When loaded, the compressor draws electrical power equal to the full-load shaft power ( \(P_{motor}\)) converted to kilowatts via \(k_{BHP}\), divided by the motor efficiency to account for electrical-to-mechanical conversion losses. This power is paid for over the fraction of annual hours that the machine actually runs loaded.

\begin{equation}\label{eq:operating-cost-loaded} C_{loaded} = \frac{P_{motor} \cdot k_{BHP} \cdot H_{annual} \cdot f_{loaded} \cdot c_{elec}}{\eta_{loaded}} \end{equation}

Symbols

\(C_{loaded}\)	Annual electricity cost in the loaded condition \([\unit{ \dollar\per\year}]\)
\(P_{motor}\)	Full-load motor brake horsepower \([\unit{ \bhp}]\)
\(k_{BHP}\)	BHP-to-kilowatt conversion factor (0.746) \([\unit{ \kilowatt\per\bhp}]\)
\(H_{annual}\)	Total compressor operating hours per year \([\unit{ \hour}]\)
\(f_{loaded}\)	Run-time loaded fraction (input ÷ 100) \([\unit{ \percent}]\)
\(c_{elec}\)	Electricity unit cost \([\unit{ \dollar\per\kilowatt\hour}]\)
\(\eta_{loaded}\)	Motor efficiency in the loaded condition (input ÷ 100) \([\unit{ \percent}]\)

Note: The run-time and efficiency inputs are entered on a 0–100 percentage scale and divided by 100 internally to obtain dimensionless fractions. Because both the loaded run-time fraction and the loaded efficiency fraction are expressed the same way (percent ÷ 100), the formula is algebraically equivalent to using dimensionless fractions throughout. The result is in dollars per year.