Interactive Lecture
Steam Tables, Quality, and Wet–Dry Vapour Properties
by Dr Arnaldo Delli Carri
Water can exist in three phases (solid, liquid, gas) depending on pressure and temperature. The T-v diagram shows how these phases relate at constant pressure (isobars).
Phase changes occur when a substance transitions between solid, liquid, and gas states. For water, these transitions are characterised by:
Two special thermodynamic states are particularly important:
Triple Point: The unique condition where all three phases coexist in equilibrium.
For water: \(T_{tp} = 273.16\) K (0.01°C), \(p_{tp} = 611.657\) Pa (0.00611 bar)
Critical Point: Beyond this point, liquid and gas phases become indistinguishable (supercritical fluid).
For water: \(T_c = 647.1\) K (373.95°C), \(p_c = 22.064\) MPa (220.64 bar)
\(v_c = 0.003155\) m³/kg
The temperature-specific volume (T-v) diagram reveals important behaviour:
Quality (Dryness Fraction): In the two-phase region, quality \(x\) represents the vapour mass fraction:
\[x = \frac{m_{vapour}}{m_{total}} = \frac{m_{vapour}}{m_{liquid} + m_{vapour}}\]
Where: \(0 \leq x \leq 1\) (0 = saturated liquid, 1 = saturated vapour)
\[v = v_f + x(v_g - v_f) = v_f + x \cdot v_{fg}\]
\(v_f\) = specific volume of saturated liquid, \(v_g\) = specific volume of saturated vapour
Heat water at constant pressure and watch the phase change unfold. Drag the Heat Input slider to add energy — observe the piston rise, bubbles form, and quality increase. Then change pressure and see how the boiling point and latent heat shift.
| Concept | Equation | Description |
|---|---|---|
| Quality (Dryness Fraction) | \(x = \frac{m_{vapour}}{m_{total}}\) | Fraction of mass in vapour phase (0 ≤ x ≤ 1) |
| Specific Volume (Two-Phase) | \(v = v_f + x \cdot v_{fg}\) | Linear interpolation between liquid and vapour |
| Enthalpy (Two-Phase) | \(h = h_f + x \cdot h_{fg}\) | Energy content in two-phase region |
| Entropy (Two-Phase) | \(s = s_f + x \cdot s_{fg}\) | Entropy in two-phase region |
| Clausius-Clapeyron Equation | \(\frac{dp}{dT} = \frac{h_{fg}}{T \cdot v_{fg}}\) | Slope of phase boundary curve |
| Ideal Gas (Superheated) | \(pv = RT\) | Approximate behaviour far from saturation |
| Latent Heat of Vaporisation | \(h_{fg} = h_g - h_f\) | Energy required for phase change at constant p, T |
| Latent Entropy of Vaporisation | \(s_{fg} = s_g - s_f\) | Entropy change during phase change at constant p, T |
Important Constants for Water:
A rigid container holds 2 kg of water at 150°C. The specific volume is measured to be 0.5 m³/kg. At this temperature, the saturation properties are: \(v_f = 0.001091\) m³/kg and \(v_g = 0.3928\) m³/kg. Calculate the quality (dryness fraction) of the mixture.
Show Solution
Solution:
Given:
Step 1: Determine the phase
Quality is only defined in the two-phase region where \(v_f \leq v \leq v_g\). We must check this first:
\[v = 0.5 \;\text{m}^3/\text{kg} \;>\; v_g = 0.3928 \;\text{m}^3/\text{kg}\]
Since \(v > v_g\), the state is superheated vapour, not a two-phase mixture.
Step 2: Conclusion
The quality (dryness fraction) concept does not apply in the superheated vapour region. Always verify that the state lies within the saturation dome before calculating quality.
Follow-up example: If instead \(v = 0.2\) m³/kg at the same temperature:
Since \(v_f = 0.001091 < 0.2 < v_g = 0.3928\), the state is in the two-phase region.
\[x = \frac{v - v_f}{v_g - v_f} = \frac{0.2 - 0.001091}{0.3928 - 0.001091} = \frac{0.198909}{0.391709} = 0.508\]
Answer: x = 0.508 (50.8% vapour, 49.2% liquid by mass)
A 0.5 m³ rigid tank contains saturated water mixture at 200 kPa with a quality of 0.6. Determine: (a) the mass of liquid, (b) the mass of vapour, and (c) the total mass in the tank. Use steam tables: at 200 kPa, \(v_f = 0.001061\) m³/kg, \(v_g = 0.8857\) m³/kg.
Show Solution
Solution:
Given:
Step 1: Calculate specific volume of mixture
\[v = v_f + x \cdot v_{fg} = 0.001061 + 0.6(0.8857 - 0.001061)\]
\[v = 0.001061 + 0.6(0.884639) = 0.001061 + 0.530783 = 0.531844 \;\text{m}^3/\text{kg}\]
Step 2: Calculate total mass
\[m_{total} = \frac{V}{v} = \frac{0.5}{0.531844} = 0.940 \text{ kg}\]
Step 3: Calculate vapour mass
\[m_{vapour} = x \cdot m_{total} = 0.6 \times 0.940 = 0.564 \text{ kg}\]
Step 4: Calculate liquid mass
\[m_{liquid} = (1-x) \cdot m_{total} = 0.4 \times 0.940 = 0.376 \text{ kg}\]
Answers:
(a) Mass of liquid = 0.376 kg
(b) Mass of vapour = 0.564 kg
(c) Total mass = 0.940 kg
Water is at a temperature of 300°C and has a specific volume of 0.02 m³/kg. At T = 300°C the saturation properties are: \(p_{sat} = 8.581\) MPa, \(v_f = 0.001404\) m³/kg, \(v_g = 0.02167\) m³/kg. Determine: (a) the phase of the water, and (b) if it is a two-phase mixture, calculate its quality x.
Show Solution
Given:
Step 1: Identify the phase region
Compare v with the saturation boundaries:
\[v_f = 0.001404 \;\text{m}^3/\text{kg} \;<\; v = 0.02 \;\text{m}^3/\text{kg} \;<\; v_g = 0.02167 \;\text{m}^3/\text{kg}\]
Since v lies between \(v_f\) and \(v_g\), the water is a two-phase (liquid–vapour) mixture at the saturation pressure of 8.581 MPa.
Step 2: Calculate quality
\[x = \frac{v - v_f}{v_g - v_f} = \frac{0.02 - 0.001404}{0.02167 - 0.001404} = \frac{0.018596}{0.020266} = \mathbf{0.918}\]
Answers:
(a) Phase: Two-phase liquid–vapour mixture at p = 8.581 MPa
(b) Quality: x = 0.918 (91.8% vapour by mass)
Calculate the energy required to convert 5 kg of saturated liquid water at 1 bar (100 kPa) to saturated vapour at the same pressure. At 1 bar: \(h_f = 417.46\) kJ/kg, \(h_g = 2675.5\) kJ/kg.
Show Solution
Solution:
Given:
Step 1: Calculate latent heat of vaporisation
\[h_{fg} = h_g - h_f = 2675.5 - 417.46 = 2258.04 \text{ kJ/kg}\]
Step 2: Calculate total energy required
\[Q = m \cdot h_{fg} = 5 \times 2258.04 = 11,290.2 \text{ kJ}\]
Answer: Q = 11,290 kJ (or 11.29 MJ) of energy is required
This is the energy needed for the isobaric (constant pressure) phase change from liquid to vapour.