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05282010, 08:54 AM




Join Date: Apr 2007
Location: Austin, TX
Posts: 410


Quote:
Originally Posted by elippse
That must be a supercharged engine to get 25" at 8000' where standard pressure is 22.22". 25" at sealevel 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'

Elippse,
You caught me! I wasn't looking at a chart when I wrote that post.
As penance, here is an excerpt from an O540A power chart:
The highlighted line shows power vs. altitude for a setting of about 22.3"/2350 rpm.
You'll notice that the rated power is 168HP at 1000', but increases to 181HP at 6000' when the the MAP/RPM are held constant. I believe that this difference is due to the reduced exhaust backpressure at altitude, but there may be other factors as well.
The full page (PDF) is here for reference.
Blue skies,
David

05282010, 09:55 AM



Join Date: Dec 2006
Location: Arroyo Grande, CA
Posts: 938


Power
Hi, David!
First, thanks for the 540 power curve; that makes a great addition to my files! But keep in mind, that in order to drop the MAP to 22.3" at 1000', where standard pressure is 28.86", requires a large pressure drop across the throttle valve which gives pumping losses. That was the lesson Lindbergh showed to the P38 pilots in the Pacific in WWII  keep the throttle wide open and reduce propeller rpm. I'll have to go back to Taylor to see what power loss occurs from Pin to Pout. I'm planning on using carb heat at altitude to reduce rpm and increase engine efficiency on trips. Even though increasing induction temperature reduces induction density directly, an engine's efficiency increases with the squareroot of the temperature, which is why the power correction factor in the box on the upper left of the complete graph, which is not shown on your post, shows the HP vs squareroot of the absolute temperature ratio, which is what I wrote about on my original response. 'Let you know what Taylor has on the pressure ratio.

05282010, 12:08 PM



Join Date: Dec 2006
Location: Arroyo Grande, CA
Posts: 938


Taylor on pumping loss
David; I did a very brief and not at all comprehensive research of CF Taylor, Vol.1, 2nd Ed., Revised, on pumping losses. If I'm interpreting Fig. 922 correctly, the major contributor to power loss is due to throttling of the inlet.

06052010, 08:06 AM




Join Date: Jul 2005
Location: Detroit, MI
Posts: 1,546


Theory vs. Charts
Early in this thread, Paul Lipps said, in part, this:

"Here's what I use in my equations. First, for your engine, determine the MAP that the factory shows on their chart of the poweraltituderpm graph. That is the MAP that you ratio to with your engine's running MAP. For instance, on my O235, 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%. "

The rest was about correction for temperature and the correction was small.
This makes perfect sense to me. Assuming a flat torque line, the HP should be proportional to the amount of fuelair mixture burned per unit of time. RPM and density therefore should give a correct answer as per Paul's calculations. So, do they?
Ever curious, I took a look at the power chart for the Superior Vantage IO360 (very similar to my XP IO360. I tested the theory for the sealevel, best power condition on an Excel spreadsheet. The lines are pretty flat.
Here is the chart for you to double check my work.
Oops, the chart does not agree with the theory. For example:
1. Paul uses 22" by 2500 RPM and gets around 71.7%. The chart says 59.4%. Error = 20.7%
2. I calculated all the MP's in one inch intervals from 18" to 29" for 2700 RPM. Paul's method gives a difference from the chart between 1.0 and 1.32, increasing as MP declines, perfect at 29" which is where the engine is rated.
3. I calculated all the RPM's in 100 rpm increments for 18" MP. The error was from 1.32 to 1.41
I was really hoping to solve this problem because you can't get prop efficiency right without correct horsepower data. I have seen other formulas and an early version of a spreadsheet from Kevin Horton. I'm still looking for the magic bullet. I certainly don't have the answer.
__________________
H. Evan's RV7A N17HH 215 hours
"We can lift ourselves out of ignorance, we can find ourselves as creatures of excellence and intelligence and skill. We can be free! We can learn to fly!" J.L. Seagull
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06052010, 01:35 PM



