Properties of Pure Substances and Ideal Gases

Phase change, p-v and T-s diagrams, saturation, dryness fraction, using steam tables, ideal gas equation, compressibility factor and processes of ideal gases.

Power plants and refrigerators use working fluids that change phase. To analyse them you need the steam tables and the idea of dryness fraction; for gases, the ideal gas law.

Phase change of water

Heating water at constant pressure: compressed liquid → saturated liquid → wet mixture → saturated vapour → superheated vapour.

  • Saturation temperature depends on pressure (100 °C at 1.013 bar).
  • Critical point (water: about 22.06 MPa, 374 °C): above it, liquid and vapour cannot be told apart.
  • Triple point: solid, liquid and vapour coexist.

Wet steam and dryness fraction

In the wet region, the dryness fraction x = mass of vapour / total mass. Any specific property:

y = y_f + x y_fg (y = v, u, h or s; f = saturated liquid, g = saturated vapour, fg = g − f)

Degree of superheat = T − T_sat.

Measuring dryness: a throttling calorimeter (h before = h after) or a separating-and-throttling calorimeter for very wet steam.

Diagrams

  • p–v: the saturation dome; isotherms are flat inside it.
  • T–s: area under a reversible process = heat transfer. Cycles are clearest here.
  • h–s (Mollier): vertical lines are isentropic; used for turbines and nozzles.

Ideal gases

pv = RT, with R = R̄/M (R̄ = 8.314 kJ/kmol·K). For air, R = 0.287 kJ/kg·K, c_p = 1.005, c_v = 0.718, γ = 1.4.

ProcessRelationIndex n
IsobaricV/T = constant0
IsothermalpV = constant1
IsentropicpV^γ = constantγ
Isochoricp/T = constant∞
PolytropicpVⁿ = constantn

Heat in a polytropic process: Q = W (γ − n)/(γ − 1).

Real gases deviate from pv = RT; the compressibility factor Z = pv/(RT) measures this (Z = 1 for an ideal gas). Gases follow the ideal law well at low pressure and high temperature.

Worked example

At 1 MPa, h_f = 762.6 kJ/kg and h_fg = 2015.3 kJ/kg (steam table values). Find the enthalpy of steam with x = 0.9.

  • h = 762.6 + 0.9 × 2015.3 = 2576.4 kJ/kg.

Practice

GATE-styleNAT · numerical answer

Using h_f = 762.6 and h_fg = 2015.3 kJ/kg, what is the enthalpy of steam with dryness fraction 0.8? (Give 2 decimal places.)

Show answer and explanationAnswer: 2374.84

h = h_f + x h_fg = 762.6 + 1612.24 = 2374.84 kJ/kg.

GATE-styleNAT · numerical answer

What is the density of air (R = 0.287 kJ/kg·K) at 100 kPa and 300 K in kg/m³? (Give 3 decimal places.)

Show answer and explanationAnswer: 1.161

ρ = p/(RT) = 100/(0.287 × 300) = 1.161 kg/m³.

GATE-styleNAT · numerical answer

Air is compressed isentropically (γ = 1.4) from 100 kPa, 300 K to 800 kPa. What is the final temperature in K? (Give 1 decimal place.)

Show answer and explanationAnswer: 543.4

T₂ = T₁ (p₂/p₁)^((γ−1)/γ) = 300 × 8^0.2857 = 543.4 K.

GATE-styleMCQ · one correct answer

On a T–s diagram, the area under a reversible process curve represents:

Show answer and explanationAnswer: A

For a reversible process δQ = T ds, so the area under the T–s curve is the heat transferred.