Question

water at 400 kPa and 60°C while the other part is evacuated. The partition is now removed, and the water expands to fill the entire tank. Determine the entropy change of water during this process, if the final pressure in the tank is 40 kPa

Answer 1

Answer:

0.4912 kJ/K

Explanation:

Using steam table, the properties of water at stage 1 can be determined:

$P_{1}$ = 400 kPa

$T_{1}$ = 60°C

Therefore, from steam table:

specific volume $v_{1} = 0.001017 m^{3}/kg$

entropy  $s_{1} = 0.8313 kJ/kg*K$

Calculating the specific volume of water after the partition is removed, since the mass is constant and volume of water is doubled:

$v_{2} = 2*0.001017 = 0.002034 m^{3}/kg$

Final pressure of water in the tank = $P_2} = 40 kPa$

dryness factor after removing the partition

$x = \frac{0.002034 - 0.001026}{3.993 - 0.001026} = 2.524*10^{-4} kg/vapor$

Therefore:

$s_{2} = s_{f2} + x*s_{fg2} = 1.0261 + (2.524*10^{-4}*6.643) = 1.02778 kJ/kg*K$

Thus, the entropy change is:

Δs = $m*(s_{2}-s_{1}) = 2.5*(1.02778 - 0.8313) = 0.4912 kJ/K$

Therefore, the entropy change of water is 0.4912 kJ/K