Question

Steam is contained in a closed rigid container with a volume of \(1 \mathrm{~m}^{3}\). Initially, the pressure and temperature of the steam are 7 bar and \(500^{\circ} \mathrm{C}\), respectively. The temperature drops as a result of heat transfer to the surroundings. Determine the temperature at which condensation first occurs, in \({ }^{\circ} \mathrm{C}\), and the fraction of the total mass that has condensed when the pressure reaches \(0.5\) bar. What is the volume, in \(\mathrm{m}^{3}\), occupied by saturated liquid at the final state?

Short answer

Condensation starts at 81.33°C. Fraction of mass condensed is 0.315, and volume occupied by liquid is 0.315 m³.

Step-by-step solution

- Understand Initial Conditions

The initial state of the steam in the container is given: a volume of 1 m³, initial pressure of 7 bar, and initial temperature of 500°C. This represents a superheated state.

- Locate Initial State on a Steam Table

Using steam tables, find the specific volume (v) of the steam at the initial conditions (7 bar and 500°C).

- Calculate Specific Volume

From the steam tables, at 7 bar and 500°C, the specific volume of steam is approximately 0.257 m³/kg.

- Determine Transition to Saturated State

Using the specific volume (0.257 m³/kg), we identify the temperature at which condensation first occurs by comparing it to the specific volumes of saturated liquid (v_f) and saturated vapor (v_g) at decreasing pressures until the specific volume matches the calculated value.

- Identify Saturation Condition

Condensation begins when the specific volume falls between the saturated liquid and vapor values. At 0.5 bar, the specific volume of saturated vapor (v_g) is 0.3749 m³/kg. This confirms condensation can occur as the v_g > 0.257 m³/kg.

- Find Temperature at Vaporization Point

From the steam tables, the temperature corresponding to the 0.5 bar pressure is 81.33°C. This is the temperature at which condensation first starts.

- Fraction of Mass Condensed

Using the specific volumes: v_f (liquid) = 0.001030 m³/kg and v_g (vapor) = 0.3749 m³/kg at 0.5 bar. Calculate the quality (x) using the equation: \[ v = v_f + x(v_g - v_f) \]Solving for x, we get: \[ 0.257 = 0.001030 + x(0.3749 - 0.001030) \]\[ x ≈ 0.685 \]The fraction of the total mass that has condensed is then 1 - x: \[ 1 - 0.685 = 0.315 \]

- Volume Occupied by Saturated Liquid

The volume occupied by the saturated liquid can be calculated by multiplying the mass fraction of the condensed liquid by the total volume: \[ V_{liquid} = 0.315 \times 1 \text{ m}^{3} = 0.315 \text{ m}^{3} \]

Key concepts

specific volume

Specific volume is a key concept in thermodynamics. It is defined as the volume occupied by a unit mass of a substance. Its units are typically in \(m^3/kg\). For steam, specific volume indicates how much space one kilogram of steam occupies.
In the exercise, the initial specific volume of the steam helps determine the changing state as thermal conditions vary. Knowing specific volumes at different states aids in identifying when condensation begins.
For instance, the initial specific volume of steam at 7 bar and 500°C was found using steam tables to be approximately 0.257 m³/kg. This value is compared to identify the transition from a superheated to a saturated state.

saturated state

The saturated state is the condition where a substance exists simultaneously in both liquid and vapor forms at equilibrium. When steam in a closed container cools down, it approaches this state.
The temperature and pressure at which steam starts to condense (turn into liquid) marks the saturated state.
From the steam tables in the exercise, we identified that at 0.5 bar pressure, the steam’s specific volume made it possible for condensation to begin. The saturated vapor specific volume (\(v_g\)) at this pressure is crucial for this process.

condensation process

Condensation is the process of changing from a vapor state to a liquid state. In the context of the exercise, when the steam cools and reaches its saturation point at 0.5 bar, some of it starts to condense.
Initially, the specific volume helps recognize when condensation starts. For condensation at 0.5 bar, the temperature was 81.33°C. The specific volumes of saturated liquid (\(v_f\)) and vapor (\(v_g\)) at this pressure help determine how much of the steam condenses.
Calculating the fraction of the steam that has condensed involves determining the quality (x) of the steam, representing the fraction or percentage of steam still in the vapor form.

steam tables

Steam tables are vital for finding the properties of water and steam at various temperatures and pressures. They provide parameters like specific volume, temperature, pressure, enthalpy, and internal energy among others.
In the exercise, using steam tables helps locate specific volumes \((v_f\), \((v_g\)) and other important values at different states. For instance, at 7 bar and 500°C, the specific volume needed was 0.257 m³/kg.
Likewise, steam tables are used to find the temperature at which steam starts condensing at a given pressure (0.5 bar), which was 81.33°C. Proper use of steam tables is essential for solving thermodynamic problems effectively.

quality (mass fraction)

The quality of steam, represented by x, is a measure of the mass fraction that is vapor in a liquid-vapor mixture. It varies between 0 (completely liquid) and 1 (completely vapor).
In thermodynamic processes, knowing the quality helps compute how much of the steam is in the vapor state and how much has condensed.
This is key in the exercise's step 7 where the quality was calculated as approximately 0.685 at 0.5 bar using the equation:
\[ v = v_f + x(v_g - v_f) \]
Therefore, the mass fraction of the steam that has condensed is the remaining part, \((1 - x)\), which equals 0.315. This information helps compute other properties like the volume occupied by the saturated liquid at the final state.