Manifold pressure - Altitude effects

[italia458]

This is not the standard question about how altitude affects manifold pressure.

I am wondering what happens to the required manifold pressure (and why), as you increase in altitude, if you want to maintain the same power? So, if at sea level, with 65% power set at 23"/2400RPM, and 20GPH -- what do I need to set the manifold to if I wanted to keep 65% power (and 20GPH) with 2400RPM at 8,000'? Assume standard atmospheric conditions.

I would assume that as long as you are moving the same amount (density) of air into the cylinders, at roughly the same temp, then you should require the same manifold setting. But, I do know that static pressure and atmospheric density don't decrease the same with an increase in altitude so I'm assuming there would be a slight difference. At 10,000' the density ratio is .7385 and the pressure ratio is .6877. The lower pressure ratio, compared to the density ratio, I believe might indicated that you require a lower manifold pressure at altitude to create the same power.

I noticed the cruise performance charts on a certain piston airplane also showed that the manifold pressure was generally lower, with an increase in altitude, to maintain the same % power. Is my hypothesis correct?

I'm looking for a detailed technical answer as to what and why, if anyone has some info on this that would be fantastic!

[ft]

It will be lower, due to the lower exhaust backpressure. IIRC, Aerodynamics for Naval Aviators has an excellent description of this. Hopefully not edited out in later editions in spite of kerosene torches having replaced Aeroplanes.

[oggers]

Good question

Is my hypothesis correct?

I believe it is. OAT is lower so for a given MAP the manifold air is denser, therefore a somewhat lower MAP will restore the air/fuel ratio. Of course, the throttle is still being opened, but not quite as much as it would be to maintain a constant MAP.

[ft]

But... the OAT is lower as the pressure is lower. As you raise the pressure in the induction system, the temperature will rise again. Now, is the atmosphere lapse rate with pressure higher or lower than what you get as you compress the air in the induction system? That determines whether the manifold air is warmer or colder than it was at a lower altitude, for the same MAP. I'm not going to venture a guess either way, but I'm willing to guess that the temperature difference won't be enough to matter.

-------- Quick-and-dirty and possibly poorly thought through thought experiment below - do not use for operating nuclear power plants, and ignore unless willing to risk subjecting yourself to gross errors - corrections welcome -------

Let's forget those pesky non-ideal properties of the atmosphere for a bit and play with the ideal gas law.

pV=nRT

p = nRT/V

n = pV/RT

Density (n) and pressure go hand in hand, unless the temperature changes. However, the temperature does change. Using your figures to calculate the ratio of T at altitude over T at SLS,

6877 = 7385*R*T/V

while SLS gives

1 = 1*R*T/V, or R*T/V = 1.

At altitude, that'd mean R*T/V = 6877/7385, or .931

R constant and V constant gives

T ratio = .931 * (273+15) K = 268 K = -5 C

Just what my CR-3 says it should be, in the standard atmosphere, so the ideal gas law seems to be valid to use for the lapse rate. Should be equally valid, within reasonable limits, as you bring p up again through the compressor of the turbo/supercharger giving you the same T in the manifold regardless of altitude. Hence, we need to look at other reasons for the power increase with altitude with constant MAP - and back pressure is one.

[italia458]

ft... A for NA is pretty good but I have found some inaccuracies in it. For example it says this, "With the exception of near closed throttle position, an increase in engine speed will produce an increase in manifold pressure."

As far as I know, a decrease in engine speed will produce an increase in manifold pressure! They also didn't explain preignition and detonation very well. They gave the impression that detonation was more dangerous than preignition. That's actually the opposite! You can run an engine with moderate detonation for quite a while, but if you have preignition it can ruin an engine in seconds!

[italia458]

I'm assuming a non-turbo engine since at sea level with the turbo not working the intake temps will be 'normal'. But when at altitude when the turbo is working, those hot gasses will definitely be affecting the temp!

