Water at 70 kPa and 1008C is compressed isentropically in a closed system to 4 MPa. Determine the final temperature of the water and the work required, in kJ/kg, for this compression.

The final temperature of the water is 676.164 °C and the specific work required is 1171.384 kilojoules per kilogram.

Let suppose that compression occurs quasi-statically, work is done on the closed system and enthalpy is increased. By First Law of Thermodynamics, we model compression process as following:

(1)

Where:

By definition of enthalpy, in kilojoules per kilogram, and by dividing the resulting expression by the mass of the entire system, we have the following expression:

(2)

Where:

From steam tables we find that initial and final states of the water are represented by the following data:

Initial state

, , , (Superheated steam)

Final state

, , , (Superheated steam)

By (1) we have that the specific work required is:

The final temperature of the water is 676.164 °C and the specific work required is 1171.384 kilojoules per kilogram.

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The answer is "909.3928  KJ".

Explanation:

The method is isentropic since the cylinders are shielded.

Calculating the work:

answer:

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explanation:

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