1 m3 of saturated liquid water at 200°C is expanded isothermally in a closed system until its quality is 80 percent. Determine the total work produced by this expansion, in kJ.

w = 134121.2 [kJ]

Explanation:

This is a classic problem of thermodynamics, with the initial saturation conditions we can find the initial enthalpy.

We have to remember that working in a closed system that expands is equal to:

W = P * DV

where:

P = pressure of the system [Pa]

DV = volume differential (Final - initial)[m^3]

W = work done by the volume change [J]

And the enthalpy in each state is given by the following expression.

H = U + P*V

Where:

H = enthalpy [J]

U = internal energy [J]

P = pressure [Pa]

V = volume [m^3]

The work performed per unit of mass is equal to:

w = m*(h2 - h1)

Where:

m = mass of the system [kg}

h2 = enthalpy at pooint with 80% of quality [kJ]

h1 = enthalpy at saturated liquid water [kJ]

The mass can be solved using the initial data of water saturation.

In the attached image we can see how was selected the specific volume for the saturated liquid water.

And the enthalpy at the initial saturation point can be found using the tables, (see the second attached image).

h1 = 852.26[kJ/kg]

And now using the definition for the quality of a pure substance in the saturation condition we can find the enthalpy at the second point.

h2 = hf + x*(hg - hf)

where:

hf = enthalpy at saturated liquid [kJ/kg]

hg = enthalpy at saturated gas [kJ/kg]

x = quality

h2 = enthalpy at the second point [kJ/kg]

using the tables we can find the values for hf and hg

hg = 2792 [kJ/kg]

hf = 852.26 [kJ/kg]

h2 = 852.26 + 0.8*(2792 - 852.26)

h2 = 2404.05[kJ/kg]

Now replacing in the equation we have:

w = 86.43 * (2404.05 - 852.26)

w = 134121.2 [kJ]

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