You didn't use the equation properly. n is equal to the moles of the gas, not density. They're not the same thing at all. You basically calculated the relationship between temperature, density and pressure of the standard atmosphere. If you didn't round to -5 degrees your answer would be -4.8 degrees Celsius, which is precisely the temp at 10,000' for a standard atmosphere. Temp and pressure both affect density and that's all you showed.

Manifold pressure, under 'perfect' conditions, would have a maximum of 29.92" at SL. When the airplane climbs up, that maximum pressure would equal the outside atmospheric pressure. At 10,000' it would be 20.58". If we are setting 65% power with 2400RPM lets say MP has to be at 20" at SL. At 10,000', based on what I've seen in performance charts, the required MP for 65% power at 2400RPM, would be lower than 20". The only reason I can come up with for that difference is that as you climb in altitude, the pressure ratio decreases faster than the density ratio. Meaning, the density doesn't decrease directly proportional to pressure because of the lower temp at altitude, which would negate part of the pressure drop. Meaning, you don't need as high a pressure to get a certain density into the cylinders since the lower temp increases density.

I can see why one might think that lower back pressure would require less MP but I haven't seen any proof of it. It makes sense but I could only see it having a VERY small effect.

[ft]

italia458,
what hot gasses?

In a naturally aspirated engine, you have expansion affecting the temperature in the manifold. In a charged engine, you (usually) have compression. What's more 'normal', as far as temperature in the manifold goes?

In a constant volume, n is directly proportional to density.

I showed that the ideal gas law applies to the standard atmosphere as used by you, which was my intention. Knowing this, and assuming that the temperature drop or increase you see due to expansion or compression from ambient to MAP is also in accordance with the ideal gas law means the hypothesis about the density/pressure interrelationship being the reason for the lower required MAP is not an explanation for the observed phenomenon. You will end up with the same temperature in the manifold regardless of ambient.

(Disclaimer still applies though.)

As for back pressure, I can't really see that proof is needed for the engine using power when it is pumping air against a pressure. I'm not really interested in debating whether it is the explanation or not. Dig up a few more references for comparison?

Standing by for additional explanations.

Also, you're quoting Aerodynamics for Naval Aviators out of context.

Edit: Lower exhaust back pressure also means more complete evacuation, which for a given MAP means you get more of a charge into the combustion chambers = more BHP.

Cheers,
Fred

[italia458]

italia458,
what hot gasses?

In a naturally aspirated engine, you have expansion affecting the temperature in the manifold. In a charged engine, you (usually) have compression. What's more 'normal', as far as temperature in the manifold goes?

The relatively hot gasses that go into the intake manifold after passing through the intercooler. I can't see any difference between a charged or NA engine with regard to what happens in the intake manifold. In a charged engine compression happens in the turbine itself, before it goes into the manifold. I'm not sure what compression you're talking about in the manifold.

I meant 'normal' in that case, as in the same as a naturally aspirated engines since it isn't benefiting from the aid of the turbo yet.

In a constant volume, n is directly proportional to density.

It was just confusing since you made no mention of volume until later on - you had labeled (n) as Density. I understand what you did at the end.

Knowing this, and assuming that the temperature drop or increase you see due to expansion or compression from ambient to MAP is also in accordance with the ideal gas law means the hypothesis about the density/pressure interrelationship being the reason for the lower required MAP is not an explanation for the observed phenomenon. You will end up with the same temperature in the manifold regardless of ambient.


I don't understand that. If I have -20 degrees Celsius air in winter, it doesn't magically become the same temperature in the manifold as the 35 degrees Celsius air in summer! The change in temperature, I would say, would follow the ideal gas laws... but starting out at a lower temp means you will finish with a lower temp.

Maybe I'm not seeing what you're trying to say... any way of clarifying what you mean?

As for back pressure, I can't really see that proof is needed for the engine using power when it is pumping air against a pressure. I'm not really interested in debating whether it is the explanation or not. Dig up a few more references for comparison?

