Properties of pure substances and steam tables

Pure substances, the constant-pressure phase-change process, saturation, quality, critical and triple points, and how to read saturated and superheated steam tables for first-law calculations.

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Why it matters

Water, refrigerants and cryogenic propellants change phase inside boilers, condensers, evaporators and rocket feed systems, and the ideal-gas equation fails near phase change. Property tables (steam tables) and the quality concept let you find the state and its energy accurately. Reading tables correctly is a core skill for Rankine-cycle and refrigeration problems.

Key ideas

Pure substance. A substance with a fixed, uniform chemical composition throughout, even if it exists in several phases (ice–water–steam). Air is treated as a pure substance while it stays gaseous; liquid air in equilibrium with gaseous air is not, because the phases have different compositions.

Phase change at constant pressure. Heat water in a piston–cylinder at constant pressure:

  1. Compressed (subcooled) liquid — below the saturation temperature.
  2. Saturated liquid — at T_sat; any more heat starts boiling.
  3. Saturated liquid–vapour mixture (wet steam) — T stays at T_sat while the liquid evaporates; v rises sharply.
  4. Saturated (dry) vapour — the last drop has evaporated.
  5. Superheated vapour — T rises above T_sat.

Saturation. At a given pressure there is one saturation temperature (and vice versa): T_sat rises with p. For water T_sat = 100 °C at 101.325 kPa, about 120.2 °C at 200 kPa and 179.9 °C at 1 MPa. Inside the dome p and T are not independent, so another property is needed to fix the state.

Quality (dryness fraction). x = m_vapour / m_total, from 0 (saturated liquid) to 1 (saturated vapour). Any specific property of the mixture is a mass-weighted average, e.g. v = vf + x·vfg.

Critical and triple points (water). Critical point: 22.06 MPa, 373.95 °C, where vf = vg and the latent heat vanishes; above it there is no distinct boiling. Triple point: 0.01 °C, 0.6117 kPa, where solid, liquid and vapour coexist.

Property diagrams. On T–v and p–v diagrams the saturated-liquid and saturated-vapour lines meet at the critical point and enclose the wet region. Constant-pressure lines are horizontal inside the dome on a T–v diagram.

Using the tables.

  • Saturated tables (by T or by p): vf, vg, uf, ug, hf, hfg, hg, sf, sfg, sg. Subscript f = saturated liquid, g = saturated vapour, fg = g − f.
  • Deciding the phase: at the given p, compare the given v (or T, h) with the saturated values. v < vf → compressed liquid; vf < v < vg → wet; v > vg → superheated.
  • Superheated tables: entered with two of p, T; interpolate linearly between rows.
  • Compressed liquid: if no table, approximate with saturated liquid values at the same temperature: v ≈ vf(T), u ≈ uf(T), h ≈ hf(T) (better: h ≈ hf(T) + vf·(p − p_sat)).

Table values depend slightly on the edition; take them from the data book you are allowed to use. The values below are standard (IAPWS-based) steam-table values, given as data.

Formulas

x = m_g / (m_f + m_g) v = vf + x·(vg − vf) = vf + x·vfg h = hf + x·hfg; u = uf + x·ufg; s = sf + x·sfg x = (v − vf) / (vg − vf) (finding quality from a known v) h = u + p·v

  • v: specific volume, m³/kg; u: specific internal energy, kJ/kg; h: specific enthalpy, kJ/kg; s: specific entropy, kJ/(kg·K)
  • p: pressure, kPa (kPa × m³/kg = kJ/kg)
  • x: quality, dimensionless (only defined inside the dome)

Worked examples

Example 1 (standard). A rigid 0.5 m³ tank holds 2 kg of water at 200 kPa. Find the phase, the quality, the mass of vapour and the specific enthalpy. Data at 200 kPa: T_sat = 120.21 °C, vf = 0.001061 m³/kg, vg = 0.88578 m³/kg, hf = 504.70 kJ/kg, hfg = 2201.6 kJ/kg.

  1. v = V/m = 0.5 / 2 = 0.25 m³/kg
  2. vf < 0.25 < vg, so the water is a wet mixture at 120.21 °C.
  3. x = (v − vf)/(vg − vf) = (0.25 − 0.001061) / (0.88578 − 0.001061) = 0.2814
  4. Vapour mass = x·m = 0.2814 × 2 = 0.563 kg
  5. h = hf + x·hfg = 504.70 + 0.2814 × 2201.6 = 1124.2 kJ/kg

Answer: wet steam, x = 0.281, m_g = 0.563 kg, h = 1124 kJ/kg

Example 2 (GATE level). 1 kg of wet steam at 1 MPa with quality 0.8 is heated at constant pressure in a piston–cylinder until it reaches 250 °C. Find the heat added, the boundary work and the change in internal energy. Data at 1 MPa: vf = 0.001127, vg = 0.19436 m³/kg, hf = 762.51, hfg = 2014.6 kJ/kg; superheated at 1 MPa, 250 °C: v = 0.23275 m³/kg, h = 2943.1 kJ/kg.

