🌡️ Psychrometrics Calculator
Determine properties of moist air. Evaluate relative humidity, wet-bulb and dew-point temperatures, specific enthalpy, volume, and vapor pressures with standard sea-level or variable barometric pressure.
📝 Configuration
Thermodynamic Definitions:
• Relative Humidity: $RH = P_v / P_{sat}(T)$
• Saturation Vapor Pressure: Buck Formulation (kPa)
• Specific Enthalpy: $h = 1.006T + \omega(2501 + 1.86T)$ (kJ/kg)
• Specific Volume: $v = R_{air}(T + 273.15)/(P - P_v)$ (m³/kg)
• Humidity Ratio: $\omega = 0.62198 P_v/(P - P_v)$ (kg/kg)
Note: Wet-bulb temperature is solved iteratively using the adiabatic saturation equation to within 10⁻⁵ °C precision.
• Relative Humidity: $RH = P_v / P_{sat}(T)$
• Saturation Vapor Pressure: Buck Formulation (kPa)
• Specific Enthalpy: $h = 1.006T + \omega(2501 + 1.86T)$ (kJ/kg)
• Specific Volume: $v = R_{air}(T + 273.15)/(P - P_v)$ (m³/kg)
• Humidity Ratio: $\omega = 0.62198 P_v/(P - P_v)$ (kg/kg)
Note: Wet-bulb temperature is solved iteratively using the adiabatic saturation equation to within 10⁻⁵ °C precision.
📊 Results & Visualization
Configure inputs and click Solve Moist Air Properties to view results.
📘 Calculation Methodology
Mathematical Model & Theory
Psychrometrics is the study of moist air properties. Key definitions include relative humidity ($RH$), humidity ratio (specific humidity $\omega$), and dew-point temperature:
$$\omega = 0.622 \frac{P_v}{P - P_v}, \quad RH = \frac{P_v}{P_{sat}(T)}$$
$$h = 1.006 T + \omega (2501 + 1.86 T) \quad \text{[kJ/kg dry air]}$$
Worked Engineering Example
Problem Statement:
Air at 25°C and 101.325 kPa has a relative humidity $RH = 50\%$. Find the humidity ratio $\omega$ and specific enthalpy $h$. ($P_{sat}$ at 25°C is 3.169 kPa).
Step-by-step Solution:
1. Calculate vapor pressure $P_v$:
$$P_v = RH \times P_{sat} = 0.50 \times 3.169 = 1.5845 \text{ kPa}$$ 2. Calculate humidity ratio $\omega$:
$$\omega = 0.622 \times \frac{1.5845}{101.325 - 1.5845} = 0.00987 \text{ kg water / kg dry air} = 9.87 \text{ g/kg}$$ 3. Calculate specific enthalpy $h$:
$$h = 1.006 \times 25 + 0.00987 \times (2501 + 1.86 \times 25) = 25.15 + 25.14 = 50.29 \text{ kJ/kg dry air}$$
Final Result:
Humidity ratio is 9.87 g/kg and enthalpy is 50.29 kJ/kg.
Air at 25°C and 101.325 kPa has a relative humidity $RH = 50\%$. Find the humidity ratio $\omega$ and specific enthalpy $h$. ($P_{sat}$ at 25°C is 3.169 kPa).
Step-by-step Solution:
1. Calculate vapor pressure $P_v$:
$$P_v = RH \times P_{sat} = 0.50 \times 3.169 = 1.5845 \text{ kPa}$$ 2. Calculate humidity ratio $\omega$:
$$\omega = 0.622 \times \frac{1.5845}{101.325 - 1.5845} = 0.00987 \text{ kg water / kg dry air} = 9.87 \text{ g/kg}$$ 3. Calculate specific enthalpy $h$:
$$h = 1.006 \times 25 + 0.00987 \times (2501 + 1.86 \times 25) = 25.15 + 25.14 = 50.29 \text{ kJ/kg dry air}$$
Final Result:
Humidity ratio is 9.87 g/kg and enthalpy is 50.29 kJ/kg.