Reading over my post I can see I didn't clearly state my point... I agree it definitely does use power to pump against a pressure and it makes sense that it would require a lower MAP in this case. However, at SL (compared to 10,000'), the percentage of horsepower required to pump air out of the engine into an atmospheric pressure that's 4.6 PSI higher I think would be small. I do agree it would have an effect on MAP but I haven't seen any data on the amount of power required in this case so I can't really draw any definitive conclusions.

It's possible that both the lower back pressure and the fact that pressure decreases faster than density, with an increase in altitude, affect the required manifold pressure. But how much of each I'm not sure.

Also, you're quoting Aerodynamics for Naval Aviators out of context.

I didn't think I was. Here is the full paragraph. If you want to read more it's on page 138.

"Of course, the engine airflow is a function of RPM for two reasons. A higher engine speed increasesthe pumping rate and the volume flow through the engine. Also, with the engine driven supercharger or impeller, an increase in engine speed increases the supercharger pres- sure ratio. With the exception of near closed throttle position, an increase in engine speed will produce an increase in manifold pressure."

EDIT: I actually think what they were referring to was an engine outside of its governing range or one with a fixed pitch prop. If you increase the throttle, the RPM and MAP both increase. That makes sense now.

[Brian Abraham]

italia458, here is a chart from a normally aspirated engine, which is rated at 400 horsepower, that you can play "what ifs" with.

You'll note that at your nominated 23"X2400RPM the sea level output is roughly 262.5/263 horsepower, depending on how thick your pencil is. This calculates out to be 65.7% power. You will note also that 23" is only available up to about 6,800 feet, as manifold pressure drops 1" per 1,000 feet. With a standard temperature lapse rate 23"X2,400RPM at 6,800 feet will produce 295 horsepower. The 30 horsepower increase is due to reduced back pressure, as previously mentioned, but also the throttle is now wide open and the butterfly is not throttling the induction system ie pumping losses have been reduced to an absolute minimum.

Having only 22" available at 8,000 feet you would need to increase RPM to obtain your 65%. I haven't crunched the numbers, I'll leave that to you.

http://i101.photobucket.com/albums/m56/babraham227/z053-1.jpg

[bookworm]

It's not that hard to calculate the pumping loss. Expanding a volume of 720 in^3 = 1.18 x 10^-2 m^3 against a pressure of 10^5 Pa costs 1.18 kJ. At 2400 RPM, that's 80 pumps per second or about 100 kW or 130 hp. So if you reduce the ambient pressure from 1000 hPa to 800 hPa, you get 20% of that, or about 26 hp back as useful work. Close enough?

[italia458]

bookworm... that's what I'm looking for. What equations did you use for that?

[ft]

Work = Force * distance

Force = Pressure * Area

Work = Pressure * Area * distance

Swept volume = Area * distance

Work = Pressure * Swept volume

italia,
the different ratios of change of pressure and density are explained by the temperature lapse rate. That's what you see running it through the gas law as I did above. The pressure drops more than the density as you climb, but that's due to the temperature dropping as well - lapse rate, or the RT/V term with R and V kept constant. Compress it to the same MAP again and the temperature comes right back up, meaning trying to find the MAP difference in the difference between temperature and density ratios is chasing a red herring. In the manifold, conditions will be the same in the end once MAP is equal.

(Don't forget the disclaimer, not doing a lot of thinking here :))

Regarding the quote, there's no need to consider the prop for it to make sense. What the text is saying is that as RPM increases, the RPM of the supercharger (geared to the crankshaft) increases as well and delivers a higher MAP. You initially left out the bit about it applying to a supercharged engine.

[bookworm]

What equations did you use for that?

Equations? You make it sound fancier than it is! I guess it's

dU = -p dV

The trickiest bit is remembering that the cylinders expand twice per engine revolution. That's engineering, which is hard. ;)

[ft]

Hmmm... one exhaust stroke per cylinder per two revolutions though, now that you mention it.

[bookworm]

Hmmm... one exhaust stroke per cylinder per two revolutions though, now that you mention it.

You seeeee. I told you it was hard... ;) Does that mean I got my factor of two the wrong way up?

