FE Other Disciplines Thermodynamics and Heat Transfer Practice Problems

The current FE Other Disciplines specification assigns 9-14 of the 110 exam questions to Thermodynamics and Heat Transfer. Form A contains 9 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A binary-cycle geothermal plant produces 450 kW of net power while receiving 3.60 MW from a brine stream at 165 $^{\circ}$ C and rejecting heat to a condenser loop at 30 $^{\circ}$ C. The ratio of the plant thermal efficiency to the maximum efficiency attainable between these two temperatures is most nearly:

(A)0.13
(B)0.15
(C)0.31
(D)0.41

Show worked solution

Answer: (D)

An actual efficiency of 12.5 percent sits at about 41 percent of the 30.8 percent reversible ceiling set by the 438 K and 303 K reservoirs.

The reversible limit between the same two reservoirs fixes the benchmark, and both temperatures enter on the absolute scale.

η_{C a r n o t} = 1 - \frac{T _{L}}{T _{H}} = 1 - \frac{303}{438} = 0.308

Plant performance follows from the metered energy streams once the net power and the brine heat input carry the same units.

η_{t h} = \frac{W _{n e t}}{Q _{H}} = \frac{450}{3600} = 0.125

Their quotient is the second-law efficiency, which the Kelvin-Planck statement holds below unity for any real engine and which lands near two-fifths for low-temperature organic Rankine hardware.

η_{I I} = \frac{0.125}{0.308} = 0.41

FE Reference Handbook — Thermodynamics: Carnot Cycle Efficiency and Power Cycle Thermal Efficiency

Why the other choices appear

(A)Reports the plant thermal efficiency itself, 450/3600 = 0.125.
(B)Builds the ceiling from Celsius readings, 1 - 30/165 = 0.818, giving 0.125/0.818 = 0.153.
(C)Stops at the Carnot ceiling, 1 - 303/438 = 0.308, and offers it as the requested ratio.