Question

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.

Answer 1

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:

$w= u_2-u_i$

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