Properties of pure substances and steam tables

Phases of a pure substance, saturation, quality, critical and triple points, and how to read steam tables in each region, with worked numericals.

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Why it matters

Steam power plants, refrigeration and air-conditioning loops and engine cooling circuits all use a working fluid that changes phase. Steam and refrigerant tables are how engineers find the state of that fluid and read its enthalpy and entropy for every energy balance. Reading the right table, in the right region, is the single most common source of lost marks in cycle problems.

Key ideas

Pure substance. A substance of fixed, uniform chemical composition throughout. It may exist in more than one phase: a mixture of ice, liquid water and steam is a pure substance. Air is treated as a pure substance as long as no phase change occurs; liquid air in equilibrium with gaseous air is not, because the two phases have different compositions.

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

  • Compressed (subcooled) liquid: T below the saturation temperature for the pressure.
  • Saturated liquid (x = 0): about to boil.
  • Wet (saturated) mixture (0 < x < 1): liquid and vapour coexist at the same T and P. Temperature stays at T_sat while heat goes into vaporisation.
  • Saturated (dry) vapour (x = 1): last drop just evaporated.
  • Superheated vapour: T above T_sat at that pressure.

Saturation. At a given pressure, boiling occurs at one temperature, T_sat; at a given temperature, at one pressure, P_sat. In the wet region P and T are not independent, so a second property (v, h, s or x) is needed to fix the state. Water boils at 100 °C at 101.325 kPa but at about 180 °C at 1 MPa.

Dryness fraction (quality). x = m_vapour / m_total, defined only in the wet region. Any specific property in the wet region is the mass-weighted average of the saturated-liquid and saturated-vapour values.

Critical point. The top of the saturation dome, where saturated-liquid and saturated-vapour states become identical and h_fg = 0. For water: about 22.06 MPa and 373.95 °C. Above the critical pressure there is no distinct boiling; a supercritical boiler has no drum and no evaporation stage.

Triple point. The one state where solid, liquid and vapour coexist. For water: 0.01 °C and 0.6117 kPa. Below the triple-point pressure, ice sublimes directly to vapour.

Using the tables.

  1. Given P and T: compare T with T_sat(P). T < T_sat → compressed liquid; T = T_sat → need x; T > T_sat → superheated table.
  2. Given P (or T) and v, h or s: compare with the f and g values. Between them → wet, find x. Above g → superheated.
  3. Compressed liquid with no table: take properties of saturated liquid at the same temperature (v ≈ v_f(T), u ≈ u_f(T), h ≈ h_f(T); a better h is h_f(T) + v_f (P − P_sat)).

Diagrams. On T–v and P–v diagrams the dome separates the regions; on the T–s diagram the area under a reversible process is the heat transferred; the h–s (Mollier) chart is used for turbines and nozzles because enthalpy drop and isentropic lines are read directly.

Reference states. Steam tables set u = 0 and s = 0 for saturated liquid at the triple point. Refrigerant tables use other references, so never mix values from different tables in one problem; only differences matter.

Formulas

v = V / m — v: specific volume, m³/kg; V: volume, m³; m: mass, kg.

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

v = v_f + x v_fg, u = u_f + x u_fg, h = h_f + x h_fg, s = s_f + x s_fg

  • Subscripts: f = saturated liquid, g = saturated vapour, fg = g − f (e.g. h_fg = latent heat of vaporisation, kJ/kg).

x = (h − h_f) / h_fg — finding quality from a known property (same form for v, u, s).

h = u + P v — h, u in kJ/kg; P in kPa; v in m³/kg (kPa·m³/kg = kJ/kg).

h_fg = T_sat s_fg — for the phase change at constant P and T; T_sat in K.

h ≈ h_f(T) + v_f (P − P_sat(T)) — compressed liquid approximation.

Worked examples

Example 1 (standard). Find the phase and the specific volume, enthalpy and entropy of water at 1 MPa and 200 °C.

