ATPL theory questions
Joined: Apr 2004
Posts: 734
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From: London, GB
KayPam,
There is a similar question reported here and, as noted by numerous responses immediately above, the factor is completely irrelevant to the question, and the answer marked correct above is clearly wrong. Nevertheless, the figure is equivalent to the quotient of two compressibility factors f/f_0 using equations 11, 16, and 20, in NACA Report 837 (Aiken, W., 1946). The ratio depends only on pressure altitude and Mach number and is therefore independent of temperature deviation.
There is no Mach number in an incompressible flow because the speed of sound is infinite. Neither Mach meters nor airspeed indicators need to be "corrected for compressibility" so it does not necessarily follow that determining Mach number by a quantity such as `(Pt - Ps) / Ps` will result in an error. The Mach number can be related to the normalised pressure rise `(Pt - Ps) / Ps` using a simple algebraic expression without needing to adopt an incompressible flow assumption. The applicable technical standard for Mach meters (SAE AS8018 citing NASA Technical Note D-822) assumes an isentropic flow in relating pressure rise to Mach number. The same is true for airspeed indicators, although in their case the impact pressure q_c–the pressure rise–is normalised by standard sea-level pressure. Impact pressure (for M ≤ 1) is related to Mach number by:
M^2 = 5*(-1+(1+q_c/p)^(2/7)),
and to true airspeed v by:
v^2 = 7*(p/rho)*(-1+(1+q_c/p)^(2/7)), rho is ambient air density.
Because these equations are based on an isentropic model they account for variation in density as air is brought to rest in a pitot probe. The latter was first given in this form by Saint-Venant and Wantzel in Journal de l'École polytechnique, vol 16, 1839 (according to Stanton [link]).
In an incompressible flow the relationship for true airspeed squared is simply,
v^2 = 2*q/rho,
and q is dynamic pressure which is the difference between total and static pressures in an incompressible flow. This incompressible flow model is not used by Mach meters or airspeed indicators. Dynamic pressure may be thought of as a first order approximation of impact pressure - see for example the paragraph following eqn 2, and particularly figure 2, in NACA Report 247 (Zahm, A., 1926). The inverse relationships for impact and dynamic pressures are,
Isentropic flow: q_c = p*(-1+(1+(1/7)*(rho/p)*v^2)^(7/2)), and,
Incompressible flow: q = (1/2)*rho*v^2.
The Maclaurin expansion of p*(-1+(1+(1/7)*(rho/p)*v^2)^(7/2)) about v^2 = 0 is (1/2)*rho*v^2 which is the Euler-Bernoulli solution to Euler's equation for an incompressible flow.
The first use of a compressibility factor, of which I'm aware, in a circular slide rule for obtaining TAS from CAS, as in eqn 9 in NASA Technical Note D-822 or eqn 20 in NACA Report 837, is described in Ezra Kotcher's (of Bell X-1 fame) US patent 2342674 of 29th Feb 1944 for an Air Speed Computer. For on the influence of errors see NACA Technical Note 1605 (Huston, W., 1948), Accuracy of Airspeed Measurements and Flight Calibration Procedures.
And, if you compute EAS/CAS, you will find 0.94.
Mach number is measured from (Pt-Ps)/Ps.
So, say you don't have any device to correct your machmeter for compressibility (which is a problem in itself.. but let's assume that).
The machmeter will overread, due to compressibility.
So, say you don't have any device to correct your machmeter for compressibility (which is a problem in itself.. but let's assume that).
The machmeter will overread, due to compressibility.
M^2 = 5*(-1+(1+q_c/p)^(2/7)),
and to true airspeed v by:
v^2 = 7*(p/rho)*(-1+(1+q_c/p)^(2/7)), rho is ambient air density.
Because these equations are based on an isentropic model they account for variation in density as air is brought to rest in a pitot probe. The latter was first given in this form by Saint-Venant and Wantzel in Journal de l'École polytechnique, vol 16, 1839 (according to Stanton [link]).
In an incompressible flow the relationship for true airspeed squared is simply,
v^2 = 2*q/rho,
and q is dynamic pressure which is the difference between total and static pressures in an incompressible flow. This incompressible flow model is not used by Mach meters or airspeed indicators. Dynamic pressure may be thought of as a first order approximation of impact pressure - see for example the paragraph following eqn 2, and particularly figure 2, in NACA Report 247 (Zahm, A., 1926). The inverse relationships for impact and dynamic pressures are,
Isentropic flow: q_c = p*(-1+(1+(1/7)*(rho/p)*v^2)^(7/2)), and,
Incompressible flow: q = (1/2)*rho*v^2.
