Exercise set

Electrical Machines and Motor Drive Systems Exercises

Solved motor-drive exercises for torque, slip, VFD speed, RMS current, braking, protection, resonance, encoder feedback and validation.

These exercises practise electrical-machine and motor-drive calculations as engineering decisions. The goal is not only to compute current, torque, slip, voltage dip, speed, thermal loading, braking energy, regenerative power, or speed-feedback plausibility. The goal is to decide whether the motor, drive, supply, protection, cooling, control feedback, and measurement evidence are consistent with the required load case.

Assume balanced three-phase systems, RMS AC quantities, and simplified screening models unless an exercise states otherwise. Real projects should also check manufacturer torque-speed curves, overload curves, cable length, insulation stress, harmonics, protection settings, safety functions, mechanical resonance, ambient derating, and commissioning records.

Release Evidence Notes

Use these worked problems as sizing and release screens until the installed motor-drive evidence proves the same duty. A credible motor-drive release package should connect shaft load, motor nameplate, drive rating, overload curve, cable installation, protection settings, thermal environment, control mode and commissioning measurements.

For motor-drive decisions, the strongest evidence is agreement between electrical, mechanical and control data:

  • shaft torque, speed, acceleration time, inertia and process load match the selected motor and drive overload region;
  • measured current, voltage, power factor, DC-link trend, speed feedback and thermal trend describe the same operating point;
  • low-speed cooling, enclosure temperature, altitude, duty RMS current, peak current and repeated braking energy are checked against real ratings;
  • cable length, voltage drop, reflected-wave stress, common-mode current, shield termination and bearing mitigation match the installed layout;
  • commissioning records state instrument class, load condition, parameter set, protection settings, firmware, interlocks and acceptance margin.

When those evidence paths disagree, do not average the nameplate and commissioning data into a pass. Identify the wrong load case, bad parameter set, overloaded drive, weak cooling path, hidden voltage sag, cable installation issue, regenerative energy path limit or measurement mismatch before release.

How to Use These Exercises

For each problem:

  1. define the boundary: shaft output, motor terminals, drive input, DC link, or supply bus;
  2. keep shaft power, electrical active power, apparent power, and current separate;
  3. state whether the calculation is continuous, short-time, starting, braking, or degraded-duty;
  4. compare the result with a rating or acceptance criterion;
  5. identify the measurement that would validate the calculation in service.

The common mistake is selecting a motor or drive from rated kW alone. Motor-drive systems fail in practice because current, torque, thermal duty, voltage sag, braking energy, low-speed cooling, parameter settings, and protection timing do not match the real duty cycle.

Engineering Boundary Notes

These exercises use simplified electrical-machine and motor-drive screens. They do not replace manufacturer drive studies, motor thermal modelling, protection coordination, insulation coordination, torsional or vibration analysis, functional-safety review, harmonic assessment, commissioning tests or machinery risk assessment. A calculated motor-drive margin applies only to the stated shaft load, motor terminals, drive rating, cable installation, control mode, thermal environment and duty cycle.

Separate shaft, motor, drive, supply and mechanical boundaries before release. A kW match does not prove current, torque, cooling, braking, cable stress, bearing current, encoder feedback, resonance avoidance or trip timing. Release evidence should show that electrical, thermal, mechanical and control measurements describe the same operating point.

Common Release Mistakes

  • selecting motor or drive size from rated kW while ignoring RMS current, overload, starting and duty cycle;
  • accepting VFD speed control without low-speed cooling, parameter set, slip and torque evidence;
  • checking braking energy for one stop while average regenerative or resistor power fails over the cycle;
  • ignoring cable length, reflected-wave stress, common-mode current, shield termination and bearing mitigation;
  • trusting current measurements without load condition, instrument class, parameter set and firmware state;
  • releasing feedback, protection or skip-frequency settings without fault-response and resonance evidence.

Scenario Map

ScenarioExercisesEngineering decision
Steady motor loading1, 2Convert shaft power, speed, slip, input power and current into a credible operating point.
Starting and acceleration3, 4Check voltage dip, short-time torque, acceleration time and overload duration.
Drive control and thermal duty5, 6, 10, 17Decide whether VFD speed, RMS current, winding temperature and low-speed derating are acceptable.
Braking and energy recovery7, 14Verify braking energy, recovery path and resistor or line-side power limits for repeated stops.
Cable and high-frequency installation effects11, 13, 15Check cable voltage drop, reflected-wave insulation stress, common-mode current and bearing-risk evidence.
Commissioning and operating evidence8, 9, 12Validate savings, measured current consistency and guarded release under representative duty.
Protection and abnormal events16Check locked-rotor or jam trip timing against thermal exposure and timing uncertainty.
Speed feedback and fault response18Check encoder edge rate, tachometer agreement, missing-pulse detection and coast-down travel after feedback loss.
Speed avoidance and resonance19Check whether VFD skip-frequency settings keep continuous operation away from a measured mechanical mode.

Validation Package Checklist

  • shaft load, speed, inertia, duty cycle, thermal environment and process operating point are documented;
  • motor nameplate, drive rating, overload curve, parameter set, firmware, control mode and protection settings are controlled;
  • shaft power, active power, apparent power, RMS current, torque, slip, voltage sag and thermal loading are separated;
  • cable length, voltage drop, reflected-wave stress, common-mode current, shield termination and bearing mitigation are checked;
  • braking energy, regenerative path, resistor duty, low-speed cooling and ambient derating are bounded;
  • commissioning data include instrument class, load condition, speed feedback, interlocks, trend evidence and acceptance margin;
  • final release decision states accept, derate, change drive settings, add filter, improve cooling, retest, restrict speed or hold.

Exercise 1: Shaft Torque, Input Power, and Line Current

A process mixer requires 55\ \text{kW} at the shaft while running at 1460\ \text{rpm}. The selected motor is supplied at 400\ \text{V} three-phase. At the operating point, estimated motor efficiency is 0.93 and power factor is 0.86.

Find:

  1. required shaft torque;
  2. electrical input active power;
  3. input apparent power;
  4. estimated line current.

Solution

Shaft torque from power and speed:

\displaystyle T=\frac{9550P}{n}
\displaystyle T=\frac{9550(55)}{1460}=360\ \text{N m}

Input active power is shaft power divided by efficiency:

\displaystyle P_{in}=\frac{P_{out}}{\eta}=\frac{55}{0.93}=59.1\ \text{kW}

Apparent power includes power factor:

\displaystyle S=\frac{P_{out}}{\eta PF}=\frac{55}{0.93(0.86)}=68.8\ \text{kVA}

Three-phase line current:

\displaystyle I_L=\frac{S}{\sqrt{3}V_{LL}}
\displaystyle I_L=\frac{68800}{\sqrt{3}(400)}=99.3\ \text{A}

Engineering Comment

The motor is not a 55\ \text{kVA} load just because it delivers 55\ \text{kW} at the shaft. The supply and switchgear see apparent power and RMS current. The 360\ \text{N m} torque is the steady process torque at this speed, not the acceleration torque or breakaway torque.

Validation should compare measured running current, voltage, speed, and process load against these estimates. If current is much higher, check overload, low voltage, incorrect connection, low power factor, mechanical drag, or a process operating point different from the design case.

Plausibility Check

A 55\ \text{kW} shaft load near 1500\ \text{rpm} should require a few hundred newton metres of torque. Because efficiency and power factor are below 1, apparent power and line current must be higher than a direct 55\ \text{kW} division would suggest.

