percent of power calculations

[RonB]

Good evening

Does anyone know what is required to determine the percent of power a Lycoming IO360 180 HP engine may be producing at a particular setting?

I have a Dynon EMS and it's obviously showing a much higher percent of power than is actually being produced by my engne and the folks at Dynon haven't been able to offer any help.

Thanks

Ron B

[n5lp]

Originally Posted by RonB

Good evening

Does anyone know what is required to determine the percent of power a Lycoming IO360 180 HP engine may be producing at a particular setting?

I have a Dynon EMS and it's obviously showing a much higher percent of power than is actually being produced by my engne and the folks at Dynon haven't been able to offer any help.

Thanks

Ron B

Kevin Horton has that one covered. Look down the left side of the page under "Top Links."

[elippse]

Originally Posted by RonB

Good evening

Does anyone know what is required to determine the percent of power a Lycoming IO360 180 HP engine may be producing at a particular setting?

I have a Dynon EMS and it's obviously showing a much higher percent of power than is actually being produced by my engne and the folks at Dynon haven't been able to offer any help.

Thanks

Ron B

Here's what I use in my equations. First, for your engine, determine the MAP that the factory shows on their chart of the power-altitude-rpm graph. That is the MAP that you ratio to with your engine's running MAP. For instance, on my O-235, rated power occurs at 28.4", not 29.92! So if I'm cruising along at 8000' and 22", the first part of the power ratio will be 77.5%. Next, you have to ratio your actual rpm vs rated rpm. So if rated is 2700 and you're at 2500, this part is 92.6%. This asumes that these engines have a fairly flat torque curve over a range of rpm so that power is directly proportional to rpm. So now we have 77.5% (0.775) times 92.6% (.926) which multiplied together gives 71.7%. Then to this I ratio the inlet temperature plus stagnation temperature rise and minus fuel evaporation drop and convert this to absolute temperature and divide this by the sea-level absolute temperature minus fuel evaporation drop, and get the square-root of this. So on this standard day the OAT was 30.5F and at 200 mph the stagnation rise was 7F. I add 30.5+7-24+459.7, 473.2, and divide this by 459.7+59-24, 494.7, giving 0.957, and a square-root of 0.978. Multiply this times the previous 71.7% and get 70.1% power Now without MAP and all of the temperatures, I could have just gotten the density ratio relative to sea-level for 8000' on a standard day which is 0.786. Some smart people found that on average, because of how the engine reacts to the induction temperature's effect on the inlet density and power, that if you raise the density ratio to the 1.135 power that will compensate. 0.786^1.135=0.761. Multiply that by the rpm ratio, 0.926, and get 70.4% power. But that only works for WOT, so if you have the throttle reduced, you need to use the first more-complete calculation with MAP. Nothing to it, right?

[n5lp]

Originally Posted by elippse

Here's what I use in my equations. First, for your engine, determine the MAP that the factory shows on their chart of the power-altitude-rpm graph. That is the MAP that you ratio to with your engine's running MAP. For instance, on my O-235, rated power occurs at 28.4", not 29.92! So if I'm cruising along at 8000' and 22", the first part of the power ratio will be 77.5%. Next, you have to ratio your actual rpm vs rated rpm. So if rated is 2700 and you're at 2500, this part is 92.6%. This asumes that these engines have a fairly flat torque curve over a range of rpm so that power is directly proportional to rpm. So now we have 77.5% (0.775) times 92.6% (.926) which multiplied together gives 71.7%. Then to this I ratio the inlet temperature plus stagnation temperature rise and minus fuel evaporation drop and convert this to absolute temperature and divide this by the sea-level absolute temperature minus fuel evaporation drop, and get the square-root of this. So on this standard day the OAT was 30.5F and at 200 mph the stagnation rise was 7F. I add 30.5+7-24+459.7, 473.2, and divide this by 459.7+59-24, 494.7, giving 0.957, and a square-root of 0.978. Multiply this times the previous 71.7% and get 70.1% power Now without MAP and all of the temperatures, I could have just gotten the density ratio relative to sea-level for 8000' on a standard day which is 0.786. Some smart people found that on average, because of how the engine reacts to the induction temperature's effect on the inlet density and power, that if you raise the density ratio to the 1.135 power that will compensate. 0.786^1.135=0.761. Multiply that by the rpm ratio, 0.926, and get 70.4% power. But that only works for WOT, so if you have the throttle reduced, you need to use the first more-complete calculation with MAP. Nothing to it, right?

See Ron, it's simple!

[digidocs]

I think that exhaust backpressure plays a role too.

If you look at the manufacturer's power charts, it appears that 25" MAP at 8000ft yields more power than 25" MAP at SL. I believe this difference is due to there being less (absolute) exhaust backpressure at altitude.

-DC

[Finley Atherton]

For cruise at either 'Best Power" (about 150 degrees ROP) or "Best Economy" (peak EGT) mixture settings you can use the "Part Throttle Fuel Consumption" graph in your Lycoming Owners Manual. All you need to use this graph is fuel flow and rpm. Be aware this is for a standard Lycoming with mags. Higher compression, electronic ignition etc will likely result in more HP. I THINK you may be able to also use this graph for LOP operations as Specific Fuel Consumption is fairly constant from around peak to about 70 degrees LOP (see Lycomings graph; "Representative Effect of Leaning on ..........."

Fin
9A

EDIT. There are a few different Part Throttle Fuel Consumption graphs in the Owners Manual. I am referring to the appropriate one for your engine model that has "Mixture Control - Manual to Best Economy or Best Power as Indicated"

[elippse]

Originally Posted by digidocs

I think that exhaust backpressure plays a role too.

If you look at the manufacturer's power charts, it appears that 25" MAP at 8000ft yields more power than 25" MAP at SL. I believe this difference is due to there being less (absolute) exhaust backpressure at altitude.

-DC

That must be a supercharged engine to get 25" at 8000' where standard pressure is 22.22". 25" at sea-level would cause some power loss due to pressure drop across the throttle valve, whereas 25" at 8000' would be due to a MAP boost and less TV drop. 'Just wundrin'

[Brantel]

If your Dynon is far off, then there must be some error in your fuel flow transducer calibration, the OAT reading, or the MAP pressure reading or you are using something other than stock compression ratio? Do you have the HP set correctly for your engine?

Originally Posted by RonB

Good evening

Does anyone know what is required to determine the percent of power a Lycoming IO360 180 HP engine may be producing at a particular setting?

I have a Dynon EMS and it's obviously showing a much higher percent of power than is actually being produced by my engne and the folks at Dynon haven't been able to offer any help.

Thanks

Ron B

[647jc]

The engine red line parameter setting in the Dynon will also affect the percent power readout. Dynon uses the red line setting as the RPM at which the engine generates 100% power at sea level. My engine is a O320 which has a red line of 2700 RPM BUT my prop red lines at 2600. So, to get all the warnings from the Dynon when I get close or hit the 2600 RPM prop red line, I have configured the engine red line RPM in the Dynon for 2600 RPM. This affects the engine percent power readout by about 2%. When the engine is running at 75% power the Dynon reads 77%. 65% engine power reads about 67%. Not a big deal but if you set your Dynon red line to something other than engine max power RPM, your percent power readout will be affected.

[Scott Hersha]

Here's a power setting table I got from Lycoming:

Well.... I couldn't link it to this post so I'll try to email it to you. It's in a word document file.