Max Endurance, Max Range...
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Max Endurance, Max Range...
So I've been trying to get a more thorough understanding of some basic aerodynamics and in the process uncovered the following definition of L/D Max;
"At this point, the least amount of power is required for both the maximum lift and minimum drag. This will provide Max endurance, Max range and Best Glide Speed."
It is my understanding that Max Endurance comes at L/D Max.
Max range is 'slightly to the right' on the C_L, C_D graph, or, more specifically, 'at a velocity such that (C_L ^1/2) / (C_D)'.
The statement that doesn't seem quite true is the "This will provide Max Endurance, Max Range and Best Glide Speed."
Am I getting caught up in sloppy writing or is the quoted definition incorrect?
"At this point, the least amount of power is required for both the maximum lift and minimum drag. This will provide Max endurance, Max range and Best Glide Speed."
It is my understanding that Max Endurance comes at L/D Max.
Max range is 'slightly to the right' on the C_L, C_D graph, or, more specifically, 'at a velocity such that (C_L ^1/2) / (C_D)'.
The statement that doesn't seem quite true is the "This will provide Max Endurance, Max Range and Best Glide Speed."
Am I getting caught up in sloppy writing or is the quoted definition incorrect?
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It all depends on the shape of the propulsive efficiency curves for your particular kind of engine. For a modern turbofan max endurance (i.e minimum rate of fuel flow) occurs at a speed slightly lower than L/D max.
Some textbooks take a simplified approach to this issue, quoting a "rule" for propellors and another "rule" for jets, but high bypass turbofans really are somewhere between the two.
Some textbooks take a simplified approach to this issue, quoting a "rule" for propellors and another "rule" for jets, but high bypass turbofans really are somewhere between the two.
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Hey Lads,
After recently sudying ATPL aerodynamics and systems I think I may be able to help out here,
Best L/D ratio is described as the most lift for the least drag (BEST L/D), it is the same speed as BEST GLIDE SPEED and best ANGLE OF CLIMB that yields the MAX EXCESS THRUST. Becasue it is the speed where the least amount of drag occurs it means an aircraft can fly around all day at the lowest power seting to maintain level flight and so leads to MAX ENDURANCE. In other words it will be the longest time an aircraft can stay airbourne.
Best range occurs at 1.32 VIMD (velocity induced minimum drag). It is also the speed at which the aircraft will yields best excess power which means the speed flown is BEST RATE OF CLIMB and MAX RANGE. Some jet aircraft will fly slightly faster (5%) than 1.32VIMD at speeds called 99% max range speeds. They're called this because the 5% increase in speed only results in a 1% decrease in MAX RANGE.
Hope this explains things well enough for you.
Rocket
After recently sudying ATPL aerodynamics and systems I think I may be able to help out here,
Best L/D ratio is described as the most lift for the least drag (BEST L/D), it is the same speed as BEST GLIDE SPEED and best ANGLE OF CLIMB that yields the MAX EXCESS THRUST. Becasue it is the speed where the least amount of drag occurs it means an aircraft can fly around all day at the lowest power seting to maintain level flight and so leads to MAX ENDURANCE. In other words it will be the longest time an aircraft can stay airbourne.
Best range occurs at 1.32 VIMD (velocity induced minimum drag). It is also the speed at which the aircraft will yields best excess power which means the speed flown is BEST RATE OF CLIMB and MAX RANGE. Some jet aircraft will fly slightly faster (5%) than 1.32VIMD at speeds called 99% max range speeds. They're called this because the 5% increase in speed only results in a 1% decrease in MAX RANGE.
Hope this explains things well enough for you.
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Originally Posted by rocket66
Best L/D ratio is described as the most lift for the least drag (BEST L/D), it is the same speed as BEST GLIDE SPEED and best ANGLE OF CLIMB that yields the MAX EXCESS THRUST. Becasue it is the speed where the least amount of drag occurs it means an aircraft can fly around all day at the lowest power seting to maintain level flight and so leads to MAX ENDURANCE. In other words it will be the longest time an aircraft can stay airbourne.