Join Date: Dec 2006
Location: Arroyo Grande, CA
Posts: 938


Power curves
WOW! 32% to 41% error! I must have really goofed this time! I often do that when writing a posting without having set down my ideas and calculations beforehand! But somehow I'm not seeing where I went wrong on my initial analysis. Because my original posting was about an engine that produced rated power at 2700 rpm, and had a sealevel WOT MAP of 28.4", I got out my Lycoming curve #13381 on the 160 HP O320D, from Lycoming specification #2283G, and did the calledfor graphing of the lines at 8000', 22" MAP, and 2500 rpm. And without correcting for the manifold temperature difference, I got 115 HP, which is 71.9% power. That seems to be pretty close to the 71.7% I originally calculated! Did I draw my lines wrong on the graph? There must be something that I am missing. Will someone please enlighten me?

06052010, 04:45 PM




Join Date: Jul 2005
Location: Detroit, MI
Posts: 1,546


It's a puzzle
Quote:
Originally Posted by elippse
WOW! 32% to 41% error! I must have really goofed this time! I often do that when writing a posting without having set down my ideas and calculations beforehand! But somehow I'm not seeing where I went wrong on my initial analysis. Because my original posting was about an engine that produced rated power at 2700 rpm, and had a sealevel WOT MAP of 28.4", I got out my Lycoming curve #13381 on the 160 HP O320D, from Lycoming specification #2283G, and did the calledfor graphing of the lines at 8000', 22" MAP, and 2500 rpm. And without correcting for the manifold temperature difference, I got 115 HP, which is 71.9% power. That seems to be pretty close to the 71.7% I originally calculated! Did I draw my lines wrong on the graph? There must be something that I am missing. Will someone please enlighten me?

Maybe I'm in error. Your idea seemed correct to me. Try the link and see if I'm reading the Superior chart correctly. I have a hard time imagining that Superior and Lycoming could be much different. I can put the XLS on the webserver, too, if it will help. RSVP.
Remember, though, that I used the sea level chart and it is commonly held that you get more HP at higher altitudes. The Superior chart, if I'm reading it right, gives 107 HP for the IO360 (180 HP) at sea level, 22" x 2500 RPM. The 360 should give about 12.5% more HP for the same RPM and MP, right? So if 107 is right on my chart, it would be 95 HP for a 320. I'd love to find out where we are actually differing.
BTW, the Superior is rated at 29", not 28.4". If I am reading the right Lycoming chart, then sea level performance, 22 x 2500 is about 102 HP which reduces our difference to 7 HP (102  95), some of which may be the difference in the MP for rated power. 22 is a greater percentage of 28.4 than of 29 by enough to account for about 2 HP. That takes the difference down to 5 HP.
Here's the problem, though: The spreadsheet says that the Ellipse formula is most accurate at the highest power settings and the least accurate at the lowest ones. The example we are discussing is at the top end where errors would be smallest.
If a theory seems good and works in one test then the theory is supported but not proven. However, if it fails in a second, valid but different test, then it is not supported and, at a minimum, needs revision. That's where I am for this idea at this time.
I wish I had even a theory to explain why %RPM x %MP does not equal %PWR. Here is one possibility. The relationship between air density and air pressure is not linear. The table I have handy goes from sea level to 15000' but it shows that while pressure% changes from 100 to 56.43, the density% changes from 100 to 62.92. This is a 11.5% difference from linear. The trouble is, I think this goes in the wrong direction to explain the problem! Either way, though, it impinges on the theory.
__________________
H. Evan's RV7A N17HH 215 hours
"We can lift ourselves out of ignorance, we can find ourselves as creatures of excellence and intelligence and skill. We can be free! We can learn to fly!" J.L. Seagull
my website
Paid $25.00 "dues" net of PayPal cost for 2015 and 2016 and 2017.

