🌬️ Psychrometric Sizer Tool

Calculate all moist air thermodynamic properties from any 2 inputs. Supports SI/Imperial units, barometric altitude adjustments, and real-time chart tracing.

⚡ Fortran 90 Engine Double Precision (IEEE 754) ✓ ISO / ASME Validated

Tools

📊 Solver Telemetry ● ACTIVE

👁️ Views 278

⚡ Solves 217

💾 Downloads 248 📦 Fortran Code 11.4 KB

📅 Released Jun 2026

⏱️ Latency < 1 ms

📝 Configuration

Key Equations:

P_sat = 0.6105·exp(17.27T/(T+237.3))
W = 0.622·P_w/(P_atm−P_w)
h = 1.006T + W(2501+1.86T) kJ/kg_da
SHR = Q_sensible/Q_total
Mixing: W_mix = (ṁ₁W₁+ṁ₂W₂)/(ṁ₁+ṁ₂)

📊 Results

Configure inputs and click Analyze to view results.

📘 Methodology

Psychrometric Properties

Moist air properties are computed from dry-bulb temperature and relative humidity using standard correlations: Magnus formula for saturation pressure, and ASHRAE relations for humidity ratio, enthalpy, specific volume, dew point, and wet bulb.

AHU Processes

Sensible heating/cooling: W constant, RH changes
Cooling & dehumidification: below dew point, condensate formed
Adiabatic humidification: follows wet-bulb line (h ≈ const)
Mixing: properties mix linearly on psychrometric chart

Sensible Heat Ratio

SHR = Q_sensible/Q_total characterizes the process. SHR=1 for pure sensible processes. Cooling coils typically have SHR = 0.6–0.8. The SHR line on the psychrometric chart determines coil selection and sizing.

📘 Calculation Methodology: Psychrometric Moist Air Properties & Humidity Ratio

Mathematical Model & Theory

Psychrometric analysis models moist air as a mixture of dry air and water vapor, quantifying humidity ratio $\omega$, relative humidity $\phi$, enthalpy $h$, and dew point $T_{dp}$:

$$\omega = 0.622 \frac{p_v}{p_{atm} - p_v}, \quad \phi = \frac{p_v}{p_{sat}(T)}$$

$$h = 1.006 T + \omega (2501 + 1.86 T) \quad [\text{kJ/kg}_{da}]$$

Assumptions

Ideal gas mixture of dry air and water vapor at atmospheric pressure.
ASHRAE Standard formulation for water saturation pressure $p_{sat}(T)$.

Academic References

ASHRAE Handbook — Fundamentals: Psychrometrics.
Moran, M. J. et al.: Engineering Thermodynamics, Ch. 12.

Worked Engineering Example

Problem Statement:
Air at $T = 25^\circ\text{C}$ and $P = 101.325\text{ kPa}$ has relative humidity $\phi = 50\%$ ($p_{sat} = 3.169\text{ kPa}$). Find humidity ratio $\omega$ and mixture enthalpy $h$.

Step-by-step Solution:
1. $p_v = 0.50 \times 3.169 = 1.5845\text{ kPa}$.
2. $\omega = 0.622 \times 1.5845 / (101.325 - 1.5845) = 0.9856 / 99.7405 = 0.00988\text{ kg}_{w}/\text{kg}_{da}$ ($9.88\text{ g/kg}$).
3. $h = 1.006(25) + 0.00988(2501 + 1.86 \times 25) = 25.15 + 0.00988(2547.5) = 25.15 + 25.17 = 50.32\text{ kJ/kg}_{da}$.
Final Result:
Humidity ratio is $\mathbf{9.88\text{ g/kg}}$ and enthalpy is $\mathbf{50.32\text{ kJ/kg}}$.

ThermoFluidCalc Report