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The prime reason for ventilation of enclosed pool areas is to prevent build up of uncomfortable and possibly damaging humidity. Therefore, ventilation rates for pool areas can best be determined by examining the moisture removing capabilities of an

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CALCULATION OF VENTILATION REQUIREMENTS FOR INDOOR POOLSBASED ON MOISTURE EVAPORATION
The prime reason for ventilation of enclosed pool areas is to prevent build up of uncomfortable andpossibly damaging humidity. Therefore, ventilation rates for pool areas can best be determined byexamining the moisture removing capabilities of an air-to-air heat exchanger at various flow rates andthen comparing this capability with the amount of moisture introduced into the air from the pool.In general , it may be said that a mechanical ventilation rate of 1 ACH (one air change per hour) of thepool enclosure will be sufficient to maintain reasonable relative humidity levels when a pool cover isused regularly. However, the ventilation system should be capable of supplying 2 (two) ACH for main-tenance of good air quality in all operating conditions.
MOISTURE PRODUCTION OF INDOOR POOLS:
The amount of water evaporated from a pool depends on:a)the surface area of the pool;b)the water temperature;c)the indoor air temperature and relative humidity;d)the amount of air movement over the pool surface.In short, the following principles apply:1)The larger the water surface area, the greater the amount of water evaporated. Therefore,using a pool cover to reduce the surface area of water exposed reduces the amount ofwater evaporated;2)The higher the water temperature, the higher the evaporation rate;3)The lower the indoor air temperature, the higher the evaporation rate;4)The lower the indoor relative humidity the higher the evaporation rate;5)The greater the air movement over the pool area, the higher the evaporation rate. Thus,activity in the pool area will increase the evaporation rate.
CHART 1
(Evaporation Rates) gives specific rates for various conditions.
Cost-effective ventilation for people
CALCULATION OF VENTILATION REQUIREMENTS FOR INDOOR POOLSBASED ON MOISTURE EVAPORATION ContinuedMOISTURE REMOVAL CAPABILITIES:
In addition to the volume of air being exchanged, moisture removal depends largely on the moisturecontent of both the inside and outside air. The actual content can be calculated knowing the air tem-perature and its relative humidity.While CHART ll (Flowrate Factor) shows specific flowrates required for a pound of water removal, thetrends below outline the principles:1)The higher the indoor relative humidity, the larger the moisture removing capabilities;
2)
The higher the indoor temperature, the larger the moisture removing capabilities;3)The lower the outdoor temperature, the larger the moisture removing capabilities.
Detailed Calculation of Total Ventilation Requirements:
Using the CHARTS and equations provided on the worksheet, and making decisions on water tem-perature, indoor temperature and relative humidity levels based on comfort levels, energy use andwindow construction (triple glazed windows will allow higher humidity levels before condensationoccurs than will single or double glazed), calculate the required total ventilation rate.The application of the theoretical total ventilation rate, to the pool in design, should be adjusted fornatural ventilation rates, pool, and pool cover use, to give the required mechanical ventilation rate.The calculation for the minimum air required to remove water which evaporates from the surface of apool is obtained from the ASHRAE APPLICATIONS HANDBOOK.The first table contains the “evaporation rate” of water based on the water temperature, room tempera-ture and room relative humidity.The second table contains the “flowrate factor” based on the indoor temperature, outdoor temperature,and room relative humidity.The total ventilation rate per square foot of water surface area is the product of these two values.The total ventilation rate is this product multiplied led by the water surface area.(See calculation next page)
CALCULATION OF VENTILATION REQUIREMENTS FOR INDOOR POOLSBASED ON MOISTURE EVAPORATION Continued
Calculation:1.
From Table 1, select the appropriate “evaporation rate” based on the room air temperature, watertemperature and relative humidity.
Example:
From Table IAir Temp. = 30
0
CWater Temp. = 28
0
CRelative Humidity = 50%Evaporation Rate = 0.050 lb/(sq.ft-hr.)
2.
From Table ll, select the corresponding “flowrate factor” depending on the indoor air temperature,outdoor air temperature
and room reIative humidity.
Example:
From Table llIndoor Temp. = 30
0
COutdoor Temp. = -10
0
CRelative Humidity = 50%Flowrate Factor = 19.20 cfm-hr/lb.3.Multiply the values obtained from Step 1 and Step 2 to obtain the minimum required flowrate.Min. flowrate = 0.050 x 19.20 cfm/sq.ft.Min. flowrate = .96 cfm/sq.ft.4.Multiply the value in Step 3 by the area of the pool.
Example:
It the pool area is 500 sq. ft. (a standard 16' x 32' pool)The total required flowrate = 480 cfm.This total ventilation rate can now be used to calculate the mechanical ventilation rate required.Assume that the pool enclosure has a natural ventilation rate of .25 ACH (or if the actual value isknown, use the known value). Calculate the pool enclosure volume and corresponding air volumeleakage in cfm.(See example next page)
CALCULATION OF VENTILATION REQUIREMENTS FOR INDOOR POOLSBASED ON MOISTURE EVAPORATION Continued
Example:
If pool enclosure is 30' x 50' x 12'Total volume is 18,000 cubic feetWith a natural ventilation rate of .25 ACHAir volume per hour= 18,000 x .25= 4500 cu.ft./hrOR 75 cu.ft./min (cfm)Subtract the natural ventilation rate from the total ventilation rate required.
In our Example:
Mechanical ventilation rate required = 480 - 75 = 405 cfmIf a pool cover is to be used regularly, the mechanical ventilation rate can be reduced by 1/3rd.
In our Example:
Mechanical ventilation rate required with pool cover = 405 x .66 = 267 cfmThis would represent the minimum mechanical ventilation rate to be supplied.
***
SYSTEM INSTALLATION
After calculating the mechanical ventilation rate required for the pool enclosure, it is necessary toconsider the distribution network for both the supply and exhaust air streams.Proper duct design will:1.minimize air flow requirements;
2.
ensure a comfortable recreation environment;3.optimize humidity control , including elimination of condensation on windows.Please refer to Sketch I and 2, enclosed, for typical layouts.
In general, please note that:
1.The airflows and duct lengths Indicated on the sketches are meant as examples only - actualperformance may differ;
CALCULATION OF VENTILATION REQUIREMENTS FOR INDOOR POOLSBASED ON MOISTURE EVAPORATION Continued
2.Supply air should be discharged near exterior windows. If the supply air is being heated, thesupply grills may be close to the ground - otherwise, discharge height should be at leasteight feet.3.Return grills and lines should be located near the ceiling;4.Maintain a reasonable distance (at least eight feet) between supply and return grills, toprevent short circuiting.
Reheating of Supply Air:
Although the air-to-air heat exchanger will recover up to 80% of indoor air temperature , the incomingfresh air supply may be uncomfortably cooI. Therefore it may be desirable to add an electric ductheater to heat the incoming air.The accompanying chart gives examples of the power required in a duct heater to heat the incomingair at various conditions.
REHEAT REQUIRED TO BRING THE INCOMING AIRSTREAM UP TOVARIOUS TEMPERATURES BASED ON TWO AIRFLOWS
Airflow Indoor Air Temp.Reheat Required at Outdoor Air Temp. (KW)(cfm)(
0
C)O
0
C-10
0
C-20
0
C-30
0
C350201.52.53.04.0350252.53.54.05.0350303.54.55.06.0700203.55.57.08.5700255.57.59.010.5700307.59.511.012.5

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