Machmeter

[Radhaz]

Hurri6

If I'm correct, one of the questions you have is regarding the increased dynamic pressure at a higher speed?

Remember that total pressure is measured by bringing the air to rest - it is dynamic + static pressure. The dynamic pressure is so high because the air has lots of energy; it's this that is being measured (think molecules pushing against a piston). That's why it is do high in the case you quote.

On another point, even though the speed of sound is temperature dependent, we don't need it to work out Mach number. Both dynamic and static pressures are local ambient - that's the body of fluid you're in at the time of measuring.

Problems arise in the real world because measuring absolute static pressure accurately is very hard, especially with all those pesky shock waves knocking around...

I hope this helps and has not merely insulted your intelligence.

[keith williams]

If you are looking for something involving slightly less maths to explain how the mach meter works the following might help. My use of the word “proportional” is not strictly mathematically correct in a few places, but this doesn’t negate the general argument.


Pdyn = 1/2Rho V squared

Where Pdyn is dynamic pressure, Rho is air density and V is TAS

Rearranging this gives

TAS = square root ( 2 x Pdyn / Rho)

So TAS is proportional to Pdyn / Rho ……Equation 1.


Increasing Static Pressure (Pstat) compresses the air, causing its density to increase.

Increasing temperature expands the air, causing its density to decrease.

So we can say that Rho is proportional to Pstat / temperature.

Rearranging this gives

Temperature is proportional to Pstat / Rho ……Equation 2.


LSS is proportional to the square root of temperature

So using equation …2 we can say that

LSS is proportional to Pstat / Rho ……Equation 3.


We now have the following


TAS is proportional to Pdyn / Rho ……Equation 1.

LSS is proportional to Pstat / Rho ……Equation 3.


We also know that Mach number = TAS /LSS....... Equation 4




Combining equations 1, 3 and 4 we can say that

Mach Number is proportional to (Pdyn / Rho) / (Pstat / Rho)

This simplifies to give

Mach Number is proportional to Pdyn / Pstat


The Mach meter takes in total pressure and static pressure and uses two capsules and a mechanical linkage to measure the ratio of Pdyn/Pstat.

The various equations provided in previous posts detail precisely how Mach Number and Pdyn/Pstat are related. The relationship is not a simple linear one so the figures that you have quoted in your initial post are not correct.

[Hurricane6]

I appreciate the notes you make but this is the proof I've been looking for. Many thanks Mr Williams!

[Hurricane6]

I think you have made a couple of pertinent points. Thank you Radhaz

[Old Smokey]

Hurricane6,

It seems you're in the process of winding up your thread, a pity, I was enjoying it.

Before you wind it up, a word or two here which may shed light on your search for P/S, presumably P0 and Ps, as applicable to the Mach number calculation. P0 and Ps show up in all of the normal Mach number calculations, as highlighted below -

CAS to Mach Number (Given CAS and Ps for the Pressure Height)

M = SQR (5 * (( P0 / Ps * (1 + Vc^2 / a0^2 / 5)^3.5 - 1) + 1)^(2/7) - 1))

Mach Number to CAS (Given Mach No. and Ps for the Pressure Height)

Vc = a0 * SQR(5 * (( Ps / P0 * ((1 + M^2 / 5)^3.5 - 1) + 1)^(2/7) - 1))

CAS / Mach Changeover Height (Given CAS and Mach Number)

Ps / P0 = (((Vc / a0)^2 / 5 + 1)^3.5 - 1) / ((1 + M^2 / 5)^3.5 - 1)

Where -

a0 = ISA Sea Level Speed of Sound (661.4787442 Kt)
Vc = CAS in Knots
M = Mach Number
P0 = ISA Sea Level Pressure (1013.25 hPa)
Ps = Static Pressure at Flight Altitude (hPa)

So we see the Pressure Ratio (Ps/P0) in the second and third examples, and the inverse Pressure Ratio (P0/Ps) in the first.

Liked the thread Hurri6, keep them coming.

Best Regards,

Old Smokey

[galaxy flyer]

And to think I have ATPLs from two countries and wings from a well-known, if not loved, AF without knowing this stuff! I am humbled by the knowledge presented here.

GF

[Pugilistic Animus]

Ah now I see what Hurricane6 was asking

Nice equations...Old Smokey...
However the best answer to the question....
Q: The Machmeter...well how do it know?

A: Don't Worry...'They' know

[Old Smokey]

PA,

Well so long as somebody knows, I guess it's alright You seem to have a pretty good grip on these things, so I'm much assured

Check your PMs PA,

Best Regards,

Old Smokey

[Hurricane6]

Dear Pugilistic Animus,
I can see your account covers the derivation. Brilliant.
I'd like to better understand though; what is an isoentrope? The slope is on which graph?
Are P and V pressure and volume? If so of what? Can you give succinct reasoning for Sigma arriving in the developing expression?
Hurricane6

[18greens]

One thing I've always wondered is....

I understood the speed of sound is faster in denser environments so it varies with pressure not temperature. But he formula is based purely on temperature not pressure.