[ft]

By my reckoning, the factor of two should have been a factor of one half, leaving us at one quarter of the calculated power loss?

[aerobat77]

I believe might indicated that you require a lower manifold pressure at altitude to create the same power.


pretty right. thats because the exhaust back pressure is lower with lesser air density.

[Atlas Shrugged]

Take a look at this and a few of the other excellent articles on the site.

Pelican's Perch #15:<br>Manifold Pressure Sucks! (http://www.avweb.com/news/pelican/182081-1.html)

One thing worth remembering is that the ONLY thing that a manifold pressure gauge will tell you is THE AMOUNT OF AIR that is AVAILABLE TO THE ENGINE at the time....nothing more.

[ft]

And here I thought the MAP gauge indicated the pressure in the intake manifold... silly me! Now, where did I get that outrageous idea from? Would that amount be in mass or volume? :8

I have a feeling this picture will go a long way for those who got lost somewhere in the calculus:

http://www.mechadyne-int.com/vva-reference/images/pv-diagram-for-a-throttled-si-engine.png

Power delivered is the area of the power loop minus the area of the pumping loop.

Simple calculations of the pumping losses land in the ballpark when assuming the pumping pressure delta to be equal to the atmospheric pressure delta. If we included the fact that the exhaust gasses pass through an orifice on their way to atmospheric, I think we'd get close. IIRC, the combustion chamber pressure and the atmospheric pressure will be at a fixed ratio determined by the fixed mass flow through the exhaust valve, so the lowering of the exhaust back pressure will be multiplied in the calculation of the pumping loss in the exhaust cycle.

For example, lower the exhaust back pressure to 0.75 of atmospheric, 750 hPa instead of 1000. The exhaust stroke combustion chamber pressure would be, say, twice the exhaust pressure, so it'd go from 2000 hPa to 1500 hPa, giving a relatively larger reduction in pumping losses measured in kW/bhp.

I really hope someone who deals with this for a living will come along and set us straight eventually. That someone is, unfortunately, not me! Born 50 years too late for Real Engines. :(

[Atlas Shrugged]

the pressure in the intake manifold

Simply put: THE AMOUNT OF AIR THAT IS AVAILABLE TO THE ENGINE

[oggers]

:hmm: Maybe the question was not as well made as I first thought, but two of the constraints in the OP were constant fuel flow and constant RPM. Therefore I would suggest that pumping losses - changed though they may be - do not explain the need to adjust MAP whilst maintaining constant fuel flow.

Lycoming advise this on power setting:

"to maintain constant power, correct manifold pressure by approx 0.18"Hg for every 10F variation in induction air temperature from standard altitude temperature. Subtract manifold pressure for temperature below standard."

That sounds like a density correction to me.

[bookworm]

Well the OP wrote:

So, if at sea level, with 65% power set at 23"/2400RPM, and 20GPH -- what do I need to set the manifold to if I wanted to keep 65% power (and 20GPH) with 2400RPM at 8,000'?

The constraints don't actually make sense, as 20 GPH at 8000 ft with an equivalent mixture setting will give more power than 20 GPH at sea level. If you allow the mixture setting to be varied, to maintain 20 GPH by enriching the mixture, then there's a continuum of MP vs mixture settings that will meet those constraints.

[italia458]

Apologies for taking so long to reply!

Here is a 2 page copy from the Jepp JAA book on Powerplants.

Powerplant.pdf - File Shared from Box - Free Online File Storage (http://www.box.com/s/qg4lhvpxnfulbezdpnmf)

It explains how the decrease in temperature (with a constant MAP) and the reduced atmospheric pressure result in more power if a constant MAP and RPM are kept when at higher altitudes.

