Engine Load vs. Output Power
Thread Starter
Registered User
Joined: Sep 2002
Posts: 655
Engine Load vs. Output Power
This is going to be a tough one to explain from my end, but hopefully
someone understands and can clarify.
A discussion has come about regarding forced air induction and
how a turbo requires exhaust gas to spool up.
Assuming there is no load on the motor (IE: Trans. in nuetral)
and the engine is rev'd , will the amount of air/fuel processed
and exhaust gas velocity created be more; less; or somewhat
similar compared to when the engine is under load?
My thoughts are since the engine has little load in nuetral,
the pistons are free to move from TDC to BDC at a much quicker
rate.
This will cause lower pressure in the combustion chamber to allow
more air/fuel charge to enter (higher VE%). By the same comparison,
air (CFM) ingested and exhausted will be equally as high.
My conclusion is the turbo will spool up faster and exhaust gas
velocity will be high.
For all the above we assume properly tuned exhaust, intake, etc.
One more thought:
Engine output power will 'somewhat' the same at 5000 RPM in nuetral
as it would be in gear at 5000 RPM under full load. The stored
energy once a clutch grabbed would transfer instant power to
the ground.
I think the main variables that change would be BSFC...or the
efficiency required to spin the motor without a load at the crank.
The above questions are to discredit these theories:
- Turbos require engine load to produce power.
- Engines do not process the amount of air flow (CFM) in nuetral
as they would under load.
someone understands and can clarify.
A discussion has come about regarding forced air induction and
how a turbo requires exhaust gas to spool up.
Assuming there is no load on the motor (IE: Trans. in nuetral)
and the engine is rev'd , will the amount of air/fuel processed
and exhaust gas velocity created be more; less; or somewhat
similar compared to when the engine is under load?
My thoughts are since the engine has little load in nuetral,
the pistons are free to move from TDC to BDC at a much quicker
rate.
This will cause lower pressure in the combustion chamber to allow
more air/fuel charge to enter (higher VE%). By the same comparison,
air (CFM) ingested and exhausted will be equally as high.
My conclusion is the turbo will spool up faster and exhaust gas
velocity will be high.
For all the above we assume properly tuned exhaust, intake, etc.
One more thought:
Engine output power will 'somewhat' the same at 5000 RPM in nuetral
as it would be in gear at 5000 RPM under full load. The stored
energy once a clutch grabbed would transfer instant power to
the ground.
I think the main variables that change would be BSFC...or the
efficiency required to spin the motor without a load at the crank.
The above questions are to discredit these theories:
- Turbos require engine load to produce power.
- Engines do not process the amount of air flow (CFM) in nuetral
as they would under load.
Registered User
Joined: May 2002
Posts: 199
From: Seattle Area
Hmmm interesting question....
When there is no load on the engine, yes the engine will rev alot easier, just try it on your own car. But I don't think that the cylinder pressure will change between an engine with load to one without any load. The cylinder is still the same volume no matter then rpm right? So the pressure drop would be the same, irrelevent to the rpm of the engine.
But do correct me if I am wrong...
Thats all my brain can handle today.
Hunter
When there is no load on the engine, yes the engine will rev alot easier, just try it on your own car. But I don't think that the cylinder pressure will change between an engine with load to one without any load. The cylinder is still the same volume no matter then rpm right? So the pressure drop would be the same, irrelevent to the rpm of the engine.
But do correct me if I am wrong...
Thats all my brain can handle today.
Hunter
Registered User
Joined: Oct 2002
Posts: 2,931
From: Upstate NY
It's easier to look at a steady state condition, or maybe a number of them to see what's happening.
The engine is a demand device which you control with the accelerator pedal, which should more correctly be called the "throttle" pedal, because you "throttle" or choke off the intake to control either power output or rpm, or both.
Steady state 5000 rpm no load: It takes less than wide open throttle (WOT) to hold the engine at 5000...a lot less. Therefore the intake is severely restricted so very little air is flowing thru the intake, very litttle is getting burned and there's not much exhaust volume to drive the turbo. Note that the engine makes more exhaust noise at 5000 under full load than it does under no load. If you were able to measure exhaust gas temp (EGT), it would be much higher under load.
Free reving in neutral (no load) is just a number of no load examples of what I said above. The rate at which the engine accelerates is determined by throttle position.
Pistons move from TDC to BDC in a time directly proportional to rpm, not load.
With less than WOT (part throttle), VE will be lower, not higher than at WOT. Actually the turbo will spool up slower, relative to engine speed than when at WOT. The turbo doesn't know how many rpm the engine is turning, it only knows how much hot exhaust gas it is injesting. The way a 1.8L VW/Audi turbo gets max torque below 2000 rpm at WOT is that there is enough hot exhaust gas to spin the turbo to near it's max rpm.
