Properties of pure substances and steam tables

Why it matters

Steam power plants, process boilers, refrigerators and heat pumps all run on a working fluid that changes phase, and no simple equation of state describes it across liquid, wet and vapour states. Engineers therefore read properties from tables or charts. Fixing the state correctly and reading the right table is the step that every Rankine-cycle, turbine or refrigeration calculation depends on.

Key ideas

Pure substance. A substance with a fixed, uniform chemical composition throughout, even if it exists in more than one phase (water + steam, liquid + vapour R-134a). A homogeneous gas mixture such as air is treated as pure as long as no component condenses.

Phase change at constant pressure (heating water in a piston–cylinder).

1. Compressed (subcooled) liquid: T below the saturation temperature for the pressure.
2. Saturated liquid: liquid about to boil (state f).
3. Wet (saturated) mixture: liquid and vapour coexist; T stays at T_sat while heat is added.
4. Saturated (dry) vapour: the last drop has just evaporated (state g).
5. Superheated vapour: T above T_sat for the pressure.

Saturation. For a pure substance, T_sat and p_sat are linked one-to-one: water boils at 99.6 °C at 100 kPa and at 179.9 °C at 1 MPa. In the wet region p and T are therefore not independent, and a second property such as dryness fraction x, v, h or s is needed to fix the state.

Dryness fraction (quality) x = m_vapour / m_total, defined only in the wet region (0 ≤ x ≤ 1). Any specific property of the mixture is the mass-weighted mean of the f and g values.

Critical point. The top of the saturation dome, where the saturated-liquid and saturated-vapour lines meet and h_fg = 0. For water: T_c ≈ 373.95 °C, p_c ≈ 22.06 MPa, v_c ≈ 0.003106 m³/kg. Above it there is no distinct boiling; the fluid is supercritical. Triple point of water: 0.01 °C, 0.6117 kPa, where solid, liquid and vapour coexist.

Property diagrams. T–v, p–v, T–s and h–s (Mollier) diagrams show the dome with the critical point at the top. On T–s, a constant-pressure line is horizontal inside the dome; on the Mollier chart, turbine expansions are read as vertical (isentropic) drops.

Using the tables — a decision procedure.

• Given p and T: compare T with T_sat(p). T < T_sat → compressed liquid; T = T_sat → wet (need one more property); T > T_sat → superheated.
• Given p and v (or h, or s): compare with the f and g values at that p. Between them → wet, find x; above g → superheated table.
• Compressed liquid is usually approximated by saturated liquid at the same temperature: v ≈ v_f(T), u ≈ u_f(T), h ≈ h_f(T) (a better estimate is h ≈ h_f(T) + v_f (p − p_sat)).
• Steam tables take u = 0 and s = 0 for saturated liquid at the triple point. Only differences in h, u and s have physical meaning.

Throttling calorimeter. Wet steam throttled through a valve keeps its enthalpy (h₁ = h₂). If the throttled steam is superheated, its p and T give h₂, and therefore x₁. This works only for steam that is already fairly dry; very wet steam needs a separating-and-throttling calorimeter.

Formulas

v = V / m, u = U / m, h = H / m, s = S / m — specific properties (m³/kg, kJ/kg, kJ/kg, kJ/kg·K).

h = u + p v

• h, u in kJ/kg; p in kPa; v in m³/kg (kPa·m³/kg = kJ/kg).

x = m_g / (m_f + m_g) — dryness fraction, dimensionless, wet region only.

v = v_f + x v_fg = v_f + x (v_g − v_f) u = u_f + x u_fg h = h_f + x h_fg s = s_f + x s_fg

• Subscript f: saturated liquid; g: saturated vapour; fg = g − f (h_fg is the latent heat of vaporisation at that pressure). Valid only inside the dome.

x = (y − y_f) / (y_g − y_f) — dryness fraction from any known property y (v, u, h or s).

h_fg = T_sat · s_fg — follows from Q = TΔs for reversible isothermal evaporation (T_sat in K).

y = y₁ + (y₂ − y₁)(T − T₁)/(T₂ − T₁) — linear interpolation between two table entries.

Worked examples

Table values below are from standard steam tables (IAPWS-based); your data book may differ in the last digit.

