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Calculation of RO System Cleaning Tank Volume
The cleaning tank in a reverse osmosis (RO) system must be designed with corrosion resistance in mind. Common materials include fiberglass-reinforced plastic, polyvinyl chloride (PVC), or steel tanks lined with rubber. In regions with extreme temperatures, it's essential to install heating or cooling systems within the enclosure to maintain optimal conditions for RO membranes. The general cleaning temperature should not fall below 15°C, as lower temperatures can reduce the effectiveness of the cleaning process.
When determining the size of the cleaning tank, several factors must be considered: the volume of the pressure vessels, the security filter, and the piping used in the system. These elements collectively determine the total water volume that needs to be accommodated during the cleaning cycle.
For example, a factory’s RO system may use a 6-3 configuration, meaning six 8-inch membrane modules in the first stage and three 8-inch modules in the second stage. Each membrane module contains six 40-inch elements. To calculate the cleaning tank volume:
1. **Pressure Vessel Volume**:
Each pressure vessel has a volume calculated as:
$ V_1 = \frac{1}{4} \times \pi \times D^2 \times L = \frac{3.14}{4} \times \left(\frac{25.4 \times 8}{1000}\right)^2 \times 6 = 0.1884 \, \text{m}^3 $
For 9 pressure vessels:
$ V_2 = 0.1884 \times 9 = 1.6956 \, \text{m}^3 $
Assuming the membrane elements occupy 30% of the vessel volume, the effective water volume is:
$ V_3 = 1.6956 \times 70\% = 1.1869 \, \text{m}^3 $
2. **Pipeline Volume**:
A 50-meter-long pipeline with a diameter of φ89×3.5 (lining hose):
$ V_4 = \frac{1}{4} \times \pi \times D^2 \times L = \frac{3.14}{4} \times 0.082^2 \times 50 = 0.264 \, \text{m}^3 $
3. **Security Filter Volume**:
A filter with a diameter of 600 mm and an effective height of 1.3 m:
$ V_5 = \frac{1}{4} \times \pi \times D^2 \times H = \frac{3.14}{4} \times 0.6^2 \times 1.3 = 0.3674 \, \text{m}^3 $
Considering 5% of the filter volume is occupied by the filter media:
$ V_6 = 0.3674 \times 95\% = 0.349 \, \text{m}^3 $
4. **Total Volume**:
$ V = V_3 + V_4 + V_6 = 1.1869 + 0.264 + 0.349 = 1.8 \, \text{m}^3 $
Adding a safety factor of 1.2 for unexpected variables:
$ V' = 1.2 \times 1.8 = 2.16 \, \text{m}^3 $
Including a 0.5 m³ buffer for the cleaning cycle:
$ V'' = 2.16 + 0.5 = 2.66 \, \text{m}^3 $
Based on this calculation, a 3 m³ cleaning tank would be appropriate for the system. This ensures sufficient capacity for all components and accounts for operational variability.