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Psychrometrics

Relative humidity from moisture content

pw=pW621.945+Wpws=0.61094·exp(17.625·tt+243.04)RH=100pwpws
where
pw
partial pressure of water vapour (kPa)
p
atmospheric pressure (kPa)
W
moisture content (g/kg dry air)
pws
saturation vapour pressure at t, Magnus (kPa)
t
dry-bulb temperature (°C)
RH
relative humidity (%)
Explain this equation

Finds relative humidity from moisture content and temperature. Use it to see what RH a given supply or mixed air condition will give once it reaches room temperature.

p_w is the vapour pressure the water in the air exerts, found from moisture content W and total pressure p. p_ws is the saturation pressure at temperature t. RH is how full the air is, as a percentage of saturation.

This simply reverses the moisture content equation: rearranging W = 621.945 p_w/(p − p_w) gives p_w = p W/(621.945 + W).

Sense check: 10 g/kg at 26 °C is about 48 %; the same air cooled to 14 °C is near 100 %, at its dew point.

Watch out: a result over 100 % means the air cannot exist at that temperature; water will condense.

Finds relative humidity from moisture content and temperature, for example to check the room RH a supply-air condition will produce after it warms up.

°C
g/kg

Enter your values to see the result.

Good to know

Moisture content stays the same when air is heated or cooled without condensation, so RH falls as air warms and rises as it cools. That is why reheating after a cooling coil lowers RH.

Common mistake

Entering moisture content in kg/kg instead of g/kg (1000 times too small).

Limits

Magnus saturation pressure, no enhancement factor. Results above 100 % mean the state is not physically possible as a gas.

Sources

Results are calculations from the published methods and figures shown, and depend on the data you enter. Follow the equipment manufacturer’s data, CIBSE guidance, BS EN 378, the F-gas Regulations and your company procedures. Refrigerant work needs the correct F-gas certification.

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