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Homework Help: Calculating height given pressures and temperatures

  1. Dec 25, 2012 #1
    1. The problem statement, all variables and given/known data
    On a certain day the barometric pressure at sea level is 30.1in.Hg, and the temperature is 70°F. The pressure gage in an airplane in flight indicates a pressure of 10.6psia, and the temperature gage shows the air temperature to be 46°F. Estimate as accurately as possible the altitude of the airplane above sea level.

    2. Relevant equations
    Here is the solution
    https://dl.dropbox.com/u/63664351/MATS2005/Capture2.PNG [Broken]

    3. The attempt at a solution
    https://dl.dropbox.com/u/63664351/MATS2005/Capture.PNG [Broken]

    1. I derived [tex] \frac{dP}{dy} = -\frac{PM}{RT} g [/tex] But on the given solution there is no M. How come?
    2. Why is T linear?
    3. Where it gets a=530 and b=-24/h
    4. Where it gets h=9192ft?
    Given P=10.6psia, P_0=30.1in.Hg and we know 29.92in.Hg=14.7psia=1atm
    then P_0=30.1*14.7/29.92=14.78843583psia
    [tex] \ln \frac{10.6}{14.78843583}=-0.3329915114 [/tex]
    [tex] \ln \frac{506}{530}=-0.04634033726 [/tex]
    [tex] \frac{g}{24*R}=0.04914629473 [/tex]
    [tex] h = 146.2120585 \text{m}= 480 \text{ft} [/tex]

    If you have different approach (except ISA), please do share.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 25, 2012 #2


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    The atmosphere has 80% nitrogen (mainly molecules with 2 N16-atoms), 20% oxygen (mainly molecules with 2 O16-atoms) and negligible traces of other gases.
    You can use the relative pressure to avoid this.

    It is not, but that assumption will give a reasonable approximation.

    Chosen to get T(0)=530 ("Fahrenheit above absolute zero"? US-units are weird) and T(airplane)=530+24.

    In the last equation, everything apart from h is known, you can solve it for h.
  4. Dec 25, 2012 #3


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    Ambient Temperature = 70 F
    Converting F to Rankine scale (where Abs. Zero = -460 F)
    then T = 70 + 460 = 530 F
  5. Dec 25, 2012 #4
    The quantity:

    [tex] \frac{g}{24*R}=0.04914629473 [/tex]

    You did not calculate this in consistent units. You were right about the M. The R in this equation needs to be divided by a factor of M. The R that should be used here is

    [tex]R = \frac{8.31}{29}\frac{J}{(K)(gm)}[/tex]

    You also need to multiply by 1000 to convert the Joules in the numerator from kg to grams.

    You also need to divide the 24 F by 1.8 to get degrees K. Once you make these corrections, you should get the right answer. This punctuates the need to pay careful attention to units when you solve a problem.
    Last edited: Dec 25, 2012
  6. Dec 26, 2012 #5
    Thanks you all for your responses!
    Still confused with conversion.
    For example
    [tex] \ln \frac{46°F}{70°F} ≠ \ln \frac{506°R}{530°R}[/tex]
    If the units inside ln are the same, then the number inside ln is without unit (the units cancel each other).
    Thus how come the calculations above are not equal?
    What unit should I use then? And how come unit that you're suggesting work?

    Following attachment is my calculations. I calculate it as precise as possible and round it only on my final answer. That being said mine differ significantly.
    Solution = 2801.722 m
    Mine = 2686.85 m
    Difference about 200 m is unacceptable, don't you think?
    And by International Standard Atmosphere that says
    [tex] T = a - 0.0065y [/tex] [tex] y_p= \frac{280.927778K - 294.261111K}{-0.0065} = 2051.282 \text{m} [/tex]
    https://dl.dropbox.com/u/63664351/MATS2005/units.png [Broken]
    Last edited by a moderator: May 6, 2017
  7. Dec 26, 2012 #6
    You are very close now. First of all, you can't express the ratio of the temperatures in degrees F, because the ideal gas law calls the temperatures to be absolute temperatures (R or K), not temperatures relative to some arbitrary reference (F or C). Secondly, you made a mistake in arithmetic: the 23 should be a 24 (like you correctly obtained when you did the problem the first time). See what you get when you make this correction.

    Finally, the temperature profile for the atmosphere changes from day to day, hour to hour, minute to minute, etc. The International Standard Atmosphere is an approximate average over time, and over the surface of the earth. Your problem, on the other hand, gives the temperature and pressure profiles at a specific location and time.
    Last edited by a moderator: May 6, 2017
  8. Dec 28, 2012 #7
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