Psychrometrics

The question

On a summer afternoon, pour iced water into a glass and within a minute it is beaded with droplets. Nobody spilled anything; the water came out of the air. On a winter morning the same glass stays dry. Why does a glass of cold water sweat?

Mixtures of ideal gases

Air is a mixture, and so are combustion products. For mixtures of ideal gases, Dalton's model makes life simple: each component behaves as if it alone filled the whole volume at the mixture temperature, exerting its own partial pressure $p_i = y_i\,p$ , where $y_i$ is its mole fraction. The partial pressures add up to the total, and the mixture's $u$ , $h$ and $s$ are mass- or mole-weighted sums of the components'. (Entropy picks up an extra term from mixing itself — another irreversibility, as Lesson 6 would predict.)

Psychrometrics applies this to the most important mixture of all: dry air plus water vapour. The water vapour is at such a low partial pressure — a few kilopascals — that it behaves as an ideal gas (Lesson 3's quiz). But unlike nitrogen, it can condense at room temperature, and that is what makes moist air interesting.

Measuring moisture

Two measures describe how much water vapour the air holds:

The , $\omega = m_v/m_a = 0.622\,p_v/(p - p_v)$ : kilograms of water vapour per kilogram of dry air. It only changes if water is added or removed.
The , $\phi = p_v/p_g(T)$ : the vapour's partial pressure as a fraction of the saturation pressure at the air temperature. It changes whenever the temperature does.

Warm air can hold far more water than cold air: saturation pressure roughly doubles every 11 K. Cool air at constant $\omega$ and its relative humidity rises, until at $\phi = 100\%$ the vapour is saturated. That temperature is the . Cool any further, and water must condense.

There's the answer to the question. The air touching the glass is cooled to the glass temperature. If that is below the air's dew point, water condenses on the glass. On a muggy summer day the dew point may be 20 °C; on a dry winter day, below freezing.

The psychrometric chart

Engineers condense all of moist-air thermodynamics onto one chart: dry-bulb temperature along the bottom, humidity ratio up the side, and curves of relative humidity, enthalpy and wet-bulb temperature laid over it. Any state is a point; any process is a line.

ToolA psychrometric chart you can drag

Glass temperature12.0 °C

Relative humidity53%ω = 14.0 g/kg

Dew point19.2 °C

Wet bulb22.5 °C

Enthalpy65.9 kJ/kgper kg of dry air

The glass at 12.0 °C is below the dew point of the room air (19.2 °C): air touching it is cooled to saturation and water condenses. The glass sweats.

Challenges0/3

Make a glass of iced water (about 5 °C) sweat — then find room air in which it stays dry.
Use the cooling coil to pull at least 5 g of water out of each kilogram of air.
Cool desert air (35 °C or hotter, under 20% RH) below 22 °C with an evaporative cooler.

The psychrometric chart for moist air at 1 atm: dry-bulb temperature along the bottom, humidity ratio ω (kg of water vapour per kg of dry air) up the side. The top curve is saturation (100% relative humidity); purple curves are 20–80%; dashed lines are constant enthalpy, close to constant wet-bulb temperature. Drag the white point to set the air, then pick a process.

The chart makes the basic air-conditioning processes visual:

Heating at constant $\omega$ runs horizontally right. Relative humidity falls, which is why heated winter air feels so dry.
Cooling runs horizontally left until it hits saturation at the dew point, then follows the saturation curve down as water condenses. Air conditioners cool and dehumidify this way.
Evaporative cooling follows a line of nearly constant enthalpy: water evaporating into the air takes its latent heat from the air, cooling it while humidifying it. It works brilliantly in dry climates and hardly at all in humid ones.
Mixing two airstreams lands on the straight line between their states, at the mass-weighted position.

The wet-bulb temperature

Wrap a thermometer bulb in a wet wick and blow air over it. Water evaporates, cooling the bulb until the heat flowing in from the air balances the latent heat carried away. The steady reading is the . It is close to the adiabatic saturation temperature — the temperature air reaches if it is saturated by evaporating water into it with no outside heat — and it is the lowest temperature evaporative cooling can reach.

The cooling tower

Back to the plant, and the white plume rising over it. Lesson 8's condenser rejected about 1.5 W of heat for every watt of electricity. If there is no big river to take it, a cooling tower does it by evaporating some of the condenser water into a stream of air:

ExploreInside the cooling tower

Outdoor air temperature25.0 °C

Outdoor relative humidity50%

Hot water from the condenser40.0 °C

Cooling range10.0 K

Air per kg of water0.51 kg

Water evaporated1.42%0.51 t/s for the plant

Daily water loss44 thousand m³must be replaced from a river

Challenges0/2

Find a hot, humid day on which the tower cannot deliver 30 °C water.
Find conditions where under 1% of the circulating water evaporates.

A wet cooling tower cools a power plant’s condenser water by evaporating a small part of it into a stream of air. Energy and mass balances per kilogram of water fix how much air is needed and how much water is lost. The water can never be cooled below the air’s wet-bulb temperature; the gap that remains is the approach. Numbers below are scaled to a plant rejecting 1.5 GW of heat.

The analysis is a control-volume energy and mass balance with moist-air properties: the air enters with enthalpy $h_1$ and humidity $\omega_1$ , leaves nearly saturated with $h_2$ and $\omega_2$ , and the water it carries away as vapour, $\dot m_a(\omega_2 - \omega_1)$ , must be replaced. Roughly 1–2% of the circulating water evaporates for every 10 K of cooling, because almost all the heat leaves as latent heat. The wet-bulb temperature of the air sets a hard floor on the cold-water temperature — and therefore on the condenser pressure and the plant's efficiency. On a hot, humid day, power plants lose output.

Estimate first

A 1 GW power plant rejects 1.5 GW of heat through a wet cooling tower. Roughly how much water does it evaporate per day, in cubic metres?

101001,00010,000100,0001,000,00010,000,000

Drag along the line to place your guess — each tick is ten times the last.

Check your intuition

Air at 30 °C with a dew point of 18 °C passes over an air-conditioner coil at 12 °C. What happens?

Going deeper
Going deeper: the enthalpy of moist air

Per kilogram of dry air, moist air's enthalpy is

$h = c_{p,a}T + \omega\,(h_{g0} + c_{p,v}T) \approx 1.005\,T + \omega\,(2501 + 1.82\,T) \ \text{kJ/kg},$

with $T$ in °C and both components referenced to 0 °C (water as liquid). The huge 2501 kJ/kg term is the latent heat: a few grams of water vapour per kilogram carry as much energy as warming the air by tens of degrees. That is why lines of constant enthalpy and constant wet-bulb temperature on the chart nearly coincide, and why dehumidification dominates the energy use of air conditioning in humid climates.

Recap

What to carry forward

Dalton's model: each gas in an ideal mixture acts as if alone, at its partial pressure.
Moist air is described by its humidity ratio $\omega$ (mass of vapour per mass of dry air) and its relative humidity $\phi = p_v/p_g$ .
Cool air at constant $\omega$ and it saturates at the dew point; colder surfaces collect condensation.
The wet-bulb temperature is the lowest temperature evaporative cooling can reach.
On the psychrometric chart, heating, cooling and dehumidification, evaporative cooling and mixing are simple lines.
Cooling towers reject heat mostly by evaporation, about 1.5% of the water per 10 K; the wet bulb limits the cold-water temperature.

Next: the flame inside the plant. How hot can it get, and how much energy does fuel really contain?