A piston-cylinder device contains 1.2 kg of saturated water vapor at 200°C. Heat is now transferred to steam, and steam expands reversibly and isothermally to a final pressure of 800 kPa. Determine the heat transferred and the work done during this process.

Heat transferred = 175.6kJ

Explanation:

Given

m = 1.2kg

T1 = 200°C = 473K

P2 = 800kPa

From the steam table, we need u1 and u2.

u1 = 2594.2Kj/Kg, s1 = 6.4302Kj/KgK

When P2 = 800kPa and T2 = T1

u2 = 2631.1Kj/Kg, s2 = 6.8177Kj/KgK

The heat transfer is calculated as follows;

Q = T∆S

Q = T(s2 - s1)

Q = 473k (6.8177Kj/KgK - 6.4302Kj/KgK)

Q = 219.19 kJ

Calculating the Energy Balance

∆E = E, out - E,in

∆E = m(u2 - u1)

So, we have

m(u2 - u1) = Q - E

1.2(2631.1Kj/Kg -2594.2Kj/Kg) = Q - E

44.28 = Q - E Make E the subject of formula

E = Q - 44.28

E = 219.19 - 44.28

E = 174.91kJ

Hence, the work done during the process is 174.91kJ

answer: 500 wph.

explanation:

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