Question 7.8: Problem: A piston‐cylinder system contains 0.1 kg of water a...
Problem: A piston‐cylinder system contains 0.1 kg of water at 0.5 MPa and 400 °C. It is cooled at constant pressure so that 100 kJ of energy is removed from the system. What is the final state of the water?
Find: Final state of the water.
Known: Mass of the water m = 0.1 kg, initial pressure P_1 = 0.5 MPa, initial temperature T_1 = 400 °C, constant pressure so final pressure P_2 = P_1 , heat removed Q_{12} = –100 kJ.
Diagram:
Governing Equations:
For a constant pressure process q_{12}=\frac{Q_{12}}{m} =h_2 – h_1.
This is a first‐law control mass problem of the type that we have seen in Chapter 4, the only difference being that we now also consider the possibility of a phase change occurring within the fluid. To determine whether this happens, we have to fix the initial and final state of the water in the system.
To fix the state of a simple compressible system we need two independent properties. In the initial state we know the pressure ( P_1 = 0.5 MPa) and temperature ( T_1 = 400 °C). From the superheated steam tables, at a pressure of 0.5 MPa, T_{sat} = 151.86 °C. Since T_1 > T_{sat} , we know that the water is in the form of superheated steam. We can therefore mark the initial state (1) on a T‐v diagram. It is cooled at constant pressure, so we can represent the process by a line that follows the isobar of P = 0.5 MPa on T‐v axes.
How do we determine the final state (2)? We know one property, pressure, but we need to determine one more, which we can get from an energy balance. For heat transfer at constant pressure,
q_{12}=\frac{Q_{12}}{m} =h_2 – h_1.
From the superheated steam tables (Appendix 8c) at P_1 = 0.5 MPa and T_1 = 400 °C, specific enthalpy h_1 = 3271.9 kJ / kgK. Therefore,
h_2=h_1 + \frac{Q_{12}}{m} =3271.9 \ kJ/kgK + \frac{(-100 \ kJ)}{0.1 \ kg} =2271.9 \ kJ/kg.
From the saturated water tables (Appendix 8b) at P_1 = 0.5 MPa, specific enthalpy of liquid h_f = 640.23 kJ / kg and vapour h_g = 2748.7 kJ / kg. Since h_f < h_2 <h_g , the final state is a saturated mixture:
h_2 = x_2h_g + (1 – x_2) h_f=h_f + x_2(h_g – h_f),
x_2=\frac{h_2 – h_f}{h_g – h_f} =\frac{2271.9 \ kJ/kg – 640.23 \ kJ/kg}{2748.9 \ kJ/kg – 640.23 \ kJ/kg} =0.773 \ 79.
Answer: The final state will be a saturated mixture with a quality of 77.4%.