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newSchoolWorks.com › VoTechWorks › Manufacturing & Marine › Robotics / Mechatronics › Motor Drives: AC/DC Motors, Variable-Frequency Drives, and Motion Control › Lesson 11
MFG-MECH15-05 · Lesson 11 · 10 clock hours · theory

Energy Efficiency, Power Factor, and Motor Drive Economics

Objective. Students will calculate true power, reactive power, apparent power, and power factor for a motor load, explain how a VFD improves system efficiency at partial load, and estimate annual energy savings when replacing a throttled valve or damper with variable-speed drive control.

⚠ Pending professional review

This lesson, its worked examples and its practice questions has not yet been verified by a licensed engineer or certified journey-level tradesperson or by a certification or approval body. It is a study draft and may contain errors. Do not use it as instructions for work on a job site or on live equipment. Always follow your instructor, your supervising professional, and current official references and regulations.

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Energy Efficiency, Power Factor, and Motor Drive Economics

Industrial motors account for a large share of electrical energy consumed in manufacturing and marine facilities. Understanding how to measure that consumption accurately—and how variable-frequency drives (VFDs) reduce it—is a core skill for every mechatronics technician.

The Power Triangle: kW, kVAR, and kVA

In an AC circuit with reactive loads such as motors and transformers, electrical power has three related quantities.

True power (P) is the power that does real work—turning a shaft, pumping fluid, moving air. It is measured in kilowatts (kW).

Reactive power (Q) is the power that flows back and forth between the source and the magnetic fields of inductive loads. It does no useful work but must be supplied by the utility. It is measured in kilovolt-amperes reactive (kVAR).

Apparent power (S) is the total power the utility must deliver. It is the vector sum of true and reactive power, measured in kilovolt-amperes (kVA).

These three quantities form a right triangle:

where θ is the angle between voltage and current.

Power factor (PF) is the ratio of true power to apparent power:

> PF = P / S = kW / kVA

A power factor of 1.0 (or 100%) means all current delivered does useful work. A power factor of 0.80 means 20% of the apparent current produces no useful output. Utilities often charge demand penalties when power factor falls below a threshold (commonly 0.90), so improving power factor saves money directly.

Displacement Power Factor vs. True Power Factor in VFD-Fed Motors

Displacement power factor describes the phase angle between the fundamental (60 Hz) voltage and current waveforms. A lightly loaded induction motor has a poor displacement power factor because its magnetizing current is relatively large compared with its load current.

True power factor (also called total power factor) accounts for both displacement and harmonic distortion. A VFD draws current from the utility in pulses, creating harmonic currents (primarily 5th and 7th harmonics). These harmonics add to the reactive burden seen by the utility even when the displacement power factor looks acceptable. For this reason, true power factor = distortion factor × displacement power factor, and true power factor is always ≤ displacement power factor when harmonics are present.

Line-side harmonic filters or active front-end VFDs can mitigate harmonic content.

Power-Factor Correction Capacitors — A Critical Safety Point

Power-factor correction capacitors are sometimes installed at motor terminals to supply reactive power locally, reducing reactive current drawn from the utility. This is acceptable practice for motors run across the line.

However, capacitors must never be connected to the output terminals of a VFD. The VFD output is a pulse-width-modulated (PWM) waveform, not a sinusoidal voltage. Capacitors on the output create resonance, overcurrent trips, and can destroy both the capacitor bank and the VFD output transistors. Capacitors for power-factor correction belong on the line (input) side of the VFD, or at the service entrance, never between the VFD and the motor.

The Affinity Laws and Why Speed Reduction Saves Energy

For centrifugal pumps and fans, the affinity laws describe how flow, pressure (head), and power change with shaft speed:

| Quantity | Relationship |

|---|---|

| Flow (Q) | Q₂/Q₁ = N₂/N₁ |

| Head (H) | H₂/H₁ = (N₂/N₁)² |

| Power (P) | P₂/P₁ = (N₂/N₁)³ |

The cubic relationship for power is the key insight. Reducing speed to 75% of full speed reduces power demand to (0.75)³ = 0.422, or about 42% of full-load power. That means a 58% reduction in power consumption at 75% speed—a dramatic saving compared with throttling a valve or damper, which wastes energy as pressure drop while the motor still runs at full speed.

