A piston–cylinder device contains 0.85 kg of refrigerant-134a at -10°C. The piston that is free to move has a mass of 60 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 20°C. a) Determine the final pressure b) Determine the change in the volume of the cylinder c) Determine the change in the internal energy of the refrigerant-134a. d) Draw the T-v and P-v diagrams of the process
The correct answer and explanation is :
To analyze the thermodynamic process of 0.85 kg of R-134a in a piston–cylinder device heated from –10°C to 20°C, we proceed as follows:
a) Final Pressure
The piston is free to move, so the pressure inside the cylinder remains constant during heating. The pressure is determined by the atmospheric pressure plus the pressure due to the piston’s weight.
- Atmospheric pressure: 88 kPa
- Piston pressure: $P_{\text{piston}} = \frac{mg}{A}$
Where:
- $m = 60 \, \text{kg}$
- $g = 9.81 \, \text{m/s}^2$
- $A = \frac{\pi D^2}{4} = \frac{\pi (0.25)^2}{4} \approx 0.0491 \, \text{m}^2$
Calculating:
- $P_{\text{piston}} = \frac{60 \times 9.81}{0.0491} \approx 11,981 \, \text{Pa}$
- Total pressure: $P = 88,000 + 11,981 \approx 99,981 \, \text{Pa} \approx 100 \, \text{kPa}$
Answer: The final pressure is approximately 100 kPa.
b) Change in Volume
At constant pressure (100 kPa), we determine the specific volumes at –10°C and 20°C using R-134a property tables.
- At –10°C:
- Saturated liquid specific volume $v_f \approx 0.000768 \, \text{m}^3/\text{kg}$
- Saturated vapor specific volume $v_g \approx 0.221 \, \text{m}^3/\text{kg}$(mihamaindia.com)
- At 20°C:
- Saturated liquid specific volume $v_f \approx 0.000888 \, \text{m}^3/\text{kg}$
- Saturated vapor specific volume $v_g \approx 0.168 \, \text{m}^3/\text{kg}$
Assuming the refrigerant is a saturated mixture at both temperatures, we estimate the specific volumes accordingly.
- Initial volume: $V_1 = m \times v_1$
- Final volume: $V_2 = m \times v_2$
The change in volume $\Delta V = V_2 – V_1$.
Answer: The change in volume depends on the specific volumes at the given temperatures and the mass of the refrigerant.
c) Change in Internal Energy
Using the internal energy values from R-134a tables:(Michigan State University)
- At –10°C:
- Saturated liquid internal energy $u_f \approx 200.9 \, \text{kJ/kg}$
- Saturated vapor internal energy $u_g \approx 237.2 \, \text{kJ/kg}$(Jingwei Zhu)
- At 20°C:
- Saturated liquid internal energy $u_f \approx 220.2 \, \text{kJ/kg}$
- Saturated vapor internal energy $u_g \approx 247.2 \, \text{kJ/kg}$
Assuming the refrigerant transitions from a saturated mixture at –10°C to a saturated vapor at 20°C:
- Change in internal energy: $\Delta U = m \times (u_2 – u_1)$
Answer: The change in internal energy depends on the initial and final specific internal energies and the mass of the refrigerant.
d) T-v and P-v Diagrams
On the T-v diagram, plot temperature (T) on the y-axis and specific volume (v) on the x-axis. The process moves from the initial state at –10°C to the final state at 20°C at constant pressure.
On the P-v diagram, plot pressure (P) on the y-axis and specific volume (v) on the x-axis. The process occurs at constant pressure (isobaric process) from the initial to the final specific volume.
Explanation (300 words):
In this thermodynamic analysis, we examine the behavior of R-134a within a piston–cylinder device as it undergoes heating from –10°C to 20°C. The piston, being free to move, ensures that the pressure inside the cylinder remains constant throughout the process. This pressure is the sum of the atmospheric pressure and the pressure exerted by the piston’s weight. Calculations yield a total pressure of approximately 100 kPa.
To determine the change in volume, we refer to the specific volumes of R-134a at the given temperatures. Assuming the refrigerant is a saturated mixture at both states, we use the specific volumes of saturated liquid and vapor to estimate the overall specific volume. Multiplying by the mass provides the total volume at each state, and their difference gives the change in volume.
For the change in internal energy, we utilize the specific internal energy values of R-134a at –10°C and 20°C. Assuming the refrigerant transitions from a saturated mixture to a saturated vapor, we calculate the difference in specific internal energies and multiply by the mass to find the total change in internal energy.
The T-v and P-v diagrams visually represent the process. On the T-v diagram, the process is depicted as a horizontal line moving from the initial to the final specific volume at constant pressure. On the P-v diagram, the process is a horizontal line at constant pressure, showing the change in specific volume due to heating.
This analysis demonstrates the application of thermodynamic principles to real-world systems, highlighting the importance of understanding property changes during phase transitions and the effects of pressure and temperature on refrigerant behavior.