Question 7.9: Problem: An insulated piston‐cylinder device contains 2 kg o...
Problem: An insulated piston‐cylinder device contains 2 kg of saturated steam at 0.5 MPa. The steam is allowed to expand reversibly until its pressure drops to 0.1 MPa. Find the work done during this process.
Find: Work done during expansion W.
Known: Saturated steam, mass of steam m = 2 kg, initial pressure P_1 = 0.5 MPa, final pressure P_2 = 0.1 MPa, reversible and insulated (adiabatic) means isentropic.
Diagram:
Governing Equations:
Energy balance Q + W = ∆U + ∆KE + ∆PE
In this problem the initial state of the system is known, since we know the pressure (P_1 = 0.5 MPa) and the quality (x = 1). To fix the final state we know the final pressure (P_2 = 0.1 MPa). We can determine a second property from the fact that the expansion process is adiabatic and reversible, which means that it is isentropic (s_1= s_2 ). Therefore, from the saturated water tables (Appendix 8b) at P_1 = 0.5 MPa, specific enthalpy of liquid s_1 = s_g = 6.8213 kJ / kgK.
At P_2 = 0.1 MPa, specific entropy of liquid s_f = 1.3026 kJ / kgK and vapour s_g = 7.3594 kJ / kgK. Since s_f< s_2< s_g , we know that the final state is a saturated vapour–liquid mixture. We can sketch the process on a P‐ν diagram.
The final quality is
x_2= \frac{s_2-s_f}{s_g-s_f}= \frac{6.8213 \ kJ/kgK -1.3026 \ kJ/kgK}{7.3594 \ kJ/kgK-1.3026 \ kJ/kgK}=0.911158.
From an energy balance,
W+\underbrace{Q}_{=0} =\Delta U=m(u_2-u_1).
At P_1 = 0.5 MPa, specific internal energy of vapour u_1= u_g = 2561.2 kJ / kg. At P_2 = 0.1 MPa, specific internal energy of liquid u_f = 417.36 kJ / kg and vapour u_g = 2506.1 kJ / kg. Using the quality of x_2 = 0.911158 from earlier,
u_2=u_f+x_2(u_g-u_f), \\ u_2 = 417.36 \ kJ/kg+0.911158×(2506.1 \ kJ/kg -417.36 \ kJ/kg)=2320.53 \ kJ/kg. \\ W=m(u_2-u_1)=2 \ kg × (2320.2 \ kJ/kg-2561.2 \ kJ /kg), \\ W= -481.340 \ kJ.
Answer: The system does 481.3 kJ of work on the surroundings while expanding.