Exercise 2: Synchronous Speed, Slip, and Rotor Frequency

A six-pole induction motor is connected to a 50\ \text{Hz} supply. During a loaded test, an independent tachometer reads 970\ \text{rpm}.

Find:

  1. synchronous speed;
  2. slip;
  3. rotor electrical frequency;
  4. whether the result is plausible for normal loaded operation.

Solution

Synchronous speed:

\displaystyle n_s=\frac{120f}{p}=\frac{120(50)}{6}=1000\ \text{rpm}

Slip:

\displaystyle s=\frac{n_s-n_r}{n_s}=\frac{1000-970}{1000}=0.030=3.0\%

Rotor electrical frequency:

f_r=sf=0.030(50)=1.5\ \text{Hz}

Engineering Comment

A 3.0\% slip is plausible for a loaded induction motor, depending on size and design. It suggests the motor is below synchronous speed by a normal amount rather than stalled or severely overloaded.

The calculation would not prove healthy operation by itself. Validation should include phase current balance, terminal voltage, vibration, temperature rise, and comparison with the manufacturer speed-current curve. If slip increases over time at the same load, investigate supply voltage, mechanical drag, rotor damage, or bearing condition.

Plausibility Check

A six-pole motor on 50\ \text{Hz} has 1000\ \text{rpm} synchronous speed, so a measured 970\ \text{rpm} is a normal loaded value. Rotor frequency near 1.5\ \text{Hz} is consistent with low slip.

Exercise 3: Across-the-Line Starting Current and Voltage Dip

A 220\ \text{kW} pump motor is supplied from a 480\ \text{V} motor-control center through a transformer rated 1500\ \text{kVA}. At rated load the motor has \eta=0.94 and PF=0.88. Locked-rotor current is 6.0 times full-load current. The transformer and upstream source are approximated by Z_{source}=0.060\ \text{pu} on the transformer base.

Estimate:

  1. full-load current;
  2. across-the-line starting current;
  3. starting apparent power on the transformer base;
  4. approximate voltage dip;
  5. whether a sensitive load with an alarm threshold at 0.90\ \text{pu} is at risk.

Solution

Rated apparent power:

\displaystyle S_{FL}=\frac{P_{out}}{\eta PF}=\frac{220}{0.94(0.88)}=265.8\ \text{kVA}

Full-load current:

\displaystyle I_{FL}=\frac{S}{\sqrt{3}V_{LL}}=\frac{265800}{\sqrt{3}(480)}=320\ \text{A}

Starting current:

I_{start}=6.0I_{FL}=6.0(320)=1920\ \text{A}

Starting apparent power:

S_{start}=\sqrt{3}V_{LL}I_{start}
S_{start}=\sqrt{3}(480)(1920)=1.60\ \text{MVA}

Per-unit starting current on a 1500\ \text{kVA} base:

\displaystyle I_{start,pu}\approx\frac{1.60}{1.50}=1.07\ \text{pu}

Approximate voltage dip:

\Delta V_{pu}\approx I_{start,pu}Z_{source,pu}=1.07(0.060)=0.064\ \text{pu}

Estimated minimum bus voltage:

V_{min}\approx1.00-0.064=0.936\ \text{pu}

Engineering Comment

The simple screen suggests the motor start alone should not pull the bus below 0.90\ \text{pu}. The sensitive load may still be at risk if pre-existing voltage drop, feeder impedance, other coincident loads, control-power transformer drop, contactor dropout behavior, or utility source weakness are significant.

Validation should use a power-quality recorder at the motor-control center and at the sensitive load during a real or staged start. The start should also be checked for acceleration time and starting torque; reducing voltage to reduce current can fail if the pump torque demand is high.

Plausibility Check

Locked-rotor current is six times full-load current, so a start around 1.9\ \text{kA} is plausible. The voltage dip remains below 10\% because the source impedance is only 0.060\ \text{pu} on a transformer base near the starting kVA.

Exercise 4: Acceleration Torque and Drive Overload Decision

A motor drive must accelerate a machine from rest to 1500\ \text{rpm} in 5.0\ \text{s}. Total reflected inertia at the motor shaft is J=11\ \text{kg m}^2. The process load torque during acceleration is 65\ \text{N m}. The candidate drive and motor can supply 520\ \text{N m} for 10\ \text{s}.

Determine whether the short-time torque rating is adequate.

Solution

Final angular speed:

\displaystyle \omega=\frac{2\pi n}{60}=\frac{2\pi(1500)}{60}=157\ \text{rad/s}

Angular acceleration:

\displaystyle \alpha=\frac{\Delta\omega}{\Delta t}=\frac{157}{5.0}=31.4\ \text{rad/s}^2

Acceleration torque:

T_{acc}=J\alpha=11(31.4)=345\ \text{N m}

Total required shaft torque:

T_{req}=T_{acc}+T_{load}=345+65=410\ \text{N m}

Torque margin:

\displaystyle \text{margin}=\frac{T_{available}-T_{req}}{T_{req}}=\frac{520-410}{410}=0.268=26.8\%

Engineering Comment

The short-time torque rating passes this simplified check with about 27\% margin. The acceleration time is also within the 10\ \text{s} overload duration.

This is not a complete release decision. Validation should confirm current limit, DC-link voltage, motor temperature, coupling torque, brake release timing, load breakaway torque, and actual speed ramp. If the drive current saturates, the acceleration time may extend and the thermal calculation changes.

Plausibility Check

The acceleration torque dominates the process torque, which is expected for a fast 5\ \text{s} run-up with 11\ \text{kg m}^2 reflected inertia. The 26.8\% torque margin is useful but not large enough to ignore breakaway torque.

Exercise 5: VFD Frequency, Speed, and Volts per Hertz

A four-pole induction motor drives a conveyor that must run at 720\ \text{rpm} under load. Expected slip at the required torque is 3.0\%. The motor is rated for 400\ \text{V} at 50\ \text{Hz} and is controlled below base speed with approximate constant volts per hertz.

Find:

  1. required synchronous speed;
  2. VFD output frequency;
  3. approximate output voltage for constant V/f;
  4. the main low-speed limitation to check.

Solution

The rotor speed is:

n_r=n_s(1-s)

Therefore:

\displaystyle n_s=\frac{n_r}{1-s}=\frac{720}{0.970}=742\ \text{rpm}

Frequency:

\displaystyle f=\frac{n_sp}{120}=\frac{742(4)}{120}=24.7\ \text{Hz}

Rated volts per hertz:

\displaystyle \frac{V}{f}=\frac{400}{50}=8.0\ \text{V/Hz}

Approximate output voltage:

V\approx8.0(24.7)=198\ \text{V}

Engineering Comment

The VFD frequency setpoint should start near 24.7\ \text{Hz} if the slip estimate is accurate. In commissioning, actual speed should be checked with an encoder or tachometer because slip changes with torque and motor temperature.

The main low-speed limitation is cooling. A shaft-mounted motor fan may move much less air at half speed. If the conveyor needs high torque for long periods at 25\ \text{Hz}, the motor may require derating, an external blower, temperature sensors, or a different motor design.

Plausibility Check

The requested speed is roughly half a four-pole motor’s base synchronous speed, so a VFD frequency near 25\ \text{Hz} is expected. The voltage also drops near half base voltage under constant volts per hertz control.

