Steam Table Calculator

Specify the state

Directions

This calculator finds the thermodynamic properties of water and steam from any two independent properties. Select two properties from the menus, enter their values and units, and click Solve (or press Enter). You can combine temperature, pressure, specific volume, internal energy, enthalpy, entropy and quality; the only pair not allowed is internal energy with enthalpy. Note that selecting quality and one other variable yields saturation properties. A state can also be selected by clicking almost anywhere on the diagram or by dragging the red dot.

The table lists all properties of the state; the two you entered are in bold. For a saturated liquid–vapor mixture, the table also lists the saturated liquid (subscript f) and saturated vapor (subscript g) values used to calculate the mixture properties. Change the units of any result from its menu, or use the SI and English buttons to change all units at once.

Sometimes more than one state has the two values you entered, for example a temperature and an enthalpy that match both a compressed liquid and a saturated mixture. All such states are listed; click a button above the table to show one of them.

The diagram shows the saturation curve, the state (red dot) and a line of constant pressure (T–s and h–s diagrams) or constant temperature (P–h and P–v diagrams) through the state. The Examples menu loads some typical states.

Details

Properties are calculated with the IAPWS Industrial Formulation 1997 for the Thermodynamic Properties of Water and Steam (IAPWS-IF97), the standard used for steam tables in engineering. IF97 divides the range covered here into four regions, each with its own equation:

Region 1: compressed liquid up to 350 °C
Region 2: superheated vapor
Region 3: liquid, vapor and supercritical fluid near the critical point (350 °C to about 590 °C, above 16.5 MPa)
Region 4: the saturation curve (liquid–vapor equilibrium)

The constants are: $$R=0.451526 \, \text{kJ}/(\text{kg K}), \qquad T_c=647.096 \, \text{K}, \qquad p_c=22.064 \, \text{MPa}, \qquad \rho_c = 322 \, \text{kg}/\text{m}^3$$

Region 1: dimensionless Gibbs free energy

$$\frac{g(p,T)}{RT}=\gamma(\pi,\tau)=\sum_{i=1}^{34} n_i\,(7.1-\tau)^{I_i}\,(\tau-1.222)^{J_i}, \qquad \pi = \frac{p}{16.53 \, \text{MPa}}, \quad \tau = \frac{1386 \text{K}}{T}$$

The other properties come from derivatives of the Gibbs free energy g:

$$ v=\left(\frac{\partial g}{\partial T}\right)_T,\qquad s=-\left(\frac{\partial g}{\partial T}\right)_p,\qquad h=g+Ts,\qquad u=h-pv $$

Region 2: ideal-gas part plus residual part

$$ \gamma = \gamma^\circ +\gamma^r,\qquad \gamma^\circ=\ln{\pi+\sum_{i=1}^{9}n_i^\circ \tau^{J_i^\circ}} ,\qquad \gamma^r=\sum_{i=1}^{43}n_i\,\pi^{I_i}\,\left(\tau-0.5\right)^{J_i} $$

$$ \pi=\frac{p}{1\:\text{MPa}},\qquad \tau=\frac{540\,\text{K}}{T} $$

Region 3: dimensionless Helmholtz free energy

$$ \frac{f(\rho,T)}{RT}=\phi(\delta,\tau)=n_1\ln{\delta}+\sum_{i=2}^{40}n_i \delta^{I_i} \tau^{J_i},\qquad \delta=\frac{\rho}{\rho_c},\quad\tau=\frac{T_c}{T} $$

$$ p=\rho\, RT\, \delta \frac{\partial\phi}{\partial\delta},\qquad h=RT\left(\tau\frac{\partial\phi}{\partial\tau}+\delta\frac{\partial\phi}{\partial\delta}\right),\qquad s=R\left(\tau\frac{\partial\phi}{\partial\tau}-\phi\right),\qquad u=h-pv $$

The Region 3 equation has density as its independent variable, so when T and P are known the density is calculated by solving the pressure equation numerically. Below the critical temperature, the largest root is the liquid and the smallest root is the vapor.

Region 4: saturation curve

The equation is quadratic in both β and ϑ, so it can be solved directly for saturation pressure from temperature or for saturation temperature from pressure.

Saturated liquid–vapor mixtures

For quality x (mass fraction vapor), each specific property y of a saturated mixture is:

Solving for the state

When temperature and pressure are given, the properties are calculated directly from the equation for that region. For other pairs, the calculator finds the state numerically: it scans the range for sign changes and then uses a bracketed root finder (Illinois method). Temperature and pressure together cannot specify a state inside the two-phase region because they are not independent there; use quality with either T or P.

Some pairs have more than one solution. For example, the enthalpy of saturated vapor has a maximum near 235 °C, so a given h_g corresponds to two saturation temperatures. Also, at a fixed temperature, the enthalpy of a compressed liquid increases with pressure, so it can equal the enthalpy of a low-quality saturated mixture at the same temperature. Internal energy and enthalpy are not allowed as a pair: for a vapor that behaves nearly as an ideal gas, both depend almost only on temperature.

Reference state and units

Internal energy and entropy are zero for saturated liquid at the triple point (0.01 °C, 0.6117 kPa). English units are converted from SI: 1 Btu/lbm = 2.326 kJ/kg, 1 Btu/(lbm·R) = 4.1868 kJ/(kg·K), 1 ft³/lbm = 0.062428 m³/kg and 1 psia = 6.894757 kPa.

This calculator covers 0 to 600 °C and 0.1 kPa to 100 MPa; it does not include metastable states or ice.

Symbols

c_p	heat capacity at constant pressure (kJ/(kg·K))
f	specific Helmholtz free energy (kJ/kg)
g	specific Gibbs free energy (kJ/kg)
h	specific enthalpy (kJ/kg)
I_i, J_i, J_i^°	exponents in the IF97 equations
n_i, n_i^°	coefficients in the IF97 equations
P, p	pressure (kPa or MPa)
p_c	critical pressure (22.064 MPa)
p_sat	saturation pressure (MPa)
R	specific gas constant for water (0.461526 kJ/(kg·K))
s	specific entropy (kJ/(kg·K))
T	temperature (K)
T_c	critical temperature (647.096 K)
T_sat	saturation temperature (K)
u	specific internal energy (kJ/kg)
v	specific volume (m³/kg)
x	quality (mass fraction of vapor)
y	any specific property: v, u, h or s
y_f, y_g	property y of saturated liquid and of saturated vapor
Greek symbols
β	reduced saturation pressure, (p_sat/1 MPa)^1/4
γ, γ^°, γ^r	dimensionless Gibbs free energy, g/(RT), and its ideal-gas and residual parts
δ	reduced density, ρ/ρ_c
ϑ	transformed saturation temperature in the Region 4 equation
π	reduced pressure
ρ	density (kg/m³)
ρ_c	critical density (322 kg/m³)
τ	inverse reduced temperature
ϕ	dimensionless Helmholtz free energy, f/(RT)

About

This simulation was generated by Professor David L. Silverstein and Dr. Loyal Murphy of the University of Mississippi using Google Gemini. It was modified in the Department of Chemical and Biological Engineering at University of Colorado Boulder for LearnChemE.com by John L. Falconer using Claude AI. It was prepared with financial support from the National Science Foundation (DUE 2336987 and 2336988). Address any questions or comments to LearnChemE@gmail.com.