Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2021 / 3

Before condensation, assume a closed parcel with fixed atmospheric mixing ratio, constant specific heat capacities, and negligible irreversible heating. The first law of thermodynamics for a dry adiabatic process isThe parcel remains in approximate pressure equilibrium with a hydrostatic environment. With the standard approximation that parcel and environmental densities are close, $dp/dz\simeq -\rho g$, henceThis is the dry adiabatic lapse rate, about $9.8\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ for air. More precisely the environmental hydrostatic balance gives $dT/dz=-g{\rho }_{\mathrm{env}}/(c_p{\rho }_{\mathrm{parcel}})$; the familiar constant dry adiabatic lapse rate uses their ratio close to one.

Let $e_v$ be the parcel's water vapor partial pressure. Fixed mole fraction before condensation means $e_v/p$ is constant. Its dew point $T_D$ is defined by $E(T_D)=e_v$, where $E$ is the saturation pressure. The Clausius-Clapeyron relation givesThe last approximation uses hydrostatic balance, the ideal gas law, and an environmental temperature close to the parcel's. Rearrangement yieldsFor a modest initial dew-point depression, $T_D/T\simeq 1$ in absolute temperature, soWith $L_v\simeq 2.5\times 10^6\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, $R_d\simeq 287\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and $R_v\simeq 461.5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, $H\simeq 1.58\times 10^5\mathrm{m}$. This dew point scale is much longer than the atmospheric pressure scale height.

The exponential function is an approximation, not the exact consequence of a dry adiabatic process. Retaining $T=T_s-{\Gamma }_{d}z$ gives $p/p_s=(T/T_s{)}^{c_p/R_d}$ and the useful comparisonIt also shows explicitly that expansion lowers the dew point even before any water vapor condenses.

At the lifting condensation level, the parcel temperature first equals its dew point. For $z/H\ll 1$, the first-order Taylor expansion gives $T_D\simeq T_{D,s}(1-z/H)$. Equating this with the dry-adiabatic process result givesNear terrestrial surface temperatures, $T_{D,s}/H\simeq 1.7$–$1.9\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$, whereas ${\Gamma }_d\simeq 9.8\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$. ConsequentlyUse absolute temperature for $T_{D,s}/H$. The temperature difference can be evaluated in kelvins or degrees Celsius, whose increments agree. The coefficient is an approximate near-surface terrestrial value, not independent of composition or ambient conditions for all atmospheres.

The dew-point depression is $6\mathrm{K}$, soKeeping $T_{D,s}=276.15\mathrm{K}$, $H\simeq 158.5\mathrm{km}$ and $c_p=1004\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ gives approximately $747\mathrm{m}$, consistent with the rounded estimate.

The real cloud base can differ because entrainment changes the parcel's atmospheric mixing ratio and temperature, surface humidity varies, and the environmental pressure profile need not match the assumed parcel profile. Finite cloud-droplet activation requires some atmospheric supersaturation, so the first visible cloud droplets need not occur exactly at the thermodynamic lifting condensation level. A parcel must also actually be lifted that far; environmental static stability can prevent this. The constant-latent heat and first-order Taylor expansion approximations produce smaller numerical differences.

A rising parcel cools, lowering the saturation pressure. Once its water vapor supply exceeds that value, atmospheric supersaturation develops. Homogeneous nucleation of pure water is strongly inhibited at the small atmospheric supersaturations typical of clouds; existing aerosol particles provide cloud condensation nuclei. Their dry sizes and composition, together with the parcel's ascent rate, determine how many cloud droplets activate.

Köhler theory combines the Kelvin equation, which raises equilibrium vapor pressure over a curved surface, with Raoult's law, which lowers it for a chemical solution. In a dilute diameter convention, with dry diameter $D_d$ and wet diameter $D$,Here $s=e_v/E(T)-1$ is atmospheric supersaturation, ${\sigma }_w$ is surface tension, and ${\kappa }_h$ measures aerosol hygroscopicity, not the dry-adiabatic process exponent. The maximum of the Köhler curve is the critical supersaturation. Larger or more hygroscopic particles generally have lower thresholds. Cloud-droplet activation requires overcoming that maximum; total aerosol number is not the activated number.

Given an observed aerosol number size distribution $n_j(D_d)=d{N}_j/d\ln D_d$ for composition classes $j$, first assign each class a hygroscopicity and compute its critical supersaturation. If the parcel reaches maximum atmospheric supersaturation $s_{\max }$, the equilibrium-threshold estimate isFor each fitted lognormal distribution, this is its upper-tail integral; it can be evaluated analytically with an error function or numerically from measured bins.

The value $s_{\max }$ must be calculated, not guessed from the aerosol distribution alone. Prescribe parcel upward velocity $w$, initial pressure, temperature and humidity, then integrate expansion cooling, latent heat release, conservation of total water, and diffusional cloud droplet growth. Schematically,where $a_s,b_s>0$, $G$ includes water vapor diffusion and thermal conduction, and $q_l$ is liquid mass fraction. Rising motion produces atmospheric supersaturation; growing particles consume it. Their competition selects the peak and which particles cross their activation barriers. Assume a well-mixed closed parcel, liquid water, soluble spherical particles, specified mixing state, and negligible collision-coalescence and fallout during activation. Slow kinetic growth, weakly soluble particles or entrainment require a more detailed treatment than the threshold integral.