Formula sheet

Electrical Machines and Motor Drive Systems Formula Sheet

Machine and motor-drive formulas for torque, speed, slip, input current, efficiency, starting current, voltage dip, VFD speed, RMS duty current, thermal limits, and validation.

Branch: Electrical Engineering
Content: Formula sheet
Updated: Jun 23, 2026
Revision: v1.0.0 · reviewed

This formula sheet collects first-pass calculations for electrical machines and motor drive systems. Use it to connect machine nameplate data, shaft torque, speed, slip, starting current, source voltage dip, variable-frequency drive operation, duty-cycle RMS current, braking energy, thermal limits, protection, and validation evidence.

The equations are screening and design-review tools. They do not replace a manufacturer torque-speed curve, finite-element electromagnetic model, thermal test, harmonic study, protection coordination study, safety assessment, or commissioning measurement. They are useful only when the machine boundary, load case, RMS and peak quantities, duty cycle, cooling assumption, supply impedance, drive current limit, protection setting, and measurement method are stated.

Symbols and Units

Symbol	Meaning	Common unit
$P$	active or mechanical power, depending on context	W or kW
$S$	apparent power	VA or kVA
$Q$	reactive power	var or kvar
$V_{LL}$	line-to-line RMS voltage	V
$I_L$	line RMS current	A
$PF$	power factor	dimensionless
$\eta$	efficiency	dimensionless
$T$	shaft torque	N m
$\omega$	angular speed	rad/s
$n$	rotational speed	rpm
$n_s$	synchronous speed	rpm
$f$	electrical frequency	Hz
$p$	number of machine poles	dimensionless
$s$	induction-machine slip	dimensionless
$J$	reflected inertia at the motor shaft	kg m^2
$\alpha$	angular acceleration	rad/s^2
$I_{FL}$	full-load current	A
$I_{start}$	starting current	A
$Z_{source,pu}$	source impedance on the selected base	pu

Use RMS values for AC current and voltage unless a formula explicitly states peak, instantaneous, or DC-link quantities. Do not mix shaft output power, electrical input power, apparent power, and converter DC-link power without defining the boundary.

Mechanical Power, Torque, and Speed

Angular speed from rpm is:

\displaystyle \omega=\frac{2\pi n}{60}

Mechanical shaft power is:

P_{mech}=T\omega

Therefore:

\displaystyle T=\frac{P_{mech}}{\omega}

In the common kW/rpm form:

\displaystyle T\ [\text{N m}]=\frac{9550P\ [\text{kW}]}{n\ [\text{rpm}]}

This relation is exact for power, torque, and speed at the same shaft. If a gearbox, belt stage, coupling, or reflected load is present, move power and torque through that transmission before comparing to the motor nameplate.

Worked Example: Shaft Torque from Rated Power

A machine must deliver $75\ \text{kW}$ at $1480\ \text{rpm}$ . Estimate required shaft torque.

Use:

\displaystyle T=\frac{9550P}{n}

Substitute:

\displaystyle T=\frac{9550(75)}{1480}=484\ \text{N m}

The value is the average shaft torque at that operating point. It does not prove that the motor can start the load, accelerate it within the required time, handle cyclic overload, or remain within its thermal limit. Those checks require torque-speed and duty-cycle information.

Three-Phase Input Power and Full-Load Current

For balanced three-phase input:

S=\sqrt{3}V_{LL}I_L

Active input power is:

P_{in}=\sqrt{3}V_{LL}I_LPF

If shaft output power, efficiency, and power factor are known:

\displaystyle P_{in}=\frac{P_{out}}{\eta}

\displaystyle S=\frac{P_{out}}{\eta PF}

\displaystyle I_L=\frac{S}{\sqrt{3}V_{LL}}

These formulas assume balanced sinusoidal fundamental quantities. Inverter input currents, distorted currents, unbalanced loads, and harmonic heating require the AC power and converter formulas linked from this page.

Worked Example: Motor Input Current from Shaft Power

A $75\ \text{kW}$ motor operates from $480\ \text{V}$ three-phase. At this load, estimate $\eta=0.94$ and $PF=0.88$ .

Input active power:

\displaystyle P_{in}=\frac{75}{0.94}=79.8\ \text{kW}

Input apparent power:

\displaystyle S=\frac{75}{0.94(0.88)}=90.7\ \text{kVA}

Line current:

\displaystyle I_L=\frac{90700}{\sqrt{3}(480)}=109\ \text{A}

This is a normal running-current estimate, not a starting-current estimate. Conductors, overloads, drive current ratings, transformer loading, and short-time current limits must be checked against the project rules and the actual machine data sheet.