The constraints don't actually make sense, as 20 GPH at 8000 ft with an equivalent mixture setting will give more power than 20 GPH at sea level. If you allow the mixture setting to be varied, to maintain 20 GPH by enriching the mixture, then there's a continuum of MP vs mixture settings that will meet those constraints.

bookworm... 20 GPH is the mixture setting! Creating a specific power (ie. 45% 65% 75%, etc) requires a specific fuel flow for each. If you create 65% power at 2000', 4000', 12000', etc you will be burning the same fuel at each altitude. You will have a higher TAS at altitude because of less drag due to reduced density. I'm not sure how the constraints on my original post don't make sense.

[bookworm]

bookworm... 20 GPH is the mixture setting!

No, it's not the mixture setting, it's the fuel flow. While in many circumstances the fuel flow is a reasonable surrogate, it isn't here.

Creating a specific power (ie. 45% 65% 75%, etc) requires a specific fuel flow for each. If you create 65% power at 2000', 4000', 12000', etc you will be burning the same fuel at each altitude.

Not so. The pumping losses, or exhaust backpressure as it puts it in your extract on the second page, is a real effect that changes the efficiency of the engine. At higher altitudes, you get more power for the same fuel.

I'm not sure how the constraints on my original post don't make sense.

But I now agree with you on this, I think. The constraints are OK, they're just hard to deal with.

Start with "at sea level, with 65% power set at 23"/2400RPM, and 20 GPH" and climb. At 8000 ft there are two effects that change the power to give you more than 65%: one is the lower temperature, which means that more air is going in at the same MP. The other is the lower exhaust backpressure, which means that the engine is more efficient.

So you can reduce the power in two ways: you can reduce the MP to maintain 65% power, but that will reduce the fuel flow below 20 GPH. So to restore it to 20 GPH, you creep the mixture control forward at the same time, to make it less efficient (assume we're on the rich side of peak). That will increase the power, so you keep reducing the MP and enrichening the mixture until you get that 65% power and 20 GPH. You're correct in that only one combination of mixture and throttle will give both 65% power and 20 GPH. But it's not easy to calculate what the corresponding MP is.

[italia458]

No, it's not the mixture setting, it's the fuel flow. While in many circumstances the fuel flow is a reasonable surrogate, it isn't here.

Sorry, I'm not sure why I just wrote that! haha. Of course it isn't the mixture setting as mixture is a fuel/air ratio.

Not so. The pumping losses, or exhaust backpressure as it puts it in your extract on the second page, is a real effect that changes the efficiency of the engine. At higher altitudes, you get more power for the same fuel.

If you're feeding your engine a certain amount of fuel, you will be able to do a certain amount of work due to the energy contained in the fuel. When the engine is creating 65% power on the ground, it is putting say 50% power to the propeller and 15% for pumping losses. Those number are made up. When you go to 8000', if you're at the same fuel flow (20GPH) and mixture (0.08), you will create the same power. Now, two things to discuss:

1) Due to the decreased temperature, keeping the same manifold pressure will provide increased density air to the engine. Depending where the mixture setting currently is, that could create more power or less power. By reducing the MAP, you create the same power from the quantity of fuel.

2) Due to decreased back pressure at altitude, there are less pumping losses, resulting in more power available for the propeller. So now, it would be say 55% power to the propeller and 10% for pumping losses.

The following two reasons contribute to a higher TAS at altitude in this situation. 1) You are pushing more power to the propeller which is able to take a bigger bite of the air. 2) The decreased atmospheric density at altitude will create less drag for a given TAS.

The manufacturer could have chosen to relate the charts to what the actual power going to the propeller is but relating it to the quantity of fuel the pilot/hard-working-owner is pumping into the engine is probably easier to calculate and simpler for the pilot to understand.

So you can reduce the power in two ways: you can reduce the MP to maintain 65% power, but that will reduce the fuel flow below 20 GPH.

I don't believe so. Say, at SL, 20GPH, 0.08 fuel/air and 20" MAP you create 65% power. If at 8000', 0.08 fuel/air and 20" MAP you will be creating more power and to do that, you need more fuel flow. Both the power and fuel flow will be higher in that case. You could lean it back to 20GPH or lower, but that would change the fuel/air mixture. To get back to 65% power, you adjust the throttle (decreasing it to lower the MAP) and adjust the mixture so that when the mixture is equal to the one at SL, the fuel flow also equals 20GPH. At 8000' it would look something like this: 20GPH, 0.08 fuel/air, 19.1" MAP and 65% power.

