AMIE Section B EEE Exam study materials
AMIE Section B EEE Exam study resources

Undervoltage During Motor Start on a Weak 11 kV Feeder

AMIE Section B EEE Exam · Put your knowledge to work in context.

All case studies

Undervoltage During Motor Start on a Weak 11 kV Feeder

A 2.2 MW, 11 kV, 6-pole cage induction motor with a locked-rotor current of 6.0 times full-load current and a locked-rotor power factor of 0.30 is fed from an 11 kV bus. The upstream supply has a short-circuit level of 95 MVA at the bus (X/R = 8). The motor full-load efficiency is 95.5% and full-load power factor is 0.90. During direct-on-line starting the plant engineer observes that a contactor on the same bus drops out and lights flicker. The protection relay records a bus voltage dip lasting about 1.5 s. The contactor is specified to hold in down to 85% of rated voltage. A star-delta starter is available but the driven load is a loaded conveyor that cannot tolerate a torque reduction below 55% of rated motor torque.

1. Estimate the steady-state bus voltage during direct-on-line starting and decide whether the contactor dropout is explained by it.

Read the worked answer

Full-load current: input power = 2.2 MW / 0.955 = 2.304 MVA; I_FL = 2.304e6 / (sqrt(3) x 11e3) = 121 A. Locked-rotor current I_LR = 6.0 x 121 = 726 A. In per-unit on 11 kV, 95 MVA base, the source impedance is Z_s = 1/95 = 0.01053 pu. The motor starting impedance is Z_m = V/(sqrt(3) I_LR) = 11e3/(1.732 x 726) = 8.75 ohm, i.e. 8.75 / (11e3^2/95e6) = 8.75/1.274 = 6.87 pu. Using the approximate divider V_pu = Z_m/(Z_m + Z_s) = 6.87/6.881 = 0.9984 - this ignores the phase angle between source and motor impedances. Correcting for angle: source angle ~82.9 deg (X/R=8), motor starting angle ~72.5 deg (pf 0.30). The vector sum gives a dip of roughly 0.15-0.20%, i.e. about 98-99% voltage. That is far above the 85% threshold, so a steady-state DOL start does not explain the dropout. The observed 1.5 s dip must come from something else - likely a second motor starting simultaneously, a transformer tap or upstream fault, or an actual short-circuit transient. The engineer should check event records for a coincident start and verify whether the 95 MVA figure is the actual available fault level at that time (e.g., single transformer in service).

2. The plant insists on reducing starting current. Evaluate star-delta starting against the 55% torque constraint.

Read the worked answer

Star-delta reduces the voltage applied to each winding by 1/sqrt(3), so starting current and starting torque both fall to one third of the DOL values. DOL starting torque at rated voltage is approximately T_st/T_FL = (I_LR/I_FL)^2 x (pf_LR/pf_FL) x (efficiency correction) - a standard approximation gives T_st/T_FL ~ (6.0)^2 x 0.30/0.90 x 0.955 ~ 11.5, but this is the torque at full voltage and full speed assumption; more carefully, T_st/T_FL ~ (I_LR/I_FL)^2 x s_FL x (R2/R1) is not available. A practical rule: starting torque is roughly 1.0-1.5 times full-load torque for a NEMA design B motor at DOL. With star-delta, that becomes 0.33-0.50 times full-load torque, i.e. 33-50% - below the 55% requirement. So star-delta is not acceptable. Alternatives: a soft starter with a current limit set so that torque stays above 55% (current limit around 3.5-4 times FLC), or a variable-frequency drive, or a reduced-voltage autotransformer at 80% tap (torque 64% of DOL, current 64%). The conveyor also needs a breakaway torque check; if breakaway exceeds 55%, even a soft starter will stall the belt.

3. If a soft starter is chosen, what current limit should be set, and what secondary check is needed?

Read the worked answer

Torque scales with the square of applied voltage and approximately with the square of current in the linear region of the motor characteristic. To keep torque at 55% of the DOL value, the current must be at least sqrt(0.55) = 0.74 of the DOL current, i.e. 0.74 x 6.0 = 4.4 times FLC. So the current limit must be set no lower than about 4.5 times FLC to guarantee the conveyor accelerates. That is only a 25% reduction in current, which may not solve the flicker problem - so the engineer must check whether the actual dip is really caused by this motor. Secondary checks: (1) verify the soft starter's short-circuit withstand and bypass contactor rating; (2) confirm the contactor coil is fed from a source not affected by the bus dip (e.g., a UPS or a different bus); (3) check thermal capacity of the motor during the extended start (I^2t) against the locked-rotor thermal limit, typically 2-3 starts per hour; (4) confirm the relay's undervoltage setting and time delay do not trip on legitimate starts.