Answer:
a. W = 194.11 KJ/kg
b. Q = 1713.5 KJ/kg
c. Q = 1713.41 KJ/kg
Explanation:
a.
Since, this is a constant pressure process. Therefore, the work done will be:
W = PΔV
where,
W = work done per unit mass
P = constant pressure = 40 bar = 4000 KPa
ΔV = change in volume = Vg - Vf
At 4000 KPa, from saturated water table:
Vf = 0.001252 m³/kg
Vg = 0.049779 m³/kg
Therefore,
ΔV = (0.049779 - 0.001252) m³/kg
ΔV = 0.048527 m³/kg
Now, the work done will be:
W = (4000 KPa)(0.048527 m³/kg)
<u>W = 194.11 KJ/kg</u>
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b.
Now for heat transfer, using the relation:
Q = ΔH
where,
Q = Heat Transfer per unit mass
ΔH = change in enthalpy of water = Hg - Hf
At 4000 KPa, from saturated water table:
Hf = 1087.4 KJ/kg
Hg = 2800.8 KJ/kg
Therefore,
Q = (2800.8 - 1087.4) KJ/kg
<u>Q = 1713.5 KJ/kg</u>
<u></u>
c.
Heat transfer per unit mass with the help of internal energy can be found out by using first law of thermodynamics:
Q = ΔU + W
where,
Q = Heat Transfer per unit mass
ΔU = change in internal energy of water = Ug - Uf
W = work done per unit mass = 194.11 KJ/kg
At 4000 KPa, from saturated water table:
Uf = 1082.4 KJ/kg
Ug = 2601.7 KJ/kg
Therefore,
ΔU = (2601.7 - 1082.4) KJ/kg = 1519.3 KJ/kg
Hence,
Q = (1519.3 + 194.11) KJ/kg
<u>Q = 1713.41 KJ/kg</u>