Exercise 6: Duty-Cycle RMS Current and Peak Overload

A hoist drive repeats the following 120\ \text{s} cycle:

SegmentCurrentDuration
lift acceleration210\ \text{A}6\ \text{s}
lift steady load145\ \text{A}34\ \text{s}
hold80\ \text{A}20\ \text{s}
lowering with braking165\ \text{A}22\ \text{s}
idle35\ \text{A}38\ \text{s}

The drive is derated to 130\ \text{A} continuous in the installed enclosure and allows 220\ \text{A} for 10\ \text{s}.

Check:

  1. RMS current for the cycle;
  2. continuous thermal screen;
  3. peak overload screen.

Solution

RMS current:

\displaystyle I_{RMS}=\sqrt{\frac{\sum I_i^2t_i}{\sum t_i}}
\displaystyle I_{RMS}=\sqrt{\frac{210^2(6)+145^2(34)+80^2(20)+165^2(22)+35^2(38)}{120}}
I_{RMS}=121\ \text{A}

Continuous thermal screen:

121\ \text{A}<130\ \text{A}

Peak overload screen:

210\ \text{A}<220\ \text{A}

and the high-current acceleration interval is 6\ \text{s}, which is below the allowed 10\ \text{s} duration.

Engineering Comment

The simplified drive current screens pass. The margin is not large, so the design should still check ambient temperature, enclosure ventilation, motor cooling, braking energy, and whether a heavier load case exists.

Validation should log drive current and temperature over several complete cycles after the enclosure reaches normal operating temperature. A single unloaded test would not validate the RMS current calculation.

Plausibility Check

The RMS current is much lower than the 210\ \text{A} peak because the high-current segment lasts only 6\ \text{s}. The peak overload check still matters separately because RMS thermal pass does not prove short-time current capability.

Exercise 7: Braking Energy and Repeated Stops

A test stand has reflected inertia J=24\ \text{kg m}^2 at the motor shaft. It decelerates from 1200\ \text{rpm} to rest in 8.0\ \text{s} using dynamic braking. The braking resistor is rated for 300\ \text{kJ} per stop and 12\ \text{kW} average over a repeated test sequence. The stop is repeated every 45\ \text{s}.

Determine whether the resistor passes the energy and average-power checks.

Solution

Initial angular speed:

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

Energy removed:

\displaystyle \Delta E=\frac{1}{2}J\omega^2=\frac{1}{2}(24)(126^2)=190000\ \text{J}=190\ \text{kJ}

The energy per stop is below 300\ \text{kJ}, so the single-stop energy check passes.

Average power over the repeated sequence:

\displaystyle P_{avg}=\frac{190000}{45}=4.22\ \text{kW}

This is below the 12\ \text{kW} average rating.

Engineering Comment

The resistor passes both simplified checks. The braking system still needs DC-link voltage protection, resistor temperature monitoring or thermal model, enclosure heat rejection, wiring and contactor ratings, and a fault response if the resistor circuit is open.

Validation should capture DC-link voltage, stop time, resistor temperature, and fault logs over repeated stops. If process inertia is increased later, this calculation must be repeated because braking energy scales directly with inertia and with speed squared.

Plausibility Check

The average braking power is much lower than the stop power because each stop is followed by a 45\ \text{s} interval. The energy per stop remains the controlling single-event check.

Exercise 8: Fan Speed Reduction with Affinity Laws

A ventilation fan consumes 38\ \text{kW} at full speed. The process can accept operation at 80\% of full speed for a low-demand period. Assume the same fan and system curve region so the affinity-law approximation is valid.

Estimate:

  1. new power;
  2. power reduction;
  3. a validation check before accepting the operating mode.

Solution

For a fan in the same operating region:

\displaystyle P_2=P_1\left(\frac{n_2}{n_1}\right)^3

Substitute:

P_2=38(0.80)^3=19.5\ \text{kW}

Power reduction:

\Delta P=38-19.5=18.5\ \text{kW}

Percentage reduction:

\displaystyle \frac{18.5}{38}=48.7\%

Engineering Comment

A modest speed reduction can produce a large power reduction because fan power is approximately cubic with speed. The approximation can be wrong if dampers, static pressure, stall, filter loading, or process constraints change the operating point.

Validation should measure airflow or process pressure, motor current, drive output frequency, temperature, and any minimum ventilation requirement. Energy savings are useful only if the process remains inside its safety and performance limits.

Plausibility Check

Power at 80\% speed is about half of full-speed power because 0.8^3=0.512. A small speed reduction giving a large power reduction is therefore expected for a fan in the same operating region.

Exercise 9: Commissioning Data Consistency Check

A newly commissioned motor-drive system has the following design estimate at a specific load case:

QuantityExpected
shaft power55\ \text{kW}
line voltage400\ \text{V}
efficiency0.93
power factor0.86
expected current99\ \text{A}

During commissioning, the measured current is 124\ \text{A} at approximately the same speed and process setting. Voltage is 397\ \text{V} line-to-line. The drive reports no current sensor fault.

Estimate the apparent power implied by the measured current and identify what the engineer should investigate.

Solution

Measured apparent power:

S_{meas}=\sqrt{3}V_{LL}I_L
S_{meas}=\sqrt{3}(397)(124)=85.3\ \text{kVA}

Expected apparent power from the design estimate was:

\displaystyle S_{exp}=\frac{55}{0.93(0.86)}=68.8\ \text{kVA}

Difference:

\Delta S=85.3-68.8=16.5\ \text{kVA}

Relative increase:

\displaystyle \frac{16.5}{68.8}=24.0\%

Engineering Comment

The measured apparent power is about 24\% higher than expected. The voltage is only slightly below nominal, so voltage alone does not explain the current increase.

The engineer should investigate actual shaft load, process condition, motor connection, motor parameters entered in the drive, power factor, harmonics, current imbalance, mechanical drag, bearing or coupling condition, and whether the expected 55\ \text{kW} load case was correctly reproduced. A release decision should wait for an active-power measurement or drive power trend, a speed check, and a thermal trend under representative duty.

Plausibility Check

A 24\% apparent-power increase is too large to dismiss as the small voltage difference between 400\ \text{V} and 397\ \text{V}. The measured current points to a load, parameter, power-factor or mechanical issue that needs investigation before release.

Exercise 10: Low-Speed Continuous Torque Derating

A self-cooled induction motor has rated continuous torque:

T_{rated}=220\ \text{N m}

At a VFD output frequency of:

f=20\ \text{Hz}

the manufacturer derating curve allows only:

k_T=0.70

of rated continuous torque because the shaft-mounted fan provides less cooling. The driven machine requires:

T_{load}=180\ \text{N m}

continuously at this speed. Check whether the operating point can be released without an external blower or derating.

Solution

Allowed continuous torque is:

T_{allow}=k_TT_{rated}
T_{allow}=0.70(220)=154\ \text{N m}

Torque utilization is:

\displaystyle U_T=\frac{T_{load}}{T_{allow}}
\displaystyle U_T=\frac{180}{154}=1.17

Torque shortfall is:

\Delta T=T_{load}-T_{allow}
\Delta T=180-154=26\ \text{N m}

The continuous low-speed operating point fails the thermal derating screen.

Engineering Comment

The motor may be able to produce the torque electrically for a short time, but continuous operation is limited by cooling. The release options are lower torque, shorter duty, an external blower, a motor rated for inverter duty at low speed, temperature feedback or a different gear ratio. Drive current alone is not enough evidence when motor cooling changes with speed.

