Pressure in atmosphere

Introduction

ISO2533:1975

The case in Toposhere (<10 km).

• Lapse rate +6.5 °C/km
• Base temp 19.0 °C
• Base atmospheric pressure 108,900 Pa equals 1.075 atm
• Base atmospheric density 1.2985 kg/m3

International Standard Atmosphere

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

Consider the function in GY91's BMP280-3.3, eg at https://startingelectronics.org/tutorials/arduino/modules/pressure-sensor/

Earth's atmosphere's changes in

• Pressure $P$
• Temperature $T$
• Density $\rho$
• Viscosity $\eta$ or $\mu$

Hydrostatic balance ${\frac {dP}{dh}}=-\rho g$ The ideal gas law $P=\rho RT$

Reference atmospheric model

How the ideal gas properties change (mainly) as a function of altitude (etc).

Static atmospheric model

$\rho ={\frac {MP}{RT}}$ and $dP=-g\rho dh$ (see above).

Isothermal-barotropic approximation and scale height

Temperature and molecular weight are constant: density and pressure are exponential functions of altitude.

The US standard atmosphere

More realistic temperature function, consisting of eight data points connected by straight lines, which is---of course---an approximation.

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

Barometric formula

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

Ideal gas law $P={\frac {\rho }{M}}RT$, where pressure $P$ is a function of $n,T,V$, thus $P=P(n,T,V)$ and the hydrostatic assumption $dP=-\rho gdz$ are needed to derive this.

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

${\frac {dp}{dh}}=-g{\frac {pM}{RT}}$

Actually, both $p$ and $T$ are dependent on altitude $h$. Thus, we will assume linear dependency on temperature $T=T_{0}-Lh$, and we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac {dp}{dh}&=- g \frac{pM}{R(T_0 - Lh)} \\ \frac {dp}{p}&=- g \frac{M}{R(T_0 - Lh)} dh \\ \ln p &= \frac{gM}{RL} \ln(T_0 - hL) + C \\ p &= p_0 \left( T_0 - hL \right)^{\frac{gM}{RL}} \end{align} }

Simplified model from Weather.gov

https://www.weather.gov/media/epz/wxcalc/pressureAltitude.pdf

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h = 44307.70\left( 1 - \left( \frac{p}{1013.25} \right)^{0.190284} \right) }