Traditional flow control with a throttle valve or discharge damper keeps the motor at full speed and full (or near-full) power while restricting flow mechanically. A VFD matches motor speed to the actual flow requirement, so only the energy needed is consumed.

Annual Energy Savings Calculation

To estimate annual savings, compare full-speed power (with throttling) to reduced-speed VFD power, then multiply by operating hours and the utility rate:

> Annual savings (kWh) = (P_full − P_VFD) × hours_per_year

> Annual savings ($) = annual savings (kWh) × $/kWh

Marine Application: HVAC Fans Running 24/7

On a ship, HVAC fans often run continuously to maintain air quality and thermal comfort for crew and cargo. At full speed around the clock the energy cost is significant. Installing a VFD allows fan speed to be modulated with demand—lower at night, higher in tropical ports. Even a modest average speed reduction to 80% yields (0.80)³ = 0.512, saving roughly 49% of fan energy. Over a year at sea this translates to measurable fuel savings, reduced engine hours, lower maintenance costs, and reduced emissions—a compelling business case for retrofit.

Safety Reminders

Before working on any motor or drive:

Key terms9

True power (P)The portion of electrical power that performs useful work, measured in watts (W) or kilowatts (kW).
Reactive power (Q)Power that oscillates between the source and inductive or capacitive loads without doing useful work; measured in kilovolt-amperes reactive (kVAR).
Apparent power (S)The total power a utility must supply, equal to the vector sum of true and reactive power; measured in kilovolt-amperes (kVA).
Power factor (PF)The ratio of true power to apparent power (PF = kW / kVA); a dimensionless number between 0 and 1 indicating how efficiently current is converted to useful work.
Displacement power factorThe power factor calculated from the phase angle between the fundamental voltage and current waveforms only, not accounting for harmonic distortion.
True power factorPower factor that accounts for both phase displacement and harmonic distortion; always less than or equal to displacement power factor when harmonics are present.
Affinity lawsA set of relationships for centrifugal pumps and fans stating that flow varies directly with speed, head varies with speed squared, and power varies with speed cubed.
DC bus capacitorsCapacitors inside a VFD that store energy to maintain a smooth DC voltage; they retain a dangerous charge after AC power is removed and must be allowed to discharge before internal servicing.
Power-factor correction capacitorsCapacitors installed on the line side of equipment to supply reactive power locally, reducing reactive current drawn from the utility; must never be installed on the output terminals of a VFD.