But it's not easy to calculate what the corresponding MP is.

That's why the manufacturer's do it for us and put it on nice charts!

[ft]

As pointed out earlier, lower ambient temperature does not necessarily mean the temperature in the manifold (which is what matters) is any lower if the MAP is the same. It all comes down to the atmospheric lapse rate.

Fuel used does not equal fuel used in combustion.

[bookworm]

If you're feeding your engine a certain amount of fuel, you will be able to do a certain amount of work due to the energy contained in the fuel. When the engine is creating 65% power on the ground, it is putting say 50% power to the propeller and 15% for pumping losses. Those number are made up. When you go to 8000', if you're at the same fuel flow (20GPH) and mixture (0.08), you will create the same power....So now, it would be say 55% power to the propeller and 10% for pumping losses.

So I think this comes down to interpretation of what we mean by the "same power". If we're talking about "brake horse power", the pumping losses must be taken into account before measuring the power -- it's power at the brake, after all. But you're saying that the constraint is that the nominal power, before taking into account such losses, is the same at altitude as at sea level.

If you look at a typical engine nomogram (as Brian Abraham posted earlier in the thread), the process of finding the (brake horse) power consists of 4 steps:

1) Find the power at full-throttle altitude
2) Find the power at sea-level
3) Interpolate between the two
4) Then correct for non-standard temperature as a density correction formula

That implies to me that steps 1 to 3 include both a correction for temperature in the ISA and for pumping losses. If the latter were not included, why bother with the complex nomogram? You'd just apply the density correction formula as in step 4.

What is not clear to me is how to work out fuel consumption as altitude changes. Lycoming, for example, publishes a chart of fuel consumption as a function of "actual brake horsepower", RPM and "mixture strength" (i.e. a line for best power and a line for best economy). I assume that this is measured at sea level. No guidance seems to be given as to how that varies with altitude. The temperature will not affect the fuel consumption for a given BHP, since denser air just means more fuel for the same mixture setting. But the pumping efficiency will mean that more BHP are produced for the same fuel at altitude compared to sea level.

[italia458]

As pointed out earlier, lower ambient temperature does not necessarily mean the temperature in the manifold (which is what matters) is any lower if the MAP is the same. It all comes down to the atmospheric lapse rate.

Fred, I read over the thread again and I understand what you're saying. It makes sense that if you bring it back to the same MAP, then the temperature should be the same. And like you said, there could be a possibility of a difference in lapse rate but I agree, if there is a difference, it would be small. Not sure why I didn't see it before. If we're all on the same page, so far it seems that a reduced back pressure at altitude is the only reason for the increased power or lower MAP required?

It seems then that the JAA book is wrong on that matter?

Fuel used does not equal fuel used in combustion.

Can you expand on that. I think under most conditions the fuel would be completely burnt, however, not always inside the cylinder and not always providing an increase in power.

[oggers]

italia:

Fred, I read over the thread again and I understand what you're saying. It makes sense that if you bring it back to the same MAP, then the temperature should be the same.

I wouldn't be so quick to back-track. The question is; if you take an MAP at sea level, and then climb to 'x' grand and set the same MAP, will the air density in the manifold be the same as before? It won’t be unless by chance the inducted air is heated by just the right amount.

The environmental lapse rate is definitely not adiabatic, whilst the process of inducting air is not strictly adiabatic either: they are different processes that exchange energy with their surroundings at a different rate per unit pressure change, are they not? ft decided “the ideal gas law seems to be valid to use for the lapse rate” by using your figures that were already derived from the standard lapse rate! I thought you'd spotted that little error.

Merely altering the throttle to set the same manifold pressure as before is unlikely to result in the same density as before. To my mind that is a very uncontroversial concept that either someone or something has to account for if our performance data is to be valid.

