Question

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.

Answer 1

Answer:

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.

$v_{specf}=0.01157[m^3/kg]$

$m=\frac{V}{v_{specf}}\\ m=\frac{1}{0.01157}\\ m=86.43[kg]$

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]