The Maclaurin expansion of p*(-1+(1+(1/7)*(rho/p)*v^2)^(7/2)) about v^2 = 0 is (1/2)*rho*v^2 which is the Euler-Bernoulli solution to Euler's equation for an incompressible flow.
The first use of a compressibility factor, of which I'm aware, in a circular slide rule for obtaining TAS from CAS, as in eqn 9 in NASA Technical Note D-822 or eqn 20 in NACA Report 837, is described in Ezra Kotcher's (of Bell X-1 fame) US patent 2342674 of 29th Feb 1944 for an Air Speed Computer. For on the influence of errors see NACA Technical Note 1605 (Huston, W., 1948), Accuracy of Airspeed Measurements and Flight Calibration Procedures.
Joined: Nov 2000
Posts: 4,328
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From: White Waltham, Prestwick & Calgary
I've lost the will to live already..... 
I've checked with a couple of other senior instructors, and they would expect the answer to be 480 knots if it was encountered in an Instruments exam.
I've checked with a couple of other senior instructors, and they would expect the answer to be 480 knots if it was encountered in an Instruments exam.
Joined: May 1999
Posts: 1,846
Likes: 4
From: Bristol, England
I, for one, am fascinated as selfin's explanation goes some way to explaining something that has worried me for some time. I have always found it difficult to reconcile the two statements:
1. On aircraft with Air Data Computers the ASI displays CAS
2. The Air Data Computer calculates CAS using variations of Saint Venant's formulae which assume compressible flow.
Because if #2 were true, the ADC would output what I know as EAS, CAS corrected for compressibility, although it still seems to be called CAS. Is this the case, selfin?
1. On aircraft with Air Data Computers the ASI displays CAS
2. The Air Data Computer calculates CAS using variations of Saint Venant's formulae which assume compressible flow.
Because if #2 were true, the ADC would output what I know as EAS, CAS corrected for compressibility, although it still seems to be called CAS. Is this the case, selfin?
Joined: Dec 2015
Posts: 507
Likes: 1
From: France
There is no Mach number in an incompressible flow because the speed of sound is infinite. Neither Mach meters nor airspeed indicators need to be "corrected for compressibility" so it does not necessarily follow that determining Mach number by a quantity such as `(Pt - Ps) / Ps` will result in an error. The Mach number can be related to the normalised pressure rise `(Pt - Ps) / Ps` using a simple algebraic expression without needing to adopt an incompressible flow assumption. The applicable technical standard for Mach meters
Nothing would prevent you from building a machmeter which would would exactly compute the mach from the following formula :
TAS=sqrt(2(Pt-Ps)/rho)
LSS=sqrt(gamma R T)
Rho=Ps/RT
TAS/LSS = sqrt(2(Pt-Ps)/Ps Gamma)
If you measure a higher Pt due to compressibility then yes you're going to need to correct it.
If you measure a CAS of 280kt at FL330, you're probably gonna want to correct something if you want to know your EAS.
In facts, you have to correct for compressibility if you use this above mentionned formula, simply because the formula is not valid for compressible flow.
For compressible flow, where a machmeter becomes useful, there are better suited formulas that you could use, and then you wouldn't need to correct for compressibility, because you avoided the need for it by choosing a better formula.
Then, I could agree that the answer marked as correct was actually wrong, i'm not in the head of the question's writers...
I was simply trying to find an explanation regarding this, without being able to tell whether I would answer 480 or 450 should this question appear on my official test (can we trust aviation exam or this other question bank when they tell us a surprising correct answer like this ?)
And from what you wrote after, I know you'll understand my message because you've written yourself the "better suited formula" I was talking about
Regarding the little technical details of real macheters, I really have no idea wheter early mach meters were able to mechanically reproduce this sort of complicated formula as written in wikipedia :
https://wikimedia.org/api/rest_v1/me...782ff6024667fa
To begin with I don't understand how you could divide a number by another with a system of mechanical gears and hands, so I don't see how you would elevate this ratio to the power 2/7
It is however very easy for modern ADC to use formulas as complicated as required thanks to IT.
Joined: Apr 2004
Posts: 734
Likes: 10
From: London, GB
Alex,
Those are equivalent statements and are, so far as I'm aware, required by SAE AS 8019 (and its successor 8019A) which is referenced by ETSO-C2d and US TSO-C2d. I don't have a copy of the SAE Aerospace Standard but I believe the USAF counterpart is MIL-PRF-27197E; table II uses impact pressure.
