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A piston-cylinder assembly contains a two-phase liquid-vapor mixture of H20 at 220 lbf/in^2 with a quality of 75%. The mixture is heated and expands at constant pressure until a final temperature of 475°F is reached. Determine the work for the process, in Btu per lb of H2O present.

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klefenz

Answer:

222.8 kJ

Explanation:

220 psi is 15.17 bar

I use an online steam calculator to obtain the steam values.

At 15.17 bar and a quality of 75%:

v1 = 0.0021 m^3/kg

If I heat it to 475 F causing it to expand at constant pressure:

475 F = 246 C

v2 = 0.149 m^3/kg

Work is:

L = p2 * v2 - p1 * v1

p1 = p2

L = p1 * (v2 - v1)

15.17 bar = 1517 kPa

L = 1517 * (0.149 - 0.0021) = 222.8 kJ

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