1.2 Properties of Pure Substances

Pure substances are the foundation of thermodynamic analysis. Every working fluid in a power plant, refrigeration cycle, or heat exchanger is modeled as a pure substance, so being comfortable with their properties and phase behavior is essential before tackling the more advanced topics in Thermo II.

This review covers phase diagrams, property tables, phase-change behavior, and two-phase mixture calculations.

Phase diagrams and property tables

Phase diagram features

A P-T phase diagram plots pressure (vertical axis) against temperature (horizontal axis) and maps out where each phase is stable. Three curves divide the diagram into solid, liquid, and gas regions, and two special points anchor the whole picture:

• Triple point: All three phases coexist in equilibrium. For water this occurs at 0.01°C and 0.6117 kPa. Below the triple-point pressure, a liquid phase cannot exist.
• Critical point: The liquid and vapor phases become indistinguishable. For water: 373.95°C and 22.064 MPa. Above this point the substance is a supercritical fluid.
• Sublimation curve: Equilibrium between solid and gas. Dry ice (solid $CO_2$) sublimes at atmospheric pressure because 1 atm is below $CO_2$'s triple-point pressure.
• Fusion curve: Equilibrium between solid and liquid. For most substances this curve slopes to the right (higher pressure raises the melting point). Water is an anomaly: its fusion curve slopes slightly to the left.
• Vaporization curve: Equilibrium between liquid and gas. It runs from the triple point up to the critical point. Water boils at 100°C only at 101.325 kPa; at higher pressures the saturation temperature rises.

Property tables and interpolation

Property tables list thermodynamic data (specific volume $v$, internal energy $u$, enthalpy $h$, and entropy $s$) at discrete temperatures and pressures. They're organized into distinct sub-tables:

• Saturated tables (indexed by temperature or pressure) give $v_f, v_g, u_f, u_g, h_f, h_{fg}, h_g, s_f, s_g$ along the saturation curve.
• Superheated vapor tables cover states to the right of the saturation dome on a T-v diagram.
• Compressed (subcooled) liquid tables cover states to the left. When compressed-liquid data aren't available, a common approximation is to use the saturated-liquid values at the same temperature.

When your state falls between two table entries, you need to interpolate.

Linear interpolation (the standard approach for $v$, $u$, $h$, and $s$):

$y = y_1 + \frac{x - x_1}{x_2 - x_1}(y_2 - y_1)$

where $x$ is the known independent variable (T or P) and $y$ is the property you're solving for. Linear interpolation works well for all common properties across small intervals. Logarithmic interpolation is sometimes mentioned for entropy, but in practice linear interpolation is used for entropy in standard steam tables and introduces negligible error over typical table spacing.

Using phase diagrams and property tables together

To find the state and properties of a substance at a given T and P:

1. Fix the state. Compare the given T and P to the saturation data. If $T < T_{sat}$ at the given P (or equivalently $P > P_{sat}$ at the given T), the substance is a compressed liquid. If $T > T_{sat}$, it's superheated vapor. If both match the saturation curve, you're in the two-phase region and need one more property (like $v$ or $x$) to pin down the state.
2. Go to the correct table. Compressed liquid, saturated, or superheated.
3. Read or interpolate. If the exact T and P appear in the table, read directly. Otherwise, interpolate between the nearest entries.
4. Two-phase case. If the substance is a liquid-vapor mixture, you can't use the superheated or compressed-liquid tables. Instead, use quality $x$ and the saturated properties (covered below).

Thermodynamic property calculations

Steam tables

Steam tables are simply the property tables for water/steam. They're the most common tables you'll encounter because water is the working fluid in most power cycles. The organization is the same as described above: saturated tables (by T and by P), superheated vapor tables, and compressed liquid tables.

A quick reference for water at 1 atm (101.325 kPa):

Notice the enormous jump in specific volume between liquid and vapor. That ratio (roughly 1:1600) is why phase identification matters so much for engineering calculations.