06062010, 10:53 PM



Join Date: Dec 2006
Location: Arroyo Grande, CA
Posts: 938


Power
Howard, I see now where you are coming from; you want a formula which will give engine power over the full range of operation, not just at WOT. WOT is what I always use in my equations because I design a propeller for the maximum power, rpm, and speed at a given density altituder, and that is what I thought was what people want. But I can see where those who have a CS prop might want to see what their power is at various combinations of MAP and rpm. If their avionics are giving them percentage of power, they must have a formula with the elements of Ps, OAT, rpm, MAP, and density. On the engine power graphs put out by the engine makers, the righthand graph shows power from rpm and MAP vs altitude. whereas the lefthand page shows power at sealevel conditions from rpm vs MAP. On this one is shown the effect of operating the engine at less than WOT where the effect of throttling the engine and thus reducing its MAP vs sealevel pressure is shown. This shows the HP loss that occurs from the pumping loss across the throttle valve. I'm going to see if I can work up an allencompassing equation that will give the power with the differential pressure across the TV. I have a curvefit presently for my O235 with four coefficients that I made years ago that matches this left side, and I'll see if I can refine that to work with all engines. It goes: HP = (rpm X a + b)(MAP + c) + d where a=1.213E3, b=2.287, c=4.91, and d=71.05

06092010, 04:11 PM



Join Date: Dec 2006
Location: Arroyo Grande, CA
Posts: 938


Well, I think I may have come up with a reasonably simple and fairly accurate method of determining engine power from rpm and MAP. I did these calculations using engine data sheets from an O235 L2C, an O320B, and an O320D. Keep in mind that these engine power graphs are not exact, they are based on estimates from various tests. Look at the lines on the graph from an angle and on some data sheets most of the lines are parallel and some converge at a distant point, and some even have some lines parallel and one or two others angled! Go figure! But they never measured the engine horsepower in flight at all of these altitudes and at all of these rpm and MAP values. They were based on engineering estimates. The sealevel numbers, of course, were probably obtained from tests on engines in a test cell operated at these various rpm and MAP, and then corrected to sealevel MAP and temperature. Do you really think they waited for a 59F day at 29.9213" pressure, or had a sealed test cell where they maintained the temperature and pressure inside the cell to a very high accuracy? No, they used correction factors with the measured torque, rpm, pressure, and temperature just as do all of the engine makers with their dynos except the aircraft engine data was probably reduced with slide rules whereas all of the new dynos do it with builtin computers. As far as the formula for power, here's what you do: first subtract the MAP from the engine's sealevel MAP at rated power, then multiply this remainder by an enginerelated constant. Subtract this new value from the SLMAP and divide this by the SLMAP. Multiply this by the actual rpm and divide by the rated rpm. Multiply this by 100 to get percentage of power. You can further refine this for inlet temperature as given previously. For my O235, SLMAP is 28.4", rated rpm is 2800, and the constant is approximately 1.4. This constant can be calculated for each engine from the data sheet. This probably isn't the untarnished silver bullet (Lone Ranger's?) that Howard was looking for, but it probably comes very close, maybe as close as the original engineering estimates!

06092010, 04:53 PM




Join Date: Jul 2005
Location: Detroit, MI
Posts: 1,546


Thanks, I'll try that!
Paul, we may be the last two guys on earth willing to go that deeply into the subject, but I will, for certain, test this with the charts for my Superior 360. More to come. Thanks.
btw  this must apply only to "best power" mixture. And LOP works a whole different way as Walter Atkinson has posted on this forum before.
__________________
H. Evan's RV7A N17HH 215 hours
"We can lift ourselves out of ignorance, we can find ourselves as creatures of excellence and intelligence and skill. We can be free! We can learn to fly!" J.L. Seagull
my website
Paid $25.00 "dues" net of PayPal cost for 2015 and 2016 and 2017.

06092010, 11:26 PM




Join Date: Mar 2005
Location: Calgary, Canada
Posts: 3,594


Most of the WW2 engines were tested on a dyno and altitude cells as it was clear that altitude affected power output through reduced exhaust back pressure. You can add that to your list of variables of rpm, MAP, IAT, humidity and AFR, not to mention throttle angle (pumping loss variations and mixture distribution changes in OXXX engines), ignition timing differences and exhaust system differences.
With enough assumptions made, inaccuracies invariably creep in. While a tidy math model would be nice, there is no way to verify the theory without the two items listed above. Interesting discussion nevertheless.
I believe I read a long time ago that Lycoming did test some of their large engines in altitude cells, not sure about the atmo fours. Does anyone know for sure?

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