Atmospheric pressure, density and temperature variation with altitude - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-06-20T02:58:41Z https://physics.stackexchange.com/feeds/question/433840 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/433840 2 Atmospheric pressure, density and temperature variation with altitude user1936752 https://physics.stackexchange.com/users/52363 2018-10-11T09:11:00Z 2018-10-11T20:12:50Z <p>I'm trying to understand how one can calculate pressure, density and temperature of the atmosphere as a function of altitude. </p> <p>My assumptions are mostly sourced from <a href="https://en.wikipedia.org/wiki/Lapse_rate" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Lapse_rate</a> and <a href="https://en.wikipedia.org/wiki/Barometric_formula#Derivation" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Barometric_formula#Derivation</a>. However, on these pages, there seems to be a little vagueness regarding what parameters are being held constant so I shall write them out explicitly here with dependence on height <span class="math-container">$z$</span> where appropriate:</p> <p>1) Air is an ideal gas so <span class="math-container">$P(z)M = \rho(z)RT(z)$</span>.</p> <p>2) The pressure is hydrostatic i.e. <span class="math-container">$dP(z) = -\rho(z) g dz$</span> </p> <p>3) There is some temperature lapse rate as a function of altitude and density of air <span class="math-container">$T(z) = f(z, \rho(z))$</span>. This allows me to take into account radiation and convection. Now, the Wikipedia page (<a href="https://en.wikipedia.org/wiki/Lapse_rate" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Lapse_rate</a>) treats <span class="math-container">$\rho$</span> as a constant and then assumes that air behaves like an adiabatic gas when it expands due to heat to obtain a valid expression for T(z). That seems incorrect though as density clearly does change with altitude.</p> <p>4) It's not clear if I can obtain <span class="math-container">$\rho(z)$</span> from some other consideration independently. </p> <p>Are there good tricks/reasonable physical assumptions to solve and obtain all three variables as a function of <span class="math-container">$z$</span>?</p> <p>EDIT: The constant density assumption is what I'm having trouble with. Why should this be true and if not, what is the way to obtain it (at least to some first order where we ignore temperature lapse)? </p> https://physics.stackexchange.com/questions/433840/-/433852#433852 1 Answer by user1476176 for Atmospheric pressure, density and temperature variation with altitude user1476176 https://physics.stackexchange.com/users/35879 2018-10-11T10:32:53Z 2018-10-11T10:32:53Z <p>The formula wikipedia quotes, <span class="math-container">\begin{align} P(z) = P_0 \exp \left( - \int_0^z \frac{M(z^*) g(z^*)}{R_\text{u} T(z^*)} \text{d} z^* \right) \end{align}</span> provides a way to calculate <span class="math-container">$P(z)$</span> when <span class="math-container">$M(z)$</span>, <span class="math-container">$g(z)$</span>, and <span class="math-container">$T(z)$</span> are known. </p> <p>Unfortunately, <span class="math-container">$T(z)$</span> is complicated: in some layers, temperature increases with altitude, while in others it decreases with altitude. There isn't really a simple model to describe this variation because the effects that drive it (absorbtion of radiation by gas, convection) are not simple. On top of this, the composition of the atmosphere varies with elevation, so <span class="math-container">$M(z)$</span> is not a constant function. The answer to your question is therefore (sadly) that there is no simple analytical derivation of the type you seek. </p> <p>If you're interested in the variation of <span class="math-container">$P(z)$</span>, I'd suggest you either look up the empirical trend or consider the simpler case of pure air at uniform <span class="math-container">$T$</span>.