An insulated pistoncylinder arrangement has an initial volum

An insulated piston/cylinder arrangement has an initial volume of 0.15m³ and contains steam at 400 kPa and 200 degrees C. The steam is expanded reversibly and adiabatically, and the work output is measured very carefully to be 30 kJ.

(a) Calculate the change in internal energy of this process

(b) Determine U and S of the steam after expansion

(c) Evaluate if the final state of steam is in the two-phase (liquid/vapor) region

Solution

Vi = 0.15 m^3
Pi = 400 kPa
Ti = 200 deg. C
Q = 0 (adiabatic)
W = 30 kJ
Mr = 18

Mass of steam,
PiVi = m(R/Mr)Ti
m = Pi*Vi*Mr / RTi
m = 400*0.15*18 / 8.314*473
m = 0.2746 kg

Energy balance for closed system,
E = Q - W, E = U
U = -W
U1 - U2 = W
U2 = U1 - W
U2 = m*u1 - W
U2 = 0.2746*2646.6761 - 30
m2u2 = 696.8538 kJ
u2 = 2537.705 kJ/kg

Evaluate internal energy, u2 = 2537.705 kJ/kg with P2 < 400 kPa
Internal energy at initial pressure, 400 kPa, at saturated vapor condition,
uif = u(Pi) = 2553.6 kJ/kg
It can be seen that u(Pi) > u2, thus there is internal energy of two-phase region at P < 4bar(400 kPa). The claim is valid with final pressure around 300 kPa. But this condition may not be reached if the entropy is less than initial entropy. However based on the data the claim is possible because entropy final may higher, than initial entropy, s2 > si.