Pressure in atmosphere - Revision history

Mol: /* Barometric formula */

2023-08-30T15:14:24Z

Barometric formula

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	== Barometric formula ==		== Barometric formula ==

			https://en.wikipedia.org/wiki/Barometric_formula

	Models how the pressure of the air changes with altitude with linear temperature change.		Models how the pressure of the air changes with altitude with linear temperature change.

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= NRLMSISE-00

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	https://en.wikipedia.org/wiki/U.S._Standard_Atmosphere		https://en.wikipedia.org/wiki/U.S._Standard_Atmosphere

	=== NRLMSISE-00 ==		=== NRLMSISE-00 ===

	Is an empirical, global reference atmospheric model of the Earth from ground to space.		Is an empirical, global reference atmospheric model of the Earth from ground to space.

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The US standard atmosphere

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	https://en.wikipedia.org/wiki/U.S._Standard_Atmosphere		https://en.wikipedia.org/wiki/U.S._Standard_Atmosphere

			=== NRLMSISE-00 ==

			Is an empirical, global reference atmospheric model of the Earth from ground to space.

	=== NASA Global Reference Atmospheric Models GRAM ===		=== NASA Global Reference Atmospheric Models GRAM ===

2023-08-30T15:10:54Z

Barometric formula

← Older revision		Revision as of 18:10, 30 August 2023
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	\end{align}		\end{align}
	</math>		</math>

	and this gives for altitude <math>h</math>		and this gives for altitude <math>h</math>

	<math>		<math>
	h = \frac{T_0 - \left( \frac{p}{p_0}\right)^{\frac{LR}{gM} }}{L}		h = \frac{T_0 - \left( \frac{p}{p_0}\right)^{\frac{LR}{gM} }}{L}

2023-08-30T15:10:38Z

Barometric formula

← Older revision		Revision as of 18:10, 30 August 2023
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	\end{align}		\end{align}
	</math>		</math>
			and this gives for altitude <math>h</math>
			<math>
			h = \frac{T_0 - \left( \frac{p}{p_0}\right)^{\frac{LR}{gM} }}{L}
			</math>

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Barometric formula

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Barometric formula

2023-08-30T14:58:53Z

Barometric formula

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	</math>		</math>

	Actually, both <math>p</math> and <math>T</math> are dependent on altitude <math>h</math>.		Actually, both <math>p</math> and <math>T</math> are dependent on altitude <math>h</math>. Thus, we will assume linear dependency on temperature <math>T = T_0 - Lh</math>, and we have

			<math>
			\begin{align}
			\frac {dp}{dh}&=- g \frac{pM}{R(T_0 - Lh)} \\
			\frac {dp}{p}&=- g \frac{M}{R(T_0 - Lh)} dh

			\end{align}
			</math>

2023-08-30T14:49:38Z

Barometric formula

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	Ideal gas law		Ideal gas law
	<math>P = \frac{\rho}{M} R T</math>, where pressure <math>P</math> is a function of <math>n, T, V</math>, thus <math>P=P(n,T,V)</math> and the hydrostatic assumption <math>dP = - \rho g dz</math> are needed to derive this.		<math>P = \frac{\rho}{M} R T</math>, where pressure <math>P</math> is a function of <math>n, T, V</math>, thus <math>P=P(n,T,V)</math> and the hydrostatic assumption <math>dP = - \rho g dz</math> are needed to derive this.


	Assume <math>\frac {dp}{dh}=-\rho(h) g = - \frac{N(h)m}{V}g</math> where <math>N(h)</math> is the number of particles at altitude <math>h</math>. The ideal gas law states that <math>pV = N R T</math>, and thus <math>N = N(h) = \frac{ p(h) V }{RT} </math>. Finally, we have

	~~<math>~~
	~~\frac{dp}{dh} = -\frac{N(h)m}{V}g = -\frac{p(h) V}{RT}~~
	~~</math>~~


	~~----~~

	Assume <math>\frac {dp}{dh}=-\rho(h) g </math> and <math>p = \rho \frac{R}{M}T</math> which gives <math>\rho = \frac{pM}{RT}</math> and we have		Assume <math>\frac {dp}{dh}=-\rho(h) g </math> and <math>p = \rho \frac{R}{M}T</math> which gives <math>\rho = \frac{pM}{RT}</math> and we have
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	\frac {dp}{dh}=- g \frac{pM}{RT}		\frac {dp}{dh}=- g \frac{pM}{RT}
	</math>		</math>

			Actually, both <math>p</math> and <math>T</math> are dependent on altitude <math>h</math>.

2023-08-30T14:48:26Z

Barometric formula

← Older revision		Revision as of 17:48, 30 August 2023
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	----		----

	Assume <math>\frac {dp}{dh}=-\rho(h) g </math> and <math>p = \rho \frac{R}{M}T</math> which gives <math>\rho = \frac{~~pTM~~}{R}</math> and we have		Assume <math>\frac {dp}{dh}=-\rho(h) g </math> and <math>p = \rho \frac{R}{M}T</math> which gives <math>\rho = \frac{pM}{RT}</math> and we have

	<math>		<math>
	\frac {dp}{dh}=- g \frac{~~pTM~~}{R}		\frac {dp}{dh}=- g \frac{pM}{RT}
	</math>		</math>