A piston–cylinder device contains steam that undergoes a reversible thermodynami

Question

A piston–cylinder device contains steam that undergoes
a reversible thermodynamic cycle. Initially the steam is
at 400 kPa and 350°C with a volume of 0.3 m3. The steam is
first expanded isothermally to 150 kPa, then compressed adiabatically
to the initial pressure, and finally compressed at
the constant pressure to the initial state. Determine the net
work and heat transfer for the cycle after you calculate the
work and heat interaction for each process.
7–161 Determine the work input and entropy generation

Explanation / Answer

p1=400kpa

t1=350 C u1=2884.5kj/kg v1=0.71396 s1=7.7399 kj/kg

p2=150 kpa t2=350 C u2=2888 kj/kg s2=8.19 kj/kg

p3=400kpa s3=s2=8.1983 u3=3132.9 kj/kg v3=.89148 m3/kg

The mass of the steam in the cylinder and the volume at state 3 are

m=V1/v1= .3/.71396=0.4202 kg

V3=mv3=.4202* .89148=0.3746 m3

process 1-2: isothermal expansion

entropy change=.4202(8.1983-7.7399)=.1926

heat =t*s=(350+273)*.1926=120kJ

work= 120-.4202(2888-2884.5)=118.5 kj

process 2-3:

isentropic

work=m(u3-u2)=.4202(3132.9-2888)=102.9 kj

hwat=0

process 3-1:

isobaric

work=400kpa(.3746-.3)=29.8 kj

heat=29.8-.4202(2884.5-3131.9)=134.2 kj

therefore net work=29.8+102.9-118.5=14.2 kj

net heat=120-134.2=-14.2

therefor heat lost=14.2 kj

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