Browse the full thermodynamic tables as in the textbook —saturation by temperature and by pressure, superheated vapor and compressed liquid— for water, 10 refrigerants and 6 gases (nitrogen, oxygen, hydrogen, helium, argon and methane), plus air and N₂, O₂, CO₂, CO, H₂ and H₂O as ideal gases and a table of critical constants, in SI and English units. The calculator gives the state from two properties, with the exact value from the equation of state and the table interpolation step by step.
With linear interpolation between the two neighboring rows: y = y₁ + (x − x₁)/(x₂ − x₁)·(y₂ − y₁). In superheated vapor, if the pressure is not in the table, interpolate first in T in the two pressure blocks and then between them (double interpolation).
Compare with saturation: at a given temperature, P > Psat is compressed liquid and P < Psat is superheated vapor; with P and h, hf < h < hg is a mixture with quality x = (h − hf)/hfg.
From the reference equations of state (IAPWS-95 for water and those of each refrigerant), computed with CoolProp, free MIT-licensed software.
Almost every thermodynamics problem starts the same way: fix the state of a substance with two properties and read the others (specific volume, internal energy, enthalpy, entropy). For water and refrigerants this is done with saturation, superheated vapor and compressed liquid tables. This guide explains how to pick the right table and how to use quality in the mixture region, and works a turbine example.
The practical rule: with T and P, compare T with Tsat(P). With P and another property (h, s, v), compare it with the saturated liquid and vapor values at that pressure: if it falls between them, it is a mixture.
y = yf + x · (yg − yf) = yf + x · yfg
for y = v, u, h or s, where f is saturated liquid and g saturated vapor. Solving for x gives the quality from a known property: x = (y − yf)/yfg.
Psat = 101.42 kPa; hf = 419.17 kJ/kg and hg = 2675.57 kJ/kg, so hfg = 2256.40 kJ/kg, the heat needed to evaporate 1 kg of water at that temperature. With x = 0.90: h = 419.17 + 0.90 · 2256.40 ≈ 2449.9 kJ/kg, and v = 0.001043 + 0.90 · (1.6718 − 0.001043) ≈ 1.505 m³/kg.
Steam enters at 1 MPa and 300 °C and expands without losses (at constant entropy) down to 100 °C.
All these values are the ones the tool's state calculator gives, from CoolProp's reference equations of state.
On paper, when the state falls between two rows of the table, you interpolate linearly. The tool shows both: the interpolated table value, as in class, and the exact value from the equation of state, so you can see how far the interpolation is off.
Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach (McGraw-Hill); M. J. Moran, H. N. Shapiro et al., Fundamentals of Engineering Thermodynamics (Wiley). The tool's properties are computed with CoolProp's equations (I. H. Bell and contributors, MIT license). Simuveliq is not affiliated with these authors, publishers or the CoolProp project.
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