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Worksheet: star-delta starter

Name:Class:Date:

Material: the free star-delta starter simulator at examrobotic.com (examrobotic.com/en/tools/star-delta-starter-simulator). Time: about 40 minutes.

Aim: check why starting in star cuts the current to a third, choose the changeover time, see when this starter does not work and understand the interlock between KM2 and KM3.

1. Motor data

  1. The simulator's motor is 11 kW and 21 A in delta at 400 V. What voltages must its nameplate show for it to be started star-delta on a 400 V supply?
  2. How much current would it draw direct on line? (see the peak boxes) ________ A, which is ________ times In.

2. Starting a fan

  1. Load «fan», timer at 6 s. Press Start and write down:
    Peak in star: ________ A    Peak at the change to delta: ________ A
  2. What fraction of the direct-on-line start is the star peak? Why that fraction?
  3. Repeat with the timer at 1 s. What happens to the peak at the change to delta? Explain it.
  4. Repeat with 15 s. Does the motor gain anything? How is the time set on a real installation?

3. With a load and without the interlock

  1. Load «loaded conveyor». Does the motor start in star? Why? Which starter would you use instead?
  2. Untick the electrical interlock and start the fan. Look at the millisecond chart. What happens at the changeover and why?
  3. Overload relay F1 is just after KM1. What current would you set it to? ________ A. Why not 21 A?

4. Find the fault

Press «Surprise fault» three times and write down what the starter does and what the fault was each time.

Answers (for the teacher)

  1. 400/690 V Δ/Y: it must run in delta at 400 V. A 230/400 V motor will not do.
  2. Just over 6 times In (about 130 A in the simulator).
  3. With the fan and 6 s, the star peak is about a third of direct on line and the changeover peak is moderate.
  4. A third: each winding gets √3 times less voltage (230 V) and the line current in star is √3 times smaller than in delta. √3 × √3 = 3. Torque also drops to a third.
  5. The changeover peak is close to a direct-on-line start: the motor has not accelerated yet.
  6. Nothing: once it stops accelerating, staying in star gives nothing and, with a load, it overheats. It is set with a clamp meter: when the current stops falling, it is time to change over.
  7. No: in star it gives a third of the torque and cannot cope with the 60 % the conveyor needs. Soft starter or variable speed drive.
  8. Short circuit and Q1 trips: KM3 closes while KM2 is still conducting through its arc.
  9. To 0.58 × In ≈ 12.2 A, because in that position it carries the winding current (In/√3), not the line current.