Refrigerant tables and charts

Refrigerant property data (for substances like R-134a, R-410A, or ammonia) are structured identically to steam tables. The key difference is the temperature and pressure ranges: refrigerants are chosen specifically because they change phase at temperatures useful for cooling applications.

For example, R-134a has a saturation temperature of about −26.1°C at 1 atm, which makes it practical for household refrigerators. You read and interpolate refrigerant tables exactly the same way you do steam tables.

Calculating properties using tables

The process is the same regardless of the substance:

1. Identify the substance and locate the correct table set.
2. Determine the phase region (compressed liquid, saturated mixture, or superheated vapor) by comparing given conditions to saturation data.
3. If conditions match a table entry exactly, read the properties directly.
4. If conditions fall between entries, interpolate using the linear formula above.
5. For two-phase states, use quality and the mixture equations from the next section.

Pure substance phase changes

Types of phase changes

Phase changes occur when a substance crosses one of the equilibrium curves on the P-T diagram:

• Melting / Freezing (solid ↔ liquid): Occurs along the fusion curve. Ice melts at 0°C and 1 atm.
• Vaporization / Condensation (liquid ↔ gas): Occurs along the vaporization curve. Boiling specifically refers to vaporization at the saturation temperature for a given pressure. Evaporation, by contrast, can occur below the boiling point at a free surface.
• Sublimation / Deposition (solid ↔ gas): Occurs along the sublimation curve. Dry ice sublimes at −78.5°C and 1 atm.

Behavior during phase changes

During any phase change at constant pressure, the temperature stays constant while energy is added or removed. On a T-v diagram, this appears as a horizontal line across the two-phase dome.

What does change dramatically is specific volume. During vaporization, $v$ increases by orders of magnitude (for water at 1 atm, $v$ goes from about 0.001 m³/kg to about 1.694 m³/kg). During condensation, the reverse happens.

The energy required to drive a phase change at constant T and P is called latent heat:

• Latent heat of fusion for water: 333.5 kJ/kg at 0°C
• Latent heat of vaporization for water: 2257 kJ/kg at 100°C

The latent heat of vaporization is nearly seven times larger than the latent heat of fusion. This is why boiling a pot of water takes so much more energy than melting the same mass of ice.

Two-phase mixture properties

Quality and the lever rule

When a substance exists as a liquid-vapor mixture (anywhere under the saturation dome), you need one additional property beyond T or P to fix the state. That property is usually quality.

Quality ($x$) is the mass fraction of vapor in the mixture:

$x = \frac{m_{vapor}}{m_{total}}$

• $x = 0$: saturated liquid (all liquid, no vapor)
• $x = 1$: saturated vapor (all vapor, no liquid)
• $0 < x < 1$: two-phase mixture

You can calculate quality from any specific property using the same pattern. For specific volume:

$x = \frac{v - v_f}{v_g - v_f} = \frac{v - v_f}{v_{fg}}$

where $v_{fg} = v_g - v_f$.

Calculating mixture properties

Once you know quality, every other specific property follows the same linear relationship:

$v = v_f + x \cdot v_{fg}$

$u = u_f + x \cdot u_{fg}$

$h = h_f + x \cdot h_{fg}$

$s = s_f + x \cdot s_{fg}$

These are equivalent to writing $v = (1 - x) v_f + x \, v_g$, just rearranged using $v_{fg} = v_g - v_f$.

Worked example: Water at 100°C (1 atm) with quality $x = 0.5$.

From the saturated table at 100°C: $v_f = 0.001043$ m³/kg, $v_g = 1.694$ m³/kg, $h_f = 419.0$ kJ/kg, $h_g = 2676.0$ kJ/kg.

$v = 0.001043 + 0.5(1.694 - 0.001043) = 0.848 \text{ m³/kg}$

$h = 419.0 + 0.5(2676.0 - 419.0) = 1547.5 \text{ kJ/kg}$

The same approach applies to $u$ and $s$. Just pull the $f$ and $g$ values from the table and plug in.