Answer:
0.4912 kJ/K
Explanation:
Using steam table, the properties of water at stage 1 can be determined:
[tex]P_{1}[/tex] = 400 kPa
[tex]T_{1}[/tex] = 60°C
Therefore, from steam table:
specific volume [tex]v_{1} = 0.001017 m^{3}/kg[/tex]
entropy [tex]s_{1} = 0.8313 kJ/kg*K[/tex]
Calculating the specific volume of water after the partition is removed, since the mass is constant and volume of water is doubled:
[tex]v_{2} = 2*0.001017 = 0.002034 m^{3}/kg[/tex]
Final pressure of water in the tank = [tex]P_2} = 40 kPa[/tex]
dryness factor after removing the partition
[tex]x = \frac{0.002034 - 0.001026}{3.993 - 0.001026} = 2.524*10^{-4} kg/vapor[/tex]
Therefore:
[tex]s_{2} = s_{f2} + x*s_{fg2} = 1.0261 + (2.524*10^{-4}*6.643) = 1.02778 kJ/kg*K[/tex]
Thus, the entropy change is:
Δs = [tex]m*(s_{2}-s_{1}) = 2.5*(1.02778 - 0.8313) = 0.4912 kJ/K[/tex]
Therefore, the entropy change of water is 0.4912 kJ/K
