's exact amount of air with each rotation...a vacuum on the inlet side. This vacuum actuates the bypass valve which then opens and creates a tunnel between the discharge and inlet sides to equalize the pressure so that the blower isn't trying to gulp air that isn't there.

Trevor and shotgun were also correct in that the only effects on performance of intake tract's characteristics are on efficiencies and to a much smaller scale than the actual increases and characteristics of the blower's discharge. As I had explained in the other thread. The blower DOES put out a fixed amount of air with each rotation and if that air didn't make it past the valves and into the combustion chamber it would still be in the manifold and with the next rotation the manifold pressure would increase and escalate with each rotation. Significant restriction would produce the observable symptom of increased manifold pressure whouch would take more power to pump against. It is at that time that the characteristics of valvetrain would come back into play and you would need larger valves, more lift, or longer duration to overcome the restriction.

I hope that helps clear things up for you.

Originally Posted by mbtoloczko

Welcome back Trevor. You picked a fine thread to make an appearance! :-)



Seems to me that this was the crux of the discussion on the SC thread. I am getting the impression that people think that since the Whipple SC is a fixed displacement blower, the amount of air being forced into the motor is increaing at a rate proportional the the rpm of the SC. This isn't the situation though. We are controlling boost pressure and air flow is the dependent variable. As the boost pressure rises, air flow into the motor will increase. The Whipple SC will make peak boost at very low rpm, and then control it at the setpoint with the bypass valve. The flow is whatever the engine can do with that increase in pressure. Its really not much different than raising the atmospheric pressure by 7 psi.

[mbtoloczko]

Originally Posted by longassname

I hope that helps clear things up for you.

Yep, I totally understand now. Thanks for tolerating my lack of understanding. I am very curious though to see what happens at higher rpm. As you are saying, I think that at some point the engine will not be able to efficiently receive the air that the SC is pumping, and manifold pressure will begin to rise. Wondering what rpm where that will occur...

[longassname]

Hey, no problem man. I'm on your side that's why I bother explaining the stuff.

Mike may well find out now that he will have a manual transmission. The reasonably low 6500 rpms we get with the automatic are a a big part of what makes so much of this stuff work for us. It lets stage 2 work for turbos despite the little fuel injectors, keeps our valves under control despite the light valve springs, and lets us pump more air before we have to address the valve train. I didn't even realize how high we went on the "boost" on Mike's car when we did it. I just cut the pulley to the smallest size I could to get hood clearance. We had gone more hours than i care to remember without sleep at that point. For all I know it might be too much already. We didn't get to drive it very far. I can cut the crank pulleys smaller easily enough t hough. I'll figure it out on my car like I had originally planned. I have extensive logging capabilities.

Hopefully the na guys who are digging into the heads will make some significant progress soon. It's allready been done. One of our customer's already has it done. He got bigger valves and springs that allow much more lift allong with a lot of head work and big cams courtesy of pantera.
I'm hoping not to have to dig into the valve train on my high boost engine but who knows?

Originally Posted by mbtoloczko

Yep, I totally understand now. Thanks for tolerating my lack of understanding. I am very curious though to see what happens at higher rpm. As you are saying, I think that at some point the engine will not be able to efficiently receive the air that the SC is pumping, and manifold pressure will begin to rise. Wondering what rpm where that will occur...