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2. A piston-cylinder device initially contains 2.4 kg saturated liquid water at 160 °C. Now heat is transferred to the water until the volume triples and the cylinder contains saturated vapor only. Determine (a) the volume of the tank, (b) the final temperature and pressure, and (c) the internal energy change of the water. H2O 2.4 kg 160 °C

2. A piston-cylinder device initially contains 2.4 kg saturated liquid water at 160 °C. Now heat is transferred to the water until the volume triples and the cylinder contains saturated vapor only. Determine (a) the volume of the tank, (b) the final temperature and pressure, and (c) the internal energy change of the water. H2O 2.4 kg 160 °C

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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### Thermodynamics Problem: Piston-Cylinder Device #### Problem Statement: A piston-cylinder device initially contains 2.4 kg of saturated liquid water at 160°C. Heat is transferred to the water until the volume triples, and the cylinder contains saturated vapor only. Determine: (a) The volume of the tank (b) The final temperature and pressure (c) The internal energy change of the water #### Diagram: The diagram depicts a simple piston-cylinder setup with the following initial conditions: - Substance: H₂O - Mass: 2.4 kg - Temperature: 160°C An energy input, denoted by Q, is transferred to the system causing a phase change. #### Solution: To solve this problem, we will follow these steps: 1. **Determine the initial volume of the saturated liquid water.** 2. **Calculate the final volume of the water after heating.** 3. **Identify the properties of the saturated vapor at the final state (Volume, Temperature, and Pressure).** 4. **Compute the internal energy change.** ##### Step-by-Step Solution: 1. **Initial Volume:** - Use the saturated water tables to find the specific volume (v_f) of saturated liquid water at 160°C. - The specific volume of water (v_f) at 160°C can be found to be approximately 0.001053 m³/kg. - Initial volume (V_i) = mass × specific volume = 2.4 kg × 0.001053 m³/kg = 0.0025272 m³. 2. **Final Volume:** - Given that the volume triples upon heating: - Final volume (V_f) = 3 × Initial volume = 3 × 0.0025272 m³ = 0.0075816 m³. 3. **Final State Properties:** - Since the water is converted to saturated vapor upon heating: - At 160°C, the specific volume of saturated vapor (v_g) is 0.39279 m³/kg. - Using this to find final volume: - V_f = mass × specific volume - 0.0075816 m³ = 2.4 kg × v_g - Since v_g for saturated vapor at 160°C is consistent and matches: - Pressure stays at the saturation pressure at 160°C, which is 618.78 k
Transcribed Image Text:### Thermodynamics Problem: Piston-Cylinder Device #### Problem Statement: A piston-cylinder device initially contains 2.4 kg of saturated liquid water at 160°C. Heat is transferred to the water until the volume triples, and the cylinder contains saturated vapor only. Determine: (a) The volume of the tank (b) The final temperature and pressure (c) The internal energy change of the water #### Diagram: The diagram depicts a simple piston-cylinder setup with the following initial conditions: - Substance: H₂O - Mass: 2.4 kg - Temperature: 160°C An energy input, denoted by Q, is transferred to the system causing a phase change. #### Solution: To solve this problem, we will follow these steps: 1. **Determine the initial volume of the saturated liquid water.** 2. **Calculate the final volume of the water after heating.** 3. **Identify the properties of the saturated vapor at the final state (Volume, Temperature, and Pressure).** 4. **Compute the internal energy change.** ##### Step-by-Step Solution: 1. **Initial Volume:** - Use the saturated water tables to find the specific volume (v_f) of saturated liquid water at 160°C. - The specific volume of water (v_f) at 160°C can be found to be approximately 0.001053 m³/kg. - Initial volume (V_i) = mass × specific volume = 2.4 kg × 0.001053 m³/kg = 0.0025272 m³. 2. **Final Volume:** - Given that the volume triples upon heating: - Final volume (V_f) = 3 × Initial volume = 3 × 0.0025272 m³ = 0.0075816 m³. 3. **Final State Properties:** - Since the water is converted to saturated vapor upon heating: - At 160°C, the specific volume of saturated vapor (v_g) is 0.39279 m³/kg. - Using this to find final volume: - V_f = mass × specific volume - 0.0075816 m³ = 2.4 kg × v_g - Since v_g for saturated vapor at 160°C is consistent and matches: - Pressure stays at the saturation pressure at 160°C, which is 618.78 k
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