Plausibility Check

The available continuous torque is 30\% below nameplate because of cooling derating. A 180\ \text{N m} load exceeding the 154\ \text{N m} low-speed allowance is therefore expected to fail.

Exercise 11: Motor Cable Voltage Drop

A drive outputs:

V_{drive}=400\ \text{V}

line-to-line to a remote motor. The motor current at the design load is:

I=99.3\ \text{A}

The three-phase cable has effective resistance per phase over the run:

R_c=0.12\ \Omega

Use the approximate resistive voltage drop:

\Delta V\approx\sqrt{3}IR_c

The minimum acceptable motor terminal voltage for this operating point is:

V_{min}=380\ \text{V}

Estimate terminal voltage and decide whether the cable run passes.

Solution

Voltage drop is:

\Delta V=\sqrt{3}(99.3)(0.12)=20.6\ \text{V}

Motor terminal voltage is:

V_{motor}=V_{drive}-\Delta V
V_{motor}=400-20.6=379.4\ \text{V}

Percentage drop is:

\displaystyle \frac{\Delta V}{V_{drive}}=\frac{20.6}{400}=5.15\%

Compare terminal voltage with the minimum:

379.4<380\ \text{V}

The cable-run screen narrowly fails the stated terminal-voltage requirement.

Engineering Comment

Voltage at the motor terminals matters for torque, current, heating and drive tuning. A real check would include cable reactance, PWM waveform, reflected-wave insulation stress, temperature-adjusted conductor resistance, shield grounding, EMC, protective device coordination and measured voltage at the motor under load.

Plausibility Check

Nearly 100\ \text{A} through 0.12\ \Omega per phase creates a voltage drop of about 20\ \text{V}, which is a few percent of a 400\ \text{V} system. A borderline terminal-voltage failure is therefore plausible.

Exercise 12: Guarded Current-release Decision

A commissioning run for the same motor-drive package measures cycle RMS current:

I_{RMS,m}=121\ \text{A}

The derated continuous current limit in the installed enclosure is:

I_{limit}=130\ \text{A}

The measurement and duty-cycle reconstruction have expanded uncertainty:

U_I=5\ \text{A}

The enclosure-rating review applies an additional release allowance of:

A_I=3\ \text{A}

against the published limit. Use:

I_{guard}=I_{RMS,m}+U_I

and:

I_{release}=I_{limit}-A_I

Decide whether the drive can be released on current margin.

Solution

Guarded RMS current is:

I_{guard}=121+5=126\ \text{A}

Release current limit is:

I_{release}=130-3=127\ \text{A}

Guarded margin is:

M_I=I_{release}-I_{guard}
M_I=127-126=1\ \text{A}

The current-release screen passes, but only by:

1\ \text{A}

Engineering Comment

This is a thin release. The current screen passes only after the duty cycle, enclosure derating and measurement uncertainty are all stated explicitly. A release package should still include thermal trend, ambient condition, drive heat-sink temperature, overload history, current imbalance and confirmation that the tested duty cycle is the worst representative duty.

Plausibility Check

The unguarded margin is 9\ \text{A}, but uncertainty and allowance consume 8\ \text{A}. A remaining margin of 1\ \text{A} is plausible and should be treated as fragile.

Exercise 13: Long-Cable Reflected-Wave Insulation Stress

A VFD feeds a remote motor through a long cable. At the operating bus voltage, the drive DC link is estimated as:

V_{dc}=650\ \text{V}

The long cable and motor terminal mismatch give an effective reflection coefficient:

\Gamma_0=0.60

The PWM voltage rise time at the drive terminals is:

t_{r0}=0.12\ \mu\text{s}

The motor insulation system is limited to repetitive terminal peaks below:

V_{pk,lim}=1000\ \text{V}

The installation standard also limits repetitive voltage slew rate at the motor terminals to:

\displaystyle \left(\frac{dV}{dt}\right)_{lim}=3000\ \text{V}/\mu\text{s}

Use the screening estimates:

V_{pk}=V_{dc}(1+\Gamma)
\displaystyle \frac{dV}{dt}\approx\frac{V_{dc}}{t_r}

A proposed output filter reduces the effective reflection coefficient to:

\Gamma_f=0.35

and increases the rise time to:

t_{rf}=0.35\ \mu\text{s}

Check the unfiltered and filtered cases against both insulation peak voltage and dV/dt limits.

Solution

Unfiltered reflected-wave peak:

V_{pk,0}=650(1+0.60)=1040\ \text{V}

Peak-voltage margin:

M_{V,0}=1000-1040=-40\ \text{V}

Unfiltered slew rate:

\displaystyle \left(\frac{dV}{dt}\right)_0=\frac{650}{0.12}=5417\ \text{V}/\mu\text{s}

Slew-rate margin:

M_{dV/dt,0}=3000-5417=-2417\ \text{V}/\mu\text{s}

The unfiltered installation fails both the peak-voltage screen and the dV/dt screen.

Filtered reflected-wave peak:

V_{pk,f}=650(1+0.35)=878\ \text{V}

Filtered peak-voltage margin:

M_{V,f}=1000-878=122\ \text{V}

Filtered slew rate:

\displaystyle \left(\frac{dV}{dt}\right)_f=\frac{650}{0.35}=1857\ \text{V}/\mu\text{s}

Filtered slew-rate margin:

M_{dV/dt,f}=3000-1857=1143\ \text{V}/\mu\text{s}

The filtered case passes both simplified release screens.

Engineering Comment

The cable voltage-drop calculation in Exercise 11 does not protect the motor insulation. A drive can deliver acceptable RMS voltage while fast PWM edges create terminal overvoltage and high dV/dt stress at the motor. A real release should check cable length, cable surge impedance, pulse width, carrier frequency, motor impulse rating, filter rating, grounding, shield termination, bearing-current mitigation and measured terminal waveform.

Plausibility Check

The unfiltered peak is only slightly above the 1000\ \text{V} limit, but the slew rate is far above the 3000\ \text{V}/\mu\text{s} limit, so the failure is driven by both peak stress and edge speed. Increasing rise time from 0.12\ \mu\text{s} to 0.35\ \mu\text{s} should reduce dV/dt by about a factor of three, which matches the calculated drop from 5417 to 1857\ \text{V}/\mu\text{s}.

Exercise 14: Regenerative Braking Front-End Limit

A motor drive with an active front end decelerates a high-inertia machine from:

n_1=1500\ \text{rpm}

to:

n_2=300\ \text{rpm}

The reflected inertia at the motor shaft is:

J=32\ \text{kg m}^2

The commanded deceleration time is:

t_d=6.0\ \text{s}

Assume 88\% of the removed kinetic energy is returned through the line-side converter:

\eta_{regen}=0.88

The active front end is rated for:

P_{regen,short}=65\ \text{kW}

for the deceleration interval and:

P_{regen,cont}=18\ \text{kW}

over the repeated production sequence. The stop repeats every:

t_{cycle}=25\ \text{s}

Check the recovered energy, average regenerative power during deceleration, sequence-average regenerative power and whether a faster 15\ \text{s} cycle would pass the continuous line-side rating.