Efficiency and Loss Balance

Machine or drive efficiency is:

\displaystyle \eta=\frac{P_{out}}{P_{in}}

Loss is:

P_{loss}=P_{in}-P_{out}

For a motor-drive package:

\eta_{system}=\eta_{drive}\eta_{motor}\eta_{mechanical}

Approximate heat that must be removed from the electrical package is the sum of converter loss, motor electrical and magnetic loss, cable/filter loss, and enclosure loss:

P_{heat}\approx P_{drive,loss}+P_{motor,loss}+P_{cable,loss}+P_{filter,loss}

The loss split matters because the thermal paths are different. Semiconductor junction temperature, motor winding temperature, bearing temperature, cable temperature, and enclosure air temperature can all have different limits.

Synchronous Speed and Slip

For an AC rotating field:

\displaystyle n_s=\frac{120f}{p}

where $p$ is the number of poles. For an induction motor:

\displaystyle s=\frac{n_s-n_r}{n_s}

Rotor electrical frequency is:

f_r=sf

Slip is required for induction-machine torque in motoring operation. A low slip near rated load is normal. A slip that is much higher than expected can indicate overload, low voltage, incorrect frequency, damaged rotor bars, excessive load torque, or incorrect measurement.

Worked Example: Slip and Rotor Frequency

A four-pole induction motor is supplied at $60\ \text{Hz}$ and runs at $1765\ \text{rpm}$ under load.

Synchronous speed:

\displaystyle n_s=\frac{120(60)}{4}=1800\ \text{rpm}

Slip:

\displaystyle s=\frac{1800-1765}{1800}=0.0194=1.94\%

Rotor electrical frequency:

f_r=sf=0.0194(60)=1.16\ \text{Hz}

The result is plausible for a loaded induction motor. If a tachometer reports a much lower speed at the same supply frequency, the engineer should check mechanical load, voltage, phase balance, rotor condition, and whether the machine is actually connected as assumed.

Load Acceleration and Required Torque

Angular acceleration is:

\displaystyle \alpha=\frac{\Delta\omega}{\Delta t}

Acceleration torque for reflected inertia is:

T_{acc}=J\alpha

Required motor shaft torque during acceleration is approximately:

T_{req}=T_{load}+T_{acc}+T_{friction}

If a gearbox is present, reflect load inertia to the motor shaft:

\displaystyle J_{ref}=J_{load}\left(\frac{\omega_{load}}{\omega_{motor}}\right)^2

This equation assumes ideal rigid gearing for the inertia transformation. Backlash, compliance, belt slip, torsional resonance, and clutch behavior require a dynamic model.

Worked Example: Acceleration Torque and Short-Time Power

A drive must accelerate a load with reflected inertia $J=8.5\ \text{kg m}^2$ from rest to $1200\ \text{rpm}$ in $4.0\ \text{s}$ . The process load torque during acceleration is $55\ \text{N m}$ .

Final angular speed:

\displaystyle \omega=\frac{2\pi(1200)}{60}=126\ \text{rad/s}

Angular acceleration:

\displaystyle \alpha=\frac{126}{4.0}=31.5\ \text{rad/s}^2

Acceleration torque:

T_{acc}=8.5(31.5)=268\ \text{N m}

Required shaft torque:

T_{req}=268+55=323\ \text{N m}

Near final speed, the mechanical power associated with this torque is:

P=T\omega=323(126)=40.7\ \text{kW}

The conclusion is not automatically “select a 40.7 kW motor.” The conclusion is that the motor and drive must supply about $323\ \text{N m}$ for the specified acceleration without violating current, voltage, thermal, mechanical, or protection limits. Continuous rating and short-time overload rating must both be checked.

Load Torque Laws

Common first-pass load models are:

Load type	Approximate torque law	Approximate power law
conveyor, hoist, positive-displacement pump	$T\approx\text{constant}$	$P\propto\omega$
fan, centrifugal pump in the same system	$T\propto\omega^2$	$P\propto\omega^3$
winder or constant-tension process	$T\propto r$ at the roll	depends on diameter and speed
machine tool spindle	$P\approx\text{constant}$ above base speed	$T\propto1/\omega$

Affinity laws for fans and pumps are valid only when the same machine, same fluid, same geometry, and similar operating region apply. Throttled systems, cavitation, static head, compressibility, control valves, and changed impellers can break the simple cubic power law.

