From here
Snoopy wrote:
LOP – fuel controls power
HP = 14.85 * FF
%power = HP/200
Where does the 14.85 come from? And why /200 and not 100?
EDIT: Nvm, I figured the 200 out myself. But still, where does the 14.85 come from?
EDIT2: Nvm, I got it, but shouldn’t it be 14.7?
hazek wrote:
but shouldn’t it be 14.7
I read 13.9 somewhere from Mike Bush. But I won’t look it up now, just that I put that in my phone as multiplicator. As far as I understand this depends mostly on the compression ratio and surely some other efficiency values. Standard compression ratio out of my head should be 8.5:1, isn’t it?
UdoR wrote:
As far as I understand this depends mostly on the compression ratio
I thought it was the stoichiometric air-fuel mixture for AVGAS?
Btw I apologize for the extremely off topic posts, perhaps this can be moved to another thread?
It’s not directly a physics law and depends purely on the engine’s efficiency (as well as the fuel used : specific energy and density).
HP = efficiency x fuel power = efficiency x mass flow x specific energy = efficiency x fuel flow x specific energy x volumetric mass.
100LL is 43.5 MJ/kg and has a density of 0.769 kg / L :
HP = efficiency x FF x 33.45 (MW/(L/s))
HP = efficiency x FF x 33.45 × 1341 (HP/MW) / 951 (gph/(L/s))
HP = efficiency x FF x 47.17
HP = 14.85 x FF if your engine is 31.5% efficient (which is close to the Carnot limit at 34%, the theoretical limit with a single stage thermodynamic machine).
There is also much variation in all of this :
So, take these exact numbers with a pinch of salt… but the order of magnitude sure looks correct and can be roughly applied to different engines (provided the proper operating conditions are met).
Isn’t 31.5% very optimistic as engine efficiency?
I would say from my old souvenirs from schools that it is a theoritical value, and reality is a bit less than 30%. It won’t be far either, but in cars we are usually far lesser, because engine speed is varying a lot and this value is ok in steady condition at optimal richness. Optimal richness is not always the case for plane as we may be in the red box…
My rule of a thumb is 1/10 of engine cubic inch size to lph… xO320 → 32l/h, xO360 —> 36 lph… you get the simple trick. I have to say the planes i’m flying doesn’t have FF indicators, some have faulty fuel gauges…
Thank you for the summary maxbc
greg_mp wrote:
My rule of a thumb is 1/10 of engine cubic inch size to lph
Another one with identical result is 1/5 of hp: 160hp = 32lph; 180hp = 36lph.
@greg_mp and @Capitaine, I think it’s possible the way the title was phrased by the mods when they moved my off topic post to a separate thread is leading you to completely misunderstand what this thread is about. At least your comments make no sense to me in relation to what this thread is about.
It’s about engine operation on the lean side of peak EGT, i.e. on the fuel scarce side of the stochastic air-fuel ratio, where the amount of power the engine produces is limited by fuel flow and not by available air as it is on the ROP. And @maxbc beautifully explained how that calculation is made.
The tittle of the thread should really be something like this: “Fuel flow to engine power formula when operating LOP”
UdoR wrote:
I read 13.9 somewhere from Mike Bush
Just listened to that presentation, he says for 7.5:1 compression ratio engines it’s 13.7 (most turbo engines) and for 8.5:1 compression ratio engines it’s 14.9. Higher compression ratio makes engines more efficient so it makes sense.
greg_mp wrote:
O320 → 32l/h, xO360 —> 36 lph… you get the simple trick
That’s funny. I feed an IO540 in front and a typical ROP power setting comes in fact close to 54 lph. Not that I use it that way, I seldom feed more than 40 lph to it (LOP).