Example 1 (standard). Steam at 1 MPa has a dryness fraction of 0.9. Find T, v, h and s. At 1 MPa: T_sat = 179.88 °C, v_f = 0.001127, v_g = 0.19436 m³/kg, h_f = 762.5, h_fg = 2014.6 kJ/kg, s_f = 2.1381, s_fg = 4.4469 kJ/kg·K.

1. Wet steam is at the saturation temperature: T = 179.88 °C.
2. v = v_f + x (v_g − v_f) = 0.001127 + 0.9 × (0.19436 − 0.001127) = 0.001127 + 0.17391 = 0.1750 m³/kg.
3. h = h_f + x h_fg = 762.5 + 0.9 × 2014.6 = 762.5 + 1813.1 = 2575.6 kJ/kg.
4. s = s_f + x s_fg = 2.1381 + 0.9 × 4.4469 = 2.1381 + 4.0022 = 6.1403 kJ/kg·K.

T = 179.9 °C, v ≈ 0.175 m³/kg, h ≈ 2575.6 kJ/kg, s ≈ 6.140 kJ/kg·K

Example 2 (GATE level). Steam from a 1 MPa main is sampled with a throttling calorimeter. After the throttle the steam is at 100 kPa and 120 °C. Find the dryness fraction in the main. Data: at 100 kPa, T_sat = 99.6 °C, so 120 °C is superheated; h(100 kPa, 120 °C) ≈ 2716.6 kJ/kg (superheated table). At 1 MPa: h_f = 762.5, h_fg = 2014.6 kJ/kg.

1. Throttling is adiabatic with no work and negligible change in kinetic energy: h₁ = h₂.
2. h₂ = 2716.6 kJ/kg, so h₁ = 2716.6 kJ/kg.
3. x₁ = (h₁ − h_f) / h_fg = (2716.6 − 762.5) / 2014.6 = 1954.1 / 2014.6 = 0.970.

x₁ ≈ 0.97 (97 % dry)

Example 3 (state identification). Water at 2 MPa and 300 °C. T_sat(2 MPa) = 212.4 °C < 300 °C, so the steam is superheated. From the superheated table: v = 0.1255 m³/kg, h = 3024.2 kJ/kg, s = 6.7684 kJ/kg·K. Check: u = h − pv = 3024.2 − 2000 × 0.1255 = 2773.2 kJ/kg, which matches the tabulated u. h ≈ 3024 kJ/kg

Common mistakes

• Using the dryness-fraction formulas for a superheated or compressed-liquid state; x has no meaning outside the dome.
• Treating p and T as two independent properties in the wet region.
• Reading the saturation table by temperature when the pressure is given (or the reverse) and picking the wrong row.
• Using v_g instead of v_g − v_f, which matters at high pressures where v_f is not negligible.
• Forgetting to convert MPa to kPa in h = u + pv; the pv term then comes out 1000 times too small.
• Assuming a throttling calorimeter works for very wet steam; it works only if the throttled steam ends up superheated.
• Linear interpolation over a wide temperature gap near the saturation line, where properties change non-linearly.

For GATE ME

Expect state identification (compressed, wet or superheated) from a given pair of properties, dryness-fraction calculations from v, h or s, throttling-calorimeter problems, and property look-ups that feed Rankine-cycle and turbine questions. GATE supplies the needed table values in the question, so practise using them quickly: deciding the region, computing x, interpolating and checking with h = u + pv.

Quick check

1. What is the dryness fraction of saturated liquid, and of dry saturated vapour?
2. Water at 1 MPa and 150 °C: what is its phase?
3. At 1 MPa, v_f = 0.001127 and v_g = 0.19436 m³/kg. A mixture has v = 0.1 m³/kg. Find x.
4. What is h_fg at the critical point?
5. Why does a throttling calorimeter need the steam to become superheated after throttling?

Answers: 1. 0 and 1. 2. Compressed liquid, since 150 °C < T_sat = 179.9 °C. 3. x = (0.1 − 0.001127)/0.193233 ≈ 0.512. 4. Zero. 5. Only then are p and T independent, so they fix h₂ (= h₁) from the superheated table.