  1. From the saturation table, T_sat at 1 MPa = 179.9 °C.
  2. Given T = 200 °C > 179.9 °C, so the water is superheated vapour. (Equivalently, P_sat at 200 °C = 1.554 MPa > 1 MPa.)
  3. From the superheated table at 1 MPa, 200 °C: v = 0.2060 m³/kg, h = 2828.3 kJ/kg, s = 6.6955 kJ/kg·K.
  4. Check with h = u + Pv: u from the table is 2622.3 kJ/kg; 2622.3 + 1000 × 0.2060 = 2828.3 kJ/kg. Consistent.

Example 2 (GATE level). A rigid vessel of volume 0.1 m³ contains 2 kg of wet steam at 1 MPa. It is heated until the contents are just dry saturated vapour. Find the initial quality, the final pressure and the heat added. Use at 1 MPa: v_f = 0.001127, v_g = 0.1943 m³/kg, u_f = 761.6, u_g = 2582.8 kJ/kg.

  1. Specific volume (constant, rigid vessel, fixed mass): v = V/m = 0.1/2 = 0.05 m³/kg.
  2. Initial quality: x₁ = (v − v_f)/(v_g − v_f) = (0.05 − 0.001127)/(0.1943 − 0.001127) = 0.04887/0.19317 = 0.253.
  3. Initial internal energy: u₁ = u_f + x₁ (u_g − u_f) = 761.6 + 0.253 × 1821.2 = 1222.2 kJ/kg.
  4. Final state: saturated vapour with v_g = 0.05 m³/kg. From the saturation table, v_g = 0.0498 m³/kg at 4 MPa, so P₂ ≈ 4 MPa (T ≈ 250 °C), u₂ = u_g ≈ 2601.8 kJ/kg.
  5. Rigid vessel, no work: Q = m (u₂ − u₁) = 2 × (2601.8 − 1222.2) = ≈ 2759 kJ.

Note that pressure rose from 1 to about 4 MPa with no volume change. Had the specific volume been smaller than v_c (0.00311 m³/kg), heating would have filled the vessel with liquid instead.

Common mistakes

  • Deciding the phase from temperature alone. Always compare with T_sat at the given pressure (or P_sat at the given temperature).
  • Using quality for a superheated or compressed state; x has no meaning outside the dome.
  • Taking h from the saturated-vapour column for wet steam instead of h_f + x h_fg.
  • Unit slip in h = u + Pv: with P in kPa and v in m³/kg, Pv is already in kJ/kg. With P in MPa you must multiply by 1000.
  • Looking up compressed liquid in the superheated table or by pressure instead of temperature.
  • Mixing steam tables from different editions or refrigerant tables with different reference states. Small differences between data books are normal; use one consistently.

For GATE ME

Expect property look-ups embedded in larger problems: quality from enthalpy or entropy after an isentropic turbine expansion, heat to a rigid tank or piston–cylinder, and throttling calorimeter questions where h is constant. Conceptual MCQs test the critical and triple points, the shape of constant-pressure lines on T–s and h–s diagrams, and where a state lies. GATE supplies steam-table data in the question; practise interpolation and identifying the region quickly.

Quick check

  1. Water is at 500 kPa and 120 °C (T_sat = 151.8 °C). What is its phase?
  2. Wet steam has h = 2000 kJ/kg at a pressure where h_f = 500 kJ/kg and h_fg = 2000 kJ/kg. What is x?
  3. What is h_fg at the critical point?
  4. Is a mixture of liquid water and steam a pure substance?

Answers: 1. Compressed (subcooled) liquid. 2. x = (2000 − 500)/2000 = 0.75. 3. Zero. 4. Yes; the chemical composition is the same in both phases.

Interview questions

All Thermodynamics and Heat Transfer 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 has a fixed, uniform chemical composition throughout, though it may exist in more than one phase. A mixture of ice, water and steam is a pure substance because every phase is H₂O. Air is treated as a pure substance as long as it stays gaseous, but liquid air in equilibrium with gaseous air is not, because the phases have different compositions. The point matters because only for a pure substance can two independent properties fix the state and be read from a single property table.

