Sol. A force balance on the piston allows us to calculate the initial pressure. Including the atmospheric pressure, which is assumed to be 100 kPa, we have

P 1 A = W + P atm A

P 1 = (16) (9.81) + (100000)

∴ P 1 = 120000 Pa or 120 kPa

To find the mass, we need the specific volume.

v 1 = v f x (v g – v f ) = 0.001 + (0.2)(1.428 – 1.001) = 0.286 m 3 /kg

The mass is then

M = V 1 /v 1 = = 0.001373 kg

When the piston just hits the stops, the pressure is still 120 kPa. The specific volume increases to

v 2 = V 2 /m = = 0.458 m 3 /kg

The quality is then found as follows, using the entries at 120ºC:

0.458 = 0.001 + x 2 (1.428 – 0.001)

∴ x 2 = 0.320 or 32.0%

After the piston hits the stops, the specific volume ceases to change since the volume remains constant. Using T 3 = 400ºC and v 3 = 0.458,

P 3 = (0.8 – 0.6) + 0.6 = 0.686 Mpa

(e) There is zero work done on the piston after it hits the stops. From the initial state until the piston hits the stops, the pressure is constant at 120 kPa; the work is then

W = P (v 2 – v 1 ) m = (120) (0.458 – 0.286) (0.001373) = 0.0283 kJ

or 28.3 J