💧 Steam Tables Advanced (IAPWS-IF97)
High-accuracy industrial steam tables solver conforming strictly to the international IAPWS-IF97 standard. Supports five input pairs with real-time T-s diagram visual plotting.
Thermodynamic State Boundaries
Water properties change non-linearly across liquid, vapor, and supercritical phases. Choose your input pair to locate the state point on the Temperature-Entropy ($T$-$s$) diagram.
Liquid (Region 1)
Saturated Mixture (Region 4)
Vapor (Region 2)
Supercritical (Region 3)
📝 Configuration
IAPWS-IF97 Standard formulation:
Calculates certified industrial properties for water and steam based on the five regions and boundary equations.
Standard Region Limits:
• Region 1: Liquid phase up to 350 °C ($623.15$ K)
• Region 2: Vapor phase up to 800 °C ($1073.15$ K)
• Region 3: Near-critical / supercritical states
• Region 4: Saturated mixture (wet dome)
Calculates certified industrial properties for water and steam based on the five regions and boundary equations.
Standard Region Limits:
• Region 1: Liquid phase up to 350 °C ($623.15$ K)
• Region 2: Vapor phase up to 800 °C ($1073.15$ K)
• Region 3: Near-critical / supercritical states
• Region 4: Saturated mixture (wet dome)
📊 Results & Visualization
Configure inputs and click Calculate Steam Properties to view results.
📘 Calculation Methodology
Mathematical Model & Theory
Thermodynamic properties of water and steam are modeled using the formulation IAPWS-IF97, which divides the thermodynamic state space of water into 5 distinct regions based on temperature and pressure boundaries:
- Region 1: Subcooled liquid (Gibbs free energy formulation $g(P,T)$)
- Region 2: Superheated vapor (Gibbs free energy formulation $g(P,T)$)
- Region 3: Near-critical and supercritical fluid (Helmholtz free energy formulation $a(\rho,T)$)
- Region 4: Saturated liquid-vapor mixture ($P_{sat}(T)$ or $T_{sat}(P)$ boundary curves)
Academic References & Standards
- IAPWS-IF97: International Association for the Properties of Water and Steam, "Revised Release on the IAPWS Formulation 1997 for the Thermodynamic Properties of Water and Steam for Industrial Use" (2014).
- ASME: ASME International Steam Tables for Industrial Use, WT. Parry et al., 3rd Edition.
Worked Engineering Example
Problem Statement:
Find the enthalpy and quality of water at $P = 10\text{ bar}$ ($1\text{ MPa}$) and $T = 179.88$°C (saturation conditions) if its entropy is $s = 5.0$ kJ/(kg·K). (At 10 bar, $s_f = 2.138$, $s_g = 6.585$ kJ/(kg·K), $h_f = 762.6$, $h_g = 2777.1$ kJ/kg).
Step-by-step Solution:
1. Calculate quality $x$ from entropy:
$$x = \frac{s - s_f}{s_{fg}} = \frac{5.0 - 2.138}{6.585 - 2.138} = \frac{2.862}{4.447} = 0.6436$$ 2. Calculate mixture enthalpy $h$:
$$h = h_f + x(h_g - h_f) = 762.6 + 0.6436 \times (2777.1 - 762.6) = 2059.2 \text{ kJ/kg}$$
Final Result:
Steam quality is 0.644 and enthalpy is 2059.2 kJ/kg.
Find the enthalpy and quality of water at $P = 10\text{ bar}$ ($1\text{ MPa}$) and $T = 179.88$°C (saturation conditions) if its entropy is $s = 5.0$ kJ/(kg·K). (At 10 bar, $s_f = 2.138$, $s_g = 6.585$ kJ/(kg·K), $h_f = 762.6$, $h_g = 2777.1$ kJ/kg).
Step-by-step Solution:
1. Calculate quality $x$ from entropy:
$$x = \frac{s - s_f}{s_{fg}} = \frac{5.0 - 2.138}{6.585 - 2.138} = \frac{2.862}{4.447} = 0.6436$$ 2. Calculate mixture enthalpy $h$:
$$h = h_f + x(h_g - h_f) = 762.6 + 0.6436 \times (2777.1 - 762.6) = 2059.2 \text{ kJ/kg}$$
Final Result:
Steam quality is 0.644 and enthalpy is 2059.2 kJ/kg.