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A pressure cooker has a volume of \(0.011 \mathrm{~m}^{3}\) and initially contains a two-phase liquid-vapor mixture of \(\mathrm{H}_{2} \mathrm{O}\) at a temperature of \(100^{\circ} \mathrm{C}\) and a quality of \(10 \%\). As the water is heated at constant volume, the pressure rises to 2 bar and the quality becomes \(18.9 \%\). With further heating a pressureregulating valve keeps the pressure constant in the cooker at 2 bar by allowing saturated vapor at 2 bar to escape. Neglecting kinetic and potential energy effects, (a) determine the quality of the \(\mathrm{H}_{2} \mathrm{O}\) at the initial onset of vapor escape (state 2 ) and the amount of heat transfer, in \(\mathrm{kJ}\), to reach this state. (b) determine the final mass in the cooker, in \(\mathrm{kg}\), and the additional amount of heat transfer, in \(\mathrm{kJ}\), if heating continues from state 2 until the final quality is \(1.0\). (c) plot the quantities of part (b) versus quality increasing from the value at state 2 to \(100 \%\).

Step by step solution

01

Understand Initial Conditions

The pressure cooker has a volume of 0.011 m鲁, and it initially contains a mixture of liquid and vapor water (H鈧侽) at a temperature of 100掳C with a quality of 10%. Quality (x) is the fraction of the mass that is in vapor form. Therefore, at the initial state, x鈧� = 0.10.

02

Determine Specific Volumes at Initial State

For water at 100掳C, the saturated liquid specific volume (v_f) and saturated vapor specific volume (v_g) can be found in the steam tables: v_f = 0.001043 m鲁/kg v_g = 1.6720 m鲁/kg Using these and the initial quality (x鈧� = 0.10), calculate the specific volume (v鈧�): v鈧� = (1 - x鈧�) * v_f + x鈧� * v_g = (1 - 0.10) * 0.001043 + 0.10 * 1.6720 v鈧� = 0.168 m鲁/kg

03

Calculate Initial Mass

Using the initial volume and the specific volume: mass (m鈧�) = Volume / specific volume = 0.011 m鲁 / 0.168 m鲁/kg m鈧� = 0.0655 kg

04

State Update to 2 bar and Determine Quality

At 2 bar (200 kPa), find v_f and v_g for water at this pressure: v_f = 0.001061 m鲁/kg v_g = 0.8857 m鲁/kg Given that the mixture's volume remains constant and the updated quality is 18.9% (x鈧� = 0.189), calculate the specific volume (v鈧�): v鈧� = (1 - x鈧�) * v_f + x鈧� * v_g = (1 - 0.189) * 0.001061 + 0.189 * 0.8857 v鈧� 鈮� 0.167 m鲁/kg

05

Amount of Heat Transfer to Reach State 2

Using steam tables, find enthalpy of liquid and vapor at 2 bar: h_f = 504.7 kJ/kg h_fg = 1904.9 kJ/kg Calculate initial and final specific enthalpies (h鈧� and h鈧�): h鈧� = h_f (at 100掳C) + x鈧� * h_fg (at 100掳C) = 419 kJ/kg + 0.10 * 2087.9 kJ/kg h鈧� = 627.79 kJ/kg h鈧� = h_f + x鈧� * h_fg = 504.7 kJ/kg + 0.189 * 1904.9 kJ/kg h鈧� = 864.47 kJ/kg Amount of heat transfer (Q鈧佲倐) to reach state 2: Q鈧佲倐 = m鈧� * (h鈧� - h鈧�) = 0.0655 kg * (864.47 kJ/kg - 627.79 kJ/kg) Q鈧佲倐 鈮� 15.53 kJ

06

Final Mass After Vapor Release

Since the pressure is constant at 2 bar and saturated vapor escapes, the final quality x鈧� is 1.0. Using the final volume and specific volume of vapor (v_g): v_g = 0.8857 m鲁/kg Final mass (m鈧�) = Volume / specific volume = 0.011 m鲁 / 0.8857 m鲁/kg m鈧� 鈮� 0.012 kg

07

Additional Heat Transfer for Full Vaporization

Enthalpy at 100% quality (x鈧� = 1.0) at 2 bar: h_g = h_f + h_fg = 504.7 kJ/kg + 1904.9 kJ/kg h_g = 2409.6 kJ/kg Additional heat transfer from state 2 to fully vaporized state: Q鈧傗們 = m鈧� * (h_g - h鈧�) Q鈧傗們 = 0.012 kg * (2409.6 kJ/kg - 864.47 kJ/kg) Q鈧傗們 鈮� 18.51 kJ

08

Plot Quality vs Heat Transfer

Create a plot with 'Quality (x)' on the x-axis ranging from 0.189 to 1.0 and 'Heat Transfer (Q in kJ)' on the y-axis. Note that the plot shows a continuous increase in heat transfer required as quality increases, starting from 15.53 kJ at x=0.189 and ending at Q鈧傗們 at x=1.0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pressure cooker thermodynamics

Pressure cookers operate based on thermodynamic principles. When the water inside is heated, it forms a liquid-vapor mixture that increases the pressure. Understanding the behavior of this system requires analyzing properties like temperature, pressure, and volume. As the water temperature rises at constant volume inside the pressure cooker, the pressure also increases because the volume of gas (steam) expands while the volume is fixed. This pressure rise can then be controlled using a pressure-regulating valve, which allows excess steam to escape and maintain a specific pressure. This helps in achieving desired cooking conditions efficiently.

specific volume calculations

Specific volume is a measure of the volume occupied by a unit mass of a substance and is crucial in thermodynamics. In the given problem, it helps calculate the specific volume of a mixture using the steam tables. For example, at the initial state when the quality is 10%, the specific volume can be calculated using: \[ v = (1 - x) \times v_f + x \times v_g \] where \( v_f \) is the specific volume of the saturated liquid, \( v_g \) is the specific volume of the saturated vapor, and \( x \) is the quality of the mixture. A correct understanding and computation of this property are essential to determining the mass and behavior of the substance in different states.

quality of vapor-liquid mixtures

Quality or dryness fraction is the ratio of the mass of vapor to the total mass of the liquid-vapor mixture. It ranges from 0 (only liquid) to 1 (only vapor). At state 1, with quality 10%, it means the mixture is largely liquid. As heating continues at constant volume, quality increases. When pressure reaches 2 bar with the new quality of 18.9%, the steam content in the mixture represents a higher fraction, indicating a larger portion of vapor. The quality changes dynamically as heat is added or removed, affecting the thermodynamic properties and equilibrium of the system.

heat transfer in phase changes

Heat transfer during phase changes plays a critical role in thermodynamics, especially in processes involving latent heat. To determine the amount of heat added to the system, we use the relation: \[ Q = m \times (h_2-h_1) \] where \( m \) is the mass, \( h_1 \) is the specific enthalpy at the initial state, and \( h_2 \) is the specific enthalpy at the final state. Latent heat is the energy required for the phase change without changing temperature. When the water in the pressure cooker heats from state 1 to state 2, a certain amount of heat is transferred causing water to evaporate raising the quality.

steam tables usage

Steam tables are critical tools in solving thermodynamic problems involving water and steam. These tables provide properties like temperature, pressure, specific volume, enthalpy, and entropy for both the liquid and vapor phases. For example, from the steam tables, at 100掳C the specific volumes \( v_f \) and \( v_g \) and the enthalpies \( h_f \) and \( h_{fg} \) for water can be identified. Correct usage of these values is crucial to finding specific volume and the total heat transfer required in the phase change process. Also, steam tables help identify the state of steam including sub-cooled, superheated or saturated conditions.