---
title: "vapour_press"
output: rmarkdown::html_vignette
vignette: >
  %\VignetteIndexEntry{vapour_press}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---

<style type="text/css">
.main-container {
  max-width: 1800px;
  margin-left: auto;
  margin-right: auto;
}
</style>

```{r, include = FALSE}
knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>", 
  fig.width = 8, 
  fig.height = 5, 
  dev = "svg"
)
Sys.setlocale("LC_ALL", "English.utf8")
```

# 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，$R_v = R^* / M_v = 8.314 / 18 = 0.4615 \;(J ·g^{-1}K^{-1})$
  
  $$
  \epsilon = \frac{R_d}{R_v} = \frac{M_v}{M_d} ≈ 0.622 \\
  \frac{R_v}{R_d} = \frac{1}{\epsilon} ≈ 1.608
  $$

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

---

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

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

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

---

# 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）

---

# 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$e$

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

---

# 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

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

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

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