Solution

Initial angular speed:

\displaystyle \omega_1=\frac{2\pi(1500)}{60}=157\ \text{rad/s}

Final angular speed:

\displaystyle \omega_2=\frac{2\pi(300)}{60}=31.4\ \text{rad/s}

Removed kinetic energy:

\displaystyle \Delta E=\frac{1}{2}J(\omega_1^2-\omega_2^2)
\displaystyle \Delta E=\frac{1}{2}(32)(157^2-31.4^2)=379\ \text{kJ}

Energy returned through the regenerative path:

E_{regen}=0.88(379)=333.5\ \text{kJ}

Average regenerative power during the deceleration interval:

\displaystyle P_{regen,d}=\frac{333.5}{6.0}=55.6\ \text{kW}

Short-time margin:

M_{short}=65-55.6=9.4\ \text{kW}

The short-time regenerative power check passes.

Sequence-average regenerative power:

\displaystyle P_{regen,avg}=\frac{333.5}{25}=13.3\ \text{kW}

Continuous line-side margin:

M_{cont}=18-13.3=4.7\ \text{kW}

The 25\ \text{s} cycle passes the continuous regenerative rating.

For a 15\ \text{s} cycle:

\displaystyle P_{regen,avg,15}=\frac{333.5}{15}=22.2\ \text{kW}

Continuous margin at the faster cycle:

M_{cont,15}=18-22.2=-4.2\ \text{kW}

The faster cycle fails the continuous line-side regenerative rating even though the single deceleration event still passes the short-time limit.

Engineering Comment

Regeneration changes the braking question from “can the resistor absorb the energy?” to “where does the recovered energy go, and what rating limits that path?” The active front end, DC link, grid connection, protection settings, line reactor, harmonic filter and any shared DC bus must be rated for both the short deceleration power and the average production sequence. A commissioning test should log DC-link voltage, line-side power, stop time, current limits, grid alarms and thermal trend over the real cycle.

Plausibility Check

The speed reduction removes most of the stored kinetic energy because energy scales with speed squared. Returning 333.5\ \text{kJ} over 6\ \text{s} gives a short-time power near 56\ \text{kW}, below the 65\ \text{kW} limit. Repeating the same event every 25\ \text{s} averages to 13.3\ \text{kW}, but shortening the cycle to 15\ \text{s} raises the average above the 18\ \text{kW} continuous rating.

Exercise 15: Common-Mode Current and Bearing-Risk Screen

A VFD feeds a motor through a shielded output cable. The installed cable length is:

L=85\ \text{m}

The effective cable capacitance from the drive output conductors to ground is estimated as:

c'=0.18\ \text{nF/m}

The motor winding, frame and installation parasitics add:

C_m=4.7\ \text{nF}

During a switching transition, the relevant common-mode voltage step is approximated as:

\Delta V_{cm}=280\ \text{V}

with rise time:

t_r=0.16\ \mu\text{s}

The plant release criterion requires the screened common-mode current pulse scale, including measurement uncertainty, to stay below:

I_{cm,lim}=18\ \text{A}

The current-probe and reconstruction uncertainty is:

U_I=2.0\ \text{A}

A proposed drive-approved common-mode filter changes the effective capacitance to:

C_{cm,f}=23.0\ \text{nF}

and changes the switching edge to:

\Delta V_{cm,f}=190\ \text{V}
t_{r,f}=0.85\ \mu\text{s}

Use:

\displaystyle i_{cm}\approx C_{cm}\frac{\Delta V_{cm}}{t_r}

Check the unfiltered and filtered cases and decide whether bearing-current mitigation and grounding validation can be released.

Solution

Unfiltered common-mode capacitance is:

C_{cm,0}=c'L+C_m
C_{cm,0}=0.18(85)+4.7=20.0\ \text{nF}

Unfiltered edge rate is:

\displaystyle \frac{\Delta V_{cm}}{t_r}=\frac{280}{0.16}=1750\ \text{V}/\mu\text{s}

Because 1\ \text{nF}\cdot1\ \text{V}/\mu\text{s}=0.001\ \text{A}:

i_{cm,0}=20.0(1750)(0.001)=35.0\ \text{A}

Guarded unfiltered current is:

I_{guard,0}=i_{cm,0}+U_I=35.0+2.0=37.0\ \text{A}

Compare with the release limit:

37.0>18

The unfiltered installation fails the common-mode current screen.

For the filtered case, the edge rate is:

\displaystyle \frac{\Delta V_{cm,f}}{t_{r,f}}=\frac{190}{0.85}=224\ \text{V}/\mu\text{s}

Filtered common-mode current is:

i_{cm,f}=23.0(224)(0.001)=5.15\ \text{A}

Guarded filtered current is:

I_{guard,f}=5.15+2.0=7.15\ \text{A}

Filtered margin is:

M_I=18-7.15=10.9\ \text{A}

Current reduction is:

\displaystyle \text{reduction}=1-\frac{5.15}{35.0}=0.853=85.3\%

The filtered installation passes the simplified common-mode current release screen.

Engineering Comment

This calculation does not say that 35.0\ \text{A} flows through a bearing. It estimates the pulse scale driven by cable and motor capacitance during fast common-mode voltage transitions. The actual paths divide through the cable shield, motor frame, grounding conductor, filters, bearings, shaft grounding device and nearby structure.

The unfiltered result is still useful because it is too large to ignore. A release package should include the selected filter or common-mode choke, shield termination, shaft-grounding or insulated-bearing decision, ground-fault protection review, current-probe measurement, vibration or bearing-noise baseline and a controlled switching-frequency setting. After the filter, the guarded screen passes, but only for the documented cable length, grounding arrangement, drive firmware, carrier frequency and measurement method.

Plausibility Check

The filter increases capacitance slightly, but it reduces the edge rate from 1750 to about 224\ \text{V}/\mu\text{s}. Since displacement current scales directly with C dV/dt, slowing the edge dominates the capacitance increase and cuts the screened current by about 85\%. A large improvement from a common-mode or output filter is therefore plausible.

Exercise 16: Locked-Rotor Thermal Protection and Jam Trip Time

A conveyor motor-drive package can see a process jam while the drive is still enabled. The motor full-load current is:

I_{FLA}=96\ \text{A}

The motor manufacturer gives a simplified cold locked-rotor thermal withstand basis of:

I_{LR}=6.0I_{FLA}

for:

t_{LR}=10\ \text{s}

During a jam test, the VFD current limit holds the measured phase current near:

I_{jam}=410\ \text{A}

The configured jam-protection logic trips after:

t_{trip}=19.0\ \text{s}

The drive output-disable and contactor-opening delay is:

t_d=0.6\ \text{s}

The validation team applies timing uncertainty:

U_t=1.2\ \text{s}

Use an I^2t thermal exposure screen. Decide whether the installed jam trip can be released, then check a revised trip setting of:

t_{trip,r}=16.0\ \text{s}

Solution

Locked-rotor thermal exposure limit in full-load-current squared seconds is:

\displaystyle H_{lim}=\left(\frac{I_{LR}}{I_{FLA}}\right)^2t_{LR}
H_{lim}=(6.0)^2(10)=360\ \text{pu}^2\text{s}

Jam current in per unit of full-load current:

\displaystyle i_{jam}=\frac{I_{jam}}{I_{FLA}}
\displaystyle i_{jam}=\frac{410}{96}=4.27\ \text{pu}

Allowable jam duration from the same thermal exposure limit:

\displaystyle t_{allow}=\frac{H_{lim}}{i_{jam}^2}
\displaystyle t_{allow}=\frac{360}{4.27^2}=19.7\ \text{s}

Guarded installed clearing time:

t_{guard}=t_{trip}+t_d+U_t
t_{guard}=19.0+0.6+1.2=20.8\ \text{s}

Timing margin:

M_t=t_{allow}-t_{guard}
M_t=19.7-20.8=-1.1\ \text{s}

Thermal utilization of the installed setting:

\displaystyle U_H=\frac{t_{guard}}{t_{allow}}
\displaystyle U_H=\frac{20.8}{19.7}=1.06

The installed jam trip fails the guarded thermal exposure screen.

