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Earth and Space Sciences · 术语表

什么是 Virtual temperature?

Definition 21.3 University Earth and Space Sciences — Year 2 · 第 21 章 — Atmospheric Thermodynamics

The virtual temperature TvT_v of moist air is the temperature that dry air would need, at the same pressure, to have the same density:

Tv=T 1+r/ε1+r≈T (1+0.61 r).T_v = T\,\frac{1 + r/\varepsilon}{1 + r} \approx T\,(1 + 0.61\,r) .

示例

Example 21.4 (The air at Norman)

At the ground the sounding gives p=971.1 hPap = 971.1\,\mathrm{hPa}, T=19.0 ∘CT = 19.0\,{}^{\circ}\mathrm{C} and a dew point of 18.5 ∘C18.5\,{}^{\circ}\mathrm{C}. The vapour pressure is the saturation vapour pressure at the dew point, e=es(18.5 ∘C)=21.5 hPae = e_s(18.5\,{}^{\circ}\mathrm{C}) = 21.5\,\mathrm{hPa} (Proposition 21.5 below). Then r=0.622×21.5/(971.1−21.5)=14.1 g/kgr = 0.622 \times 21.5/(971.1 - 21.5) = 14.1\,\mathrm{g}/\mathrm{kg}, q=13.9 g/kgq = 13.9\,\mathrm{g}/\mathrm{kg}, and Tv−T=0.61×0.0141×292.15 K=2.5 KT_v - T = 0.61 \times 0.0141 \times 292.15\,\mathrm{K} = 2.5\,\mathrm{K}: this air is as light as dry air two and a half degrees warmer.

Example 21.20 (The energy in the Norman sounding)

Lift the surface air of Example 21.4. It saturates at 964 hPa964\,\mathrm{hPa}, 64 m64\,\mathrm{m} up, and follows a moist adiabat. Below 900 hPa900\,\mathrm{hPa} the sounding is itself nearly saturated and the parcel stays a few tenths of a degree colder than its surroundings; then comes the inversion between 908 hPa908\,\mathrm{hPa} and 889 hPa889\,\mathrm{hPa}, where the temperature rises by 4.1 K4.1\,\mathrm{K} in 178 m178\,\mathrm{m} and the parcel finds itself 5 K5\,\mathrm{K} too cold. It becomes warmer than its surroundings only at 732 hPa732\,\mathrm{hPa}, 2.4 km2.4\,\mathrm{km} above the ground, its LFC, and stays warmer up to its EL at 236 hPa236\,\mathrm{hPa}, 10.9 km10.9\,\mathrm{km} above sea level, just below the tropopause the sounding reports at 235 hPa235\,\mathrm{hPa}. On a fine grid of pressures the areas of the skew-T figure are CAPE =2100 J/kg= 2100\,\mathrm{J}/\mathrm{kg} and CIN =−200 J/kg= -200\,\mathrm{J}/\mathrm{kg} with the temperature, and 2230 J/kg2230\,\mathrm{J}/\mathrm{kg} and −130 J/kg-130\,\mathrm{J}/\mathrm{kg} with the virtual temperature: the moisture of the parcel adds its own buoyancy. Parcel theory gives wmax⁡=2×2100=65 m/sw_{\max} = \sqrt{2 \times 2100} = 65\,\mathrm{m}/\mathrm{s}.

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