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Mach Number / Temperature Rise  
User currently offlineBlackbird From , joined Dec 1969, posts, RR:
Posted (9 years 4 months 4 weeks 1 day 9 hours ago) and read 6622 times:

Isn't there a formula for determining basically how the temperature will rise as the mach number increases?

If I recall, there was a 100 in the formula...

Anyone know the formula?

Andrea V. Kamarov

14 replies: All unread, jump to last
 
User currently offlineTimz From United States of America, joined Sep 1999, 6902 posts, RR: 7
Reply 1, posted (9 years 4 months 4 weeks 1 day 9 hours ago) and read 6617 times:

Temperature of what?

User currently offlineLeanOfPeak From United States of America, joined Oct 2004, 509 posts, RR: 1
Reply 2, posted (9 years 4 months 4 weeks 1 day 8 hours ago) and read 6607 times:

Do you mean stagnation temperature?

T0 = T [1 + (g+1)/2 M2]

The temperature that would result if the flow were slowed to zero velocity adiabatically?


User currently offlineNewark777 From United States of America, joined Dec 2004, 9348 posts, RR: 29
Reply 3, posted (9 years 4 months 4 weeks 1 day 8 hours ago) and read 6604 times:

ad·i·a·bat·ic Audio pronunciation of "adiabatically" ( P ) Pronunciation Key (d--btk, d--)
adj.

Of, relating to, or being a reversible thermodynamic process that occurs without gain or loss of heat and without a change in entropy.


 biggrin 

Harry

[Edited 2005-07-27 04:46:46]


Why grab a Heine when you can grab a Busch?
User currently offlineHT1000 From French Polynesia, joined Jun 2005, 40 posts, RR: 0
Reply 4, posted (9 years 4 months 4 weeks 1 day 8 hours ago) and read 6589 times:

Might be this one :

T = (TAS/100)² - (1/10).(X) with X = (TAS/100)²

T is not TAT but the difference between TAT and SAT. (In °C)
Works for TAS < 240 KT





 scratchchin 



Few Were Born With It. Even Fewer Know What To Do With It.
User currently offlineVikkyvik From United States of America, joined Jul 2003, 10335 posts, RR: 26
Reply 5, posted (9 years 4 months 4 weeks 1 day 8 hours ago) and read 6588 times:

I'm not really sure what you mean, but I'll give you the generic formula for speed of sound (which depends on the temp):

a = sqrt(gamma*R*T)

where:
a = speed of sound (in meters per second)
T = temp in Kelvin (zero degrees Celsius = 273.15 degrees Kelvin)

and specific to air:
gamma = adiabatic constant = 1.4
R = specific gas constant = 287


so for a temperature of 20 deg. C in air (68 deg F):

a = sqrt(1.4*287*293.15) = 343.2 m/s (about 773 mph)

~Vik



How can I be an admiral without my cap??!
User currently offlineLeanOfPeak From United States of America, joined Oct 2004, 509 posts, RR: 1
Reply 6, posted (9 years 4 months 4 weeks 1 day 7 hours ago) and read 6570 times:

Aaaagh...

g is a ratio of specific heats, so it is indeed unit-less.

PLEASE put J/kgK after R = 287.


User currently offlineBlackbird From , joined Dec 1969, posts, RR:
Reply 7, posted (9 years 4 months 4 weeks 19 hours ago) and read 6427 times:

I was basically getting at stagnation temps...

AVKam


User currently offlineVikkyvik From United States of America, joined Jul 2003, 10335 posts, RR: 26
Reply 8, posted (9 years 4 months 3 weeks 6 days 19 hours ago) and read 6381 times:

LeanOfPeak,

My fault - didn't mean to offend your sensibilities  Smile

I knew it wasn't unitless, but just didn't notice the omission (it's been awhile since I went over this stuff).

~Vik



How can I be an admiral without my cap??!
User currently offlineLeanOfPeak From United States of America, joined Oct 2004, 509 posts, RR: 1
Reply 9, posted (9 years 4 months 3 weeks 6 days 18 hours ago) and read 6375 times:

It's not so much my sensibilities. With the commonality of Imperial units in the aerospace sector, the omission of units introduces the possibility for substantial error, as it leaves the impression of a dimensionless constant applicable regardless of the set of units being used (i.e. Mach number, Reynolds number, Froude number, Fibonnacci number [That one's a joke, so no one take it seriously as a dimensionless constant, though it is a real term], Poisson's ratio, and lift, drag, and frictional coefficients).