Engine power @ 5000 in neutral or no-load will only be what it takes to spin the engine and flywheel 5000 rpm. This is approximately the friction hp. There will be no useful hp aka "brake hp" being produced. In gear under full load and WOT, the engine will produce the most hp it can at that rpm with a resultant higher inlet airflow, exhaust flow, turbo boost, heat, and noise.
BSFC is BRAKE Specific Fuel Consumption, and by definition it would be meaningless or infinite, if you prefer. Brake means a load measured at the flywheel, so dividing fuel (lbs/hr) by zero flywheel hp gives a terribly big number.
Turbochargers don't actually produce power, they merely pump air. They absorb heat energy and therefore power from the engine.
Turbosuperchargers are demand devices: little exhaust input = little compressor output. Positive displacement superchargers like Magnacharger (built by Eaton) pump the same amount of air per rev irrespective of the engine load. The Magnacharger has a butterfly valve at the output which can direct the air back into the inlet under light load conditions like cruising or free revving. That's why Magnacharger can claim less than one hp to drive the blower at cruise rpm and throttle opening. It's sorta lifting itself up by its own bootstraps.
Sorry, but I can't discredit the theories you mentioned. It's just the opposite; they are essentially correct.
Reading this over it sounds like I'm flaming you. That's not the intent. I think you are mistaken. You did ask.
My $.02
The engine is a demand device which you control with the accelerator pedal, which should more correctly be called the "throttle" pedal, because you "throttle" or choke off the intake to control either power output or rpm, or both.
Steady state 5000 rpm no load: It takes less than wide open throttle (WOT) to hold the engine at 5000...a lot less. Therefore the intake is severely restricted so very little air is flowing thru the intake, very litttle is getting burned and there's not much exhaust volume to drive the turbo. Note that the engine makes more exhaust noise at 5000 under full load than it does under no load. If you were able to measure exhaust gas temp (EGT), it would be much higher under load.
Free reving in neutral (no load) is just a number of no load examples of what I said above. The rate at which the engine accelerates is determined by throttle position.
Pistons move from TDC to BDC in a time directly proportional to rpm, not load.
With less than WOT (part throttle), VE will be lower, not higher than at WOT. Actually the turbo will spool up slower, relative to engine speed than when at WOT. The turbo doesn't know how many rpm the engine is turning, it only knows how much hot exhaust gas it is injesting. The way a 1.8L VW/Audi turbo gets max torque below 2000 rpm at WOT is that there is enough hot exhaust gas to spin the turbo to near it's max rpm.
Engine power @ 5000 in neutral or no-load will only be what it takes to spin the engine and flywheel 5000 rpm. This is approximately the friction hp. There will be no useful hp aka "brake hp" being produced. In gear under full load and WOT, the engine will produce the most hp it can at that rpm with a resultant higher inlet airflow, exhaust flow, turbo boost, heat, and noise.
BSFC is BRAKE Specific Fuel Consumption, and by definition it would be meaningless or infinite, if you prefer. Brake means a load measured at the flywheel, so dividing fuel (lbs/hr) by zero flywheel hp gives a terribly big number.
Turbochargers don't actually produce power, they merely pump air. They absorb heat energy and therefore power from the engine.
Turbosuperchargers are demand devices: little exhaust input = little compressor output. Positive displacement superchargers like Magnacharger (built by Eaton) pump the same amount of air per rev irrespective of the engine load. The Magnacharger has a butterfly valve at the output which can direct the air back into the inlet under light load conditions like cruising or free revving. That's why Magnacharger can claim less than one hp to drive the blower at cruise rpm and throttle opening. It's sorta lifting itself up by its own bootstraps.
Sorry, but I can't discredit the theories you mentioned. It's just the opposite; they are essentially correct.
Reading this over it sounds like I'm flaming you. That's not the intent. I think you are mistaken. You did ask.
My $.02
Last edited by OldSStroker; Sep 21, 2003 at 08:51 PM.
Thread Starter
Registered User
Joined: Sep 2002
Posts: 655
No flame taken!
That's great info, except for one concept I'd like to debate a little
further: Piston movement from TDC to BDC.
If the piston is shooting down to BDC at let's say 200 milliseconds
at 4000 RPM under moderate load, it *may* take the engine
one second to reach 5000 RPM.
If the engine was under higher load, wouldn't the piston slow
down causing higher MAP and take the engine longer to reach
5000 RPM?
This would also create a higher pressure in the exhaust system
slowing down exhaust velocity and preventing the turbo from
spooling quicker?
I don't think I fully understand the quote, "Pistons move from TDC to BDC in a time directly proportional to rpm, not load."
