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2.5: Problems

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    Exercise \(\PageIndex{1}\) International Standard Atmosphere

    After the launch of a spatial probe into a planetary atmosphere, data about the temperature of the atmosphere have been collected. Its variation with altitude (\(h\)) can be approximated as follows:

    \[T = \dfrac{A}{1 + e^{\tfrac{h}{B}}},\label{eq2.5.1}\]

    where \(A\) and \(B\) are constants to be determined.

    Assuming the gas behaves as a perfect gas and the atmosphere is at rest, using the following data:

    • Temperature at \(h = 1000\), \(T_{1000} = 250\ K\);
    • \(p_0 = 100000 \dfrac{N}{m^2}\);
    • \(\rho_0 = 1 \dfrac{Kg}{m^3}\);
    • \(T_0 = 300\ K\);
    • \(g = 10 \dfrac{m}{s^2}\).

    determine:

    1. The values of \(A\) and \(B\), including their unities.
    2. Variation law of density and pressure with altitude, respectively \(\rho (h)\) and \(p (h)\) (do not substitute any value).
    3. The value of density and pressure at \(h = 1000 m\).
    Answer

    We assume the following hypotheses:

    (a) The gas is a perfect gas.

    (b) It fulfills the fluidostatic equation.

    Based on hypothesis (a):

    \[P = \rho RT.\label{eq2.5.2}\]

    Based on hypothesis (b):

    \[dP = -\rho gdh.\label{eq2.5.3}\]

    Based on the data given in the statement, and using Equation (\(\ref{eq2.5.2}\)):

    \[R = \dfrac{P_0}{\rho_0 T_0} = 333.3 \dfrac{J}{(Kg \cdot K)}\]

    1. The values of \(A\) and \(B\):
      Using the given temperature at an altitude \(h = 0\) \((T_0 = 300\ K)\), and Equation (\(\ref{eq2.5.1}\)):
      \[300 = \dfrac{A}{1 + e^0} = \dfrac{A}{2} \to A = 600 \ K.\]
      Using the given temperature at an altitude \(h = 1000\) (\(T_{1000} = 250\ K\)), and Equation (\(\ref{eq2.5.1}\)):
      \[250 = \dfrac{A}{1 + e^{\tfrac{1000}{B}}} = \dfrac{600}{1 + e^{\tfrac{1000}{B}}} \to B = 2972\ m.\]
    2. Variation law of density and pressure with altitude:
      Using Equation (\(\ref{eq2.5.2}\)) and Equation (\(\ref{eq2.5.3}\)):
      \[dP = -\dfrac{P}{RT} gdh.\label{eq2.5.7}\]
      Integrating the differential Equation (\(\ref{eq2.5.7}\)) between \(P(h = 0)\) and \(P, h = 0\) and \(h\):
      \[\int_{P_0}^{P} \dfrac{dP}{P} = \int_{h = 0}^{h} -\dfrac{g}{RT} dh.\label{eq2.5.8}\]
      Introducing Equation (\(\ref{eq2.5.1}\)) in Equation (\(\ref{eq2.5.8}\)):
      \[\int_{P_0}^{P} \dfrac{dP}{P} = \int_{h = 0}^{h} -\dfrac{g(1 + e^{\tfrac{h}{B}})}{RA} dh.\label{eq2.5.9}\]
      Integrating Equation (\(\ref{eq2.5.9}\)):
      \[Ln \dfrac{P}{P_0} = -\dfrac{g}{RA} (h + Be^{\tfrac{h}{B}} - B) \to P = P_0 e^{-\tfrac{g}{RA} (h + Be^{\tfrac{h}{B}} - B)}.\label{eq2.5.10}\]
      Using Equation (\(\ref{eq2.5.2}\)), Equation (\(\ref{eq2.5.1}\)), and Equation (\(\ref{eq2.5.10}\)):
      \[\rho = \dfrac{P}{RT} = \dfrac{P_0 e^{-\tfrac{g}{RA} (h + Be^{\tfrac{h}{B}} - B)}}{R \tfrac{A}{1 + e^{\tfrac{h}{B}}}}\label{eq2.5.11}\]
    3. Pressure and density at an altitude of 1000 m:
      Using Equation (\(\ref{eq2.5.10}\)) and Equation (\(\ref{eq2.5.11}\)), the given data for \(P_0\) and \(g\), and the values obtained for \(R, A\), and \(B\):
    • \(\rho (h = 1000) = 1.0756 \tfrac{kg}{m^3}.\)
    • \(P(h = 1000) = 89632.5 Pa.\)

    2.5: Problems is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by Manuel Soler Arnedo via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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