You can ask the folks who set up the Mars Climate Orbiter regarding the consequences of that one.  Sad


User currently offlineAirmansv From United States of America, joined Apr 2005, 12 posts, RR: 0
Reply 10, posted (9 years 4 months 2 weeks 3 days 22 hours ago) and read 6273 times:

LeanOffPeak has the correct equation.

Total (stagnation) to Static Temperature equals
1+ (1/2 *(GAMMA-1))* Mn**2, where gamma equals Cp / Cv, and is 1.4 under standard conditions.

You cannot measure static temperature by the way when you are moving. It is the total temperature that you measure. Hence one calculates Mn from total to static pressure ratio equation, and use it in the above equation having measured total temperature to calculate static temperature.


User currently offlineLehpron From United States of America, joined Jul 2001, 7028 posts, RR: 21
Reply 11, posted (9 years 1 month 3 weeks 6 days 14 hours ago) and read 6004 times:

This has been buggin me since I learned of this equation:

Quoting LeanOfPeak (Reply 2):
Do you mean stagnation temperature?

T0 = T [1 + (g+1)/2 M2]

The temperature that would result if the flow were slowed to zero velocity adiabatically?

Regarding that, say a plane were moving at M5 at 80km (somehow) where the temperature is 165 Kelvin. Accordingly, the stagnation temperature should be 990 Kelvin. But the local density of air is a million times smaller than sea level. How does it get hot with almost no air out there?

Even the space just above the Earth has been measured to have an average of 4K, to the space shuttle, the stagnation average would be 104K @ M25 even when its almost a vacume out there!



The meaning of life is curiosity; we were put on this planet to explore opportunities.
User currently offlineVikkyvik From United States of America, joined Jul 2003, 10335 posts, RR: 26
Reply 12, posted (9 years 1 month 3 weeks 6 days 6 hours ago) and read 5955 times:

Quoting Lehpron (Reply 11):
Regarding that, say a plane were moving at M5 at 80km (somehow) where the temperature is 165 Kelvin. Accordingly, the stagnation temperature should be 990 Kelvin. But the local density of air is a million times smaller than sea level. How does it get hot with almost no air out there?

Even the space just above the Earth has been measured to have an average of 4K, to the space shuttle, the stagnation average would be 104K @ M25 even when its almost a vacume out there!

I'm not totally sure, but I'll take an educated guess. Remember that temperature is the average kinetic energy of the molecules, while heat is the total energy. So a very sparse atmosphere can have a high temperature, but very little stored heat energy as compared to a dense atmosphere. Also, with a sparse atmosphere, one needs to add very little heat to increase the temperature. So the overall kinetic energy that is lost when the flow is stagnated should be more than enough to raise the average kinetic energy (temperature) of a sparse atmosphere.

~Vik



How can I be an admiral without my cap??!
User currently offlineLeanOfPeak From United States of America, joined Oct 2004, 509 posts, RR: 1
Reply 13, posted (9 years 1 month 3 weeks 6 days 6 hours ago) and read 5949 times:

For starters, neither a shock wave nor friction is an adiabatic process.  Smile

User currently offlinePhollingsworth From United Kingdom, joined Mar 2004, 825 posts, RR: 5
Reply 14, posted (9 years 1 month 3 weeks 5 days 21 hours ago) and read 5934 times:

Quoting Newark777 (Reply 3):
ad·i·a·bat·ic Audio pronunciation of "adiabatically" ( P ) Pronunciation Key (d--btk, d--)
adj.

Of, relating to, or being a reversible thermodynamic process that occurs without gain or loss of heat and without a change in entropy.

Except that isn't the correct definition. That would be an isentropic process (entropy remains the same). Adiabatic is only the first part. All isentropic processes are adiabatic, but not all adiabatic processes are isentropic, think standard treatment of shock-waves (i.e. frozen flow). Shock-waves are inherently highly viscous phenomena and are often not truly adiabatic; however, in may cases treating them adiabatically results in estimations that are well within the acceptable range of error.


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