Question

One kilogram of water at 300°C expands against a piston in a cylinder until it reaches ambient pressure, 100 kPa, at which point the water has a quality of 90.2%. It may be assumed that the expansion is reversible and adiabatic. What was the initial pressure in the cylinder and how much work is done by the water?

Answer 1

C.V. Water. Process: Rev., Q = 0

Energy Eq. : m(u2 − u1) = 1Q2 − 1W2 = − 1W2

Entropy Eq. m(s2 − s1) = ∫ dQ/T

Process: Adiabatic Q = 0 and reversible => s2= s1

State 2: P2 = 100 kPa, x2 = 0.902

s2 = 1.3026 + 0.902 × 6.0568 = 6.7658 kJ/kg K

u2 = 417.36 + 0.902 × 2088.7 = 2301.4 kJ/kg

State 1 At T1 = 300°C, s1= 6.7658

⇒ P1 = 2000 kPa, u1 = 2772.6 kJ/kg

From the energy equation

1W2 = m(u1 - u2) = 1(2772.6 – 2301.4) = 471.2 kJ