  1. State 1: h₁ = hf + x·hfg = 762.51 + 0.8 × 2014.6 = 2374.19 kJ/kg; v₁ = vf + x·(vg − vf) = 0.001127 + 0.8 × 0.193233 = 0.15571 m³/kg
  2. Constant-pressure closed process: Q = m·(h₂ − h₁) = 2943.1 − 2374.19 = 568.9 kJ
  3. W = m·p·(v₂ − v₁) = 1000 × (0.23275 − 0.15571) = 77.0 kJ
  4. ΔU = Q − W = 568.9 − 77.0 = 491.9 kJ

Answer: Q = 568.9 kJ, W = 77.0 kJ, ΔU = 491.9 kJ

Common mistakes

  • Using the ideal-gas equation for steam near saturation.
  • Applying x outside the dome — quality has no meaning for superheated vapour or compressed liquid.
  • Entering the tables with T and p inside the wet region: they are dependent, so they cannot fix the state.
  • Approximating compressed liquid with saturated values at the same pressure instead of the same temperature.
  • Writing h = x·hg (forgetting the liquid part); the correct form is h = hf + x·hfg.
  • Mixing kPa and MPa when computing p·v.

For GATE AE

Expect questions that ask you to locate a state from two properties, find quality from v or h, and do first-law balances for heating, mixing or throttling of steam with table data supplied in the question. Conceptual questions cover the critical and triple points, the shape of the vapour dome on T–v and p–v diagrams, and the meaning of quality. Practise phase identification quickly — it decides which table you use.

Quick check

  1. What is the quality of a saturated vapour?
  2. At 500 kPa, vf = 0.001093 and vg = 0.3749 m³/kg. Is water with v = 0.2 m³/kg wet or superheated?
  3. Why can p and T not fix a state inside the wet region?
  4. What happens to hfg at the critical point?
  5. Given hf = 504.7 kJ/kg, hfg = 2201.6 kJ/kg, x = 0.9, find h.

Answers: 1. x = 1. 2. Wet (vf < v < vg). 3. Saturation ties T to p, so they are not independent. 4. It becomes zero. 5. 2486.1 kJ/kg.

Interview questions

All Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What is a pure substance in the context of thermodynamics?Concept

    A pure substance in thermodynamics is a material with a uniform and invariable chemical composition. It can exist in more than one phase, but its chemical composition must remain the same in each phase. Examples include water, nitrogen, and carbon dioxide.

  2. 2.Explain the significance of steam tables in thermodynamics.Concept

    Steam tables provide the thermodynamic properties of water and steam, which are essential for engineering calculations. They include data on properties such as temperature, pressure, specific volume, enthalpy, and entropy at various states. Engineers use these tables to design and analyze systems like boilers, turbines, and heat exchangers.

  3. 3.Why is water commonly used as a working fluid in thermodynamic cycles?Application

    Water is commonly used as a working fluid because it has a high specific heat capacity, which allows it to store and transfer large amounts of energy. It also has a high latent heat of vaporization, making it efficient for phase change processes. Additionally, water is non-toxic, readily available, and inexpensive.

  4. 4.What happens to the specific volume of steam as it transitions from saturated liquid to saturated vapor?Application

    As steam transitions from saturated liquid to saturated vapor, its specific volume increases significantly. This is because the molecules move further apart during the phase change from liquid to vapor, resulting in a larger volume occupied by the same mass.

  5. 5.Explain the difference between superheated steam and saturated steam.Concept

    Saturated (dry) steam is vapour exactly at the saturation temperature for its pressure, on the saturated-vapour line; it can coexist with liquid, and removing any heat at constant pressure starts condensation. Superheated steam is at a temperature above the saturation temperature for its pressure, so it can lose some heat without condensing; the difference T − T_sat is the degree of superheat. In the superheated region p and T are independent and fix the state, whereas saturated steam needs only one of them.

  6. 6.How does pressure affect the boiling point of a pure substance?Application

    The boiling (saturation) temperature rises with pressure: a liquid boils when its vapour pressure equals the surrounding pressure, and vapour pressure increases with temperature, so a higher pressure needs a higher temperature. For water T_sat is 100 °C at 101.325 kPa, about 120 °C at 200 kPa and about 180 °C at 1 MPa, and the relation is quantified by the Clausius–Clapeyron equation. At the same time the latent heat hfg decreases with pressure and vanishes at the critical point.

  7. 7.What is the critical point of a substance, and why is it important?Concept

    The critical point of a substance is the temperature and pressure at which the liquid and vapor phases become indistinguishable. Beyond this point, the substance exists as a supercritical fluid. It is important because it defines the upper limit for the liquid-vapor phase boundary and is crucial for designing equipment that operates at high temperatures and pressures.

  8. 8.Calculate the enthalpy change when 1 kg of water at 100°C is converted to steam at the same temperature. Assume the latent heat of vaporization is 2260 kJ/kg.Numerical

    The enthalpy change (ΔH) can be calculated using the formula: ΔH = m * L, where m is the mass and L is the latent heat of vaporization. For 1 kg of water, ΔH = 1 kg * 2260 kJ/kg = 2260 kJ.

  9. 9.If the pressure of steam is increased while keeping the temperature constant, what phase change might occur?Application

    If the pressure of steam is increased while keeping the temperature constant, the steam may condense into liquid water. This is because increasing pressure at constant temperature can push the steam into the saturated liquid region, causing condensation.

  10. 10.Determine the specific volume of steam at 1 MPa and 200°C using steam tables.Numerical

    First identify the phase: at 1 MPa the saturation temperature is about 179.9 °C, so steam at 200 °C is superheated and must be read from the superheated table, not the saturated one. The superheated table at 1 MPa and 200 °C gives v ≈ 0.2060 m³/kg (with h ≈ 2828 kJ/kg). Table values can differ slightly between editions, so quote the data book you are using.

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