Starting Current and Voltage Dip

For across-the-line induction-motor starting:

I_{start}=k_{LR}I_{FL}

where $k_{LR}$ is the locked-rotor current multiple, often taken from manufacturer data or a motor code letter.

A simple per-unit voltage-dip screen is:

\Delta V_{pu}\approx I_{start,pu}Z_{source,pu}

If existing load current is significant, it should be included in the voltage profile, but the starting-current transient is usually the dominant new disturbance.

Starting current is not the same as starting torque. For many induction motors, starting torque is roughly proportional to voltage squared:

T_{start}\propto V^2

Reducing voltage can reduce current and voltage dip, but it can also prevent acceleration if load breakaway torque is high.

Worked Example: Starting Current and Voltage-Dip Screen

A $150\ \text{kW}$ motor on a $480\ \text{V}$ bus has estimated $\eta=0.95$ , $PF=0.89$ , and locked-rotor current equal to $6.2I_{FL}$ . The source transformer base is $1000\ \text{kVA}$ with $Z_{source}=0.055\ \text{pu}$ .

Rated apparent power:

\displaystyle S_{FL}=\frac{150}{0.95(0.89)}=177.4\ \text{kVA}

Full-load current:

\displaystyle I_{FL}=\frac{177400}{\sqrt{3}(480)}=213\ \text{A}

Starting current:

I_{start}=6.2(213)=1320\ \text{A}

Starting apparent power:

S_{start}=\sqrt{3}(480)(1320)=1.10\ \text{MVA}

On the transformer base:

\displaystyle I_{start,pu}\approx\frac{1.10}{1.00}=1.10\ \text{pu}

Approximate voltage dip:

\Delta V_{pu}\approx1.10(0.055)=0.061\ \text{pu}

The bus may dip to roughly $0.94\ \text{pu}$ before existing-load drop and feeder impedance are added. That may be acceptable for some equipment and unacceptable for sensitive controls. The result should be validated with a motor-starting study or captured RMS voltage trend during commissioning.

Variable-Frequency Drive Speed and Volts per Hertz

Ideal synchronous speed under VFD control is still:

\displaystyle n_s=\frac{120f_{out}}{p}

Induction motor shaft speed is approximately:

n_r=n_s(1-s)

Below base speed, simple scalar control often keeps:

\displaystyle \frac{V}{f}\approx\text{constant}

At very low speed, stator resistance voltage drop, cooling reduction, sensorless-estimation limits, and load friction can make the simple volts-per-hertz rule inaccurate. Above base speed, voltage is usually limited and the drive enters field weakening; torque capability falls as speed rises.

Worked Example: VFD Output Frequency for Required Speed

A four-pole induction motor must run a conveyor at about $900\ \text{rpm}$ . Expected slip at the required torque is $2.5\%$ .

From $n_r=n_s(1-s)$ :

\displaystyle n_s=\frac{n_r}{1-s}=\frac{900}{0.975}=923\ \text{rpm}

Required output frequency:

\displaystyle f=\frac{n_sp}{120}=\frac{923(4)}{120}=30.8\ \text{Hz}

This is a starting setpoint, not proof of delivered speed. Commissioning should verify tachometer speed, motor current, torque margin, cooling at reduced speed, and whether conveyor breakaway torque changes the slip estimate.

Duty-Cycle RMS Current and Thermal Screening

For a repeating duty cycle, a first-pass RMS current is:

\displaystyle I_{RMS}=\sqrt{\frac{\sum I_i^2t_i}{\sum t_i}}

The same form can be used for RMS torque when torque is proportional to current in the valid operating region:

\displaystyle T_{RMS}=\sqrt{\frac{\sum T_i^2t_i}{\sum t_i}}

Thermal acceptability requires:

I_{RMS}\le I_{continuous,derated}

and each short-time segment must also satisfy:

I_{peak}\le I_{overload,allowed}

RMS screening is not a complete thermal model. It ignores thermal time constants, hotspot location, speed-dependent cooling, ambient temperature, altitude, enclosure airflow, harmonic loss, and repeated braking energy unless those effects are added explicitly.

