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Gelneren [198K]
3 years ago
6

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.
Engineering
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

The answer is "909.3928  KJ".

Explanation:

70 \ kPa  \ \ and \ \ 100^{\circ}C \\\\s_i= 7.56162\ \frac{kJ}{kgK}\\\\u_i= 2509.39 \ \frac{kJ}{kg}\\\\

The method is isentropic since the cylinders are shielded.

Calculating the work:

w= u_2-u_i \\\\

   = 3418.7728-2509.38 \\\\=909.3928 \ KJ

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