The second statement does not mean the ADC outputs EAS. If EAS is displayed as the primary airspeed by some ADC (fitted in a civil aircraft) then please provide a reference. The Saint-Venant solutions for compressible flow use impact pressure only. Constant CAS is equivalent to constant impact pressure. The Euler-Bernoulli solution for incompressible flow uses dynamic pressure only. Constant EAS is equivalent to constant dynamic pressure. The term "compressibility correction" is contentious in my opinion because it may suggest that EAS is more accurate than CAS. EAS clearly makes flight mechanics equations more tractable but it should be used with caution in the transonic regime.
KayPam,
The equation you would use was derived for a compressible medium and the only value Mach number can take in an incompressible flow is zero. Full understanding of the theory underlying the speed of sound proved surprisingly challenging in the history of science, defying such brilliant minds as Newton and Euler (it will be 2033 before all of Euler's work has been translated to English), and credit ultimately went to Laplace (Sur la Vitesse du Son dans l'air et dans l'eau, Annales des Chimie et de Physique 3, 238–241 (1816)). An excellent summary of the struggle to resolve this problem is given in Laplace and the speed of sound by Bernard Finn who was, like John D Anderson, a curator at the Smithsonian Institution. Anderson provides a straight-forward derivation in Fundamentals of Aerodynamics, section 8.3, which is very similar to this this two-page derivation plucked liberally from a search for "derivation of the speed of sound."
An incompressible flow would prevent it. It isn't clear how you would propose to "correct" a zero Mach number. Such a Mach meter belongs in Eric Laithwaite's Multiplcation of Bananas by Umbrellas.
... then you've used an airspeed indicator using Saint-Venant's solution for an isentropic flow. That is, dynamic pressure cannot be measured in flight, and any incompressible flow solutions using a measured isentropic total pressure will be in error (including the fabled Mach meter you've proposed). The proposal amounts to misinterpreting useful information (impact pressure) as an approximation (dynamic pressure), followed by a "correction" to recover the original information. Such a scheme may work at low speed and low altitude but it is inconsistent and unnecessarily restrictive. Indeed it would have been simpler to have used Saint-Venant's solution directly in the first place to obtain CAS and Mach number, which is what modern ASIs and Mach meters do.
That has already been achieved in purely mechanical instruments. There are numerous US patents outlining basic operations such as subtraction, division, and exponents can be dealt with by taking logarithms. You raise a very good question on precisely how Saint-Venant's solution can be implemented in analog ASIs and Mach meters.
1. On aircraft with Air Data Computers the ASI displays CAS
2. The Air Data Computer calculates CAS using variations of Saint Venant's formulae which assume compressible flow.
2. The Air Data Computer calculates CAS using variations of Saint Venant's formulae which assume compressible flow.
Because if #2 were true, the ADC would output what I know as EAS, CAS corrected for compressibility, although it still seems to be called CAS.
KayPam,
Nothing prevents you from calculating a mach number, from speed and temperature, even in an incompressible flow.
Nothing would prevent you from building a machmeter which would would exactly compute the mach from the following formula
If you measure a CAS of ...
It is however very easy for modern ADC to use formulas as complicated as required thanks to IT.
Joined: Dec 2015
Posts: 507
Likes: 1
From: France
Sorry for having overcomplicated things in the following messages...
First of all, I don't know why you're saying air would be uncompressible ? It is very much compressible when you squeeze it into a syringe or any kind of air pump or even a balloon. It also suffers waves of compression when sound travels through it. But even if we admit that LSS would be infinite in an incompressible flow :
Mach number is speed/speed of sound (TAS/LSS)
You're saying LSS is infinite in uncompressible flow... And you're probably saying that at low speeds the air is uncompressible.
Well, problem is sound does take time to travel even in still air. Even in water, a fluid that's almost uncompressible, sound takes time to travel.
So basically we have air at low speed : it's almost uncompressible so we can use Bernoulli's theorem. It is a tiny bit compressible so we can use the sqrt(gamma R T) = LSS formula. If speed increases, the LSS formula becomes "more correct" (according to you, you still have to explain why it would less correct at low speeds), but Bernoulli becomes less correct. And that's where the correction factor is needed.
But in any case, nothing prevents the engineer from performing computations that do not make perfect physical sense. One wants to compute a mach number as TAS/sqrt(gamma R T) ? No one can forbid him to do so.. Maybe tell them they're wrong but nothing more.
