FE Mechanical Thermodynamics Practice Problems

The current FE Mechanical specification assigns 10-15 of the 110 exam questions to Thermodynamics. Form A contains 11 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A rigid vessel of volume $0.30$ m $^{3}$ contains 2.0 kg of water at 1.0 MPa. At 1.0 MPa the saturation properties are $v_{f} = 0.001127$ m $^{3}$ /kg, $v_{g} = 0.19444$ m $^{3}$ /kg, $u_{f} = 761.7$ kJ/kg, and $u_{f g} = 1, 822.0$ kJ/kg. The total internal energy of the contents is most nearly:

(A)1,520 kJ
(B)2,810 kJ
(C)4,330 kJ
(D)5,170 kJ

Show worked solution

Answer: (C)

The specific volume of $0.150$ m $^{3}$ /kg places the water at 77% quality, which carries about 4,330 kJ of internal energy.

Dividing the vessel volume by the charge mass fixes the specific volume, and its position between $v_{f}$ and $v_{g}$ confirms a two-phase state.

v = \frac{0.30}{2.0} = 0.150 m^{3} / kg

Quality measures how far that specific volume has advanced across the saturation span.

x = \frac{0.150 - 0.001127}{0.19444 - 0.001127} = 0.770

Internal energy follows the same lever rule between the liquid and vapor values, and the vessel holds 2.0 kg.

u = 761.7 + 0.770 (1822.0) = 2165 kJ/kg

U = 2.0 (2165) = 4330 kJ

FE Reference Handbook — Thermodynamics: Properties for Two-Phase (Saturated) Systems, Quality

Why the other choices appear

(A)Treats the contents as saturated liquid, $U = 2.0 (761.7) = 1523$ kJ.
(B)Drops $u_{f}$ and keeps only the evaporation term, $U = 2.0 (0.770) (1822.0) = 2806$ kJ.
(D)Treats the contents as saturated vapor, $U = 2.0 (761.7 + 1822.0) = 5167$ kJ.