Worked Example: Intermittent Duty Current

A drive cycle repeats every $100\ \text{s}$ :

Segment	Current	Duration
acceleration	$180\ \text{A}$	$8\ \text{s}$
running	$95\ \text{A}$	$70\ \text{s}$
braking	$140\ \text{A}$	$6\ \text{s}$
idle	$25\ \text{A}$	$16\ \text{s}$

RMS current:

\displaystyle I_{RMS}=\sqrt{\frac{180^2(8)+95^2(70)+140^2(6)+25^2(16)}{100}}

I_{RMS}=101\ \text{A}

If the drive is derated to $110\ \text{A}$ continuous in the enclosure, the RMS current screen passes. The engineer still has to check whether $180\ \text{A}$ for $8\ \text{s}$ is inside the drive overload curve, whether braking energy is acceptable, and whether low-speed cooling is adequate.

Regeneration and Braking Energy

Rotational kinetic energy is:

\displaystyle E_k=\frac{1}{2}J\omega^2

Energy removed when slowing from $\omega_1$ to $\omega_2$ is:

\displaystyle \Delta E=\frac{1}{2}J(\omega_1^2-\omega_2^2)

Average braking power over time $\Delta t$ is:

\displaystyle P_{brake,avg}=\frac{\Delta E}{\Delta t}

The energy must go somewhere: a braking resistor, a regenerative front end, a DC bus shared with another load, storage, mechanical friction, or heat inside the machine and process. The energy path must be designed, protected, and validated.

Worked Example: Braking Resistor Energy Screen

A centrifuge has reflected inertia $J=18\ \text{kg m}^2$ at the motor shaft and slows from $1800\ \text{rpm}$ to rest in $12\ \text{s}$ .

Initial angular speed:

\displaystyle \omega_1=\frac{2\pi(1800)}{60}=188.5\ \text{rad/s}

Energy removed:

\displaystyle \Delta E=\frac{1}{2}(18)(188.5^2)=320000\ \text{J}=320\ \text{kJ}

Average braking power:

\displaystyle P_{brake,avg}=\frac{320000}{12}=26.7\ \text{kW}

The resistor or regenerative system must handle the energy per stop and the average power over repeated stops. A resistor that survives one stop may overheat during a production sequence with short intervals between stops.

Protection and Derating Checks

Useful screening ratios are:

\displaystyle \text{current margin}=\frac{I_{limit}-I_{case}}{I_{limit}}

\displaystyle \text{torque margin}=\frac{T_{available}-T_{required}}{T_{required}}

\displaystyle \text{thermal loading}=\frac{I_{RMS}}{I_{continuous,derated}}

\displaystyle \text{voltage margin}=\frac{V_{min,allowed}-V_{min,case}}{V_{nominal}}

These ratios should be calculated for the actual load cases: starting, acceleration, continuous duty, overload, braking, regeneration, stall, low-speed operation, loss of cooling, and restart after a trip.

Common derating drivers include ambient temperature, altitude, enclosure type, switching frequency, output frequency, low-speed fan cooling, cable length, harmonic current, dirty filters, weak source impedance, unbalanced supply voltage, and corrosive or dusty environment.

Validation Measurements

The calculation file should name the measurement that validates each important assumption.

Calculation	Validation evidence
full-load current	clamp meter or drive current trend at known load
speed and slip	tachometer, encoder count, or drive speed estimate checked against independent measurement
starting current	event capture or power-quality recorder
voltage dip	RMS voltage trend at the affected bus
acceleration torque	acceleration time, current trend, and speed trend
duty-cycle RMS current	logged current over complete production cycle
braking energy	DC-link voltage, resistor temperature, regenerative power, or stop-time trend
thermal margin	winding, bearing, drive heatsink, and enclosure temperature trend
protection	injected test, staged fault, parameter review, trip log, or commissioning record

Validation should capture both value and context: load condition, ambient temperature, supply voltage, parameter set, firmware version, enclosure state, cooling state, cable configuration, and measurement uncertainty.

Common Mistakes

Common mistakes include using shaft kW as if it were input kVA, treating VFD output as a sinusoidal utility source, ignoring locked-rotor current, selecting a motor by steady power while ignoring acceleration torque, using volts-per-hertz control below its valid speed range, and assuming that reduced voltage always solves starting problems.

Other mistakes are thermal and operational: checking RMS current but not peak overload, validating one stop but not repeated braking, ignoring low-speed cooling, changing switching frequency without revisiting thermal and EMC effects, applying long motor cables without insulation and leakage-current checks, and preserving no record of the final drive parameter set.

The safe engineering habit is to keep electrical, magnetic, mechanical, thermal, control, protection, and measurement assumptions together. A motor-drive calculation is useful only when it predicts the load case that will actually be commissioned.

REF

Disciplines