# Manifold pressure - Altitude effects

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**Manifold pressure - Altitude effects**

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!

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!

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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.

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Good question

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.

Is my hypothesis correct?

Join Date: Oct 2000

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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.

-------- 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.

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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!

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!

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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.

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.

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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

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

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

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

*Last edited by ft; 28th Jan 2012 at 20:23. Reason: Tone*

Thread Starter

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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?

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?

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.

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.

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?

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.

*"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.

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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.

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.

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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?

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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.

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.

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What equations did you use for that?

dU = -p dV

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

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

Pelican's Perch #15:<br>Manifold Pressure Sucks!

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.

Pelican's Perch #15:<br>Manifold Pressure Sucks!

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.

Join Date: Oct 2000

Location: N. Europe

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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?

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

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.

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

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.

*Last edited by ft; 30th Jan 2012 at 19:47.*