You then go on to say that this speed, at which max C_L/C_D is achieved, is the speed for maximum endurance.
You are also assuming that having the engine at "lowest power setting" for level flight leads to maximum endurance. Maximum endurance is obviously related to lowest fuel burn per time period, which indeed is minimal power setting. For a propellor-driven airplane, that occurs at a maximal value of ((C_L)^(3/2))/C_D (Anderson, Introduction to Flight, eqn 6.27).
So, which is it? Max endurance at max C_L/C_D, or max endurance at max ((C_L)^(3/2))/C_D? You can't have both. Or can you?
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EW73 your correct mate, apologies for leaving that part out.
PBL...I'm not too sure what your asking, but I'll see if I can clarify a little.
Best or (MAX L/D) occurs always at a certain angle of attack. The angle of attack it occurs depends on many different factors dependant on aircraft type. If we are talking about Jet aircraft it gets a little more complex because engine efficiency becomes a factor.
A Jet engine works at its most efficient when it operates in whats called 'Design RPM'. Design RPM is found usually in the top 15% or the power range. If it operates anywhere outside this range, be it above or below, it will be less efficient.
So if a Jet aircraft needs to fly for max endurance, it needs to fly at its best/max L/D speed (angle) as high as possible, with the engines operating in the 'Design RPM' band.
If the same Jet aircraft wants to fly for best range it must also fly as high as possible, with the engines operating in the 'Design RPM' band and at a speed that is 1.32 times as fast as its best L/D speed. E.G if the best L/D speed is 100kts the best range speed will be 1.32 x 100 = 132Kts
If we are talking about Light piston aeroplanes then the principle is the same except the aircraft should be flown at its best L/D speed as high as possible (considering the heights they operate in) with what ever power setting can maintain that speed in level flight.
Hope that helps a little
rocket
PBL...I'm not too sure what your asking, but I'll see if I can clarify a little.
Best or (MAX L/D) occurs always at a certain angle of attack. The angle of attack it occurs depends on many different factors dependant on aircraft type. If we are talking about Jet aircraft it gets a little more complex because engine efficiency becomes a factor.
A Jet engine works at its most efficient when it operates in whats called 'Design RPM'. Design RPM is found usually in the top 15% or the power range. If it operates anywhere outside this range, be it above or below, it will be less efficient.
So if a Jet aircraft needs to fly for max endurance, it needs to fly at its best/max L/D speed (angle) as high as possible, with the engines operating in the 'Design RPM' band.
If the same Jet aircraft wants to fly for best range it must also fly as high as possible, with the engines operating in the 'Design RPM' band and at a speed that is 1.32 times as fast as its best L/D speed. E.G if the best L/D speed is 100kts the best range speed will be 1.32 x 100 = 132Kts
If we are talking about Light piston aeroplanes then the principle is the same except the aircraft should be flown at its best L/D speed as high as possible (considering the heights they operate in) with what ever power setting can maintain that speed in level flight.
Hope that helps a little
rocket
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Originally Posted by rocket66
PBL...I'm not too sure what your asking, but I'll see if I can clarify a little.
Originally Posted by Anderson, Introduction to Flight, Section 1.16.2
Maximum endurance for a propellor-driven airplane occurs when the airplane is flying at a velocity such that (C_L)^(3/2)/C_D is at its maximum
not C_L/C_D as you proposed. (Unless C_L is 1!!!!)
That means that your reasoning to your conclusion is mistaken (for you gave that reasoning for a generic airplane, so for example a propellor-driven one). It might be worth while for you to inquire where the mistake lies. My questions gave some hints.
PBL
PBL is in fact correct here Rocket, Max L/D will not necessarily giove yu Max endurence, though it will be in the same ballpark.
The simplest explanation I can recall is Max Endurence will occur at Max excess Thrust (and full throttle height for that condition), max range occurs at max excess power.
Is that more or less where the formula lead PBL?
The simplest explanation I can recall is Max Endurence will occur at Max excess Thrust (and full throttle height for that condition), max range occurs at max excess power.
Is that more or less where the formula lead PBL?
