Thermodynamics 1

Question

A piston/cylinder assembly contains 0.1 kg R-410A of quality x = 0.25 and at −20°C. The piston is initially free-floating. Stops are mounted so that the volume at the stops is three times greater than the initial volume V_final=3V_initial. Heat is added to the control volume by a 50°C temperature source so that the final temperature is 20°C. Kinetic and potential energy changes are negligible.

1. Calculate the magnitude of the total entropy generation.
2. Qualitatively draw the process in a p-v diagram. You must explain step-by-step the process line.

Answer 1

To calculate the total entropy generation for the process you've described, we need to follow a series of steps. Let's start with the calculation and then move on to the qualitative drawing of the process in a p-v (pressure-volume) diagram.

### Calculation of Total Entropy Generation

1. Determine Initial and Final States:

- Initial State: R-410A with a quality of x = 0.25% at T_initial = -20^oC So ν_initial=0.013627 m³/Kg H_initial=76.97 KJ/Kg and S_initial=0.3079 KJ/kg-K and considering 0.1kg for the mass, the initial volume will be 0.001363 m³.

- Final State: R-410A with T_final = 20^oC and the volume is 3 times the initial volume, so the final volume is 0.004088 m³ and specific volume of final will be 0.040881 m³/kg which is more than saturated specific volume, so R-410A is at superheat condition. Now, we need to interpolate and find the fluid condition by interpolation and find the final pressure and final entropy and final enthalpy of the fluid.

At saturation, pressure is 1444.2 kPa at T=20^oC. and specific volume of saturated vapor is 0.01758 m³/kg. at T=20^oC and pressure of 600kPa, the specific volume is 0.05106 m³/kg.

Our final specific volume is 0.040881 m³/kg and by linear interpolation we can find

P_final=(0.040881-0.01758)*((600-1444.2)/(0.05106-0.01758))+1444.2=856.664 kPa

This way we can find the final state enthalpy and entropy as

H_final=297.922

S_final=1.1069

So change in entropy is Mass*(S_final-S_initial)=0.1*(1.1069-0.3079)=0.0799 kJ/kg

The total heat transferred is the change in enthalpy which is m*(H_final-H_initial)=0.1*(297.922-76.97)=22.0952 kJ and the temperature of the heat source is 50^oC or 323.15K.

so the generated entropy is delta S-Q/T_source=0.0799-22.0952/323.15=0.011526 kJ/K

### Qualitative p-v Diagram