For the revised setting:

t_{guard,r}=16.0+0.6+1.2=17.8\ \text{s}

Revised timing margin:

M_{t,r}=19.7-17.8=1.9\ \text{s}

Revised thermal utilization:

\displaystyle U_{H,r}=\frac{17.8}{19.7}=0.904

The revised setting passes the simplified guarded screen, provided the shorter trip does not create nuisance trips during normal acceleration or short overload.

Engineering Comment

Locked-rotor and process-jam protection is not the same as normal overload sizing. A current limit can reduce stall current below nameplate locked-rotor current, but it also lets the drive keep feeding the motor for longer. The release decision must include the motor thermal withstand basis, VFD current limit, trip logic, contactor or safe-torque-off response, restart inhibit and the worst credible jam load.

The revised trip setting is acceptable only after acceleration logs, normal transient load peaks and jam-test traces prove separation between real jams and healthy starts. Validation evidence should include current trend, speed feedback or tachometer evidence, drive fault timestamp, output-disable timestamp, motor temperature trend, reset workflow and operator handover.

Plausibility Check

The jam current is about 4.3 times full-load current, so every second of jam adds about 18 full-load-current squared seconds of thermal exposure. A 20.8\ \text{s} guarded clearing time slightly exceeds the 19.7\ \text{s} allowable time, while shortening the logic delay by 3\ \text{s} creates about 1.9\ \text{s} of guarded margin. That scale is plausible for a borderline protection setting.

Exercise 17: Winding Temperature Rise and Insulation-Class Gate

A motor-drive package finishes a loaded commissioning run. The stator winding resistance was measured cold as:

R_c=0.420\ \Omega

at:

T_c=24^\circ\text{C}

Immediately after the nominal-load test, the measured hot resistance is:

R_h=0.585\ \Omega

Use the copper resistance-temperature relation:

\displaystyle T_h=\frac{R_h}{R_c}(234.5+T_c)-234.5

The motor uses an insulation system with absolute winding-temperature limit:

T_{lim}=155^\circ\text{C}

The release rule adds:

T_{hotspot}=10^\circ\text{C}

for hot-spot allowance and:

U_T=4^\circ\text{C}

for measurement and cooldown uncertainty. Check the guarded winding-temperature margin. Then evaluate a service-factor run with:

R_{h,SF}=0.645\ \Omega

Solution

Nominal-load average winding temperature from the resistance method is:

\displaystyle T_h=\frac{0.585}{0.420}(234.5+24)-234.5
T_h=125.5^\circ\text{C}

Temperature rise above the cold ambient measurement is:

\Delta T_h=125.5-24=101.5^\circ\text{C}

The guarded winding temperature is:

T_{guard}=T_h+T_{hotspot}+U_T
T_{guard}=125.5+10+4=139.5^\circ\text{C}

Insulation-class margin is:

M_T=T_{lim}-T_{guard}
M_T=155-139.5=15.5^\circ\text{C}

The nominal-load test passes this simplified insulation-temperature screen.

For the service-factor run:

\displaystyle T_{h,SF}=\frac{0.645}{0.420}(234.5+24)-234.5
T_{h,SF}=162.5^\circ\text{C}

Guarded service-factor temperature:

T_{guard,SF}=162.5+10+4=176.5^\circ\text{C}

Service-factor margin:

M_{T,SF}=155-176.5=-21.5^\circ\text{C}

The service-factor run fails the guarded insulation-temperature gate. The drive should not be released for that overload duty without derating, improving cooling, changing the duty cycle, selecting a different motor or obtaining manufacturer-backed thermal evidence.

Engineering Comment

Current screening is not the same as winding-temperature validation. A motor can pass a current or torque calculation while its thermal path, ambient condition, enclosure, fan speed, harmonic loss or duty cycle pushes the winding insulation beyond its release limit. The resistance method estimates average winding temperature, so the hot-spot allowance matters.

Release evidence should include cold and hot resistance measurement timing, ambient temperature, cooldown correction if needed, load state, drive waveform, switching frequency, motor cooling condition, enclosure airflow, RTD or thermistor trends if available, insulation class, service factor, restart limits and the duty cycle being authorized. A service-factor nameplate value should not be treated as continuous release permission without thermal evidence.

Plausibility Check

The hot resistance at nominal load is about 39\% higher than the cold resistance, which is consistent with a winding temperature near 125^\circ\text{C}. Adding a 14^\circ\text{C} combined hot-spot and uncertainty allowance still leaves about 15^\circ\text{C} margin to a 155^\circ\text{C} limit. The service-factor resistance is much higher, so the guarded failure is expected rather than a rounding artifact.

Exercise 18: Encoder Feedback Loss and Coast-Down Release Gate

A VFD-controlled conveyor motor uses a 1024 pulse per revolution quadrature encoder. The controller counts all four A/B edges, so the effective count is:

N_e=4096\ \text{edges/rev}

The commanded operating speed is:

n_{cmd}=1450\ \text{rpm}

The controller input has a reliable edge-rate limit of:

f_{e,lim}=110\ \text{kHz}

During commissioning, an independent tachometer reads:

n_{tach}=1428\ \text{rpm}

while the encoder-derived speed reports:

n_{enc}=1472\ \text{rpm}

The release rule limits guarded speed disagreement to:

D_{lim}=2.0\%

and applies a speed-reference uncertainty allowance of:

U_D=0.4\ \text{percentage points}

The drive declares feedback loss if no valid encoder edge is received for:

t_{loss}=45\ \text{ms}

After the feedback fault, safe torque off is confirmed after:

t_{STO}=80\ \text{ms}

and the mechanical brake reaches effective torque after:

t_b=250\ \text{ms}

Timing uncertainty is:

U_t=40\ \text{ms}

Once the brake is effective, measured deceleration is approximately:

\alpha_b=180\ \text{rad/s}^2

The release limit for post-fault shaft travel is:

N_{allow}=18\ \text{rev}

Check the encoder edge rate, speed-plausibility gate and guarded post-fault travel. Then evaluate a revised configuration with:

n_{enc,r}=1438\ \text{rpm}
t_{loss,r}=20\ \text{ms}
t_{STO,r}=50\ \text{ms}
t_{b,r}=180\ \text{ms}

Solution

The commanded encoder edge rate is:

\displaystyle f_e=\frac{n_{cmd}N_e}{60}
\displaystyle f_e=\frac{1450(4096)}{60}=98{,}987\ \text{Hz}=99.0\ \text{kHz}

Input utilization is:

\displaystyle U_f=\frac{f_e}{f_{e,lim}}
\displaystyle U_f=\frac{99.0}{110}=0.900

Edge-rate margin is:

M_f=110-99.0=11.0\ \text{kHz}

The input edge-rate screen passes, but it does not prove that the feedback value is correct.

