DOL Start on a Weak 400 V Bus
A 15 kW, 400 V, 50 Hz, four-pole, star-connected squirrel-cage induction motor drives a loaded conveyor. At rated voltage the nameplate data give full-load slip 4.0%, full-load current 29 A, direct-on-line starting current 6.0 times full-load current, starting torque 1.5 times full-load torque, and breakdown torque 2.5 times full-load torque. The conveyor presents a constant torque equal to the motor’s rated torque and cannot be unloaded during starting. The site is fed by a transformer that is already heavily loaded; during a DOL start the motor terminal voltage falls to 80% of rated value. The electrician proposes a star-delta starter, arguing that it will reduce the starting current enough to prevent the voltage dip and let the motor accelerate. The plant engineer is not convinced because the conveyor is fully loaded at standstill. They ask for a quantitative check before changing the starter.
1. Using the square-law voltage dependence of induction-motor torque, determine whether the motor can accelerate the conveyor from standstill with a DOL start at 80% terminal voltage. Support the answer with standstill torque and the stable operating slip.
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At rated voltage the standstill torque is T_st = 1.5 T_FL. At 80% terminal voltage, torque scales as V², so T_st(0.8) = 1.5 × 0.8² T_FL = 1.5 × 0.64 T_FL = 0.96 T_FL.
The load is a constant 1.0 T_FL. At standstill the motor torque is less than the load torque, so the net accelerating torque is negative and the motor cannot start. It will remain stalled and draw a locked-rotor current of about 6 × 0.8 = 4.8 times full-load current.
The breakdown torque at 80% voltage is 2.5 × 0.64 = 1.6 T_FL, which is above the load torque, so a stable operating point does exist on the torque–speed curve if the motor could be brought up to speed. Using the approximate torque expression T = 2T_max/(s/s_m + s_m/s) with s_m ≈ 0.2 gives an intersection near s ≈ 7%. But the motor never reaches that region because it cannot accelerate through the low-speed range. The DOL start fails.
2. Evaluate the star-delta proposal quantitatively at the same 80% terminal voltage. Give the starting torque and line current in per-unit of full-load values, and explain why it does not solve the problem.
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In star, the phase voltage is 1/√3 of the delta value, so the starting torque is one-third of the DOL value at the same supply voltage. Including the 80% terminal voltage:
T_st(star-delta) = 1.5 × 0.8² / 3 T_FL = 0.32 T_FL.
The line current becomes 6 × 0.8 / 3 = 1.6 times full-load current. The current reduction is real, but the starting torque is only about one-third of the load torque. The conveyor cannot be accelerated; the starter merely reduces the current while the motor remains stalled.
Star-delta is a reduced-voltage starter, so it always reduces torque along with current. It is useful when the load torque during acceleration is low, not for a constant rated-torque load that must start fully loaded.
3. Recommend a practical remedy, and state the load condition under which the star-delta starter would be acceptable.
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The practical options are to unload or reduce the conveyor during starting, reinforce the supply so the terminal voltage stays near rated, use a variable-frequency drive that can produce high starting torque within a controlled current limit, or use a wound-rotor motor with slip-ring resistance starting. A soft starter alone would also reduce voltage and torque, so it is not a solution for this constant-torque load at 80% terminal voltage.
Star-delta would be acceptable if the load torque during starting were below about 0.32 T_FL, for example a centrifugal pump or fan with a quadratic torque–speed characteristic, or a conveyor that can be started empty and loaded afterwards. The key check is always the torque margin at every speed from standstill to the operating point, not just the current reduction.