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rudderrudderrat Hi Ted,
Sometimes a couple of graphs are worth a thousand words.
experimentalaircraft.info
Sometimes a couple of graphs are worth a thousand words.
experimentalaircraft.info
have a look!
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Hi FE Hoppy,
I always thought you were a genius, but I didn't realise you were a Time Lord also.
The post is dated Jan 2010
I always thought you were a genius, but I didn't realise you were a Time Lord also.
Those swine have plagiarised my post on here from a few years ago!
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Originally Posted by Wizofoz
The simplest explanation I can recall is Max Endurence will occur at Max excess Thrust (and full throttle height for that condition), max range occurs at max excess power.
Is that more or less where the formula lead PBL?
Is that more or less where the formula lead PBL?
You have a given amount of energy in your fuel tank. Energy is force x distance, and power is energy per unit time.
Let's apply a constant thrust to attain a constant velocity. Then
Energy = Force x distance = Force x Velocity x Time = constant
because that energy represents a fixed amount of fuel sitting in the tank.
If you want to maximise Time for this given energy, you need to minimise Force x Velocity, which is power. So maximising endurance is minimising power (P) required to hold level flight.
If you want to maximise distance for this given energy, the distance is Velocity x Time, so you need to minimise Force. So maximising range is minimising Force required to hold level flight.
Force is required thrust, T, and since you are constant velocity, this must equal drag, D. Since you are level, lift L must equal weight W. So T/W = D/L, trivially, which means T = W / (L/D). W is given constant (except for fuel burn, which we can ignore here), so minimising T means maximising L/D. So maximum range is achieved at maximum L/D.
Minimising P is a little trickier, although it is still easy algebra:
P = Thrust x Velocity = (W / (C_L/C_D) ) x V,
using the expression for T above and the relation L/D = C_L/C_D. But now
L = 1/2 x rho x V^2 x S x C_L = W
where rho and V are the freestream density and velocity, and S the wetted area. This is direct from the definition of C_L. So, rearranging,
V = sqrt( (2 x W) / (rho x S x C_L) )
and substituting into the above expression for P gives
P = (W / (C_L/C_D) ) x sqrt( (2 x W) / (rho x S x C_L)
= sqrt( (2 x W^2 x (C_D)^2) / (rho x S x (C_L)^3)
of which the variable part is C_D / (C_L)^(3/2). So minimising this is maximising (C_L)^(3/2) / C_D.
I think the reason this result is proposed for "propellor-driven" aircraft is that, obviously, incompressible dynamics is being used here; no compressibility corrections.
PBL
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Gday PBL
The way I have explained it is the way it had been explained in the text I used to sit the ATPL exam. I'll go back and review it at some stage and see if I can see where I may have made a mistake.
I must put in I have not been lucky enough to fly jets yet but tech logs like this one will certainly help in the interviews getting there.
Will let you know
rocket
The way I have explained it is the way it had been explained in the text I used to sit the ATPL exam. I'll go back and review it at some stage and see if I can see where I may have made a mistake.
I must put in I have not been lucky enough to fly jets yet but tech logs like this one will certainly help in the interviews getting there.
Will let you know
rocket
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The problem with "theoretical" approaches to this question is that they have to make simplifying assumptions about engine and propulsive efficiency.
For example, the "power out" of a propulsion unit is thrust*TAS. Fuel flow corresponds to "power in". If the efficiency of converting fuel flow to power out (i.e combination of engine efficiency and propulsive efficiency) was constant over a range of speeds then you could say that minimum fuel flow would correspond to minimum (drag*TAS). But can you be sure that efficiency is constant even for a piston engine - prop?
For jets the common assumption is that they follow actuator disc theory and therefore propulsive efficiency increases with speed. This leads to the assumption that fuel flow is proportional to thrust and therefore min fuel flow corresponds to min drag. But again, is this an accurate assumption for a modern turbofan?
The only accurate solution is to use the graphs supplied by the manufacturer.
For example, the "power out" of a propulsion unit is thrust*TAS. Fuel flow corresponds to "power in". If the efficiency of converting fuel flow to power out (i.e combination of engine efficiency and propulsive efficiency) was constant over a range of speeds then you could say that minimum fuel flow would correspond to minimum (drag*TAS). But can you be sure that efficiency is constant even for a piston engine - prop?
