Machine Efficiency

For a machine that converts one form of energy into another, no — that would violate the first law of thermodynamics. But for a heat pump or air conditioner, the useful output is heat moved from a cold reservoir to a hot one, not energy converted. A heat pump with a COP of 3 delivers 3 kW of heat for every 1 kW of electricity consumed; the extra 2 kW comes from the outdoor air (or ground, or water), not from nowhere. The energy balance is preserved at the system level. The reason we do not call this "efficiency > 100 %" is that efficiency is defined for conversion, not for transport. The correct metric for a heat pump is the coefficient of performance, or COP, which is expected to be well above 1. Regulators sometimes express this as a "seasonal performance factor" (SPF) or "primary energy ratio" to avoid confusion with efficiency.

Efficiency is not a constant. For an electric motor, η rises from near zero at no load to a peak at around 75 % of rated load, then falls slightly as load increases further toward 100 %. The reason is that losses are split into two kinds: fixed losses (iron losses, friction, windage) that do not depend on load, and variable losses (copper losses, which go as load squared) that do. At light load, fixed losses dominate and efficiency is low. At rated load, the two are roughly balanced and efficiency is at its peak. Beyond rated load, the copper losses dominate and efficiency falls again. The same shape applies to transformers, pumps, fans and most rotating machines. This is why running a large motor at light load wastes energy, and why proper sizing matters as much as nominal efficiency class.

Efficiency (η) measures how much of the input power is converted into useful output power, regardless of whether the input is real (kW) or apparent (kVA). Power factor (PF) measures how much of the apparent power (kVA) drawn from the supply is real power (kW) that actually does work: PF = kW / kVA. A motor running at 90 % efficiency and 0.85 PF draws 100 kW of electrical input if it delivers 90 kW of shaft power, but the apparent power it presents to the supply is 100/0.85 = 117.6 kVA. The extra apparent power does not consume energy — it just circulates between the supply and the motor's magnetic field — but it does load the supply conductors and transformers. Both metrics matter for industrial plants: efficiency determines the energy bill, power factor determines whether the utility applies a penalty or requires a capacitor bank.

Because efficiency depends on the ratio of volume to surface area, and that ratio gets worse as machines get smaller. A large machine has a lot of volume — where the useful work happens — and comparatively little surface area — where losses escape. A small machine has proportionally more surface and less volume, so losses are a larger fraction of the total. This is why a 10 MW generator can reach 98.5 % efficiency while a 1 kW motor struggles to reach 88 %. The same principle applies across technologies: a large transformer is more efficient than a small one, a utility-scale gas turbine more efficient than a small one, a big cargo ship's diesel engine more efficient than a lawnmower's petrol engine. The practical consequence is that centralising generation and industrial processes generally improves efficiency, while distributed small-scale generation and appliances are typically less efficient per unit of output.

Start by understanding where the losses are. For a motor, the losses split roughly into copper (resistive) losses, iron (magnetic) losses, friction and windage, and stray load losses. Copper losses fall when the motor runs at lower current; iron losses are roughly constant; friction and windage rise with speed. The three highest-leverage actions are usually: (1) run the machine closer to its design point — a pump running at 25 % flow should be trimmed or speed-controlled rather than throttled; (2) reduce oversizing — replace a 100 kW motor running at 20 kW with a properly sized 25 kW unit; (3) upgrade to a higher efficiency class — IE4 motors are typically 2–4 percentage points more efficient than IE2 at the same size. Beyond these, maintenance matters enormously: a pump with a worn impeller or a motor with clogged cooling passages loses efficiency rapidly, so regular inspection and cleaning often delivers more savings than a capital replacement.

The lifetime cost of a machine is dominated by the energy it consumes, not by its purchase price. For an industrial motor running continuously, the purchase price is typically 1–3 % of the twenty-year total cost of ownership, while energy is 95 % or more. A 1 % improvement in efficiency on a 100 kW motor running 8,000 hours per year at €0.15/kWh saves 8,000 × 0.15 × 0.01 × 100 = €1,200 per year, or €24,000 over a 20-year life — more than the cost of the motor itself. This is why energy efficiency regulations focus on lifetime cost rather than purchase price, and why premium-efficiency equipment often pays for itself in a few years even with a significant price premium. The same logic applies to compressors, pumps, fans, boilers and transformers, and it is the economic basis of every industrial energy audit.

It depends on the machine type and size. For an industrial motor above 50 kW, target at least 95 % (IE3) and preferably 96 % (IE4). For a small motor below 1 kW, 85 % is realistic and 90 % is excellent. For a transformer, 98 % is the minimum and 99 % is achievable above 100 kVA. For a pump, target within 5 percentage points of the best-efficiency point (BEP) at the intended operating flow. For a combustion engine, the peak efficiency is what matters but the actual operating efficiency depends on the duty cycle. In every case, be realistic about the losses you cannot eliminate — the manufacturer's data sheet is a starting point, but the actual installed efficiency depends on the load profile, the power quality, the temperature and the maintenance. Design for the operating point, not for the nameplate.