Star-Delta Starting Judgement for a Compressor Motor
A garment factory's compressed-air supply is driven by a 415 V, 50 Hz, 11 kW, 4-pole squirrel-cage motor running at 1440 rpm at full load, with efficiency 88% and power factor 0.84. The nameplate gives Ist/In = 6.5, Tst/Tfl = 2.2 and Tmax/Tfl = 2.4. The motor winding is delta-connected for 415 V and is at present started direct-on-line from a 63 kVA, 4.2%-impedance transformer that also feeds the factory lighting; every start produces a visible flicker and the office UPS drops out on roughly one start in five. The compressor is a reciprocating type with a mechanical unloader valve that can be arranged to close only after the motor has run up. A star-delta starter is already in the store, and the works manager wants to use it without changing the motor, the compressor or the transformer. The motor drives the compressor through a V-belt with a small overload margin.
1. Calculate the motor full-load current, the DOL starting current, the full-load torque, the starting torque and the slip.
Read the worked answer
In = P/(√3 × V × η × cosφ) = 11,000/(1.732 × 415 × 0.88 × 0.84) = 20.7 A. DOL starting current Ist = 6.5 × 20.7 = 135 A. Synchronous speed Ns = 120f/p = 120 × 50/4 = 1500 rpm; slip s = (1500 − 1440)/1500 = 0.04, i.e. 4%. Full-load torque Tfl = P/ω = 11,000/(2π × 1440/60) = 11,000/150.8 = 73 N·m. DOL starting torque = 2.2 × 73 = 161 N·m, and pull-out torque = 2.4 × 73 = 175 N·m. So the motor has ample starting torque when started direct-on-line; the problem is the 135 A current drawn from a 63 kVA transformer. Also note the pull-out margin: Tmax is only about 2.4 × Tfl, so any voltage dip during running sharply reduces the motor's ability to ride through a load surge, since torque varies as V².
2. Calculate the line current and starting torque with a star-delta starter on the same motor, and judge whether it can start this compressor if the compressor needs 60% of full-load torque.
Read the worked answer
In star-delta starting the winding sees 1/√3 of the line voltage, so the phase current falls to one-third and the line current also falls to one-third of the DOL value: Ist(star-delta) = 135/3 = 45 A. Torque falls with the square of the winding voltage, i.e. to one-third: Tst(star-delta) = 161/3 ≈ 54 N·m = 73% of Tfl. The compressor needs 60% of Tfl = 0.60 × 73 = 44 N·m. Since 54 N·m > 44 N·m, the starter will accelerate the compressor, but the margin is only about 22%. Judgement: use star-delta only together with the unloader, so the motor starts against no compression load; if a start occurs with the receiver already pressurised, the torque margin disappears and the motor will stall in star and never reach changeover. Set the overload relay according to where it is connected - to In if it is in the line, or to In/√3 ≈ 12 A if it is in the delta leg - and set the star-to-delta timer long enough for the motor to reach near full speed.
3. Compare the supply voltage dip for DOL and star-delta starting and state your final recommendation to the works manager.
Read the worked answer
Starting kVA at DOL = √3 × 415 × 135 = 97 kVA. Voltage dip ≈ (97/63) × 4.2% = 6.5%, which is enough to make lamps flicker and drop the UPS. With star-delta: starting kVA = √3 × 415 × 45 = 32 kVA, giving dip ≈ (32/63) × 4.2% = 2.2%, which the lighting and UPS will tolerate. Torque check under the reduced voltage: at 6.5% dip the DOL torque would be 0.87 × 161 = 140 N·m; at 2.2% dip the star-delta torque is 0.957 × 54 = 51 N·m, still above the 44 N·m required. Recommendation: fit the star-delta starter, interlock it with the unloader valve so the compressor always starts unloaded, verify that the changeover to delta happens after the motor is near full speed, and retest the flicker during a start. If heavy-load starting is unavoidable, a soft starter or a larger transformer is the correct answer - a star-delta starter cannot be used on a motor that is star-connected for 415 V, so confirm the winding is delta-rated before installation.