If we're all on the same page, so far it seems that a reduced back pressure at altitude is the only reason for the increased power or lower MAP required? It seems then that the JAA book is wrong on that matter?


It doesn’t seem that way to me. That JAA book is certainly correct if taken in context. I’ve got the same thing in two different sets of official study material. Variation of exhaust back pressure and change of density in the manifold are both going on at the same time along with variation in throttle losses - the real variation in pumping losses.

I would certainly be hesitant to question credible texts on the basis of various ramblings in this thread.

[italia458]

I'm not sure any of us really have a good idea of what is happening or else we would have this solved by now! I am entertaining the thought that it could be different from what I initially thought.

I understand ELR to be the actual lapse rate that you would measure. I do agree that it varies with altitude but if we assume that the lapse rate is actually constant all the way up to the tropopause, what else would change? Dewpoint is also changing - usually dropping, which means that the quantity of water per cubic metre in the air is dropping. But is the quantity of water per cubic metre dropping because of the expansion - or is it something else? If it was exactly because of expansion, then the water quantity per cubic metre should remain constant at the same absolute pressure, thus not affecting the density.

to maintain constant power, correct manifold pressure by approx 0.18"Hg for every 10F variation in induction air temperature from standard altitude temperature. Subtract manifold pressure for temperature below standard.

The charts would be based on a standard atmosphere so this would correct for a non-standard atmosphere. Assuming the lapse rate remains constant (and all other factors constant), wouldn’t ft be correct in saying that if you keep the same absolute pressure, the temperature should be the same?

ft decided “the ideal gas law seems to be valid to use for the lapse rate” by using your figures that were already derived from the standard lapse rate! I thought you'd spotted that little error.

I didn't really show he was wrong, I just said that it didn't necessarily prove something. The standard lapse rate is based on that ideal gas law and that's what ft showed.

I'm no expert with this stuff, but I don't believe air moves 100% adiabatically in the atmosphere and I don't believe air will go from the dry adiabatic lapse rate, immediately to the saturated (moist/wet) adiabatic lapse rate as soon as any moisture is present. If you want examples, look at a METAR and then calculate the cloud base based on the dry adiabatic lapse rate and dewpoint lapse rate (2.5/1000') and you'll find that most of the time the cloud bases will be different than calculated. I’m currently studying meteorology in more depth now so I’m hoping to find more answers that might help answer this MAP scenario. There are articles out there that will explain why the SALR is not actually constant.

I would certainly be hesitant to question credible texts on the basis of various ramblings in this thread.

I'm not hesitant to question texts - they're only credible if they can stand up to questioning and prove they are correct. However, I am hesitant to discredit what a 'credible' text says. I do think the JAA books are credible from what I've seen so far...

[oggers]

italia:

I'm not sure any of us really have a good idea of what is happening or else we would have this solved by now!

Engineers and meteorologists have solved this. There is a variation in density (even for a given MAP), back pressure and throttle losses all going on. The principles are well described. You had the density bit right to begin with, it's just that you hadn't considered the other two things at that stage.

The standard lapse rate is based on that ideal gas law and that's what ft showed.

No. What ft showed was that if you take pressure and density from sea level and apply the ISA derived correction for a given altitude (which accounts for gravity, radiation, convection, humidity etc), when you put them back into the gas law, lo and behold you can get the ISA correction for temperature at that altitude!

As I said before: the environmental lapse rate and the inlet tract are different processes that exchange energy with their surroundings at a different rate. It doesn't matter whether you use standard conditions or real time conditions.

[italia458]

Engineers and meteorologists have solved this.

I meant the people in this thread!

There is a variation in density (even for a given MAP), back pressure and throttle losses all going on. The principles are well described.

Where are these principles 'well described'? If you know, please share!

You had the density bit right to begin with, it's just that you hadn't considered the other two things at that stage.

What are the two other things I had to consider, other than the density?

No. What ft showed was that if you take pressure and density from sea level and apply the ISA derived correction for a given altitude (which accounts for gravity, radiation, convection, humidity etc), when you put them back into the gas law, lo and behold you can get the ISA correction for temperature at that altitude!