That's great info, except for one concept I'd like to debate a little
further: Piston movement from TDC to BDC.
If the piston is shooting down to BDC at let's say 200 milliseconds
at 4000 RPM under moderate load, it *may* take the engine
one second to reach 5000 RPM.
If the engine was under higher load, wouldn't the piston slow
down causing higher MAP and take the engine longer to reach
5000 RPM?
This would also create a higher pressure in the exhaust system
slowing down exhaust velocity and preventing the turbo from
spooling quicker?
I don't think I fully understand the quote, "Pistons move from TDC to BDC in a time directly proportional to rpm, not load."
Registered User
Joined: Oct 2002
Posts: 2,931
From: Upstate NY
Originally posted by Zero_to_69
Piston movement from TDC to BDC.
If the piston is shooting down to BDC at let's say 200 milliseconds
at 4000 RPM under moderate load, it *may* take the engine
one second to reach 5000 RPM.
Well it's closer to 8 milliseconds for a half rev, but you are talking about engine acceleration from one rpm to another, usually expressed in rpm per second. Don't confuse rpm, or engine "speed" or velocity with rpm per second, which is the change of rpm or engine acceleration.
In first gear it could take about 1 second under full load to rev from 4-5000 rpm. Free revving it could be faster, sure, but with no load, there is very little airflow, as I discussed earlier.
If the engine was under higher load, wouldn't the piston slow
down causing higher MAP and take the engine longer to reach
5000 RPM?
This would also create a higher pressure in the exhaust system
slowing down exhaust velocity and preventing the turbo from
spooling quicker?
More load = more exhaust volume, and yes more exhaust back pressure because the turbo can only eat it so fast. Think of the energy in the exhaust as being proportional to the volume and heat in the gas. Velocity depends on flow and resistance. The turbo continues to speed up and produces more boost which makes more exhaust gas, and so it continues.... BTW, Isn't MAP an intake measurement?
I don't think I fully understand the quote, "Pistons move from TDC to BDC in a time directly proportional to rpm, not load."
All I meant is that the piston moves mechanically thru it's stroke in the time it takes the engine to rotate 1/2 a rev. Load doesn't matter, unless it is changing the rod length considerably.
Again, don't confuse rpm with rpm per second.
Piston movement from TDC to BDC.
If the piston is shooting down to BDC at let's say 200 milliseconds
at 4000 RPM under moderate load, it *may* take the engine
one second to reach 5000 RPM.
Well it's closer to 8 milliseconds for a half rev, but you are talking about engine acceleration from one rpm to another, usually expressed in rpm per second. Don't confuse rpm, or engine "speed" or velocity with rpm per second, which is the change of rpm or engine acceleration.
In first gear it could take about 1 second under full load to rev from 4-5000 rpm. Free revving it could be faster, sure, but with no load, there is very little airflow, as I discussed earlier.
If the engine was under higher load, wouldn't the piston slow
down causing higher MAP and take the engine longer to reach
5000 RPM?
This would also create a higher pressure in the exhaust system
slowing down exhaust velocity and preventing the turbo from
spooling quicker?
More load = more exhaust volume, and yes more exhaust back pressure because the turbo can only eat it so fast. Think of the energy in the exhaust as being proportional to the volume and heat in the gas. Velocity depends on flow and resistance. The turbo continues to speed up and produces more boost which makes more exhaust gas, and so it continues.... BTW, Isn't MAP an intake measurement?
I don't think I fully understand the quote, "Pistons move from TDC to BDC in a time directly proportional to rpm, not load."
All I meant is that the piston moves mechanically thru it's stroke in the time it takes the engine to rotate 1/2 a rev. Load doesn't matter, unless it is changing the rod length considerably.
Again, don't confuse rpm with rpm per second.
Thread Starter
Registered User
Joined: Sep 2002
Posts: 655
Yes, MAP is an Intake measurement... my thought was that
the manifold pressure would rise, and the turbo would continue
to force air, but the "slower moving pistons" couldn't process
the extra charge (creating a higher MAP).
the manifold pressure would rise, and the turbo would continue
to force air, but the "slower moving pistons" couldn't process
the extra charge (creating a higher MAP).
Registered User
Joined: Jan 2001
Posts: 293
From: houston, Tx
Exhaust output is pretty much rpm and load dependent. The throttle basically controls manifold pressure so without higher manifold pressure or density even at higher rpm there is little power or exhaust produced to spin the turbo. That's why turbo cars have to load the engine against the converter for instance to build boost up before launch or shoot a small shot of NOS.
Thread Starter
Registered User
Joined: Sep 2002
Posts: 655
So what you're saying is the boost created by loading up
against the torque converter is "stored boost" in the manifold?
