humidity and psychrometric measurement

Absolute humidity is the density $\rho$ of water vapour in air at a given temperature, while relative humidity shows how close that vapour is to saturation:

$$\varphi=\frac{\rho}{\rho_0}\cdot100\%=\frac{p}{p_0}\cdot100\%,$$

where $\rho_0$ and $p_0$ are the density and partial pressure of saturated vapour at the same temperature. Replacing the density ratio by the pressure ratio is valid because water vapour in air obeys the ideal-gas equation of state $p=\frac{\rho}{M}RT$, so at constant temperature $p\propto\rho$. The dew point is the temperature at which the vapour becomes saturated at constant pressure; cooling below the dew point condenses the excess vapour. Psychrometric measurement relies on evaporation: the wet thermometer cools more the drier the air is, and from the difference $\Delta t=t_1-t_2$ the vapour partial pressure is found with the psychrometric formula $p=p_2-A\,p_{\text{atm}}\,\Delta t$, where $p_2$ is the saturated vapour pressure at the wet-bulb temperature, $p_{\text{atm}}$ is atmospheric pressure, and $A\approx6.6\times10^{-4}\ \text{K}^{-1}$ for an aspiration psychrometer.

At 20 °C the saturated water-vapour density is 17.3 g/m³. The absolute humidity of the air is 10.4 g/m³. Find the relative humidity.

A psychrometer reads $t_1=22$ °C (dry bulb) and $t_2=16$ °C (wet bulb). Atmospheric pressure is 101.3 kPa. The saturated vapour pressure is 1.82 kPa at 16 °C and 2.64 kPa at 22 °C. Find the relative humidity.

A room of volume 60 m³ is at 20 °C with relative humidity 60 %. The saturated vapour density is 17.3 g/m³ at 20 °C and 9.4 g/m³ at 10 °C. How much water condenses if the air is cooled to 10 °C?

Home psychrometer (safe independent measurement).

1. Take two identical thermometers. Wrap the bulb of one in a piece of gauze or cambric moistened with room-temperature water — this is the wet-bulb thermometer.
2. Fan air over the wet bulb (with a fan or a book) for 3–4 minutes until its reading stops changing, and read the dry-bulb thermometer at the same time. Record $t_1$ and $t_2$.
3. Compute $\Delta t=t_1-t_2$. From a table take $p_2$ at $t_2$ and $p_1^{\text{sat}}$ at $t_1$; then compute $p=p_2-A\,p_{\text{atm}}\,\Delta t$ and $\varphi=\frac{p}{p_1^{\text{sat}}}\cdot100\%$.
4. Repeat 5 times in different places in the room, compare with a household hygrometer, and explain the differences.
5. In a spreadsheet, plot $\varphi(\Delta t)$ for $t_1=22$ °C, taking $p_2$ from the table.

Safety: do not heat thermometers, do not use an open flame, and do not break the instruments. If temperatures above 40 °C or below 0 °C, fog, frost, or other hazardous conditions are required, work only with a simulation or tabulated data.

Show how the dew point and the dry-bulb and wet-bulb readings relate to the saturation-vapour-pressure curve, and how the temperature difference yields relative humidity.

Graph: temperature on the horizontal axis, saturated vapour pressure $p_0(T)$ on the vertical axis (an increasing curve). Three points are marked on the curve: $A$ — dry bulb $t_1$ (height $p_1^{\text{sat}}$), $B$ — wet bulb $t_2$ (height $p_2$), $C$ — dew point $t_d$ (height $p$). A dashed horizontal line drawn from $p$ on the pressure axis meets the curve at the dew point; the segment between $A$ and $B$ on the temperature axis is labelled $\Delta t$.

The air on a July afternoon is at $T = 30\,^\circ\text{C}$ with relative humidity $\varphi = 70\%$. The saturated vapour pressure at that temperature is $p_{\text{sat}}(30\,^\circ\text{C}) = 4.24\ \text{kPa}$, so the partial pressure of the water vapour is $e = 0.70 \times 4.24\ \text{kPa} = 2.97\ \text{kPa}$. At $24\,^\circ\text{C}$ saturated vapour has the same pressure, so the dew point is $24\,^\circ\text{C}$. The can is taken from the fridge at $4\,^\circ\text{C}$: its surface is far colder than the dew point, water vapour condenses on it, and within a minute a film of drops appears.

Uses the psychrometric definitions: relative humidity $\varphi = e/p_{\text{sat}}(T)$ compares the real vapour partial pressure with the saturated value at the same temperature, and the dew point is the temperature $T_d$ at which $e = p_{\text{sat}}(T_d)$. Any surface colder than $T_d$ forces condensation.

A coach measures the air with a psychrometer: dry bulb $T = 30\,^\circ\text{C}$, wet bulb $T_w = 26\,^\circ\text{C}$, depression $\Delta T = 4\,^\circ\text{C}$. With $p_{\text{sat}}(26\,^\circ\text{C}) = 3.36\ \text{kPa}$ and the psychrometric coefficient $A = 0.00068\ \text{K}^{-1}$, $e = 3.36 - 0.00068 \times 101.3 \times 4 = 3.08\ \text{kPa}$. Since $p_{\text{sat}}(30\,^\circ\text{C}) = 4.24\ \text{kPa}$, $\varphi = 3.08/4.24 \approx 73\%$. Sweat on the skin evaporates from a vapour-pressure deficit of only $4.24 - 3.08 = 1.16\ \text{kPa}$, so the players overheat faster than in dry air at the same $30\,^\circ\text{C}$.

Uses the psychrometric measurement: the wet-bulb thermometer loses heat by evaporation, and the rate of that cooling is set by the vapour-pressure deficit $p_{\text{sat}}(T_w) - e$; the pair of dry and wet readings gives $\varphi$, which controls evaporation from the skin.

In a kitchen the thermometer reads $T = 22\,^\circ\text{C}$ and a hygrometer reads $\varphi = 60\%$. From $p_{\text{sat}}(22\,^\circ\text{C}) = 2.64\ \text{kPa}$ the vapour partial pressure is $e = 0.60 \times 2.64 = 1.59\ \text{kPa}$, and the dew point is about $14\,^\circ\text{C}$, because $p_{\text{sat}}(14\,^\circ\text{C}) = 1.60\ \text{kPa}$. A $2\ \text{kg}$ load of wet clothes holds about $1\ \text{kg}$ of water; it leaves only while $e < p_{\text{sat}}$, and the deficit here is $2.64 - 1.59 = 1.05\ \text{kPa}$. The single-glazed window cools to $10\,^\circ\text{C}$, below the dew point, so drops of condensate run down the pane.

Uses the same two psychrometric quantities — relative humidity $\varphi = e/p_{\text{sat}}(T)$ and the dew point — to decide both how fast water evaporates from the cloth and where it condenses on the cold glass.