Worked examples4

1. A motor draws 45 kVA from the line and delivers 36 kW of true power. Find the reactive power and power factor.
  1. Identify known values: S = 45 kVA, P = 36 kW.
  2. Calculate power factor: PF = P / S = 36 / 45 = 0.80.
  3. Use the power triangle to find reactive power: Q = √(S² − P²) = √(45² − 36²) = √(2025 − 1296) = √729 = 27 kVAR.
Answer: Power factor = 0.80 (80%); Reactive power Q = 27 kVAR.
2. A pump motor runs at full speed (1800 RPM) drawing 30 kW while a throttle valve controls flow. The same pump is switched to VFD control and runs at 75% speed (1350 RPM) to deliver the same flow demand. Calculate (a) power at reduced speed using the affinity laws and (b) annual energy savings in kWh, assuming 8,000 operating hours per year.
  1. Apply the affinity law for power: P₂/P₁ = (N₂/N₁)³ = (0.75)³ = 0.4219.
  2. Calculate VFD power: P₂ = 30 kW × 0.4219 = 12.66 kW.
  3. Calculate power saved per hour: ΔP = 30 kW − 12.66 kW = 17.34 kW.
  4. Calculate annual energy savings: Savings = 17.34 kW × 8,000 h = 138,720 kWh per year.
Answer: VFD power at 75% speed ≈ 12.66 kW; annual energy savings ≈ 138,720 kWh.
3. Using the result from the previous example, calculate the annual dollar savings if the utility charges $0.12 per kWh.
  1. Annual energy savings = 138,720 kWh (from previous example).
  2. Annual dollar savings = 138,720 kWh × $0.12/kWh = $16,646.40.
Answer: Annual dollar savings ≈ $16,646 per year.
4. A ship's HVAC fan motor runs 24 hours a day, 365 days a year at full speed drawing 15 kW. After VFD installation the average operating speed is 80% of full speed. Calculate (a) average power with VFD, (b) annual energy savings in kWh, and (c) annual dollar savings at $0.15/kWh.
  1. Apply affinity law for power: P₂ = P₁ × (0.80)³ = 15 kW × 0.512 = 7.68 kW.
  2. Power saved: ΔP = 15 kW − 7.68 kW = 7.32 kW.
  3. Annual operating hours: 24 h/day × 365 days = 8,760 h/year.
  4. Annual energy savings: 7.32 kW × 8,760 h = 64,123 kWh/year.
  5. Annual dollar savings: 64,123 kWh × $0.15/kWh = $9,618.45.
Answer: VFD average power ≈ 7.68 kW; annual savings ≈ 64,123 kWh; annual dollar savings ≈ $9,618.

The math in this lesson3

Learners convert kW saved to annual kWh and dollar savings at $0.12 per kWh.
7.RP.3F-IF.2G-SRT.8
Brush up in the math halls at newSchoolWorks.com:
In class: Lecture: define kW, kVAR, kVA, and power factor; derive the power triangle; explain displacement power factor vs. true power factor in VFD-fed motors; discuss power-factor correction capacitors and why they must not be placed on VFD output terminals. · Calculation exercise: given a pump motor running at full speed with a throttle valve vs. the same motor on a VFD at 75% speed, students use the affinity laws (flow ∝ speed, power ∝ speed³) to calculate the percentage power savings and annual kWh savings at a given cents-per-kWh rate. · Group discussion: marine application case — a ship's HVAC fan running 24/7; students calculate potential fuel savings by installing a VFD and present a one-paragraph business case justifying the investment.
Materials: Power-triangle diagram posters, scientific calculators, affinity-laws reference sheet, sample utility bill with kWh rate, marine HVAC case-study handout, whiteboard
Check: Power-triangle and affinity-laws calculation worksheet graded. Pass = 80% with power factor calculated correctly and affinity-law power reduction calculation correct.

Practice check8

Answer each question, then check it. Nothing is recorded; this is your own practice.

Industrial Electricity
1. A motor circuit has an apparent power of 50 kVA and a true power of 40 kW. What is the power factor?
Industrial Electricity
2. A motor draws 60 kW of true power and operates at a power factor of 0.75. What is the apparent power the utility must supply?
Motor and Motor Controls
3. A centrifugal fan runs at full speed drawing 20 kW. Using the affinity laws, what is the approximate power drawn when speed is reduced to 60% of full speed?
Motor and Motor Controls
4. Why must power-factor correction capacitors never be installed on the output terminals of a VFD?
Motor and Motor Controls
5. What is the primary reason a throttle valve wastes energy compared with a VFD for flow control on a centrifugal pump?
Industrial Electricity
6. Before opening a VFD enclosure for internal servicing after removing AC power, a technician must:
Industrial Electricity
7. True power factor differs from displacement power factor in VFD-fed motor circuits because:
Motor and Motor Controls
8. A pump motor draws 25 kW at full speed. The VFD reduces speed to 80% of full speed. Using the affinity laws, what are the approximate annual energy savings in kWh over 8,760 hours of operation?
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