Energy efficiency is not about turning off lights. For industrial and commercial facilities, lighting is typically 10% to 15% of the electric bill. The real money is in motors (60% to 70% of industrial electricity), compressed air systems (20% to 30% of motor energy in many plants), and demand charges that penalize peak usage regardless of total consumption. A facility spending $20,000 per month on electricity can often reduce that by $3,000 to $5,000 per month through measures that cost nothing (fixing compressed air leaks, adjusting motor schedules) or that pay for themselves within 2 years (VFDs on variable-load fans and pumps, power factor correction).
The first step in any energy efficiency effort is understanding your electric bill. Most commercial and industrial customers pay two separate charges: a usage charge ($/kWh for the total energy consumed) and a demand charge ($/kW for the peak power draw during the billing period). The demand charge can be 30% to 50% of the total bill, and it is set by the single highest 15-minute or 30-minute average demand in the entire billing period. One bad peak — starting two large motors simultaneously, running the air compressor and the chiller at the same time — sets the demand charge for the entire month. This guide covers how to read your bill, identify the biggest opportunities, and calculate the payback on efficiency investments.
Demand Charges vs Usage Charges
Your electric bill has two main components. The usage charge (also called energy charge or consumption charge) is based on the total kilowatt-hours (kWh) consumed during the billing period. It represents the actual energy used. Typical commercial rates range from $0.06 to $0.15 per kWh depending on location, rate class, and time of use. The usage charge is like paying for the total gallons of water you use.
The demand charge is based on the peak kilowatt (kW) demand during the billing period, typically measured as the highest 15-minute average demand. Typical demand charges range from $8 to $25 per kW per month. If your peak demand is 500 kW, you pay $4,000 to $12,500 per month in demand charges alone, regardless of how much total energy you consumed. The demand charge is like paying for the size of the pipe, not the water that flows through it.
Demand charges exist because the utility must build and maintain generation, transmission, and distribution infrastructure sized for the peak load, not the average load. A facility that draws 500 kW for one hour and 100 kW for the other 719 hours in a month forces the utility to maintain 500 kW of capacity for that customer. The demand charge recovers that capacity cost. Reducing peak demand by even 50 kW — through load staggering, soft starters, or demand controllers — saves $400 to $1,250 per month every month.
Load factor (average demand ÷ peak demand) indicates how well you use your electrical capacity. A load factor of 0.80 means you are using 80% of your peak capacity on average. Industrial facilities typically run 0.50 to 0.80. Below 0.50 suggests significant demand reduction opportunity. Above 0.80 is excellent. Many utilities offer lower rates or rebates for high load factor customers. Track your load factor monthly; a sudden drop means something changed in your operation that is costing you demand charges.
Load Factor = Total kWh ÷ (Peak kW × Hours in billing period)
Example: 150,000 kWh used, 400 kW peak, 720 hours (30 days)
LF = 150,000 ÷ (400 × 720) = 0.52 (52%)
Low load factor = high demand charges relative to usage. Target: ≥0.65
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Compressed Air: The Most Expensive Utility
Compressed air is the most expensive utility in most industrial facilities. Only 10% to 15% of the electrical energy input to an air compressor is converted to useful work at the point of use. The rest is lost as heat. A 100 HP compressor running 6,000 hours per year at $0.10/kWh costs about $45,000 per year in electricity. If 30% of the compressed air is lost to leaks (a typical finding in plants that have not performed a leak audit), that is $13,500 per year blowing through holes in fittings, hoses, and connections.
Leak detection and repair is the single highest-ROI energy efficiency measure in most plants. An ultrasonic leak detector ($300 to $3,000 for the instrument) can identify leaks by their high-frequency sound signature, even in noisy environments. A typical leak audit in a 50,000-square-foot plant finds 30 to 100 leaks. Repairing them (tightening fittings, replacing hoses, installing couplings) costs $500 to $2,000 in labor and materials and saves $5,000 to $15,000 per year in electricity. The payback is usually measured in weeks, not years.
System pressure reduction is the second opportunity. Every 2 psi reduction in system pressure reduces compressor energy consumption by approximately 1%. Many plants run their compressors at 110 to 120 psi because one machine 500 feet from the compressor room needs 90 psi and the pressure drop in the distribution piping eats up the difference. Fixing the piping bottleneck (upsizing header pipe, adding a receiver at the point of use, or eliminating unnecessary restrictions) allows the compressor pressure to drop to 95 to 100 psi, saving 5% to 10% of compressor energy.
