impossible question
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impossible question
I'm dealing with the JAA questionare for the ATPL and I don't succeed to answer this question:
At an altitude of 25000ft, the t° is -40C° and pressure is 375mb. The interval corresponding to 1 mb decrease in pressure is:...?
I'd like to know if there's a special formula to solve this question.
Thank you!!
At an altitude of 25000ft, the t° is -40C° and pressure is 375mb. The interval corresponding to 1 mb decrease in pressure is:...?
I'd like to know if there's a special formula to solve this question.
Thank you!!
Join Date: Oct 2000
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The pressure rise per feet is proportional to the air density. With 1 mb/30 ft at ground level, all you need is the air density in the given situation.
No, I'm nowhere near certified in any way but that's how I'd do it in my little engineering mind.
Cheers,
/Fred
No, I'm nowhere near certified in any way but that's how I'd do it in my little engineering mind.
Cheers,
/Fred
Join Date: May 2000
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suggestion
1mb is 30ft approx up to around 5000ft
after that it increases up to around 70ft
and the altitudes concorde flies at 55plus 1mb equates to around
200ft+!
So I eyeball the question!
of course I am wrong more often nowadays.
after that it increases up to around 70ft
and the altitudes concorde flies at 55plus 1mb equates to around
200ft+!
So I eyeball the question!
of course I am wrong more often nowadays.
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There is a formula. To solve this problem, two steps are necessary:
1. Finding air density (rho)
rho = p/(R x T), where
p= air pressure in mb
R=Gas-constant (=287,04 m^2 / s^2 x K)
T= Temperature in Kelvin (=°C + 273)
2. Applying the formula:
dh= -1 / ( g x rho), where
dh=interval corresponding to 1mb press.change expressed in METERS
g=acceleration (=9,80665 m/s^2)
rho=air density as solved above
If you would be very picky, you could also account for decreasing acceleration at 25000ft! Nevertheless this should yield quite accurate results.
1. Finding air density (rho)
rho = p/(R x T), where
p= air pressure in mb
R=Gas-constant (=287,04 m^2 / s^2 x K)
T= Temperature in Kelvin (=°C + 273)
2. Applying the formula:
dh= -1 / ( g x rho), where
dh=interval corresponding to 1mb press.change expressed in METERS
g=acceleration (=9,80665 m/s^2)
rho=air density as solved above
If you would be very picky, you could also account for decreasing acceleration at 25000ft! Nevertheless this should yield quite accurate results.
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Your FTO is using an out-of-date worksheet. This is a question that used to be required for the old CAA Instruments exam, but has been deleted from the syllabus of the JAA (Flight Instruments) part of the combined Instrumentation paper. The JAA do expect you know ball park figures for how many feet to a hectopascal, but only by remembering - ie, about 30 feet/hectopascal at 5000 feet, about 110 at 40,000 feet, etc.
The old CAA syllabus required you to know the following formula:
No of feet to a millibar at any altitude = 96 x T/ pressure in millilbars at that pressure level
where T = OAT in degrees Absolute (or Kelvin, if you prefer).
So, to take your example:
No of feet to a miilibar at 25,000 feet = 96 x 233/375
which, my trusty calculator tells me, is 59.648 feet per millibar.
Well, I must say, common sense and experience tends to suggest 60 is about right, so the formula must work.
The FTO where I work stopped teaching this about 2 years ago, when we realised from the feedback that the JAA were no longer asking the question in this form and we have dropped it from our worksheets. Also, it's no longer in the Learning Objectives. The CAA actually specifically sent out a circular saying that it had been dropped.
The old CAA syllabus required you to know the following formula:
No of feet to a millibar at any altitude = 96 x T/ pressure in millilbars at that pressure level
where T = OAT in degrees Absolute (or Kelvin, if you prefer).
So, to take your example:
No of feet to a miilibar at 25,000 feet = 96 x 233/375
which, my trusty calculator tells me, is 59.648 feet per millibar.
Well, I must say, common sense and experience tends to suggest 60 is about right, so the formula must work.
The FTO where I work stopped teaching this about 2 years ago, when we realised from the feedback that the JAA were no longer asking the question in this form and we have dropped it from our worksheets. Also, it's no longer in the Learning Objectives. The CAA actually specifically sent out a circular saying that it had been dropped.
Last edited by oxford blue; 23rd May 2002 at 20:01.
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These questions were designed to test your knowledge of the physics behind the mechanisms. Instrument Makers are now Instrument Technicians, (or Avionics LAEs!) and no longer repair and adjust altimeter mechanisms for a living and our own last true instrument basher retired a couple of years ago. Now there's no-one in the company who can repair mechanical instruments; all our pitot statics and gyros are returned to the manufacturer these days. Its still interesting to reflect on all the old formulae and details of how temperature and density changes were corrected, but there's really not much practical use for such knowledge any longer. Most instruments use force feedback mechanisms and corrections are made digitally in the associated computer. Does anyone recall the last time a pilot snagged an altimeter for moving too many feet when he changed the millibar setting?
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Through difficulties to the cinema
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Through difficulties to the cinema
