Starting a 90 kW conveyor on a 250 kVA transformer
A plant is fed by a dedicated 250 kVA, 415 V, 50 Hz transformer of 4.5% impedance. A new 90 kW, 415 V, 4-pole induction motor is to be added on the same board; its nameplate gives 158 A full-load current, 6.0 pu locked-rotor current, 1.9 pu locked-rotor torque and 0.89 full-load power factor. It drives a belt conveyor whose breakaway torque was measured at 0.70 pu of the motor's rated torque. The utility contract limits the voltage dip at the point of common coupling to 8%. A PLC on the same board drops out if its supply falls below 85% of nominal for more than 100 ms. A star–delta starter is already in stores; a soft starter with an adjustable current limit costs extra and needs a two-week lead time; the mechanical installation is already behind schedule. The existing load on the transformer is about 60 kVA at 0.85 power factor.
1. Compute the point-of-common-coupling voltage dip for direct-on-line and star–delta starting, and determine whether the star–delta starter in stores can actually break the conveyor away.
Read the worked answer
Short-circuit capacity at the PCC.
I_sc = S_rated / (√3 · V · z_pu) = 250,000 / (√3 × 415 × 0.045) ≈ 7.73 kA
Direct-on-line.
I_start = 6.0 × 158 = 948 A
Dip ≈ I_start / I_sc = 948 / 7727 = 12.3%
The contract limit is 8%, so DOL is non-compliant. Note that the residual 87.7% is still above the PLC's 85% threshold, so the PLC would probably ride through - the binding constraint is the utility contract, not the PLC.
Star–delta.
Phase current and torque both fall by a factor of 3 (the winding voltage is 1/√3, and torque goes as V²):
- I_start = 948 / 3 = 316 A → dip = 316 / 7727 = 4.1% ✓ (well inside 8%)
- T_start = 1.9 / 3 = 0.633 pu ✓ dip, but ✗ torque
At 4.1% dip the motor terminal voltage is 95.9%, so the available breakaway torque is 0.633 × 0.959² ≈ 0.58 pu, against a required 0.70 pu. The conveyor will stall or, worse, sit in a stalled condition drawing locked-rotor current until the overload trips. The star–delta starter in stores is not usable, and this is the kind of judgement that must be made before installation, not after.
2. Size a soft-starter current limit that satisfies both the 8% dip limit and the 0.70 pu breakaway torque, and state whether the resulting margin is adequate.
Read the worked answer
Current allowed by the dip limit.
I_max = 0.08 × 7727 = 618 A = 3.91 pu
Torque available at a given current limit. For a soft starter, reducing the voltage reduces current and torque together, and around locked rotor T ∝ I² (equivalently T ∝ V²):
T/T_LR = (I/I_LR)² ⇒ T_pu = 1.9 × (I_limit/6.0)²
At I_limit = 3.9 pu: T = 1.9 × (0.65)² = 0.803 pu
That exceeds the 0.70 pu requirement - but only by 15%. Applying the usual 1.15–1.2 safety factor for voltage variation, belt tension drift and temperature gives a required 0.81–0.84 pu, which is right at or slightly beyond the capability. The design is marginal, not comfortable.
Recommended setting. 3.8 pu (600 A): dip = 7.8% ✓, torque = 1.9 × (0.633)² = 0.762 pu ✓ with a 9% margin over breakaway. Set the ramp to reach full voltage within about 10 s after breakaway, and confirm the motor's thermal duty: a 10 s start at 3.8 pu is roughly 10 × 3.8² ≈ 144 pu·s of I²t, which is inside the thermal withstand of a 90 kW machine but should be checked against the manufacturer's hot-start curve.
Judgement. The margin is thin enough that the recommendation should be conditional: verify the measured 0.70 pu breakaway torque after final belt tensioning, and consider starting the conveyor empty. If the torque has to rise further, a variable-frequency drive is the correct answer - it delivers rated torque at 1.5–2 pu current and removes the trade-off entirely.
3. The project manager wants to install the star–delta starter and 'fix it later if the conveyor stalls'. List the specific field measurements and calculations you would require before energising, and what each would change.
Read the worked answer
1. Actual transformer impedance, not the nameplate value. Impedance tolerances of ±10% are normal, and a 5.0% transformer instead of 4.5% reduces I_sc from 7.73 kA to 6.96 kA, raising the DOL dip to 13.6% and the 3.8 pu dip to 8.6% - non-compliant. A short-circuit test (or the works test certificate) settles it.
2. Measured breakaway torque of the loaded conveyor. The 0.70 pu figure is the single number the whole decision rests on. Measure it with a torque transducer on the drive shaft, or start the conveyor unloaded with a VFD on hire and record the torque at the moment of breakaway. If it is really 0.45 pu, star–delta works and the soft starter is unnecessary; if it is 0.85 pu, nothing short of a VFD works.
3. Actual PCC dip during the first start. Log the voltage at the PCC with a power-quality analyser, not at the motor terminals - the 8% limit is contractual and applies at the PCC.
4. PLC ride-through. Confirm that the 85%/100 ms threshold is real, and add a small UPS or a wide-input 24 V supply with hold-up so the PLC is not the weak link if the transformer is later loaded more heavily.
5. Protection coordination. Check that the motor's overload relay is set for the soft-start duty (not the DOL curve) and that the upstream breaker does not trip on the 3.8 pu current for 10 s.
On the manager's proposal. Star–delta is a fixed 1/3 torque reduction that no setting can change, so 'fix it later' means replacing the starter after a failed start - two weeks of lead time anyway, plus a stalled motor on a loaded conveyor. The decision should be made now, not deferred.