Misuse of compressed air is the third opportunity. Compressed air for blowoff, cooling, and drying is common in manufacturing but wildly inefficient. A single open 1/4-inch blow-off nozzle at 100 psi uses about 100 SCFM of compressed air, equivalent to 25 HP of compressor capacity running continuously. Engineered air nozzles (Venturi nozzles) reduce air consumption by 50% to 80% while maintaining the same cleaning force. Replacing open blow-offs with engineered nozzles across a plant can eliminate 20% to 40% of compressed air demand.
Motor Efficiency and VFD Savings
Motors consume 60% to 70% of industrial electricity. The efficiency of a motor is the ratio of mechanical output power to electrical input power. A standard-efficiency 50 HP motor might be 91% efficient, meaning it draws 55 HP (41 kW) of electrical power to produce 50 HP (37.3 kW) of mechanical power. A premium-efficiency motor of the same size might be 94% efficient, drawing only 53.2 HP (39.7 kW). The 1.3 kW difference, running 6,000 hours per year at $0.10/kWh, saves $780 per year. The premium motor costs perhaps $500 more than the standard motor. Payback: 8 months.
The bigger opportunity is in variable-load applications. Many motors in HVAC, pumping, and fan applications run at full speed and use throttling valves or dampers to control output. A pump that needs to deliver 70% of its maximum flow runs at full speed with a throttle valve partially closed, wasting 30% or more of the pump energy as pressure drop across the valve. A Variable Frequency Drive (VFD) slows the motor to match the actual demand, reducing energy consumption by 50% or more at partial load. The affinity laws for pumps and fans show that power varies with the cube of speed: at 70% speed, the power draw is only 34% of full-speed power.
VFDs are most effective on centrifugal loads (fans, pumps, blowers) where the torque requirement decreases with speed. They are less effective on constant-torque loads (conveyors, positive-displacement pumps, extruders) where the torque remains the same regardless of speed. For centrifugal loads with variable demand, a VFD typically saves 30% to 50% of annual energy cost with a payback of 1 to 3 years, often with utility rebates covering 30% to 50% of the VFD cost.
Motor sizing is another efficiency factor. An oversized motor runs at less than full load, where efficiency drops. A 50 HP motor running at 25% load (12.5 HP) might be only 85% efficient instead of 94%. It also has a lower power factor at part load, which can incur power factor penalties. Right-sizing motors during replacement (installing a 25 HP motor where a 50 HP motor was running at half load) improves both efficiency and power factor at essentially zero incremental cost.
Power Factor Correction
Power factor is the ratio of real power (kW, the power that does useful work) to apparent power (kVA, the total power the utility must deliver). Motors, transformers, and inductive loads draw reactive power (kVAR) that oscillates between the load and the source without doing useful work. A power factor of 0.85 means that for every 100 kVA of apparent power delivered, only 85 kW is doing useful work. The other 15 kVAR is reactive power that loads the utility's distribution system without producing revenue.
Most utilities penalize power factor below 0.85 or 0.90 through one of three mechanisms: a direct power factor penalty surcharge (typically 1% increase in bill for each 0.01 below the target), billing demand based on kVA instead of kW (which inflates the demand charge when power factor is low), or a reactive power charge ($/kVAR for the reactive component). A facility with 500 kW demand and 0.80 power factor has an apparent power of 625 kVA. If billed on kVA at $15/kVA, the demand charge is $9,375 vs $7,500 if the power factor were 1.00. That is $1,875 per month in avoidable cost.
Power factor correction is achieved by adding capacitors (or capacitor banks) that supply reactive power locally, reducing the reactive power drawn from the utility. The capacitors can be installed at the service entrance (correcting the whole facility), at individual motor control centers (correcting groups of motors), or at individual motors (correcting specific loads). Service entrance correction is cheapest but does not reduce losses in the facility's internal wiring. Motor-level correction provides the best overall benefit but costs more to install.
Sizing the capacitor bank: kVAR needed = kW × (tan(θ1) − tan(θ2)), where θ1 is the existing power factor angle and θ2 is the target power factor angle. For 500 kW at 0.80 PF corrected to 0.95 PF: kVAR = 500 × (0.75 − 0.33) = 210 kVAR. A 225 kVAR automatic capacitor bank costs approximately $8,000 to $15,000 installed and saves $1,500 to $2,000 per month in power factor penalties. Payback: 4 to 10 months. This is one of the fastest-payback investments in energy efficiency.
kVAR needed = kW × (tan θ₁ − tan θ₂)
Where θ = arccos(PF)
PF 0.80: tan θ = 0.75
PF 0.85: tan θ = 0.62
PF 0.90: tan θ = 0.48
PF 0.95: tan θ = 0.33
PF 1.00: tan θ = 0.00