An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and a quality of 0.25. An electric heater inside the tank is turned on to heat this H2O until the pressure increases to 200 kPa. Please determine the change in total entropy of water during this process. Hint: See if you can find the electrical work consumed during this process.

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rejkjavik rejkjavik · Answer 1 · 2019-11-27T12:16:06+01:00

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

[tex]v_1 = vf + x(vg - vf)[/tex]

[tex]= 0.001053 + 0.25(1.1594 - 0.001053) [/tex]

[tex]v_1 = 0.20964 m^3/kg[/tex]

[tex]s_1 = Sf + x(Sfg)[/tex]

[tex]= 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K[/tex]

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume so[tex] v_1 - v_2 = 0.29064 m^3/kg[/tex]

[tex]v_2 = v_1 = vf + x(vg - vf)[/tex]

[tex]=0.29064 = x_2(0.88578 - 0.001061)[/tex]

[tex]x_2 = 0.327[/tex]

[tex]s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K[/tex]

Change in entropy [tex]\Delta s = m(s_2 - s_1)[/tex]

=3( 3.36035 - 2.88) = 1.44 kJ/kg