This is probably the sort of reasons why EASA ends up asking surprising questions like the one we saw.
It is probably the the same sort of reason that could lead to an EASA question like the one we saw yesterday.
Because except by using wrong formulas, I don't see why this 0.94 thingy would be required.
Correct equations allow us to forget about this kind of correction factor.
I was simply explaining the wrong physical reasoning leading to require a 0.94 correction factor.
The equation you would use was derived for a compressible medium and the only value Mach number can take in an incompressible flow is zero.
Mach number is speed/speed of sound (TAS/LSS)
You're saying LSS is infinite in uncompressible flow... And you're probably saying that at low speeds the air is uncompressible.
Well, problem is sound does take time to travel even in still air. Even in water, a fluid that's almost uncompressible, sound takes time to travel.
So basically we have air at low speed : it's almost uncompressible so we can use Bernoulli's theorem. It is a tiny bit compressible so we can use the sqrt(gamma R T) = LSS formula. If speed increases, the LSS formula becomes "more correct" (according to you, you still have to explain why it would less correct at low speeds), but Bernoulli becomes less correct. And that's where the correction factor is needed.
But in any case, nothing prevents the engineer from performing computations that do not make perfect physical sense. One wants to compute a mach number as TAS/sqrt(gamma R T) ? No one can forbid him to do so.. Maybe tell them they're wrong but nothing more.
This is probably the sort of reasons why EASA ends up asking surprising questions like the one we saw.
Such a scheme may work at low speed and low altitude but it is inconsistent and unnecessarily restrictive.
Because except by using wrong formulas, I don't see why this 0.94 thingy would be required.
Correct equations allow us to forget about this kind of correction factor.
I was simply explaining the wrong physical reasoning leading to require a 0.94 correction factor.
Joined: Apr 2004
Posts: 734
Likes: 10
From: London, GB
KayPam,
That means the bulk modulus K is finite. Increasing K results in higher speeds. In other words for an acoustic wave to have a finite travel time the volume of the material must change. The more pressure that is required to effect that change, the higher the speed.
In an incompressible flow (and medium) K is infinite which means the propagation speed is infinite, in turn reducing the Mach number to zero for all speeds.
Assuming an incompressible flow at low speeds is reasonable because compressibility effects are negligible while more accessible equations describe flow behaviour very well. From a rational choice perspective the fidelity reduction is rewarded by more tractable mathematics so that the simplifying assumption is rational. In the same vein it would be irrational to introduce a Mach number to most incompressible flow treatments.
To ensure consistency such a Mach number would need to be qualified as an artificial one. What would be the practical application of such a number?
The factor in the question was irrelevant but there is insufficient information for us to determine why it was given.
Even in water, a fluid that's almost uncompressible, sound takes time to travel.
In an incompressible flow (and medium) K is infinite which means the propagation speed is infinite, in turn reducing the Mach number to zero for all speeds.
So basically we have air at low speed : it's almost uncompressible so we can use Bernoulli's theorem. It is a tiny bit compressible so we can use the sqrt(gamma R T) = LSS formula. If speed increases, the LSS formula becomes "more correct" (according to you, you still have to explain why it would less correct at low speeds), but Bernoulli becomes less correct.
But in any case, nothing prevents the engineer from performing computations that do not make perfect physical sense. One wants to compute a mach number as TAS/sqrt(gamma R T) ? No one can forbid him to do so.. Maybe tell them they're wrong but nothing more.
Because except by using wrong formulas, I don't see why this 0.94 thingy would be required.
Joined: Dec 2015
Posts: 507
Likes: 1
From: France
I pass 33, 81 and 21 this morning.
However, there was an inconsistency in flight planning.
It said that contingency fuel should be 5% of the trip fuel, in the question.
However, we were given the cruise consumption of the aircraft and I was able to calculate that 5% of the trip fuel corresponded to 3 minutes of cruise consumption.
I thought there was a minimum of 5' fuel consumption for contingency ?
In this case, should we believe the question and consider that contingency is 5%, or apply the rules and consider it's 5' ?
Thanks
However, there was an inconsistency in flight planning.
It said that contingency fuel should be 5% of the trip fuel, in the question.
However, we were given the cruise consumption of the aircraft and I was able to calculate that 5% of the trip fuel corresponded to 3 minutes of cruise consumption.
I thought there was a minimum of 5' fuel consumption for contingency ?
In this case, should we believe the question and consider that contingency is 5%, or apply the rules and consider it's 5' ?
