Dehumidification load calculation for a room

Calculate water removal from the change in humidity ratio, then calculate the heat moved using the same airflow and air states. A fixed temperature drop cannot establish either result without inlet moisture content and an outlet condition. Keep three quantities separate: water collected, coil cooling capacity and electrical input. For an electrical room, also determine whether the equipment rejects heat indoors or carries it outside.

Start with water mass, not temperature

Humidity ratio is the mass of water vapor divided by the mass of dry air. Its usual units are lb water/lb dry air or kg water/kg dry air. Unlike relative humidity, it directly supports a water mass balance.

Relative humidity changes with temperature even when no water leaves the air. Heating can lower relative humidity without removing moisture. Cooling initially raises relative humidity without removing moisture, until a surface becomes cold enough for condensation.

The enclosure relative humidity calculation explains that temperature effect. For a dehumidification calculation, convert each measured temperature and relative humidity into humidity ratio at the local atmospheric pressure.

Use a psychrometric chart or calculator that reports:

• Dry-bulb temperature.
• Relative humidity or dew point.
• Humidity ratio.
• Moist-air enthalpy.
• Specific volume per unit mass of dry air.

Water vapor is a gas, not suspended liquid droplets. Condensing it therefore releases latent heat, even when the air temperature is far below water's boiling point.

Define the room load and the air process

A room moisture load and a coil's moisture-removal rate answer different questions.

The room load includes moisture entering through ventilation, infiltration, wet processes and evaporation. The coil removal rate depends on airflow and the difference between its entering and leaving humidity ratios.

For a coil, use:

Water removal rate = dry-air mass flow × (inlet humidity ratio − outlet humidity ratio)

With mass flow in lb dry air/h, the result is lb water/h. Do not substitute total air mass without checking the basis.

For an initial room dry-down, a first estimate is:

Water initially removed from room air = room dry-air mass × (initial humidity ratio − target humidity ratio)

That estimate excludes moisture released from walls, stored materials and other wet surfaces. It also excludes continuing moisture entry. Those contributions can dominate the actual drying time.

Specify the final temperature alongside the target relative humidity. A target of 50% relative humidity does not define one moisture content across different temperatures.

A worked airflow calculation

Consider 100 cfm of actual entering airflow at approximately sea-level pressure, 14.7 psia [101.3 kPa]. Assume the air enters at 65°F [18.3°C] and 60% relative humidity.

For illustration, assume the air leaves the cooling coil saturated at 40°F [4.4°C]. This is an idealized outlet condition, not a prediction for a particular appliance.

The entering specific volume is approximately 13.4 ft³/lb dry air. Therefore:

Dry-air mass flow = 100 × 60 ÷ 13.4 ≈ 448 lb dry air/h

Using unrounded humidity ratios, the difference is approximately 0.00266 lb water/lb dry air:

Water removal = 448 × 0.00266 ≈ 1.19 lb/h

That equals approximately 0.54 kg/h, or 27–28 US liquid pints/day during continuous operation. One US liquid pint of water weighs approximately 1.04 lb.

The result follows from airflow and the humidity-ratio difference. It does not follow from the 25°F [13.9°C] temperature drop alone.

Add sensible and latent coil loads

The coil must cool the air and condense the collected water. Counting only air cooling misses a substantial part of the load.

For a quick check near these conditions, dry air has a specific heat of approximately 0.24 Btu/(lb·°F). The example's sensible load is therefore:

Sensible load ≈ 448 × 0.24 × 25 ≈ 2,690 Btu/h

Water vapor also contributes some sensible heat. Do not apply liquid water's specific heat to water that remains vapor in the entering air.

The latent heat of condensation near these temperatures is approximately 1,050 Btu/lb water. The latent portion is therefore:

Latent load ≈ 1.19 × 1,050 ≈ 1,250 Btu/h

The combined estimate is approximately 4,000 Btu/h, allowing for rounding and the vapor's sensible contribution.

An enthalpy calculation handles both contributions more consistently:

Total coil load ≈ dry-air mass flow × (inlet enthalpy − outlet enthalpy)

For the example:

Total coil load ≈ 448 × (24.2 − 15.2) ≈ 4,030 Btu/h

That is approximately 1.18 kW of thermal cooling. This estimate neglects the small liquid-condensate enthalpy correction. For a detailed equipment model, include the drain-water state using consistent enthalpy references.

