Working Principle of Star-Delta Starter

Digital ElectronicsElectronElectronics & Electrical

Circuit Diagram and Working Principle of Star-Delta Starter

The figure shows the connection diagram of a 3-phase induction motor with a star-delta starter. The star-delta starter is a very common type of starter and is extensively used for starting the squirrel cage induction motors. It is used for starting a squirrel cage induction motor which is designed to run normally on delta connected stator winding.

When the switch S is in the START position, the stator windings are connected in star. When the motor attains a speed about 80 % of rated speed, then the changeover switch S is thrown to the RUN position which connects the stator windings in delta.

By connecting the stator windings, first in star and then in delta, the line current drawn by the motor at starting is reduced to one-third than the starting current with the windings connected in delta. At the time starting, when the stator windings are connected in star, then each phase gets a voltage equal to 𝑉𝐿/√3, where VL is the line voltage.

Since the torque developed by an induction motor is directly proportional to the square of the applied voltage, the star-delta starter reduces the starting torque to one-third as compared to the torque obtained with the direct-delta starting.

Theory of Star-Delta Starter

At starting, the stator windings are connected in star fashion, thus the voltage across each phase winding is

$$\mathrm{𝑉_1 =\frac{𝑉_𝐿}{\sqrt{3}}}$$

Where,

  • VL = Line voltage
  • V1 = Stator phase voltage

Let,

𝐼𝑠𝑡.𝑌.𝑝ℎ = Starting current per phase with stator windings connected in star

𝐼𝑠𝑡.𝑌.𝐿 = Starting line current with stator windings connected in star

Also, for star connection, the line current is equal to the phase current, i.e.,

$$\mathrm{𝐼_{𝑠𝑡.𝑌.𝐿} = 𝐼_{𝑠𝑡.𝑌.𝑝ℎ}}$$

Again,

𝐼𝑠𝑡.∆.𝑝ℎ = Starting current per phase with stator windings connected in delta

𝐼𝑠𝑡.∆.𝐿 = Starting line current with stator windings connected in delta

𝐼𝑠𝑐.∆.𝑝ℎ = Short circuit phase current with stator windings connected in delta

𝑍𝑒10 = Equivalent impedance per phase of motor referred to stator at standstill

Therefore, the starting current per phase with starting windings connected in star is given by,

$$\mathrm{𝐼_{𝑠𝑡.𝑌.𝑝ℎ} =\frac{𝑉_1}{𝑍_{𝑒10}}=\frac{𝑉_𝐿}{\sqrt{3} 𝑍𝑒10}… (1)}$$

And the starting current per phase with starting windings connected in delta is

$$\mathrm{𝐼_{𝑠𝑡.∆.𝑝ℎ} =\frac{𝑉_𝐿}{𝑍_{𝑒10}}… (2)}$$

For the delta connection,

$$\mathrm{Line \:current = \sqrt{3} × Phase\: current}$$

$$\mathrm{⇒ 𝐼_{𝑠𝑡.∆.𝐿} = \sqrt{3} × 𝐼_{𝑠𝑡.∆.𝑝ℎ} =\frac{\sqrt{3} × 𝑉_𝐿}{𝑍_{𝑒10}}… (3)}$$

Now, taking the ratio of starting line current with star-delta starting to the starting line current with the direct delta switching is

$$\mathrm{\frac{𝐼_{𝑠𝑡.𝑌.𝐿}}{𝐼_{𝑠𝑡.∆.𝐿}}=\frac{𝐼_{𝑠𝑡.𝑌.𝑝ℎ}}{𝐼_{𝑠𝑡.∆.𝐿}}=\frac{(𝑉_𝐿/\sqrt{3}𝑍_{𝑒10})}{\sqrt{3}(𝑉_𝐿/𝑍_{𝑒10})}=\frac{1}{3}}$$

⇒ 𝐼_{𝑠𝑡.𝑌.𝐿} = 𝐼_{𝑠𝑡.𝑌.𝑝ℎ} =\frac{1}{3}× 𝐼_{𝑠𝑡.∆.𝐿} … (4)

Hence, with the star-delta starter, the starting current from the main supply is one-third of that with the direct delta switching.

Also,

$$\mathrm{\frac{Starting\:torque\:with \:Y − ∆ \:starter}{Starting torque with direct delta switching} =\frac{(𝑉_𝐿/\sqrt{3})^2}{𝑉_𝐿^2} =\frac{1}{3}… (5)}$$

Thus, with the star-delta starting, the starting torque is reduced to one-third times of the starting torque obtained with the direct delta switching.

raja
Published on 28-Aug-2021 14:30:04
Advertisements