Do you realize that's what I said? The 'ISA derived correction for a given altitude (which accounts for gravity, radiation, convection, humidity etc)', as you stated, would be the pressure and density at that altitude. If you plug that pressure and density change from SL into the ideal gas law, you'll get the temperature change at that altitude. You just expanded what I had said.

[oggers]

italia:

Do you realize that's what I said?

Er, that'll be why I wrote "I thought you'd spotted that little error" and "I wouldn't be so quick to back-track". The whole point is that since then you have changed position and said that ft's post made sense after all, and you are now questioning your own textbooks.

I meant the people in this thread!

Yes I realise that. The point I was making is that the understanding is already out there, including the reference you yourself gave. Meanwhile, two people in this thread - and you are one - are attempting to reinvent the wheel on this density topic to the extent that you are now questioning your own sources of knowledge.

The ref you linked to explains succinctly the change of density at constant MAP as one climbs. I'm at a loss to understand why you can't accept the explanation as it makes perfect sense. When one goes to altitude, for a given ambient pressure the density is higher. Ergo, when one goes to altitude, for a given MAP the density is higher.

What doesn't make sense is ft's hypothesis that if you set an MAP the density will also be set at any altitude.

So before we go any further can you just confirm which of those two makes sense to you? Because so far you have swapped between both. :ok:

[italia458]

The ref you linked to explains succinctly the change of density at constant MAP as one climbs. I'm at a loss to understand why you can't accept the explanation as it makes perfect sense. When one goes to altitude, for a given ambient pressure the density is higher. Ergo, when one goes to altitude, for a given MAP the density is higher.

Succinctly, yes but in exhaustive detail, no. just because something makes sense, does not mean that it is correct. I have seen countless times, things that are explained and 'make sense' but are either misleading or outright wrong.

I can take either stance. If I make assumptions, I can 'prove' either side. That should be obvious after reading this thread. However, it's also obvious on here that none of us really know what the answer is. You might actually have the right answer, oggers, but there isn't enough info presented here to definitely say which answer on here, if any, is correct.

My goal is to find the correct answer. If it seems that my initial 'guess' was wrong, I don't care. I'd rather find the correct answer than hold on relentlessly to my initial 'guess' and know I might not be correct.

You still didn't answer my question about where the principles are 'well described'. Or that other question I asked. Can you answer those two questions before we go on?

[oggers]

italia:

The reference you gave is correct on this. It's very uncontroversial. It requires only that you read it and understand it.

OTOH the opposing view being pushed by yourself and ft whereby the air density in the manifold will be the same at any altitude as long as you set the same MAP, is nonsense that violates one or both of the following ground truths:

a) From basic meteorology; that the atmosphere exchanges energy with its surroundings - ie ground and space - and the standard ELR is therefore not adiabatic.

b) From basic common sense; that air passing into a manifold will not (except by coincidence) exchange energy with the hot engine parts at the same rate as either the standard, or actual, environmental lapse rate - no matter where you put the throttle.

-----

I can take either stance. If I make assumptions, I can 'prove' either side. That should be obvious after reading this thread

Well, you certainly haven't proved this one:

so far it seems that a reduced back pressure at altitude is the only reason for the increased power or lower MAP required? It seems then that the JAA book is wrong on that matter?

...:ugh:

[italia458]

I don't think this conversation is helping answer any questions.

From basic common sense

Sometimes 'basic common sense' can lead you in the wrong direction. I've seen it many times. I don't disagree with what you're saying, however, I don't see definitive evidence either way. I think your definition of 'definitive evidence' is different than mine. Try and find a scientific report that has 'basic common sense' written somewhere in it.

Well, you certainly haven't proved this one:

so far it seems that a reduced back pressure at altitude is the only reason for the increased power or lower MAP required? It seems then that the JAA book is wrong on that matter?

I put question marks there for a reason.

[oggers]

What a lot of waffle.