12.2 Motors, Drives & Electrical Fundamentals

Key Takeaways

  • Ohm's law states that voltage equals current times resistance, and power in a DC or single-phase resistive circuit equals volts times amps.
  • Three-phase power is calculated as volts times amps times power factor times the square root of three, or about 1.732.
  • Single phasing occurs when one leg of a three-phase supply is lost, causing the remaining legs to draw excessive current and overheat the motor.
  • Reversing any two of the three leads on a three-phase motor reverses its direction of rotation.
  • A variable frequency drive changes motor speed by varying frequency, and the affinity laws show that power varies with the cube of speed.
Last updated: August 2026

12.2 Motors, Drives & Electrical Fundamentals

The WPI Need-to-Know Criteria list general electrical principles including troubleshooting breakers, relays, and circuits as supporting knowledge at the intermediate level across nearly every content area. Operators are not electricians, but they are expected to read a nameplate, interpret an amperage reading, and know when to stop and call one.


1. The Fundamental Relationships

Ohm's law

E=I×RE = I \times R

where E is voltage in volts, I is current in amperes, and R is resistance in ohms.

Power

CircuitFormula
DC or single-phase resistive$P = E \times I$
Single-phase with power factor$P = E \times I \times \text{PF}$
Three-phase$P = E \times I \times \text{PF} \times \sqrt{3}$, where $\sqrt{3} \approx \textbf{1.732}$

1 horsepower=746 watts1 kW=1,000 W1 \text{ horsepower} = 746 \text{ watts} \qquad 1 \text{ kW} = 1{,}000 \text{ W}

Worked example. A three-phase motor draws 42 amps at 460 volts with a power factor of 0.86.

P=460×42×0.86×1.732=28,780 W28.8 kWP = 460 \times 42 \times 0.86 \times 1.732 = 28{,}780 \text{ W} \approx \textbf{28.8 kW}

Converting to horsepower drawn: 28,780 ÷ 746 ≈ 38.6 hp of electrical input.

Power factor

Power factor is the ratio of real power (kW) to apparent power (kVA). Inductive loads such as motors lag, producing a power factor below 1.0. A low power factor means the utility must deliver more current for the same useful work, and utilities commonly impose a power factor penalty. Correction uses capacitor banks. Motors running lightly loaded have notably poor power factor, which is one reason oversizing motors is costly.


2. Three-Phase Induction Motors

The workhorse of the plant. Key nameplate data an operator must read:

Nameplate itemMeaning
HorsepowerRated mechanical output
VoltageDesign supply voltage, e.g., 230/460 V
Full load amps (FLA)Current at rated load — compare your clamp meter reading to this
RPMFull-load speed (slightly below synchronous speed; the difference is slip)
Service factor (SF)Permissible overload multiplier, e.g., 1.15 means 15% above rated hp continuously
Frame sizePhysical dimensions for replacement
Insulation classTemperature capability: A, B, F, H in ascending order
EnclosureODP (open drip proof), TEFC (totally enclosed fan cooled), explosion-proof
EfficiencyNominal efficiency; premium efficiency motors pay back quickly on continuous duty

Synchronous speed and slip

Ns=120×fnumber of polesN_s = \frac{120 \times f}{\text{number of poles}}

At 60 Hz: 2-pole = 3,600 rpm, 4-pole = 1,800 rpm, 6-pole = 1,200 rpm. Actual full-load speed is a few percent lower — that difference is slip, and it is what produces torque in an induction motor.

Reversing rotation

Swap any two of the three phase leads. This is the single most-tested electrical fact in operator exams, and it matters because a centrifugal pump running backward still moves some water and can fool an operator into chasing a hydraulic problem.