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

    Steam tables list the properties of water (P, T, v, u, h, s) in the saturated, superheated and compressed-liquid regions, because water does not behave as an ideal gas near saturation. The saturation tables give the f and g values used with the quality, and the superheated table gives properties as a function of P and T. Every Rankine-cycle, turbine, boiler or condenser energy balance needs enthalpies from them, and entropies to locate the isentropic end state. Values are relative to a reference state (u and s zero for saturated liquid at the triple point), so only differences are meaningful.

  3. 3.How do you differentiate between saturated and superheated steam?Concept

    Saturated steam is at the saturation temperature for its pressure; dry saturated steam (x = 1) has just finished evaporating, and any heat removal starts condensation at constant temperature. Superheated steam is at a temperature above T_sat at its pressure, so it can lose some heat while only its temperature falls, with no condensation. The degree of superheat is T − T_sat. To tell them apart, compare the measured temperature with T_sat for the measured pressure, or compare v, h or s with the saturated-vapour value.

  4. 4.Why is steam commonly used as a working fluid in power plants?Application

    Steam is commonly used as a working fluid in power plants because it has a high enthalpy of vaporization, which allows it to carry a large amount of energy. Additionally, steam can be easily expanded in turbines to produce mechanical work, and it can be condensed back into water for reuse, making it efficient for cyclic processes.

  5. 5.Water is at a temperature T. What phase is it in if its pressure is below, equal to or above the saturation pressure for T?Application

    If the pressure is below P_sat(T), the water is superheated vapour, because at that pressure it would boil at a lower temperature than T. If the pressure is above P_sat(T), it is compressed (subcooled) liquid. Only when P equals P_sat(T) can liquid and vapour coexist, and then a second property such as v, h or the quality is needed to fix the state. This comparison, or the equivalent one of T against T_sat(P), is the first step before choosing which table to use.

  6. 6.Explain how the properties of steam change as it goes through a turbine.Application

    As steam passes through a turbine, it expands and does work on the turbine blades, causing a drop in pressure and temperature. The steam's specific volume increases, and its enthalpy decreases as energy is extracted. Depending on the initial conditions, the steam may become saturated or even partially condense by the time it exits the turbine.

  7. 7.Why is it important to avoid moisture in steam entering a turbine?Application

    Water droplets in the last stages of a turbine travel more slowly than the vapour, so the moving blades strike them at high relative velocity, eroding the leading edges. The liquid also does little useful work and adds friction and drag losses, lowering stage efficiency. For this reason designers keep the exhaust dryness fraction at roughly 0.88 to 0.9 or higher, by superheating, reheating or using moisture separators.

  8. 8.Calculate the specific enthalpy of steam at 2 MPa and 250°C using steam tables.Numerical

    First locate the state: T_sat at 2 MPa is about 212.4 °C, and 250 °C is above it, so the steam is superheated. From the superheated table at 2 MPa and 250 °C, h ≈ 2903 kJ/kg (with v ≈ 0.1115 m³/kg and s ≈ 6.55 kJ/kg·K). Small differences in the last digit between data books are normal.

  9. 9.Determine the quality of steam at 1 MPa with a specific enthalpy of 2500 kJ/kg.Numerical

    At 1 MPa the saturation table gives h_f ≈ 762.5 kJ/kg and h_g ≈ 2777 kJ/kg, so h_fg ≈ 2014.6 kJ/kg. Since 2500 kJ/kg lies between h_f and h_g, the steam is wet. Quality x = (h − h_f)/h_fg = (2500 − 762.5)/2014.6 ≈ 0.86, so about 86% of the mass is vapour.

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

    The critical point is the top of the saturation dome, where saturated liquid and saturated vapour have identical properties and the latent heat h_fg becomes zero; for water it is about 22.06 MPa and 373.95 °C. Above the critical pressure, heating takes the fluid from liquid-like to vapour-like continuously, with no boiling and no two-phase region. That is why supercritical boilers have no steam drum and use once-through designs, and why refrigerants are chosen with critical temperatures well above the condensing temperature.

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