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Steam is one of the most important working fluids in engineering thermodynamics, especially in power generation, heating, and various industrial processes. Understanding the properties of steam is essential for analyzing thermodynamic cycles, designing boilers, and optimizing energy systems. The state of steam at any moment can be described by its thermodynamic properties such as pressure, temperature, and specific volume.
To visualize how these properties relate to each other, engineers use the p-v-T surface, a three-dimensional representation showing the relationship between pressure (p), specific volume (v), and temperature (T) of steam. This surface helps us understand phase changes, such as boiling and condensation, and distinguishes different steam regions like saturated, wet, and superheated steam.
In this section, we will explore these properties from first principles, learn how to interpret the p-v-T surface, and use steam tables and Mollier diagrams to solve practical engineering problems.
Before diving into the p-v-T surface, let's define the fundamental properties that describe the state of steam:
Each property is independent but related through the thermodynamic state of steam. Knowing any two properties allows us to determine the third, thanks to the steam's equation of state and phase behavior.
The p-v-T surface is a three-dimensional plot showing how pressure, specific volume, and temperature of steam vary together. It is crucial for understanding steam behavior, especially during phase changes.
Key features of the p-v-T surface include:
Inside the saturation dome, steam is a mixture of liquid and vapor. The quality or dryness fraction (x) indicates the proportion of vapor in this mixture. Outside the dome, steam is either compressed liquid (left side) or superheated vapor (right side).
Understanding this surface helps engineers predict how steam will behave in boilers, turbines, and condensers, which is essential for efficient power plant operation.
Because the p-v-T surface is complex, engineers rely on steam tables and the Mollier diagram to find steam properties quickly and accurately.
Steam Tables provide numerical values of pressure, temperature, specific volume, enthalpy, and entropy for saturated and superheated steam at various conditions.
| Pressure (MPa) | Temp. Sat. (°C) | vf (m³/kg) | vg (m³/kg) | hf (kJ/kg) | hg (kJ/kg) | sf (kJ/kg·K) | sg (kJ/kg·K) |
|---|---|---|---|---|---|---|---|
| 0.1 | 99.61 | 0.00104 | 1.6720 | 419.04 | 2675.5 | 1.3036 | 7.3549 |
| 0.5 | 151.83 | 0.00109 | 0.3749 | 640.09 | 2748.7 | 1.8710 | 6.5922 |
| 1.0 | 179.91 | 0.00105 | 0.1944 | 762.81 | 2776.2 | 2.1384 | 6.4460 |
Mollier Diagram (h-s diagram) plots enthalpy (h) against entropy (s) for steam at various pressures and temperatures. It is a valuable graphical tool for quick estimation of steam properties during thermodynamic processes, especially in turbines and compressors.
Note: Steam tables and Mollier diagrams are indispensable for engineers working with steam, as they provide accurate data beyond the ideal gas approximations.
Given steam at a pressure of 0.5 MPa with a specific volume of 0.5 m³/kg, determine whether the steam is saturated, wet, or superheated. Also, find the corresponding temperature.
Step 1: From the saturated steam table at 0.5 MPa, note the specific volumes:
Step 2: Compare the given specific volume \( v = 0.5 \, m^3/kg \) with \( v_g \).
Since \( v > v_g \), the steam is not saturated or wet; it is in the superheated region.
Step 3: To find the temperature, refer to the superheated steam tables at 0.5 MPa and look for specific volume close to 0.5 m³/kg. For example, at 0.5 MPa:
By interpolation, the temperature corresponding to \( v = 0.5 \, m^3/kg \) is approximately 220°C.
Answer: The steam is superheated at about 220°C and 0.5 MPa.
Steam at 0.1 MPa has a specific volume of 1.0 m³/kg. Calculate the dryness fraction (quality) of the steam.
Step 1: From the saturated steam table at 0.1 MPa:
Step 2: Use the formula for quality \( x \):
\[ x = \frac{v - v_f}{v_g - v_f} \]
Substitute values:
\[ x = \frac{1.0 - 0.00104}{1.6720 - 0.00104} = \frac{0.99896}{1.67096} \approx 0.598 \]
Answer: The dryness fraction of the steam is approximately 0.60, meaning 60% vapor and 40% liquid by mass.
Find the enthalpy and entropy of superheated steam at 0.8 MPa and 300°C using the Mollier diagram.
Step 1: Locate the pressure line for 0.8 MPa on the Mollier diagram.
Step 2: Move along the 0.8 MPa line to the point corresponding to 300°C.
Step 3: Read the enthalpy (h) and entropy (s) values from the diagram.
Typical values from Mollier diagram or superheated steam tables:
Answer: Enthalpy is approximately 3050 kJ/kg and entropy is 7.0 kJ/kg·K at 0.8 MPa and 300°C.
Steam expands isentropically from 3 MPa and 400°C to 0.1 MPa. Calculate the work done per kg of steam during this expansion.
Step 1: Identify initial state (State 1):
From superheated steam tables at 3 MPa and 400°C:
Step 2: Since the process is isentropic, entropy at final state (State 2) is \( s_2 = s_1 = 6.7 \, kJ/kg \cdot K \).
Step 3: At \( p_2 = 0.1 \, MPa \), find \( h_2 \) corresponding to \( s_2 = 6.7 \) from steam tables or Mollier diagram.
At 0.1 MPa:
Since \( s_2 = 6.7 < s_g \), steam is superheated.
From superheated steam tables at 0.1 MPa and \( s = 6.7 \), approximate \( h_2 \approx 2600 \, kJ/kg \).
Step 4: Calculate work done during expansion (assuming no heat transfer):
\[ W = h_1 - h_2 = 3215 - 2600 = 615 \, kJ/kg \]
Answer: The work done during isentropic expansion is approximately 615 kJ/kg of steam.
Steam initially at 0.2 MPa and 120°C is compressed isobarically to saturated liquid at the same pressure. Using the p-v-T surface, determine the change in specific volume and temperature.
Step 1: Identify initial state (superheated steam):
From superheated steam tables at 0.2 MPa and 120°C, specific volume \( v_1 \approx 1.2 \, m^3/kg \).
Step 2: Final state is saturated liquid at 0.2 MPa:
Step 3: Calculate change in specific volume:
\[ \Delta v = v_f - v_1 = 0.00116 - 1.2 = -1.19884 \, m^3/kg \]
The negative sign indicates a large decrease in volume during compression.
Step 4: Temperature change is from 120°C (superheated) to 120.2°C (saturated liquid), essentially constant pressure but temperature approaches saturation temperature.
Answer: The specific volume decreases drastically from 1.2 to 0.00116 m³/kg, and temperature approaches the saturation temperature of 120.2°C during isobaric compression.
When to use: Quickly identifying steam phase regions on the p-v-T surface during problem solving.
When to use: When exact pressure or temperature values are unavailable in tables to improve accuracy.
When to use: To avoid errors when calculating properties of wet steam and interpreting results.
When to use: During time-constrained entrance exams to save time and reduce calculation errors.
When to use: To prevent unit conversion errors in calculations and ensure consistency.
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