Indoor Pool Hall Dehumidification and Ventilation Calculator

How much water an indoor pool evaporates into the hall, and the air or dehumidifier needed to hold the humidity.

Calculated from standard textbook formulas, with every step shown. The formulas are fixed, so the working does not need checking: what matters is the numbers you type in. Use these calculators only if you are competent to choose the inputs (loads, material strength, dimensions, supports) and check your units and assumptions. For learning and checking your own working: design and sign-off are for a qualified engineer. Drafting Central does not store the numbers you type.

How this calculator works

An indoor pool puts a lot of water vapour into the hall air. Left alone the humidity rises, the air becomes uncomfortable, and moisture condenses on cold surfaces such as windows, frames and the roof, which corrodes and rots the building. The hall air must be kept at a humidity you choose, commonly in the range of about 50 to 60 percent, and a degree or two warmer than the water.

There are two ways to remove the moisture. Ventilation brings in outdoor air that is drier than the hall and exhausts the humid air: the air flow needed is the moisture load divided by the difference in humidity ratio, but all the heat in that air is thrown away. A heat-pump dehumidifier cools the hall air to condense the moisture and puts the heat back into the pool or the air, which uses far less energy. Heat recovery on the exhaust air also cuts the heat lost.

The calculator finds the evaporation from the pool with the ASHRAE equation, the outdoor air flow and air changes needed, the heat load of that air with and without heat recovery, the dehumidifier electrical power, and checks whether the coldest surface in the hall is above the dew point of the air.

Formulas used

  • \dot m_w = q_e / h_{fg}
  • \dot m_a = \frac{\dot m_w}{W_r - W_o}
  • Q = \dot m_a (h_r - h_o)
  • T_{surface} > T_{dew\ point}

Worked example

With the default numbers in the calculator above:

  1. Vapour pressure difference driving the evaporation: 1.578 kPa
  2. Evaporation (latent load on the hall): 31.78 kW
  3. Water evaporated into the hall (the moisture to remove): 46.97 kg/h
  4. Humidity ratio of the hall air and of the outdoor air: 0.013833 and 0.006345 kg/kg
  5. Dew point of the hall air: 19.05 °C
  6. Outdoor air needed to carry the moisture out (ventilation only): 1525 L/s
  7. Air changes per hour in the hall: 2.196
  8. Heat needed to bring that outdoor air to the hall condition (no heat recovery): 58.17 kW
  9. With sensible heat recovery: 40.55 kW
  10. Dehumidifier alternative: latent heat to remove: 31.78 kW
  11. Dehumidifier compressor electrical power: 7.946 kW
  12. Condensation check: coldest surface against the dew point: surface BELOW the dew point (condensation)

Change the numbers in the calculator to see the full working, every step, for your own case.

Common questions

What humidity should I hold in an indoor pool hall?

It is a design choice, often about 50 to 60 percent. Lower humidity makes swimmers colder and raises evaporation, higher humidity risks condensation. The limits that apply to your pool are set by the designer and the standard.

Why should the air be warmer than the water?

Air colder than the water increases the evaporation and chills swimmers when they get out. Common practice is air 1 to 2 degrees above the water temperature.

Why does the hall need to be at a slightly lower pressure?

So that moist air does not leak out into the rooms around the pool, where it condenses inside the walls.

Related calculators

Further reading: ASHRAE Handbook, HVAC Applications (natatoriums); Cengel and Boles, Thermodynamics: An Engineering Approach; Moran and Shapiro, Fundamentals of Engineering Thermodynamics.

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