A dehumidifier formula calculator helps you figure out how much water a dehumidifier needs to remove from the air to reach a desired level of humidity. This is crucial for maintaining comfortable and healthy indoor air quality. By using this calculator, you can ensure that your dehumidifier is effective for your specific space, whether it’s a room, a basement, or any other area.
Purpose and Functionality of the Calculator
The primary purpose of this calculator is to determine the amount of water that needs to be removed from the air to achieve a specific humidity level. This calculation depends on several factors:
- Volume of the Room (V): The total cubic volume of the space, measured in cubic meters (m³).
- Current Relative Humidity (RH1): The current humidity level in the room, expressed as a percentage (%).
- Desired Relative Humidity (RH2): The target humidity level after using the dehumidifier, expressed as a percentage (%).
- Temperature (T): The temperature of the room in degrees Celsius (°C), which affects the moisture content of the air.
How the Calculator Works
The calculator uses a series of formulas to determine the amount of water to be removed:
- Calculate the Saturation Vapor Pressure (Pws) at the given temperature (T) using the Magnus formula: Pws=0.61094×e(17.625×TT+243.04)P_{ws} = 0.61094 \times e^{\left(\frac{17.625 \times T}{T + 243.04}\right)}Pws=0.61094×e(T+243.0417.625×T) This gives the saturation vapor pressure in kilopascals (kPa).
- Calculate the Actual Vapor Pressure (Pw1 and Pw2) for both the current and desired humidity levels: Pw1=RH1100×PwsP_{w1} = \frac{RH1}{100} \times P_{ws}Pw1=100RH1×Pws Pw2=RH2100×PwsP_{w2} = \frac{RH2}{100} \times P_{ws}Pw2=100RH2×Pws
- Convert Vapor Pressure to Absolute Humidity (AH1 and AH2), which is the mass of water vapor per unit volume of air, using the ideal gas law: AH=Pw×1000R×(T+273.15)AH = \frac{P_w \times 1000}{R \times (T + 273.15)}AH=R×(T+273.15)Pw×1000 Where RRR is the specific gas constant for water vapor (approximately 461.5 J/kg/K).
- Calculate the Amount of Water to Remove per Cubic Meter of Air: Water_to_Remove=(AH1−AH2)×VWater\_to\_Remove = (AH1 – AH2) \times VWater_to_Remove=(AH1−AH2)×V
Step-by-Step Example
Let’s calculate the amount of water a dehumidifier needs to remove for a room with the following specifications:
- Volume (V) = 50 m³
- Current Humidity (RH1) = 70%
- Desired Humidity (RH2) = 50%
- Temperature (T) = 22°C
Calculation Steps
- Calculate the Saturation Vapor Pressure (Pws) at 22°C: Pws=0.61094×e(17.625×2222+243.04)P_{ws} = 0.61094 \times e^{\left(\frac{17.625 \times 22}{22 + 243.04}\right)}Pws=0.61094×e(22+243.0417.625×22) Pws≈0.61094×e(0.94004)P_{ws} \approx 0.61094 \times e^{(0.94004)}Pws≈0.61094×e(0.94004) Pws≈0.61094×2.5602P_{ws} \approx 0.61094 \times 2.5602Pws≈0.61094×2.5602 Pws≈1.564 kPaP_{ws} \approx 1.564 \, \text{kPa}Pws≈1.564kPa
- Calculate the Actual Vapor Pressure: Pw1=70100×1.564P_{w1} = \frac{70}{100} \times 1.564Pw1=10070×1.564 Pw1=1.0948 kPaP_{w1} = 1.0948 \, \text{kPa}Pw1=1.0948kPa Pw2=50100×1.564P_{w2} = \frac{50}{100} \times 1.564Pw2=10050×1.564 Pw2=0.782 kPaP_{w2} = 0.782 \, \text{kPa}Pw2=0.782kPa
- Convert Vapor Pressure to Absolute Humidity: AH1=1.0948×1000461.5×(22+273.15)AH1 = \frac{1.0948 \times 1000}{461.5 \times (22 + 273.15)}AH1=461.5×(22+273.15)1.0948×1000 AH1=1094.8461.5×295.15AH1 = \frac{1094.8}{461.5 \times 295.15}AH1=461.5×295.151094.8 AH1≈0.0078 kg/m³AH1 \approx 0.0078 \, \text{kg/m³}AH1≈0.0078kg/m³ AH2=0.782×1000461.5×(22+273.15)AH2 = \frac{0.782 \times 1000}{461.5 \times (22 + 273.15)}AH2=461.5×(22+273.15)0.782×1000 AH2=782461.5×295.15AH2 = \frac{782}{461.5 \times 295.15}AH2=461.5×295.15782 AH2≈0.0057 kg/m³AH2 \approx 0.0057 \, \text{kg/m³}AH2≈0.0057kg/m³
- Calculate the Amount of Water to Remove: Water_to_Remove=(0.0078−0.0057)×50Water\_to\_Remove = (0.0078 – 0.0057) \times 50Water_to_Remove=(0.0078−0.0057)×50 Water_to_Remove=0.0021×50Water\_to\_Remove = 0.0021 \times 50Water_to_Remove=0.0021×50 Water_to_Remove=0.105 kgWater\_to\_Remove = 0.105 \, \text{kg}Water_to_Remove=0.105kg Water_to_Remove≈0.105 litersWater\_to\_Remove \approx 0.105 \, \text{liters}Water_to_Remove≈0.105liters
This means the dehumidifier needs to remove approximately 0.105 liters of water from the air to achieve the desired humidity level.
Relevant Information Table
| Parameter | Symbol | Value | Unit |
|---|---|---|---|
| Volume of the Room | V | 50 | m³ |
| Current Humidity | RH1 | 70 | % |
| Desired Humidity | RH2 | 50 | % |
| Temperature | T | 22 | °C |
| Saturation Vapor Pressure | Pws | 1.564 | kPa |
| Actual Vapor Pressure 1 | Pw1 | 1.0948 | kPa |
| Actual Vapor Pressure 2 | Pw2 | 0.782 | kPa |
| Absolute Humidity 1 | AH1 | 0.0078 | kg/m³ |
| Absolute Humidity 2 | AH2 | 0.0057 | kg/m³ |
| Water to Remove | 0.105 | liters |
Conclusion
A dehumidifier formula calculator is an essential tool for determining how much water needs to be removed from the air to reach a desired humidity level. By understanding and applying the correct formulas, you can ensure that your dehumidifier is effective for your specific needs. This helps in maintaining comfortable and healthy indoor air quality, preventing issues like mold growth and damage to belongings. Whether for home, office, or industrial use, this calculator ensures you achieve optimal humidity levels efficiently.