What equipment ratings do and do not establish

A published collection capacity applies to the stated test conditions and operating procedure. Check the entering temperature, humidity, airflow basis and treatment of defrost or cycling before comparing calculations.

A collection rating does not establish a fixed evaporator temperature drop. Likewise, electrical input alone does not establish the outlet humidity ratio.

Real coils have uneven surface temperatures and finite contact effectiveness. Some air can pass through without reaching the coldest surface condition. Leaving air is therefore not automatically saturated at the coil surface temperature.

Also distinguish the evaporator outlet from the complete appliance outlet. A refrigeration dehumidifier commonly reheats the dried air across its condenser before discharge.

The entering dew point in the example is about 51°F [10.6°C]. A surface near that temperature provides little condensation potential; colder surfaces provide more. The dew point in a cool electronics chamber article explains the same surface-temperature threshold.

A 40°F [4.4°C] air outlet does not, by itself, prove that the coil freezes. Frost depends on local surface temperature, moisture conditions and operating time.

Separate refrigeration power from room heating

Thermal cooling capacity and electrical input are different quantities. Use 1 W = approximately 3.412 Btu/h when converting units.

A refrigeration system moves heat rather than converting electricity directly into cooling. Its cooling coefficient of performance is:

Cooling COP = evaporator cooling capacity ÷ electrical input

If the example coil moved 1.18 kW of heat while the complete appliance consumed 0.60 kW, its cooling COP would be approximately 2.0. That is an illustrative ratio, not an equipment prediction. Define whether the input includes the compressor, fan, pump and controls.

The condenser rejects the evaporator heat plus compressor input. If both coils operate inside the same room, the appliance does not provide net room cooling.

For a boundary around the complete appliance, the net total-energy addition is approximately its electrical input, allowing for condensate leaving and other boundary flows. However, its sensible heating effect also includes conversion of vapor's latent energy into sensible heat. Do not treat electrical input as the only sensible term in a room humidity-and-temperature calculation.

If the condenser rejects heat outside, the room balance changes. The enclosure heat calculation guide uses the same principle: define which heat sources and heat-transfer paths cross the chosen boundary.

An ice-water loop needs a stored-energy balance

A cold-water coil can remove moisture if its effective surface temperature stays below the entering dew point. Its available cooling depends on coolant temperature, flow and heat-transfer performance.

Ice melting at 32°F [0°C] absorbs approximately 144 Btu/lb [334 kJ/kg]. Supplying the example's roughly 4,000 Btu/h load through melting alone would require:

Ice melt rate ≈ 4,000 ÷ 144 ≈ 28 lb/h

That is approximately 13 kg/h under idealized conditions. Warming the meltwater absorbs additional heat, but rising coolant temperature also reduces condensation capability.

An indirect coil keeps coolant separate from the air and collected condensate. An open water-contact system needs a separate evaporation and moisture-transfer analysis.

Removing warmed water from the room carries energy away. Producing the ice requires energy elsewhere, so pump power alone is not the system's total energy requirement. Specify the storage duration, starting coolant condition and acceptable final temperature.

What this changes around electrical panels

Room dehumidification can reduce the moisture entering a cabinet during door openings or air exchange. It does not guarantee that every internal surface remains above dew point.

A cold pipe, cooling surface or overnight temperature drop can still create local condensation. Use the condensation guide for those enclosure-specific checks rather than treating room relative humidity as a complete protection strategy.

A cabinet heater and room dehumidifier also perform different jobs. Heating raises surface temperature; dehumidification removes water. The cabinet heater versus room dehumidifier comparison explains that application choice.

If room conditions become the design basis for a new cabinet, record their full operating range. Use the enclosure selector when choosing usable dimensions and environmental requirements, then check thermal suitability separately.

What to specify for a reliable calculation

A useful calculation package identifies the air states, moisture sources and heat destination. Send the following requirements rather than requesting capacity from a temperature drop alone:

Compare predicted water collection with measured condensate over a stable operating period. Account for water retained on the coil, drainage delays and defrost cycles. A defensible result closes both the water balance and the energy balance without assuming that cooling watts equal electrical watts.

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