3. Motor Failures and Protection

FailureCauseIndication
Overload / overheatingMechanical binding, excessive load, high ambient temperature, clogged cooling fins, low voltageAmps above FLA, overload relay trips, burnt smell, discolored windings
Single phasingLoss of one leg of the three-phase supply — a blown fuse, an open contact, a broken conductorMotor hums but will not start, or continues running while drawing very high current on the remaining legs. Rapidly destroys the windings
Voltage imbalanceUnequal phase voltagesA small percentage imbalance produces a much larger current imbalance and significant heating
Insulation failureMoisture, heat, age, contaminationLow megohm reading on insulation resistance test; ground fault
Bearing failureLubrication, misalignment, belt tension, shaft currents from VFDsNoise, heat, vibration
Moisture intrusionFailed seals in submersiblesSeal-fail alarm; low insulation resistance

Protective devices

DeviceProtects against
Fuses and circuit breakersShort circuits and ground faults — fast, high-current events
Overload relays (thermal or electronic)Sustained overcurrent — slower, heat-based; sized to motor FLA and service factor
Phase monitor / phase failure relaySingle phasing, phase reversal, voltage imbalance
Ground fault protectionCurrent leaking to ground
Moisture and over-temperature sensorsBuilt into submersible motors

The distinction that gets tested: a breaker or fuse protects the circuit from short circuits; an overload relay protects the motor from running too hot. They are not interchangeable, and a repeatedly tripping overload should never be "solved" by installing a larger heater or a bigger breaker.


4. Starting Methods

An induction motor draws locked rotor current of roughly 6 to 8 times full-load amps at start. On large motors that inrush causes voltage dip and mechanical shock.

MethodCharacter
Across-the-line (full voltage)Simplest and cheapest; full inrush and full starting torque shock
Reduced voltage (autotransformer, part winding, wye-delta)Lower inrush, lower starting torque
Solid-state soft starterRamps voltage smoothly; reduces both electrical inrush and mechanical/hydraulic shock (helps prevent water hammer)
Variable frequency driveRamps frequency and voltage; full speed control

5. Variable Frequency Drives and the Affinity Laws

A VFD rectifies incoming AC to DC and inverts it back to AC at a variable frequency, changing motor speed. Because a centrifugal machine's behavior follows the affinity laws, this is where the energy savings live:

Q2=Q1(N2N1)H2=H1(N2N1)2P2=P1(N2N1)3Q_2 = Q_1\left(\frac{N_2}{N_1}\right) \qquad H_2 = H_1\left(\frac{N_2}{N_1}\right)^2 \qquad P_2 = P_1\left(\frac{N_2}{N_1}\right)^3

  • Flow varies directly with speed
  • Head varies with the square of speed
  • Power varies with the CUBE of speed

Worked example. A pump delivering 1,400 gpm at 60 ft using 40 hp is slowed to 80% speed.

  • Flow: 1,400 × 0.80 = 1,120 gpm
  • Head: 60 × 0.80² = 60 × 0.64 = 38.4 ft
  • Power: 40 × 0.80³ = 40 × 0.512 = 20.5 hp

A 20% speed reduction cuts power almost in half. This is why VFDs replace throttling valves wherever flow must be varied — throttling wastes the energy as heat across the valve, while slowing the pump never generates it.

VFD cautions

  • Minimum speed limits — pumps must still overcome static head, and motors need cooling airflow. Below roughly 25–30 Hz, a TEFC motor's shaft-mounted fan may not cool it adequately.
  • Shaft currents can pit bearings; shaft grounding rings or insulated bearings are used on larger drives.
  • Harmonics feed back into the electrical system and may require line reactors or filters.
  • Heat — VFD enclosures need ventilation and clean filters.

6. Electrical Safety at the Plant

  • Lockout/tagout is mandatory before any work on electrically driven equipment — covered fully in Section 13.3.
  • Test before you touch. Verify de-energization with a meter you have proven on a known live source.
  • Arc flash is a serious hazard at motor control centers; boundaries, labeling, and appropriate arc-rated PPE are required.
  • Only qualified persons open energized enclosures. An operator's job is to observe, report, and isolate — not to troubleshoot inside a live MCC bucket.
Test Your Knowledge

A three-phase pump motor rotates backward after a rewiring job. How is rotation corrected?

A
B
C
D
Test Your Knowledge

A pump running at full speed uses 60 hp. If a variable frequency drive slows it to 70 percent speed, approximately what power will it require?

A
B
C
D
Test Your Knowledge

A three-phase motor hums but will not start, and when it does run it draws very high current on two legs. What has most likely occurred?

A
B
C
D
Test Your Knowledge

What does a motor nameplate service factor of 1.15 indicate?

A
B
C
D