Speed disagreement against the independent tachometer is:

\displaystyle D=\frac{|n_{enc}-n_{tach}|}{n_{tach}}
\displaystyle D=\frac{|1472-1428|}{1428}=0.0308=3.08\%

Guarded disagreement is:

D_g=D+U_D
D_g=3.08+0.4=3.48\%

Speed-plausibility margin is:

M_D=D_{lim}-D_g
M_D=2.0-3.48=-1.48\ \text{percentage points}

The installed encoder speed fails the guarded plausibility gate.

Guarded fault-response time before the brake is effective is:

t_g=t_{loss}+t_{STO}+t_b+U_t
t_g=0.045+0.080+0.250+0.040=0.415\ \text{s}

The shaft speed in revolutions per second is:

\displaystyle n_{rps}=\frac{1428}{60}=23.8\ \text{rev/s}

Post-fault travel before effective braking is:

N_d=n_{rps}t_g
N_d=23.8(0.415)=9.88\ \text{rev}

Angular speed at fault is:

\omega=2\pi n_{rps}
\omega=2\pi(23.8)=149.5\ \text{rad/s}

Brake stopping angle is:

\displaystyle \theta_b=\frac{\omega^2}{2\alpha_b}
\displaystyle \theta_b=\frac{149.5^2}{2(180)}=62.1\ \text{rad}

Convert brake stopping angle to revolutions:

\displaystyle N_b=\frac{62.1}{2\pi}=9.89\ \text{rev}

Total guarded post-fault travel is:

N_{tot}=N_d+N_b
N_{tot}=9.88+9.89=19.77\ \text{rev}

Travel margin is:

M_N=N_{allow}-N_{tot}
M_N=18.0-19.77=-1.77\ \text{rev}

The installed feedback-loss response fails the guarded post-fault travel gate.

For the revised encoder reading:

\displaystyle D_r=\frac{|1438-1428|}{1428}=0.700\%
D_{g,r}=0.700+0.4=1.10\%

The revised speed-plausibility gate passes:

M_{D,r}=2.0-1.10=0.90\ \text{percentage points}

Revised guarded fault-response time is:

t_{g,r}=0.020+0.050+0.180+0.040=0.290\ \text{s}

Revised pre-brake travel is:

N_{d,r}=23.8(0.290)=6.90\ \text{rev}

The same brake deceleration gives:

N_{tot,r}=6.90+9.89=16.79\ \text{rev}

Revised travel margin is:

M_{N,r}=18.0-16.79=1.21\ \text{rev}

The revised configuration passes the simplified speed-plausibility and post-fault travel gates, provided fault injection confirms the timing and the brake torque is repeatable at the worst load and temperature.

Engineering Comment

Encoder feedback is both a control input and a release risk. A high enough edge-rate margin only proves the input channel can count the expected pulses; it does not prove correct pulses-per-revolution configuration, direction decoding, shielding, target alignment, missed-edge handling or agreement with the physical shaft speed.

Safe torque off removes drive torque, but it does not stop the rotating system instantly. Post-fault travel depends on detection delay, logic delay, STO confirmation, brake delay, mechanical inertia, load torque, brake torque, speed and uncertainty. A defensible release should include encoder configuration records, tachometer comparison, edge-rate calculation, dropout fault injection, STO status timestamp, brake response trace, stop-distance evidence and a reset workflow that prevents automatic restart after feedback loss.

Plausibility Check

A 1024 pulse quadrature encoder counted on all edges produces about 99\ \text{kHz} at 1450\ \text{rpm}, which is close enough to the 110\ \text{kHz} limit to deserve a margin check. At roughly 24\ \text{rev/s}, a few hundred milliseconds of feedback-fault response easily creates several revolutions of travel before braking becomes effective. That explains why the original timing fails even though the brake deceleration itself is strong.

Exercise 19: VFD Skip-Frequency Resonance Release Gate

A VFD-driven pump showed a repeatable vibration peak during a commissioning run-up. A bump test and order-tracking review identify the relevant support mode as:

f_n=22.8\ \text{Hz}

The dominant excitation is the 1\times shaft-speed order. The release rule requires the operating excitation to stay outside the measured mode by:

M_f=1.50\ \text{Hz}

after allowing frequency uncertainty:

U_f=0.25\ \text{Hz}

The installed VFD skip-frequency band blocks continuous operation only from:

n_{skip,0}=1320\ \text{rpm}\ \text{to}\ 1410\ \text{rpm}

The process recipe requests continuous operation at:

n_{op}=1440\ \text{rpm}

A revised recipe can operate at:

n_{op,r}=1515\ \text{rpm}

if the drive output frequency remains below:

f_{drive,lim}=55\ \text{Hz}

Use a four-pole induction motor with estimated loaded slip:

s=0.020

Finally, check a revised VFD skip band of:

n_{skip,r}=1260\ \text{rpm}\ \text{to}\ 1480\ \text{rpm}

and a skip-band crossing ramp rate of:

r_n=220\ \text{rpm/s}

against a maximum allowed dwell time through the avoided region of:

t_{dwell,lim}=1.5\ \text{s}

Solution

The forbidden excitation-frequency band is the measured mode plus uncertainty and required separation:

f_{low}=f_n-U_f-M_f
f_{low}=22.8-0.25-1.50=21.05\ \text{Hz}
f_{high}=f_n+U_f+M_f
f_{high}=22.8+0.25+1.50=24.55\ \text{Hz}

Because the excitation is 1\times shaft speed:

n=60f

So the forbidden shaft-speed band is:

n_{low}=60(21.05)=1263\ \text{rpm}
n_{high}=60(24.55)=1473\ \text{rpm}

The installed skip band is:

1320\ \text{rpm}\ \text{to}\ 1410\ \text{rpm}

It does not cover the full forbidden band. The uncovered lower portion is:

1320-1263=57\ \text{rpm}

and the uncovered upper portion is:

1473-1410=63\ \text{rpm}

The requested continuous speed is:

n_{op}=1440\ \text{rpm}

This lies inside the forbidden band:

1263<1440<1473

Its distance from the high edge of the forbidden region is only:

1473-1440=33\ \text{rpm}

The installed recipe should not be released for continuous operation at 1440\ \text{rpm}.

For the revised operating speed:

n_{op,r}=1515\ \text{rpm}

The margin above the forbidden band is:

M_n=1515-1473=42\ \text{rpm}

Check the drive output frequency. For a four-pole motor:

\displaystyle n_s=\frac{120f_{drive}}{P}

and loaded speed is approximated by:

n=(1-s)n_s

Therefore:

\displaystyle f_{drive}=\frac{Pn}{120(1-s)}
\displaystyle f_{drive}=\frac{4(1515)}{120(0.98)}=51.5\ \text{Hz}

Drive-frequency margin is:

M_{drive}=55-51.5=3.5\ \text{Hz}

The revised operating point passes the drive-frequency limit and sits above the avoided speed region.

The revised skip band is:

1260\ \text{rpm}\ \text{to}\ 1480\ \text{rpm}

It covers the calculated forbidden band:

1260<1263\quad \text{and}\quad 1480>1473

with small coverage margins:

M_{low}=1263-1260=3\ \text{rpm}
M_{high}=1480-1473=7\ \text{rpm}

Time to cross the forbidden band at the specified ramp rate is:

\displaystyle t_{dwell}=\frac{1473-1263}{220}
t_{dwell}=0.955\ \text{s}

The crossing dwell check passes:

0.955<1.5

The revised release is acceptable only if the VFD parameter set enforces the full skip band, the process does not command continuous speed inside that band and the run-up data confirms no excessive vibration during the brief crossing.

