Properties of Pure Substances & Steam

Steam/pure-substance properties are a steady GATE ME topic (~1–2 marks) and the property backbone for the Rankine cycle. Quality, the saturation dome, and steam-table interpolation are the tested skills.

Phase behaviour

A pure substance has a fixed composition (water, refrigerant). On a T–v or p–v diagram, the saturation dome separates compressed liquid (left), the two-phase liquid–vapour mixture (inside), and superheated vapour (right).

• Saturation temperature/pressure: the boiling point at a given pressure (they are linked one-to-one under the dome).
• Critical point: above it, liquid and vapour are indistinguishable (for water, ~22.1 MPa, 374°C).
• Triple point: where solid, liquid and vapour coexist (water 0.01°C, 0.611 kPa).

Quality & property evaluation

Inside the dome, dryness fraction (quality) x = mass of vapour / total mass. Any property y = y_f + x·y_fg (y_fg = y_g − y_f):

• h = h_f + x·h_fg; s = s_f + x·s_fg; v = v_f + x·v_fg.
  Subscript f = saturated liquid, g = saturated vapour. Read h_f, h_fg, etc. from steam tables; the Mollier (h–s) diagram speeds turbine-expansion work.

Exam Tricks & Tips

• 🎯 Inside the dome, property = y_f + x·y_fg — the quality x linearly interpolates every property between saturated liquid and vapour.
• 🎯 h = h_f + x·h_fg — the workhorse for boiler heat and turbine work in the Rankine cycle.
• 🎯 Saturated liquid x = 0, saturated (dry) vapour x = 1; superheated steam is beyond x = 1 (use superheat tables, not quality).
• 🎯 Critical point: no distinct phase change; triple point: three phases coexist — common conceptual MCQs.
• 🎯 Mollier (h–s) chart reads turbine work directly as the enthalpy drop along a constant-entropy line (ideal expansion).
• ❌ Common mistake: applying x = h_f + x·h_fg to superheated steam — quality is defined only inside the two-phase dome; superheated states need the superheat tables.

Expected exam pattern

A 1-mark quality, saturation, or critical/triple-point MCQ/NAT, and a 2-mark steam-property problem (find h or s at a given p and x, then use it in a cycle). Table interpolation and the h = h_f + x·h_fg relation are frequently tested.

Quick recap

Pure substance on p–v/T–s: liquid | two-phase dome | vapour. Saturation T–p linked; critical point (no phase change), triple point (3 phases). Quality x = m_vapour/m_total; y = y_f + x·y_fg (h, s, v). x = 0 sat. liquid, 1 dry vapour; superheat beyond. Use steam tables/Mollier chart.

Cover the answer, recall, then check. 11 cards on pure-substance properties for GATE ME.

Q1. What is a pure substance?
A1. A substance of fixed, uniform chemical composition throughout (e.g. water, a single refrigerant), even across phase changes.

Q2. Define dryness fraction (quality).
A2. x = mass of vapour / total mass of the mixture; ranges 0 (saturated liquid) to 1 (saturated dry vapour).

Q3. How do you find any property inside the two-phase dome?
A3. y = y_f + x·y_fg, where y_fg = y_g − y_f (e.g. h = h_f + x·h_fg).

Q4. What do subscripts f, g, and fg mean?
A4. f = saturated liquid, g = saturated vapour, fg = difference (g − f), the latent change.

Q5. What are saturation temperature and pressure?
A5. The temperature and pressure at which boiling/condensation occurs; they are uniquely linked under the saturation dome.

Q6. What happens at the critical point?
A6. Liquid and vapour become indistinguishable — no distinct phase change occurs above it (water: ~22.1 MPa, 374°C).

Q7. What is the triple point?
A7. The unique state where solid, liquid, and vapour coexist (for water, 0.01°C and 0.611 kPa).

Q8. What is superheated steam?
A8. Vapour heated above its saturation temperature at a given pressure (beyond x = 1); use superheat tables, not quality.

Q9. What does the Mollier (h–s) diagram give quickly?
A9. Turbine work as the enthalpy drop along a constant-entropy (isentropic) expansion line.

Q10. Compressed (subcooled) liquid is found where on the diagram?
A10. To the left of the saturated-liquid line (below saturation temperature at that pressure).

Q11. Common mistake with the quality relation?
A11. Using y = y_f + x·y_fg for superheated steam — quality is defined only inside the two-phase region.

