Properly designing ventilation and air conditioning for a swimming pool is not a simple task. There are at least several dozen different methods that can be used to calculate the necessary amount of air to assimilate moisture gains in a pool area.
In Poland, none of these methods have been formalized into a standard or regulation, so, colloquially speaking, there is a "free-for-all" approach. However, our western neighbors have legally regulated this matter, and in Germany, the VDI 2089 standard is in force. I personally use this standard as well when sizing pool ventilation/air conditioning systems.
That said, it’s important to note that this standard does not fully address the issue of pool ventilation/air conditioning. Why? Because it only allows for the calculation of moisture gains from the pool basin, wet floors, water attractions, and wet, dry, and latent heat gains, and based on this, the required air flow to assimilate the moisture gains can be sized. However, it does not enable designing a ventilation/air conditioning system in such a way that it maintains the assumed temperature in the pool hall.
To achieve this, it is necessary to account for the thermal load of the pool hall during the summer, resulting from heat gains through transparent and opaque partitions, and in winter, the heat losses calculated using the EN 12831 standard.
Let’s proceed with a calculation example based on the VDI 2089 standard, utilizing the KHB module of the IX-CHART program, as well as the following assumptions:
Pool hall with a pool basin area of 312 m²
Wet floors accounting for 50% of the pool basin area
Water attractions: geyser, mushroom fountain, jacuzzi
Calculated summer thermal load from partitions and lighting, determined using the Teknosim 2000 program by Lindab: Qtw ~39 kW
Calculated heat losses for the pool hall, determined using the InstalSoft program according to EN 12831: Qw = 75 kW
Pool water temperature: +28°C
Pool hall air temperature: +30°C
Maximum relative humidity: RH = 55%
Alright, we select the pool hall calculator module and enter the data according to the assumptions:
We obtained the following information from the calculations:
Required ventilation airflow for moisture gain assimilation: 30,952 m³/h
Total moisture gains from the pool: wc = 97.2 kg/h
Wet heat gains (simplified here as latent heat, though not exactly the same): Ql = 68.9 kW
Dry heat gains (simplified here as sensible heat, though not exactly the same): Qs = -3.9 kW
Total heat gains from the pool basin and attractions: Qtsp=Ql+Qs=65kW
Total heat gains including all sources in summer: Qt=Qtsp+Qtw=65+39=104 kW
To maintain a pool hall temperature of +30°C in summer, it is evident that the supply air temperature must be lower than the pool hall temperature. Additionally, the calculated 97.2 kg/h of moisture must be removed. If we assume the permissible temperature difference between the supply air and the pool hall air temperature to maintain occupant comfort is no more than 5K, we must adjust the calculations for the ventilation/air conditioning airflow capable of achieving these conditions. The minimum supply air temperature we can apply is +25°C.
At this point, the Psychrometric/Mollier chart comes to our aid. Specifically, I will use the KCW module, which saves me the tedious process of manual calculations and plotting on the chart.
Thus, I enter the data into the calculator, and the results are as follows:
In step 1, we define the parameters of the intake air.
In step 2, we select the option to calculate the airflow based on the heat balance.
In step 3, we input the calculated total heat and moisture gains.
In step 4, we specify the required air temperature in the room.
In step 5, we define the air temperature after the cooling coil and the temperature of the coil surface. These two parameters can be iteratively adjusted to achieve the previously calculated ventilation airflow.
It is important to ensure that the assumed Δt = 5K is not exceeded, which is the case here, as the supply air temperature required to remove the calculated heat and moisture gains is +26.6°C, resulting in Δt = 3.4°C.
In step 6, we can read the required cooling capacity to be provided by the cooling unit and the amount of condensed water from the intake air. In this case, we need 92.6 kW of cooling capacity.
On the Mollier chart, such processes appear as follows:
Of course, if it turned out that the heat gains were greater and achieving Δt = 5K was not possible, it would then be necessary to increase the ventilation airflow. In this case, the determining factor for the required airflow would not be the amount of moisture to be removed but the amount of heat needed to maintain the temperature.
We can even test what would happen to the airflow if the heat gains increased by 50 kW, reaching 154 kW instead of 104 kW, with a minimum supply air temperature of +25°C. The result is shown below:
The required airflow would then be 50,700 m³/h, and the required cooling capacity would rise to 198 kW! So, if we only considered moisture gains and used a smaller airflow, we would indeed be able to remove the moisture gains, but we would fail to maintain the designed temperature of +30°C. :(
Now, let’s return to the previous example, but let’s also calculate the winter variant.
The heat gains in the pool hall remain the same; however, the external conditions have changed, and the heat gains from the partitions have now turned into heat losses.
Our balance now looks as follows:
Moisture gains: wc = 97.2 kg/h
Heat gains: Qt = 65 kW - 75 kW = -10 kW, so we need to supply +10 kW of heat along with the ventilation air to compensate for these losses. This will result in the need to supply air warmer by a certain Δt than the temperature of the pool hall.
Since, in winter, the ventilation air is dry, there is no need to use the cooling system to dehumidify the air. The external air is sufficient. However, during freezing conditions, the air is so dry that it may be necessary to use air recirculation to avoid overly drying the air, which would intensify evaporation. Recirculation, when possible, is the best method for heat recovery. In addition to heat recovery in the mixing chamber, we also use a rotary heat exchanger, which further helps recover energy.
We will perform the calculations for a temperature of -20°C and RH = 100%. The results from the ventilation central calculator are as follows:
To remove 97.2 kg/h of moisture from the pool hall and simultaneously supply 10 kW of heat to maintain the temperature, while keeping the same airflow as in summer, i.e. 30,888 m³/h, we would need to use an air recirculation rate of 80.315% and raise the air temperature by Δt = 1.013°C relative to the pool hall temperature. This means we would need to supply air at a temperature of 31.013°C, which would require a heating power of 34.43 kW to the heating unit.
This level of recirculation can only be used on the condition that the amount of fresh air supplied, approximately 5,000 m³/h, will be sufficient from a hygienic standpoint for the people in the pool hall. Assuming a minimum of 30 m³/h per person, the hall could accommodate about 166 people at the same time.
The processes on the Mollier chart are shown below:
To summarize, in this specific case, the air flow could be adopted according to VDI 2089, but it is always important to consider the calculated heat load and conduct a full analysis to ensure that the designed ventilation system maintains both humidity and temperature at the proper levels.
For our case, the required air supply flow is V = 30,888 m³/h, the required cooling power Qc = 92.6 kW, and the required heating power Qh = 34.4 kW.
As you can see, this is not a simple task, but by following the step-by-step procedure outlined above and having IX-CHART at hand, it becomes relatively quick to handle.
I invite you, dear reader, to download the program and explore the tool on your own.
If you would like to show your gratitude for the knowledge shared and want this portal to continue growing, feel free to buy me a coffee, for which I would be very grateful :)
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