Engineering Comment

A VFD can make a resonance problem worse because it allows continuous operation at speeds that a fixed-speed motor might only pass through during start-up. The key release question is therefore not only whether the motor can produce torque at the requested speed. It is whether the requested continuous speed, skip-frequency parameters, acceleration ramp, process control loop and vibration evidence all avoid the measured mechanical mode.

Skip-frequency settings should be treated as controlled safety and reliability parameters. A release package should include the mode-identification evidence, tachometer or encoder speed reference, vibration sensor location, operating load, VFD parameter export, operator speed limits, process recipe limits, alarm thresholds, post-change run-up trend and maintenance instruction that prevents later removal of the skip band.

Plausibility Check

A 22.8\ \text{Hz} 1\times resonance corresponds to about 1368\ \text{rpm} before adding guard bands. After uncertainty and separation are included, the avoided range grows to roughly 1260 to 1470\ \text{rpm}. The original 1440\ \text{rpm} recipe is therefore too close to the measured mode, while a 1515\ \text{rpm} recipe and a wider skip band are plausible mitigations.

Review Table

CheckResultInterpretation
Shaft torque360\ \text{N m}Steady process torque, not acceleration or breakaway torque.
Estimated line current99.3\ \text{A}Supply equipment sees apparent power and RMS current.
Loaded slip3.0\%Plausible for normal induction-motor load.
Starting voltage minimum0.936\ \text{pu}Simple screen stays above 0.90\ \text{pu}.
Acceleration torque margin26.8\%Short-time torque passes but still needs current validation.
VFD setpoint24.7\ \text{Hz}, 198\ \text{V}Speed depends on slip and low-speed cooling.
Duty RMS current121\ \text{A}Continuous screen passes installed derating.
Braking energy190\ \text{kJ} per stopBelow resistor single-stop and average-power ratings.
Fan power reduction48.7\%Affinity-law savings require airflow validation.
Commissioning current increase24.0\%Blocks release until load or parameter cause is resolved.
Low-speed torque deratingutilization 1.17Continuous low-speed duty fails without cooling change.
Motor cable terminal voltage379.4\ \text{V}Narrowly fails the 380\ \text{V} terminal limit.
Guarded RMS current126\ \text{A} versus 127\ \text{A} release limitPasses with only 1\ \text{A} margin.
Reflected-wave insulation stressunfiltered 1040\ \text{V} and 5417\ \text{V}/\mu\text{s}; filtered 878\ \text{V} and 1857\ \text{V}/\mu\text{s}Unfiltered case fails, filtered case passes the simplified screen.
Regenerative braking front end55.6\ \text{kW} short-time and 13.3\ \text{kW} average at 25\ \text{s} cyclePasses the stated limits, but a 15\ \text{s} cycle fails continuous regeneration.
Common-mode current screenunfiltered guarded 37.0\ \text{A}, filtered guarded 7.15\ \text{A}Unfiltered case fails bearing-risk release; filtered case passes with measurement evidence.
Locked-rotor jam protectionallowable 19.7\ \text{s}, installed guarded 20.8\ \text{s}, revised guarded 17.8\ \text{s}Installed trip fails; revised setting passes with guarded evidence.
Winding temperaturenominal guarded 139.5^\circ\text{C}, service-factor guarded 176.5^\circ\text{C}Nominal load passes insulation screen; service-factor run fails.
Encoder feedback faultedge-rate utilization 0.900, installed speed disagreement 3.48\% guarded, installed post-fault travel 19.77\ \text{rev}, revised travel 16.79\ \text{rev}Edge-rate passes, but installed speed plausibility and coast-down travel fail until feedback and timing are revised.
VFD resonance skip bandforbidden 1263 to 1473\ \text{rpm}, original recipe 1440\ \text{rpm}, revised recipe 1515\ \text{rpm}Original continuous speed and skip band fail; revised skip band and operating speed pass with narrow parameter margins.

Review Checklist

Before accepting a motor-drive calculation, check:

  • whether the boundary is shaft output, motor terminals, drive output, drive input or supply bus;
  • whether the load case is continuous, accelerating, starting, braking, low-speed or degraded duty;
  • whether torque, current, apparent power, voltage and thermal duty are all inside their relevant ratings;
  • whether low-speed operation includes cooling derating or external blower evidence;
  • whether cable length, voltage drop, reflected-wave stress and grounding are compatible with the drive;
  • whether long-cable PWM installations check both terminal peak voltage and dV/dt insulation stress;
  • whether common-mode current, bearing-current mitigation and ground-fault interaction are checked for long VFD cables;
  • whether RMS duty current and peak overload are both checked;
  • whether winding temperature, hot-spot allowance and insulation class are checked when thermal release depends on loaded operation;
  • whether braking energy is checked per stop and over the repeated sequence;
  • whether regenerative braking has a rated energy path through the active front end, DC bus, resistor, storage or grid connection;
  • whether locked-rotor or jam protection checks current limit, trip delay, output-disable delay, thermal exposure and restart inhibit;
  • whether encoder speed feedback has edge-rate margin, correct pulses-per-revolution configuration, independent speed agreement and a tested fault response;
  • whether safe torque off, brake delay and coast-down travel are checked before accepting a feedback-loss or overspeed response;
  • whether VFD speed commands, skip-frequency bands, acceleration ramps and operator recipes avoid measured mechanical resonance modes;
  • whether commissioning data includes current, voltage, speed, power, temperature and process load at the same operating point;
  • whether guarded release includes measurement uncertainty, enclosure derating and the worst representative duty cycle.

Common Mistakes

  • Selecting the motor or drive from rated kilowatts alone while torque, current, overload duration, thermal duty and starting or braking cases control the application.
  • Mixing shaft power, electrical input power, apparent power and line current when checking feeders, protection, drives and motor heating.
  • Treating low-speed VFD operation as thermally equivalent to rated-speed operation without external cooling or a derating curve.
  • Checking a single acceleration event but ignoring repeated starts, duty RMS current, DC-link heating, motor temperature and process jam cases.
  • Accepting a current pass without winding-temperature evidence, hot-spot allowance, ambient context and insulation-class margin.
  • Assuming regeneration is acceptable because kinetic energy is recoverable, while the active front end, resistor, shared DC bus or grid connection has a lower rating.
  • Ignoring cable length until reflected-wave voltage, terminal voltage, common-mode current, bearing current or ground-fault interaction appears during commissioning.
  • Accepting commissioning current without confirming process load, voltage, speed, parameter set, slip estimate, power factor and measurement uncertainty.
  • Using affinity-law savings for fans or pumps without checking minimum speed, airflow or flow validation, resonance, control stability and process limits.
  • Leaving a measured resonance inside the allowed VFD speed range because the motor current, torque and temperature screens pass.
  • Setting protection from nominal current only while overload, locked-rotor thermal exposure, jam trip timing, ground fault, insulation stress, interlock and safe-torque-off behavior remain unverified.
  • Trusting encoder feedback because pulses are present while edge-rate margin, pulses-per-revolution configuration, tachometer agreement, missed-pulse handling and post-fault travel remain unchecked.
  • Treating a passed calculation as a release when thermal trend, vibration, bearing noise, cable shield termination, firmware settings and operator handover are still undocumented.
REF

See also