Star-delta starting
About this app
Motor with a heavy inertial flywheel (flying shear): direct start vs. Y start compared — I(t) and n(t) curves live, the winding schematic morphs between star (230 V, ⅓ current) and delta (400 V). Switch manually or via a timer relay; switching too early reveals the current surge.
Why star-delta starting?
A three-phase motor started direct-on-line draws a multiple of its rated current, up to 8 × Iₙ in the app. With a heavy flywheel the run-up takes a long time and loads the supply, fuses and cables. In star-delta starting the motor first runs in star connection: each winding sees 230 V instead of 400 V (factor 1/√3). Line current drops to ⅓ overall, so 8 × Iₙ becomes about 2.7 × Iₙ. Since torque scales with U², it also drops to ⅓, and the run-up takes longer.
What the app shows
A motor drives a heavy flywheel, as in a mechanical shear. You choose between direct-on-line, manual and automatic star-delta starting. The curves I(t) and n(t) are drawn live, with the direct start kept as a reference. The circuit panel shows the winding diagram switching between star (star point connected) and delta (windings in a ring), along with phase voltage, current ratio and torque. A status line describes the phases: star start, changeover gap (K3 opens, K2 closes) and delta operation.
Test the changeover moment
In manual mode you press Y→Δ yourself. Try switching early and late: if you change over while the motor is still slow, current jumps almost to the direct-start value, and the app reports the surge as a multiple of Iₙ. Close to rated speed only a small residual surge remains. In automatic mode a timer relay takes over, adjustable from 3 to 15 s. After switch-off the heavy flywheel coasts for a long time. Soft starters and variable frequency drives are modern alternatives.
Frequently asked questions
- Why is the starting current in star connection one third?
- In star, each winding sees 230 V instead of 400 V, so phase current drops by a factor of 1/√3. In delta, line current is also √3 times the phase current, while in star both are equal. Together this gives 1/√3 · 1/√3 = ⅓ of the line current of a direct start in delta.
- What happens if a star-delta starter switches over too early?
- If the motor has not yet reached high speed at changeover, it draws almost the full starting current in delta, as with a direct start. The benefit of the star start is lost. That is why changeover happens near rated speed, usually via a timer relay.
- When is star-delta starting suitable?
- It suits motors that run in delta at 400 V and start against a low load torque, because starting torque drops to ⅓. For starts under heavy load or with precise speed control, soft starters or variable frequency drives are often used today.
- What can I adjust in the star-delta app?
- You choose direct, manual or automatic starting and set the timer relay between 3 and 15 s. In manual mode you decide the changeover moment yourself and see the resulting current surge in the curves.
Subject: Electrical & Control Engineering
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