FE Electrical and Computer Power Systems Practice Problems

The current FE Electrical and Computer specification assigns 8-12 of the 110 exam questions to Power Systems. Form A contains 8 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A run-of-river hydroelectric plant passes $850 ft^{3} / s$ through its turbine at a net head of $62 ft$ . The combined turbine-generator efficiency is $0.85$ , and the specific weight of water is $62.4 lbf/ft^{3}$ . The electrical power delivered to the station bus is most nearly:

(A)3,790 kW
(B)4,460 kW
(C)5,080 kW
(D)5,980 kW

Show worked solution

Answer: (A)

The 3.29 million ft-lbf/s of hydraulic power, reduced by the combined turbine-generator efficiency and then converted to watts, puts about 3.8 MW on the bus.

Hydraulic power available at the runner is the weight flow rate carried through the net head.

P_{h y d} = γ Q H = (62.4) (850) (62) = 3.288 \times 1 0^{6} ft \cdot lbf/s

Converting at $550 ft \cdot lbf/s$ per horsepower and charging the single combined efficiency against the shaft leaves the station output.

P_{h y d} = \frac{3.288 \times 1 0 ^{6}}{550} = 5979 hp

P_{o u t} = (0.85) (5979) (0.746) = 3791 kW

FE Reference Handbook — Fluid Mechanics: Pump Power Equation (hydraulic power, turbine form)

Why the other choices appear

(B)Omitting the 0.85 efficiency and converting the gross 5979 hp gives (5979)(0.746) = 4460 kW.
(C)Reporting the net shaft value 5082 hp as if the number were already kilowatts gives 5080.
(D)Reporting the gross hydraulic value 5979 hp as if the number were already kilowatts gives 5980.