Calculators
Calculate star-delta starting
Star-delta starting cuts the starting current to a third. That is the good news. The bad news sits right beside it. It also cuts the starting torque to a third. Design by current alone and you build a plant that hangs in star.
Result
- Starting current in star
- —A
- Will the motor start in star?
- —
- Starting current direct on line
- —A
- Starting torque in starRelative to the rated torque.
- —%
- Saving compared with direct on line
- —%
Please fill all fields with valid numbers.
A guide, not a design to standard. We give no warranty for the correctness of the results. The applicable standards and case-by-case verification govern.
The grey values are an example. Simply enter your own.
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Calculating is the easy part.
A formula gives you a number. Designing a plant also demands installation method, grouping, discrimination, the standards in force and a look at the installed base. That is what we take on.
Method and questions
Star-delta starting
Current and torque in star, and whether the motor will start at all.
Discuss a projectWorked example
The default case the calculator starts with, worked through once. Change the values above and it recalculates immediately.
Inputs
- Rated motor current
- 55 A
- Starting current ratio I_A / I_N
- 7
- Starting torque ratio M_A / M_N
- 2.2
- Load torque at start
- 40 %
Result
- Starting current in star
- 128 A
- Will the motor start in star?
- Yes. The starting torque in star is well above the load torque.
- Starting current direct on line
- 385 A
- Starting torque in star
- 73.3 %
- Saving compared with direct on line
- 66.7 %
Formula and standards
Formula
- I_A(delta) = I_N · (I_A / I_N)
- I_A(star) = I_A(delta) / 3
- M_A(star) = M_A(delta) / 3
- Starting condition met if M_A(star) > M_load
Assumptions and standards
- In star connection each winding sees the voltage divided by root three. Current and torque therefore fall to a third of the direct-on-line values.
- The nameplate figures apply to direct-on-line starting in delta.
- Changing over from star to delta produces a brief current peak that can reach the direct-on-line starting current. It is not shown here and belongs in the design of the protection.
When is star-delta unsuitable?
Whenever high torque is needed from the start, for example with loaded conveyors, piston compressors or agitators in viscous media. It suits loads whose torque rises with speed, such as fans and centrifugal pumps.
What happens at the changeover?
The motor is briefly disconnected, the field decays, and reconnecting in delta produces a current peak. At an unfavourable instant it reaches the direct-on-line starting current, and half the benefit is gone. Change over only once the speed is close to its final value.
Why exactly a third?
The winding voltage falls by root three, and with it the winding current. In addition, the line current in star equals the winding current, whereas in delta it is root three larger. Together that gives one third. Torque goes with the square of the voltage, so it also falls to a third.
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