Thanks
Joined: Jun 2012
Posts: 1,175
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From: -
A jet aeroplane is flying long range cruise. How does specific range / fuel flow change?
A Decrease / decrease
B Decrease / increase [marked wrong]
C Increase / decrease [marked correct]
D Increase / increase
A Decrease / decrease
B Decrease / increase [marked wrong]
C Increase / decrease [marked correct]
D Increase / increase
Joined: Jan 2011
Posts: 660
Likes: 20
From: England
This is a very old question and you, like thousands of other students over the years, are making the mistake of comparing lrc with mrc. The question does not ask you to do this.
As time passes the aircraft becomes lighter, so the drag decreases. This increases the specific range and decreases the fuel flow.
The greatest surprise to me is that after all these years some of the questions banks do not explain this.
As time passes the aircraft becomes lighter, so the drag decreases. This increases the specific range and decreases the fuel flow.
The greatest surprise to me is that after all these years some of the questions banks do not explain this.
Joined: Jan 2011
Posts: 660
Likes: 20
From: England
The fact that many students make the mistake of assuming that the question requires a comparison between lrc and mrc, may be interpreted as evidence that the question is defective. But it is also possible that the real problem is the way in which many students are addressing the questions. There is actually nothing in the question to suggest that the lrc.mrc comparison is intended.
Students often make the same error with questions along through lines of *increasing flap angle in straight and level flight will?* many student pick answers such as *increase in lift* or *the aircraft will climb*. But these options ignore the clue *in straight and level flight*.
Although the lrc question has probably been deleted from the cqb by now, I believe that it should be retained in commercial question banks in order to illustrate the danger of misinterpreting questions.
Students often make the same error with questions along through lines of *increasing flap angle in straight and level flight will?* many student pick answers such as *increase in lift* or *the aircraft will climb*. But these options ignore the clue *in straight and level flight*.
Although the lrc question has probably been deleted from the cqb by now, I believe that it should be retained in commercial question banks in order to illustrate the danger of misinterpreting questions.
Joined: May 2017
Posts: 2
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From: Bahrain
I have question that really I can't find answer for it 
which is :
Q: During a pre-flight inspection it is noticed that the nose oleo strut extension is 2" instead of the
normal 6", but there is no sign of oil leakage. This indicates:
A. an internal oil leak
B. gas charge pressure too low
C. gas charge pressure too high
D. tire pressure too low
I am not sure but I think the correct answer is "B" PLEASE HELP ME
which is : Q: During a pre-flight inspection it is noticed that the nose oleo strut extension is 2" instead of the
normal 6", but there is no sign of oil leakage. This indicates:
A. an internal oil leak
B. gas charge pressure too low
C. gas charge pressure too high
D. tire pressure too low
I am not sure but I think the correct answer is "B"
PLEASE HELP ME
Joined: Jan 2011
Posts: 660
Likes: 20
From: England
Well, Let’s look at the options and see what we can deduce from them.
The length of the oleo is determined by the amount of oil and gas within it and the weight of the aircraft pressing down upon it. Any loss of oil or gas will reduce the length. Any excess of oil or gas will increase the length.
It is also worth noting that if the low gas charge pressure has been caused by a small gas leak, this leak will not be visually detectable. But an oil leak would be evident from the oil deposits in the area of the leak.
An internal (option A) leak would allow the oil and gas to mix, but would not change their total volume, so this would not change the oleo length. So option A is incorrect.
The length of the oleo is determined by the amount of fluid and gas which is held within it. So the tyre pressure (option D) is incorrect.
If the gas charge pressure is too high (option C) the oleo length will be greater than normal. So option C is incorrect.
If the gas charge pressure is too low (option B)the oleo will be shorter than normal, so this is the correct option.
The length of the oleo is determined by the amount of oil and gas within it and the weight of the aircraft pressing down upon it. Any loss of oil or gas will reduce the length. Any excess of oil or gas will increase the length.
It is also worth noting that if the low gas charge pressure has been caused by a small gas leak, this leak will not be visually detectable. But an oil leak would be evident from the oil deposits in the area of the leak.
An internal (option A) leak would allow the oil and gas to mix, but would not change their total volume, so this would not change the oleo length. So option A is incorrect.
The length of the oleo is determined by the amount of fluid and gas which is held within it. So the tyre pressure (option D) is incorrect.
If the gas charge pressure is too high (option C) the oleo length will be greater than normal. So option C is incorrect.
If the gas charge pressure is too low (option B)the oleo will be shorter than normal, so this is the correct option.
