**Determining Water Consumption for Active humidity control Units**

### Operating Principles

Microclimate generators will automatically add moisture to, or subtract moisture from, a constantly flowing air supply to maintain constant relative humidity levels in a sealed enclosure. All units use an internal reservoir to retain condensate water from the dehumidifying process; this water is used as a source of water for humidifying, or expelled via an overflow pipe. During periods where the unit is only humidifying, it may be necessary to add low mineral content water to the reservoir of recycling units. Water will almost certainly be needed for positive pressure units.

### Determining Water Demand for Positive Pressure Units

For positive pressure feed units, the quantity of water used depends on the volume of air delivered and relative humidity of the ambient air entering the unit.

To Calculate the Water Used by a Positive Pressure System

- Determine the flow rate of the air supply to the enclosure (F)
- Determine the humidity of the air entering the unit (RH in)
- Note the humidity of the air being delivered by the unit (RH out)
- Determine the moisture content of the air entering the unit (MC in), and the air being delivered by the unit (MC out) (a simple chart is below)
- Subtract the moisture content of the air being delivered from the moisture content of the ambient air supplied to the unit to obtain the amount of moisture used to humidify each m3 (MC out – MC in = MC). Multiply the result by the flow rate and the period you wish to measure (t) (MC X F X t = water used).

An Example:

A positive a pressure unit is delivering half a cubic meter each minute. The ambient air entering the unit is at 30% RH and the unit is delivering an air stream at 50% RH. How much water will it use in a day?

- From the chart below, we see that 50% RH has a moisture content of 8.7 g/ m3 (MC out) and 30% RH (MC in) has a moisture content of 5.2 g/ m3, so we can determine the water (demand) needed to create the appropriate RH in each cubic meter of air. (MC out – MC in = demand) is (8.7 ? 5.2 = 3.5)
- Each cubic meter of air is using 3.5 g of water.
- A gram is equivalent to 1/1000 of a liter of water, so the unit is consuming 0.0035 liters of water for every cubic meter of air that is being humidified.
- Remember, the unit is delivering about 30 m3 each hour, or 720 m3 in a day. Multiply the total time by the amount of water used per m3 of air, to get the total demand each day (720 m3 X 0.0035 = 2.5 liters of water consumed each day).

### To Calculate the Water Used by a Recycling System

For recycling units, the quantity of water used or condensate generated depends primarily on leakage into the showcase of ambient air. In some recycling unit applications, it is possible that water may never have to be added to the unit. This is because a recycling unit?s reservoir may be regularly replenished with condensate. The calculations below are, at best, approximate, and do not consider changes in temperature. We suggest that experience and a data log will provide the best guidelines. However, you may use the following procedure to estimate your water usage in extreme conditions (eg. very dry ambient air and leaky casework).

- Determine the leakage rate of the air entering the case
- Determine the humidity of the air entering the case
- Determine the resulting moisture content of the air entering the case
- Note the humidity of the air being maintained in the case
- Determine the new moisture content of the air that has leaked into the case
- Subtract the moisture content of the air leaking into the case from the moisture content of the case. This will give you an approximation of the daily consumption

An Example:

A 2 m3 case is leaking at 3 air changes per day, so air is entering the case at 6 cubic meters per day

The leaking air is entering the case at 20% RH

The moisture content of the air leaking into the case is 6 m3 X 3.5 g / m3 = 21 g water.

The air in the case is being maintained at 50%.

The new moisture content of the air that has leaked into the case is (6 m3 X 8.7 g / m3 = 52.2 g)

The difference between the original moisture content and the new moisture content of the air that has leaked into the case is (52.2 g – 21 g = 31.2 g).

The unit will use about 31.2 g of water per day.

RH% | g/m3 |

10 | 2 |

20 | 3.5 |

30 | 5.2 |

40 | 6.9 |

50 | 8.7 |

60 | 10.4 |

70 | 12.3 |