For jets the common assumption is that they follow actuator disc theory and therefore propulsive efficiency increases with speed. This leads to the assumption that fuel flow is proportional to thrust and therefore min fuel flow corresponds to min drag. But again, is this an accurate assumption for a modern turbofan?
The only accurate solution is to use the graphs supplied by the manufacturer.
Last edited by Rivet gun; 14th Oct 2010 at 22:50.
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Maybe some simplification is in order here....max endurance is about staying up in the air the longest amount of time...max range is about going the farthest distance..
Both require max L/D but the dif is in the engines....if you can hang over an island, or weather, flying at say130kt going around in circles burning the least amount of fuel...that's going to be your max endurance......on the other hand, max range will be some power/speed ratio gives you most forward speed for the least amount of fuel burned...
Both require max L/D but the dif is in the engines....if you can hang over an island, or weather, flying at say130kt going around in circles burning the least amount of fuel...that's going to be your max endurance......on the other hand, max range will be some power/speed ratio gives you most forward speed for the least amount of fuel burned...
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i know i am getting something seriously wrong here, so hoping you guys can help me clarify something.
does fuel flow increase with thrust? i would assume it increases right?
then if thats the case, if SFC is the ratio of the amount of fuel used to the amount of thrust produced, then as thrust increases, fuel flow would increase also, but thrust will increase more in proportion hence the SFC ratio falls? is that right? so there would be a point where the increase in fuel used is offset by the biggest gain in thrust, hence the best SFC?.
then if thats the case, if flying for max endurance, if flying at the lowest possible thrust setting to maintain SL flight will give relatively less fuel flow, then why fly higher? isn't the engine having to work harder to produce the same amount of thrust than at low altitudes - so fuel flow will increase as the engine is trying to produce the same amount of thrust, but has to work harder due to the less dense air? isn't the optimum operating range just where SFC is achieved, but not necessarily where the least fuel flow is achieved as a given quantity?
sorry if the answer is really obvious to you guys, but i am hoping to learn more, and this is something that is getting me a bit confused!
thanks!
does fuel flow increase with thrust? i would assume it increases right?
then if thats the case, if SFC is the ratio of the amount of fuel used to the amount of thrust produced, then as thrust increases, fuel flow would increase also, but thrust will increase more in proportion hence the SFC ratio falls? is that right? so there would be a point where the increase in fuel used is offset by the biggest gain in thrust, hence the best SFC?.
then if thats the case, if flying for max endurance, if flying at the lowest possible thrust setting to maintain SL flight will give relatively less fuel flow, then why fly higher? isn't the engine having to work harder to produce the same amount of thrust than at low altitudes - so fuel flow will increase as the engine is trying to produce the same amount of thrust, but has to work harder due to the less dense air? isn't the optimum operating range just where SFC is achieved, but not necessarily where the least fuel flow is achieved as a given quantity?
sorry if the answer is really obvious to you guys, but i am hoping to learn more, and this is something that is getting me a bit confused!
thanks!
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TedUnderwood, rocket66, johns7022,
it is well to keep in mind that there appear to be texts out there with significant elementary mistakes in them.
It is usual for texts, even academic engineering texts, to have mistakes in them, but successful texts such as Anderson go through many editions (Anderson is on his sixth) and there is much more chance that the mistakes will have been pointed out by users and corrected. What I understand is that there are study texts out there for practical pilot training which have significant mistakes in them, which are not corrected over time. It is probably wise to be aware of this!
The math as I gave it is elementary, and thus acts as a good guide to reality.
As Rivet gun points out, it is not a perfect guide to the question posed. What it relates, accurately enough, is the amount of required power and thrust to the conditions which minimise them for a given quantity of energy.
What it doesn't deal with is the amount of fuel a given engine will need to burn to achieve that required thrust, respectively power. An engine can be considered as a converter, which converts latent energy (in the fuel) into, say, force. We all know that an engine will need to use more energy than comes out, because a certain amount is lost in the conversion process. And if that relation between potential energy in and force out is not relatively simple, then using force out, respectively power out, as a proxy for energy in, as my math did, could be somewhat misleading.
