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Humidity Ratio

Mass of water vapour per unit mass of dry air.

Formula

w=0.622pvp−pvw = 0.622 \frac{p_v}{p - p_v}

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Variables

wHumidity ratio (kg/kg)
p_vVapour pressure (Pa)
pTotal pressure (Pa)

Example

p_v=1700, p=101325: w ≈ 0.0106 kg/kg

How it works

The humidity ratio tells you how much water vapour is mixed into a flow of dry air, expressed on a dry-air mass basis. It comes from treating moist air as a gas mixture: the partial pressure of water vapour is related to its own mass fraction, and the factor 0.622 is the ratio of the molecular mass of water vapour to that of dry air. As the vapour pressure approaches the total pressure, the denominator shrinks and the humidity ratio rises sharply, which is why very humid air contains much more moisture per kilogram of dry air.

When to use it

Use this when analysing psychrometric processes such as heating, cooling, humidification, dehumidification, or evaporative cooling. It is also essential for converting between measured vapour pressure, relative humidity, and moisture content in HVAC design calculations.

Worked example, step by step

Given: - Vapour pressure, p_v = 2800 Pa - Total pressure, p = 95000 Pa Step 1: Write the formula w = 0.622 \times p_v / (p - p_v) Step 2: Substitute the values w = 0.622 \times 2800 / (95000 - 2800) Step 3: Calculate the denominator 95000 - 2800 = 92200 Pa Step 4: Calculate the ratio 2800 / 92200 ≈ 0.03037 Step 5: Multiply by 0.622 w ≈ 0.622 \times 0.03037 ≈ 0.0189 Final answer: w ≈ 0.0189 kg/kg This means there are about 0.0189 kg of water vapour per kg of dry air.

Common mistakes to avoid

  • Using total moist-air mass instead of dry-air mass in the denominator
  • Forgetting to subtract vapour pressure from total pressure
  • Mixing units such as kPa and Pa without converting consistently

Did You Know?

The constant 0.622 is the ratio of the molar masses of water and dry air.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026