Is the speed of sound the same if the temp is static ( eg 15c) and the pressure varies eg 990 mb vs 1013mb?

[Pugilistic Animus]

I'd like to better understand though; what is an isoentrope? The slope is on which graph?

Hurricane6

The isoentrope is a region of constant entropy , this is an involved explanation dealing with the second law of thermodynamics I'm afraid the actual explanation is very involved I could explain.
but I'm not sure of your background, in order to pitch it to the correct level I don't mean in anyway to sound discouraging but if you haven't had calculus based physics or calculus.

it may be difficult to explain correctly as it doesn't really lend itself to a verbal analysis, the story is really best told mathematically.

I wrote that because I know there are many levels of understanding on pprune and I generally write on a few levels at once in light of this...are you an engineering student or science major? and what level of math and physics have you had?

Are P and V pressure and volume? If so of what? Can you give succinct reasoning for Sigma arriving in the developing expression?

P is pressure but the 'v' in that case represents 'velocity' those are Equations of the Bernoulli form in which the pressure is corrected adiabatic compression using the adiabatic gas laws in other word since P,T,Volume and density can all be related it is possible to have a function in which adiabatic gas laws can be substituted into Bernoulli equation in order to see how pressure varies with velocity changes giving the correction for adiabatic correction...thus you end up with Bernoulli's equation for compressible flow

The 'V' in the .5rhoV^2 equation is in terms of TAS--- TAS is the Equivalent Airspeed corrected for the altitude density ratio [not density altitude-that's different]...just as a very simple conceptual explan....in order to have the same number on the airspeed indicator...say 100 knots at 50' agl as at 10000 feet the speed would need to b increased because the distance between the molecules 'path length' is greater at high altitudes due to lower density that is the TRUE AIRSPEED... the correction factor sigma is Rho/Rho0 @ ISA...

I understood the speed of sound is faster in denser environments so it varies with pressure not temperature. But he formula is based purely on temperature not pressure.

Is the speed of sound the same if the temp is static ( eg 15c) and the pressure varies eg 990 mb vs 1013mb?

18greens:
The exact answer arises from 'kinetic molecular theory'- Mathytouches upon it nicely in his post

but in brief pressure and temperature are related to one another in a fixed relation however because there's a fixed relation between P and T every time one varies the 'P' there's a corresponding change in 'T' so the the temperature will not remain static [without an external heater] if the pressure changes...


It may be a while between responses,if there are further questions, up to a couple of weeks, I'm a little busy now...

in the mean time

You Tube

[fantom]

For Firq Saiq (Crew experienced in Arab lands only),

It's a needle going round a dial.

What more do you want?

[18greens]

Sorry pugilist, I'm not letting you get away with that explanation. True in a closed environment p varies with t. But in the real world a static t can exist with many different ps at the same altitude.

Does lss remain the same?

[Pugilistic Animus]

True in a closed environment p varies with t. But in the real world a static t can exist with many different ps at the same altitude.
It is also true in a local environment...a static T does not exist without a change in 'p' since T in a WAT curve for example every pressure altitude/temp combination leads to a corresponding ISA density altitude, the temp always varies with pressure and vice versa....
makes sense?

[Hurricane6]

Machmeter formula

Nice fireworks Pugilistic!
Look, I expect to be able to follow your steps using differentiation and integration provided your terms are all defined well.. If you wade into differential equations you'll lose me but I'm not expecting that to happen. I've been confronted with the thermodynamic laws in the past, yes. So have no fear... keep the English unambiguous and the steps complete and wish me luck with grasping it all. You've said a lot already I do realise and yes, due to what you've said I have inklings now, thanks. But when I was a student the physics master or tutor wanted to see all the steps... QED.

[Pugilistic Animus]

Sonic velocity:
Small pressure disturbances within a specific control volume [a discrete region within an open thermodynamic system i.e a local environment enclosing your wave] propagate at sonic velocity 'a' in air [generally sonic velocity in any medium is 'c'] this discussion does not apply to blast waves and strong shock waves

if one were to imagine themselves as a stationary observer, observing the control volume that contains the propagating wave front within a pipe one would see a steady state process and would one see that sound waves move as a longitudinal wave comprised of a series of rarefactions and compressions....

Now if one were to travel with the wave--- assuming that wave propagates left to right then the air [fluid] at the right of the control volume would appear to be moving toward the wavefront at sonic velocity 'a' and the air to the left will b moving away from at a speed a-dv ( were 'v' is the velocity at which the wave moves to the left away from the observer)---a rarefaction

Since fluid is composed of matter and matter has mass....and matter cannot be created or destroyed.... the mass at the left of the wave must equal the mass at the right so we achieve as mass balance of ml=mr

I can write this as rhoAa=(rho+drho)A(a-dv) A is the cross sectional area of the control volume within the pipe...since it appears on both sides, it cancels and one is left with...a*d(rho)-rhodv=0 ....1