</p> https://physics.stackexchange.com/questions/433840/-/433884#433884 3 Answer by Bert Barrois for Atmospheric pressure, density and temperature variation with altitude Bert Barrois https://physics.stackexchange.com/users/178462 2018-10-11T13:24:31Z 2018-10-11T13:24:31Z <p>Heat transport in the lower atmosphere is predominantly convective, at least in daytime. (Convection abates after the ground has cooled off at night.) Up- and downdrafts do not exchange much energy, so it is a good approximation to assume that their expansion and compression are adiabatic. </p> <p>Adiabatic expansion of ideal gases is described by <span class="math-container">$P\sim {{\rho }^{\gamma }}$</span>, where <span class="math-container">$\gamma \equiv {{C}_{P}}/{{C}_{V}}=7/5$</span> for <em>dry</em> air. It follows that <span class="math-container">$P\sim {{T}^{7/2}}$</span> and <span class="math-container">$\rho \sim {{T}^{5/2}}$</span>. Combining these rules with the equation of hydrostatic equilibrium, <span class="math-container">$dP/dz=-g\rho$</span>, and the ideal gas law, <span class="math-container">$RT=PV=Pm/\rho$</span>, we find a constant lapse rate: <span class="math-container">$dT/dz=-\tfrac{\gamma -1}{\gamma }mg/R=$</span> -9.8 deg/km, where <em>m</em> = 29 g/mol. But this value exaggerates the measured average temperature profile. </p> <p>The International Standard Atmosphere use in aero engineering has <span class="math-container">$dT/dz=$</span> -6.5 deg/km, consistent with <span class="math-container">$\gamma$</span> = 1.26, from sea level up to 11 km, a good approximation at mid-latitudes. You may use <span class="math-container">\begin{align} &amp; T(z)=290K-(6.5\tfrac{\deg }{\text{km}})z \\ &amp; P(z)/P(0)={{[T(z)/T(0)]}^{1.26/0.26}} \\ &amp; \rho (z)/\rho (0)={{[T(z)/T(0)]}^{1.00/0.26}} \\ \end{align}</span> In most climes, however, the air is not so dry. On a muggy day with a dew point of 20°C, the partial pressure of water vapor will be 17.5 out of 760 torr. That’s 2.30% by molar content or 1.43% by weight. When a buoyant blob of air ascends, it will initially cool off at 9.8 deg/km, but when its temperature reaches the dew point, the water vapor will begin to condense, usually onto particulate nuclei. (Caveat: Very clean air can become supersaturated. Surface tension acts as an obstacle to the formation of fog droplets from scratch.) At this temperature, water releases about 585 cal/g as it condenses. Given the slope of its vapor pressure curve, about 1.1 torr/deg at 20°C, condensation will roughly triple the heat capacity of saturated air, reducing <span class="math-container">$\gamma$</span> and the lapse rate until the air has dried out. The effect is even greater in tropical climes, where the altitude of the tropopause can be as high as 17.5 km. </p> https://physics.stackexchange.com/questions/433840/-/433965#433965 2 Answer by Chet Miller for Atmospheric pressure, density and temperature variation with altitude Chet Miller https://physics.stackexchange.com/users/102308 2018-10-11T20:12:50Z 2018-10-11T20:12:50Z <p>The two basic equations are <span class="math-container">$$\rho=\frac{PM}{RT}$$</span> and <span class="math-container">$$\frac{dP}{dz}=-\rho g$$</span> If we eliminate the (altitude-dependent) density from these equations, we obtain: <span class="math-container">$$\frac{dP}{dz}=-\frac{PM}{RT}g\tag{1}$$</span> For the troposphere, the equation for the adiabatic reversible expansion and compression of convected air parcels is: <span class="math-container">$$\frac{P}{P_0}=\left(\frac{\rho}{\rho_0}\right)^{\gamma}$$</span>where the subscript 0 revers to the values at ground level (z = 0). If we combine this equation with the ideal gas law, we obtain: <span class="math-container">$$\frac{T}{T_0}=\left(\frac{P}{P_0}\right)^{\frac{\gamma-1}{\gamma}}\tag{2}$$</span>We can now substitute Eqn.2 into Eqn. 1 to obtain a equation (strictly valid for the troposphere) that involves only the pressure P (i.e., the density and temperature have been eliminated): <span class="math-container">$$\frac{dP}{dz}=-\frac{MgP_0^{\frac{\gamma-1}{\gamma}}}{RT_0}P^{1/\gamma}$$</span>If we integrate this equation between z = 0 and arbitrary z, we obtain: <span class="math-container">$$\left(\frac{P}{P_0}\right)^{\frac{\gamma-1}{\gamma}}=1-\frac{(\gamma-1)}{\gamma}\frac{Mgz}{RT_0}\tag{3}$$</span>Combining Eqns. 2 and 3 then yields: <span class="math-container">$$T=T_0-\frac{(\gamma-1)}{\gamma}\frac{Mg}{R}z\tag{3}$$</span>Eqn. 3 suggests that, for the assumed adiabatic convective expansion and compression of air parcels in the troposphere, the tropospheric temperature should vary linearly with altitude z. The vertical temperature gradient predicted by this equation is called the "dry adiabatic lapse rate," and has a value of 9.8 C/km. The actual temperature gradient observed in the atmosphere is less than this, with a value of 6.5 C/km.</p>