Steam power plants and refrigeration all hinge on how a pure substance changes phase. Reading property tables and the p–v, T–s and Mollier diagrams correctly is a make-or-break exam skill. GATE tests dryness fraction, property interpolation and process representation.

Core concept: a pure substance passing through vaporisation moves along a saturation dome; inside the dome it is a liquid–vapour mixture described by the dryness fraction, and properties come from tables or charts.

Deep explanation

Beginner — phases and the saturation dome

On a T–v or p–v diagram, the saturation dome separates subcooled liquid (left), the wet mixture (inside), and superheated vapour (right). At a given pressure, saturation temperature T_sat is where boiling occurs. Terms: subcooled/compressed liquid (below T_sat), saturated liquid (f), saturated vapour (g), superheated (above T_sat).

Intermediate — dryness fraction and mixture properties

Inside the dome, the dryness fraction (quality) x = mass of vapour/total mass. Any specific property of the mixture:
v = v_f + x·v_fg, h = h_f + x·h_fg, s = s_f + x·s_fg,
where subscript f = saturated liquid, g = saturated vapour, fg = difference (e.g. h_fg = latent heat of vaporisation). x = 0 is saturated liquid, x = 1 is saturated (dry) vapour. Interpolate between table entries for intermediate pressures/temperatures.

Advanced — the diagrams and their uses

• T–s diagram: area under a reversible path = heat; used to visualise Rankine/Carnot cycles.
• Mollier (h–s) diagram: enthalpy vs entropy; directly reads turbine work (vertical drop for isentropic expansion) and dryness at exhaust — the practical steam-cycle tool.
• p–h diagram: the refrigeration workhorse.
• Critical point: above it liquid and vapour are indistinguishable (water: 22.1 MPa, 374°C). The triple point is where all three phases coexist.
• Clausius–Clapeyron equation relates the saturation pressure–temperature slope to latent heat: dp/dT = h_fg/(T·v_fg).

Worked example

Wet steam at 10 bar has a dryness fraction of 0.9. Given h_f = 762.6 kJ/kg and h_fg = 2013.6 kJ/kg, find the specific enthalpy.
h = h_f + x·h_fg = 762.6 + 0.9 × 2013.6 = 762.6 + 1812.2 = 2574.8 kJ/kg. This is the enthalpy carried into a turbine; the difference to the exit enthalpy gives the work.

GATE relevance

Dryness-fraction property calculation, table interpolation, and reading T–s/Mollier diagrams are essential Thermodynamics skills tested directly and as steps inside Rankine-cycle problems. The mixture-property formula (h = h_f + x·h_fg) is one of the most-used relations in the subject.

Exam tricks & shortcuts

• Inside the dome, every property is f-value + x × fg-value — one formula for v, h, s, u.
• Turbine work on a Mollier chart is a near-vertical (isentropic) drop; read exit dryness directly.
• At the critical point h_fg → 0 (no distinct phase change).
• Mnemonic: "f plus x-fg, for any property you see."

Key formulas

• Dryness fraction: x = m_vapour/(m_vapour + m_liquid).
• Properties: h = h_f + x·h_fg; s = s_f + x·s_fg; v = v_f + x·v_fg.
• Saturation: at given P, T_sat fixed; latent heat h_fg.
• Superheated steam: T > T_sat (use steam tables); subcooled T < T_sat.

Wet steam properties interpolate between saturated liquid and vapour via the dryness fraction x.

Worked Example

Problem: Wet steam at 10 bar has a dryness fraction (quality) x = 0.9. From steam tables at 10 bar: h_f = 762.8 kJ/kg, h_fg = 2015.3 kJ/kg, and v_g = 0.1944 m³/kg. Find the specific enthalpy and the specific volume of the wet steam (neglect v_f).

Solution:
For a liquid–vapour mixture, any specific property is the saturated-liquid value plus the quality times the vaporisation term.

Specific enthalpy:
h = h_f + x·h_fg = 762.8 + 0.9 × 2015.3
= 762.8 + 1813.8 = 2576.6 kJ/kg.

Specific volume (v_f is negligible compared with v_g):
v ≈ x·v_g = 0.9 × 0.1944 = 0.175 m³/kg.

The quality x = 0.9 means the mixture is 90% vapour by mass, so its properties lie 90% of the way from saturated liquid toward saturated vapour.

Answer: h ≈ 2576.6 kJ/kg and v ≈ 0.175 m³/kg.