(Unfortunately, Rivetg phrases this using the old trope of "theoretical" versus - what? - "practical", I presume. Let me warn against taking this to have much real meaning. For most people, "theoretical" is "what I don't understand" (often math) and "practical" is "what I do understand". That is OK as long as it is regarded as a personal attribute, and not something objective. But for people such as myself who actually use these intellectual tools on a daily basis to get answers to real questions, it is a more or less meaningless distinction. When you have a nut to loosen, you go find the right spanner, and when you use it, you have to keep in mind that if the fit is too loose, you might round the edges of the nut and make it difficult to loosen even with a tighter-fit spanner. No one explains this caveat by saying "the problem with theoretical approaches is that you might round the corners of the nut". They say "watch out that you don't round the nut because of a loose fit." So, in this case, watch out that the efficiency characteristics of the engine doesn't affect the result too much. As Rivetg says, in a practical case it is best to use manufacturer's data.)
orangeboy,
it seems to me your considerations about SFC and burn are appropriate. But they do not lead you by themselves to an answer to the question originally posed. You are missing out the aerodynamic component, which I concentrated on. As Rivetg points out, you need both engine efficiency (relation between fuel-energy in and force, resp power output) and aerodynamics (relation between force, resp. power required and range, resp. endurance) to get an answer as to how far, resp. long you can go with a given amount of fuel.
PBL
it is well to keep in mind that there appear to be texts out there with significant elementary mistakes in them.
It is usual for texts, even academic engineering texts, to have mistakes in them, but successful texts such as Anderson go through many editions (Anderson is on his sixth) and there is much more chance that the mistakes will have been pointed out by users and corrected. What I understand is that there are study texts out there for practical pilot training which have significant mistakes in them, which are not corrected over time. It is probably wise to be aware of this!
The math as I gave it is elementary, and thus acts as a good guide to reality.
As Rivet gun points out, it is not a perfect guide to the question posed. What it relates, accurately enough, is the amount of required power and thrust to the conditions which minimise them for a given quantity of energy.
What it doesn't deal with is the amount of fuel a given engine will need to burn to achieve that required thrust, respectively power. An engine can be considered as a converter, which converts latent energy (in the fuel) into, say, force. We all know that an engine will need to use more energy than comes out, because a certain amount is lost in the conversion process. And if that relation between potential energy in and force out is not relatively simple, then using force out, respectively power out, as a proxy for energy in, as my math did, could be somewhat misleading.
(Unfortunately, Rivetg phrases this using the old trope of "theoretical" versus - what? - "practical", I presume. Let me warn against taking this to have much real meaning. For most people, "theoretical" is "what I don't understand" (often math) and "practical" is "what I do understand". That is OK as long as it is regarded as a personal attribute, and not something objective. But for people such as myself who actually use these intellectual tools on a daily basis to get answers to real questions, it is a more or less meaningless distinction. When you have a nut to loosen, you go find the right spanner, and when you use it, you have to keep in mind that if the fit is too loose, you might round the edges of the nut and make it difficult to loosen even with a tighter-fit spanner. No one explains this caveat by saying "the problem with theoretical approaches is that you might round the corners of the nut". They say "watch out that you don't round the nut because of a loose fit." So, in this case, watch out that the efficiency characteristics of the engine doesn't affect the result too much. As Rivetg says, in a practical case it is best to use manufacturer's data.)
orangeboy,
it seems to me your considerations about SFC and burn are appropriate. But they do not lead you by themselves to an answer to the question originally posed. You are missing out the aerodynamic component, which I concentrated on. As Rivetg points out, you need both engine efficiency (relation between fuel-energy in and force, resp power output) and aerodynamics (relation between force, resp. power required and range, resp. endurance) to get an answer as to how far, resp. long you can go with a given amount of fuel.
PBL
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That 5% increase in speed is the difference between MRC Max Range Cruise and LRC Long Range Cruise.
ew73
ew73