In other words, I have a couple of friends with manual transmission
that use turbos and they can spool their turbo up without load.
The boost gauge isn't showing as much manifold pressure (boost)
but is that because the exhaust velocity is helping to create enough
of a scavenging effect that the manifold pressure is lower?
Another anaology is my friend Jason with a GTP (supercharged).
His boost will lower if he drops the exhaust system, yet his
60ft times, 1/4 mile times and MPH all improve dramatically.
So...just because the manifold pressure shows an increase,
does it really mean the boost is being applied, or is it waiting
for exhaust manifold pressure to decrease before it can be
processed?
Are you all with me on this?
against the torque converter is "stored boost" in the manifold?
In other words, I have a couple of friends with manual transmission
that use turbos and they can spool their turbo up without load.
The boost gauge isn't showing as much manifold pressure (boost)
but is that because the exhaust velocity is helping to create enough
of a scavenging effect that the manifold pressure is lower?
Another anaology is my friend Jason with a GTP (supercharged).
His boost will lower if he drops the exhaust system, yet his
60ft times, 1/4 mile times and MPH all improve dramatically.
So...just because the manifold pressure shows an increase,
does it really mean the boost is being applied, or is it waiting
for exhaust manifold pressure to decrease before it can be
processed?
Are you all with me on this?
Registered User
Joined: Oct 2002
Posts: 2,931
From: Upstate NY
Originally posted by Zero_to_69
So what you're saying is the boost created by loading up
against the torque converter is "stored boost" in the manifold?
Or just that the engine is producing power, therefore more exhaust gas, therefore more boost. Even though the car isn't moving, the engine doesn't know that ; it just feels the load of the stalled converter which is very much like a dyno.
In other words, I have a couple of friends with manual transmission that use turbos and they can spool their turbo up without load.
The boost gauge isn't showing as much manifold pressure (boost)
but is that because the exhaust velocity is helping to create enough of a scavenging effect that the manifold pressure is lower?
Nope, not IMO. It goes back to restricted or throttled intake minimizing air in and therefore exhaust out. They can't spool up with clutch in at WOT; engine would just hit rev limiter. I'll bet an identical engine with auto will produce lots more boost against a stalled TC than free revving.
Another anaology is my friend Jason with a GTP (supercharged).
His boost will lower if he drops the exhaust system, yet his
60ft times, 1/4 mile times and MPH all improve dramatically.
So...just because the manifold pressure shows an increase,
does it really mean the boost is being applied, or is it waiting
for exhaust manifold pressure to decrease before it can be
processed?
Are you all with me on this?
My take is that exhaust part of the turbo, which does the driving is more efficient with pipe dropped, therefore it takes less of the engine's (exhaust) power to drive the turbo, so more is left at the flywheel. When running down the track is boost actually less with dropped pipe, or is that just at the line stalling the TC?
The compressor part of the turbocharger doesn't know what's driving it. It's very similar to the compressor in a Vortech or other belt driven centrifugal supercharger. Only the drive mechanism is different, and of course there isn't a fixed drive ratio between engine and compressor in the turbo.
my $.02
So what you're saying is the boost created by loading up
against the torque converter is "stored boost" in the manifold?
Or just that the engine is producing power, therefore more exhaust gas, therefore more boost. Even though the car isn't moving, the engine doesn't know that ; it just feels the load of the stalled converter which is very much like a dyno.
In other words, I have a couple of friends with manual transmission that use turbos and they can spool their turbo up without load.
The boost gauge isn't showing as much manifold pressure (boost)
but is that because the exhaust velocity is helping to create enough of a scavenging effect that the manifold pressure is lower?
Nope, not IMO. It goes back to restricted or throttled intake minimizing air in and therefore exhaust out. They can't spool up with clutch in at WOT; engine would just hit rev limiter. I'll bet an identical engine with auto will produce lots more boost against a stalled TC than free revving.
Another anaology is my friend Jason with a GTP (supercharged).
His boost will lower if he drops the exhaust system, yet his
60ft times, 1/4 mile times and MPH all improve dramatically.
So...just because the manifold pressure shows an increase,
does it really mean the boost is being applied, or is it waiting
for exhaust manifold pressure to decrease before it can be
processed?
Are you all with me on this?
My take is that exhaust part of the turbo, which does the driving is more efficient with pipe dropped, therefore it takes less of the engine's (exhaust) power to drive the turbo, so more is left at the flywheel. When running down the track is boost actually less with dropped pipe, or is that just at the line stalling the TC?
The compressor part of the turbocharger doesn't know what's driving it. It's very similar to the compressor in a Vortech or other belt driven centrifugal supercharger. Only the drive mechanism is different, and of course there isn't a fixed drive ratio between engine and compressor in the turbo.
my $.02
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