So, one assumes that no heat or work crosses the wavefront boundary so there no change in potential energy within the propagating wave... so we achieve energy balance...if you recall the quantity 'H' from thermodynamics which represents total energy (U+e) where U=internal energy e is other types of energy and work---without going into the math--- energy balance Uleft =Uright...this is represented by dH-adv...which solves to H+a^2/2 =(h+dh)+(a-dv)^2 ...2

due to negligible changes in pressure or temperature...the wave propagation is adiabatic AND isoentropic (assumed)....if you recall that dG =h-TdS where 'S' in the entropy...then you'll see that for an isoentropic flow TdS=0....one ends up with dH =dP/rh0....3...the d is a script delta indicating partial differential equations...which I can't write here


Combining relations 1,2 and3 one obtains a^2 =(dP/dRho)s the little 's' indicates an isoentropic process ....using "Maxwell relations" I can actually write this as a^2=k(dP/drho)T where k=ratio of specific heats, so and of course recalling that P=rhoRT and using that to solve the highlighted differential equation a^2 =k(dP/drho)T =k[ drhoRT/drho]T.=kRT..note because this involves partial differential equations, I did not show the intermediate steps.... solving it one finally obtains

a =kRT^0.5

this explains not only why the speed of sound depends solely on temperature but it also explains why an isoentropic flow is assumed...it really doesn't get any easier than this...QED!

here's typical thermodynamics class...the lady exclaiming at the end is the one who got the 'A'

You Tube

[Pugilistic Animus]

on a lighter note check out my thread on fireworks in Jet Blast....
http://www.pprune.org/jet-blast/4875...ml#post7294648

[jcbmack]

Matter

Originally Posted by Pugilistic Animus

Sonic velocity:
Small pressure disturbances within a specific control volume [a discrete region within an open thermodynamic system i.e a local environment enclosing your wave] propagate at sonic velocity 'a' in air [generally sonic velocity in any medium is 'c'] this discussion does not apply to blast waves and strong shock waves

if one were to imagine themselves as a stationary observer, observing the control volume that contains the propagating wave front within a pipe one would see a steady state process and would one see that sound waves move as a longitudinal wave comprised of a series of rarefactions and compressions....

Now if one were to travel with the wave--- assuming that wave propagates left to right then the air [fluid] at the right of the control volume would appear to be moving toward the wavefront at sonic velocity 'a' and the air to the left will b moving away from at a speed a-dv ( were 'v' is the velocity at which the wave moves to the left away from the observer)---a rarefaction

Since fluid is composed of matter and matter has mass....and matter cannot be created or destroyed.... the mass at the left of the wave must equal the mass at the right so we achieve as mass balance of ml=mr

I can write this as rhoAa=(rho+drho)A(a-dv) A is the cross sectional area of the control volume within the pipe...since it appears on both sides, it cancels and one is left with...a*d(rho)-rhodv=0 ....1

So, one assumes that no heat or work crosses the wavefront boundary so there no change in potential energy within the propagating wave... so we achieve energy balance...if you recall the quantity 'H' from thermodynamics which represents total energy (U+e) where U=internal energy e is other types of energy and work---without going into the math--- energy balance Uleft =Uright...this is represented by dH-adv...which solves to H+a^2/2 =(h+dh)+(a-dv)^2 ...2

due to negligible changes in pressure or temperature...the wave propagation is adiabatic AND isoentropic (assumed)....if you recall that dG =h-TdS where 'S' in the entropy...then you'll see that for an isoentropic flow TdS=0....one ends up with dH =dP/rh0....3...the d is a script delta indicating partial differential equations...which I can't write here


Combining relations 1,2 and3 one obtains a^2 =(dP/dRho)s the little 's' indicates an isoentropic process ....using "Maxwell relations" I can actually write this as a^2=k(dP/drho)T where k=ratio of specific heats, so and of course recalling that P=rhoRT and using that to solve the highlighted differential equation a^2 =k(dP/drho)T =k[ drhoRT/drho]T.=kRT..note because this involves partial differential equations, I did not show the intermediate steps.... solving it one finally obtains

a =(kRT)^.5

this explains not only why the speed of sound depends solely on temperature but it also explains why an isoentropic flow is assumed...it really doesn't get any easier than this...QED!

here's typical thermodynamics class...the lady exclaiming at the end is the one who got the 'A'

http://www.youtube.com/watch?v=ndVhgq1yHdA

PA, what would apply to strong shock waves and blast waves, then? Thanks.

[Pugilistic Animus]

Blast waves from HEX are supersonic...the math is formidable. Not amenable to writing it on PPRuNe

[wiedehopf]

Originally Posted by Pugilistic Animus

Blast waves from HEX are supersonic...the math is formidable. Not amenable to writing it on PPRuNe

So you seem to check this forum and are more likely to read it than my response in the tech log.
Also no one will mind in here as it's not really busy.

Your private messages don't work because you have too many saved. You probably deleted the incoming messages but you need to delete the outgoing messages too.

Below the red bar is a small box with "Jump to Folder" written above it, the print is quite small.

Select "Sent Items".
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Have a nice day.