2.5: Problems
- Page ID
- 77950
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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:
- The values of \(A\) and \(B\), including their unities.
- Variation law of density and pressure with altitude, respectively \(\rho (h)\) and \(p (h)\) (do not substitute any value).
- 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}:
\[R = \dfrac{P_0}{\rho_0 T_0} = 333.3 \dfrac{J}{(Kg \cdot K)} \nonumber \]
- 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}:
\[300 = \dfrac{A}{1 + e^0} = \dfrac{A}{2} \to A = 600 \ K. \nonumber \]
Using the given temperature at an altitude \(h = 1000\) (\(T_{1000} = 250\ K\)), and Equation \ref{eq2.5.1}:
\[250 = \dfrac{A}{1 + e^{\tfrac{1000}{B}}} = \dfrac{600}{1 + e^{\tfrac{1000}{B}}} \to B = 2972\ m. \nonumber \] - 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}:
\[\rho = \dfrac{P}{RT} = \dfrac{P_0 e^{-\tfrac{g}{RA} (h + Be^{\tfrac{h}{B}} - B)}}{R \tfrac{A}{1 + e^{\tfrac{h}{B}}}}\label{eq2.5.11} \] - 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 \(B\):
- \(\rho (h = 1000) = 1.0756 \tfrac{kg}{m^3}.\)
- \(P(h = 1000) = 89632.5 Pa.\)
- The values of \(A\) and \(B\):