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Volumetric efficiency and torque output. 5

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dano73327

Automotive
Aug 8, 2007
1
This post relates to a similar discussion I've had on another board.

I have been under the impression that the Volumetric efficiency of an IC engine would very closely follow the torque curve of the engine and that generally speaking the torque peak of the engine would represent the point of maximum VE.
Am I mistaken?
If so please explain.

Also what would be the proper way to determine accurately the VE of an engine?
 
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Correct. If ve drops off then less heat can be produced because less fuel can burn.
 
"Am I mistaken?" As a trend, no.

"Also what would be the proper way to determine accurately the VE of an engine? "

measure the flow rate in the intake.


Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
 
If you cannot measure air flow, measure fuel flow and a:f ratio then calculate.

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The torque peak occurs at the maximum product value of volumetric efficiency and combustion efficiency (Vol Eff x Comb. Eff). By plotting this value you can tell where the torque peak will occur but not its actual value. This parameter can also be useful if worked backwards to look at the shape of the combustion efficiency curve (Assuming that you don't have too many clever bits of kit like variable output oil pumps that put a step in the friction curve).
 
...or, you can equate two equations for HP and then backsolve for VE:

A) HP = (T*RPM)/(5252) ...from: T = (5252*HP)/(RPM)

B) HP = (AP*CR*VE)*(CID*RPM)/(5252*150.8)

C) (T*RPM)/(5252) = (AP*CR*VE)*(CID*RPM)/(5252*150.8)

where:
T = Engine Torque, ft-lbs(f)
HP = Horspower
AP = Air Pressure, 14.7 psi
CR = Compression Ratio, ie: 8=9:1, 10=10:1
VE = Volumetric Efficiency; ie: stock ~ 0.90; high-performance ~ 1.00; race ~ 1.10
CID = Cubic-Inch-Displacement
RPM = Engine Revolutions-per-Minute
 
According to historic car magazines Some historic F1 engines (BRM H16 I think) have had the reputation for excessive friction at higher rpm that actually hurt the power band. In that case VE could be high, but the torque at a particular rpm would be down due to mechanical losses.
In that case in any engine the "real" torque peak is higher than the dyno reports simply because adding a few hundred rpm "costs" available torque despite the cylinder filling, etc still being good.
 
...VE and TORQUE "track" together, ie: VE will 'peak' at exactly the same RPM that TORQUE peaks at...because TORQUE is dependent upon VE.
 
".....VE will 'peak' at exactly the same RPM that TORQUE peaks at...because TORQUE is dependent upon VE"

Not entirely true. Torque will peak when BMEP peaks. Which may, or may not, be when VE peaks. BMEP is IMEP minus FMEP. FMEP losses include pumping losses, accessory losses and also mechanical friction losses.
 
"...or, you can equate two equations for HP and then backsolve for VE:
A) HP = (T*RPM)/(5252) ...from: T = (5252*HP)/(RPM)
B) HP = (AP*CR*VE)*(CID*RPM)/(5252*150.8)
C) (T*RPM)/(5252) = (AP*CR*VE)*(CID*RPM)/(5252*150.8)"

I sure may be wrong again, but don't see any terms I recognize to capture the influence of things like windage trays, roller rockers, air density (depends on temp and pressure), if its an air-cooled engine spinning a huge cooling fan (that I don't think is part of VE) or even "correct" A/F ratio or ignition timing
 
1) the effects of engine components that INHANCE (or HINDER) VE are embedded within the cumulative VE number.

2) true, the final BMEP = IMEP-FMEP value will change in magnitude, but typically without (cavet wording) much change to RPM.
 
IMEP is strongly influenced by VE, but also by ignition timing, air/fuel ratio, charge motion, and local temperatures, the first two of which may or may not be optimized for MBT at every rpm, and the last two of which are characteristics of the base engine that will vary with rpm.
 
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