Question

A piston-cylinder contains 25 g of saturated water vapor and kept at 300 kPa. We add to the saturated water vapor an amount of electrical work of 7.2 kJ (via a resistance heater). Simultaneously, we lose 3.7 kJ from the water.

1.Show the path on a PV diagram.

2.Determine the change in internal energy, the PV work, and the enthalpy of the system.

3.Determine the final temperature of the system

Answer #1

By 1st law of thermodynamics for a closed system,

U = Qnet + W.......heat and work provided to the system are taken as positive.

U = Qin - Qout + W

U = 7.2 - 3.7 + W

U = 3.5 kJ -PdV

dH = dU + d(PV)

dH = dU + PdV + VdP

dH = 3.5 - PdV + PdV + VdP

dH = 3.5 + VdP

As the process occurs at constant process, dP = 0

.: dH = 3.5 kJ

or H = 3.5 kJ

.: m(H2 - H1) = 3.5 kJ

For saturated steam at 300 kPa,

specific entalpy of satuarted vapor Hv = 2724.9 kJ/kg

.: H1 = Hv = 2724.9 kJ/kg

.: 0.025 kg (H2 - 2724.9) kJ/kg = 3.5 kJ

.: H2 = 2864.9 kJ/kg

As H2 > Hv , steam is superheated.

Now, search for T where P = 300 kPa and H = 2864.9 kJ/kg in the superheated steam table.

we get, **T = 199.5 C**

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