In rpkgs/hydroTools: Tools for hydrological model

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前言

为解决的问题：

为何冬天比夏天干？夏天的水汽含量是冬天的多少倍？

冬天，华南地区（如广州）比武汉地区水汽含量多多少倍？

1. 水汽压$e$

干空气与水汽的相对分子质量，分别用符号$R_d$和$R_v$来表示（下标$_d$表示dry，$_v$表示water）。

根据高中化学可知，$M_d = 28.97 g/mol, M_v = 18 g/mol$。带入上述公式，可得

• 对于干空气，$R_d = R^* / M_d = 8.314 / 28.97 = 0.287\;(J ·g^{-1}K^{-1})$

• 对于水汽，$R_v = R^* / M_v = 8.314 / 18 = 0.4615 \;(J ·g^{-1}K^{-1})$

$$ \epsilon = \frac{R_d}{R_v} = \frac{M_v}{M_d} ≈ 0.622 \ \frac{R_v}{R_d} = \frac{1}{\epsilon} ≈ 1.608 $$

$$ R_v = \frac{1}{\epsilon} R_d $$

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$$ \rho_v = \frac{e}{R_v T} = \frac{\epsilon e}{R_d T} \ $$

$$ \rho_d = \frac{p - e}{R_d T} \ $$

$$ \begin{align} \rho &= \rho_d + \rho_v \ &= \frac{p - e}{R_d T} + \frac{e}{R_v T} \ &= \frac{p - e}{R_d T} + \frac{\epsilon e}{R_d T} \ &= \frac{p - (1 - \epsilon)e }{R_d T} \ &= \frac{p}{R_d T} (1 - 0.378 \frac{ e }{p}) \end{align} $$

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2. 水汽压$e$，与比湿$q$

2.1. 已知$e$，求$q$

$$ \begin{align} q &= \frac{\rho_v}{\rho_d + \rho_v} \ &= \frac{\epsilon e}{p - e + \epsilon e} \ &= \frac{\epsilon e}{p - (1 - \epsilon)e} \end{align} $$

2.2. 已知$q$，求$e$

$$ qp = (1 - \epsilon ) e q + \epsilon e \ e = \frac{qp}{ \epsilon + (1 - \epsilon) q } $$

转化的意义： - 已知饱和水汽压，求饱和比湿q - 已知相对湿度，求比湿（回答为何冬天干）

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3. 水汽压$e$，与混合比$w$

3.1. 已知$e$，求$w$

$$ \begin{align} w &= \frac{\rho_v}{\rho_d } \ &= \frac{\epsilon e}{p - e} \ \end{align} $$

3.2. 已知$w$，求$e$

$$ wp = we + \epsilon e \ e = \frac{w p}{w + \epsilon} $$

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4. 水汽压$e$，与水汽密度$\rho_v$的转换

4.1. 简化版

$$ \begin{align} \frac{e}{P} &= \frac{\rho_v}{\rho} \frac{R_v}{R}, (R_d ≈ R) \ &≈ \frac{\rho_v}{\epsilon \rho} \end{align} $$

$$ \rho_v = \epsilon \rho \frac{e}{P} $$

4.2. 完整版

$$ \frac{e}{P - e} = \frac{\rho_v}{\rho_d} \frac{R_v}{R_d} \ \frac{e}{P - e} = \frac{\rho_v}{\epsilon \rho_d} $$

$$ \frac{e}{P} = \frac{\rho_v}{\epsilon \rho_d + \rho_v} \ $$

$$ \rho_v = \epsilon \rho_d \frac{e}{P - e} $$

rpkgs/hydroTools documentation built on Oct. 8, 2024, 7:47 p.m.