Compressed-Air Leak Load and Annual Cost (Load/Unload Test)

Turns a stopwatch reading into a repair budget.

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Checked against published worked examples from: US DOE Industrial Technologies Program.

Details, formula, and sources

ScopeWith production off, the fraction of a cycle the compressor runs loaded (t_load / (t_load + t_unload), the DOE Compressed Air Challenge method) times the compressor's actual delivered capacity (not the nameplate) is the leakage flow, converted to kilowatts at the system specific power (about 16-22 kW per 100 cfm by compressor type) and to dollars at the run hours and rate. A neglected system loses 20-30 percent of its air to leaks, a well-run one under 10, and the difference is thousands a year on the plant's most expensive utility. Needs no flow meter. An estimate from a stopwatch test, not a metered audit.

Formula or tableleak_fraction = load_min / (load_min + unload_min); leak_cfm = compressor_cfm x leak_fraction; leak_kw = leak_cfm x specific_power / 100; annual_kwh = leak_kw x run_hours; annual_cost = annual_kwh x rate_kwh.

Source and editionThe US DOE Compressed Air Challenge load/unload leak test (leak fraction = t_load / (t_load + t_unload), leak flow = fraction x compressor cfm) and the compressed-air specific-power convention (about 16-22 kW per 100 cfm at 100 psig (Compressed Air Challenge Sourcebook: 16-19 single-stage lubricant-injected screw, 15-17 two-stage, 17-22 lubricant-free; DOE Tip Sheet 3 uses 18)), by name; the relations are public.

Free accessThe DOE Compressed Air Challenge load/unload method and the specific-power convention are public; the DOE compressed-air tip sheets and sourcebook are free at energy.gov.

What governsEstimate. AHJ and licensed professional govern.

API field namescompressor_cfm, load_min, unload_min, specific_power, run_hours, rate_kwh, leak_fraction, leak_cfm, leak_kw, annual_kwh, annual_cost

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