Five kg of water is contained in a piston–cylinder assembly, initially at 5 bar and 360°C. The water is slowly heated at constant pressure to a final state. The heat transfer for the process is 2660 kJ and kinetic and potential energy effects are negligible. Determine the final volume, in m3, and the work for the process, in kJ.

Explanation:

Given that,

Mass = 5kg of water

Initial pressure P•i = 5bar = 500 kPa

Pressure is constant P = 5bar

Initial temperature T•i = 360°C

Heat transferred Q = 2660 KJ

Kinetic and potential energy are negligible

a. Final volume

The initial enthalpy and specific volume can be determine from the steam table.

h•1 = 3188.83 KJ/kg

v•1 = 0.5796 m³/kg.

So, we can calculate the final enthalpy by using

Q = m(h2—h•1)

2660 = 5 (h2—3188.83)

2660/5 = h2—3188.83

h2 — 3188.83 = 532

h2 = 520 + 3188.83

h2 = 3720.83 KJ/kg.

Using steam table, we can determine the final temperature and final specific volume

T•f = 608°C

v•f = 0.8115 m³/kg

Then, specific volume = volume /mass

v•f = v2/m

v2 = vf × m

v2 = 0.8115 × 5

v2 = 4.0575m³

b. Workdone?

W = m•P•(vf —vi)

W = 5 × 500(0.8115 — 0.5796)

W = 579.75 KJ

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