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2 Diesel Generators 900KVA connected to the bus bar for
essential load as an alternative for power taken from
11/440V 3KVA Transformer supplying this bus bar. my
question here is in the event of power failure on 11KV side
can I feed the power to 11 KV bus by using the same
transformer as reverse feed, what will be the inrush
current???
- 3 Answers
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- Ashok Leyland, I also Faced
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Answer Posted / a. magesh kumar
My understanding is two 900 KVA generator, alternatively
feeds the supply to transformer (one at a time), the
transformer step down voltage 11KV/ 400V (3KVA). This 440V
feeds to connected load (seconday side of the transformer).
Ur question: In the event of Power failure 11KV side
(primary), what will be the inrush current?
whether u r asking short circuit current or else?
Inrush rush Current:
AC power may happen to be applied to the primary winding of
a power transformer at any point in the voltage cycle.
Excessive current inrush can result if the primary voltage
suddenly goes from zero to a high value rather than
increasing sinusoidally as it does during normal operation.
Excessive inrush current to a large transformer may not
only blow a fuse, trip a breaker, cause a sharp dip in
mains voltage, damage transformer windings, or damage
electrical supply circuits, it also induces a very large
voltage spike in the transformer secondary winding. Large
secondary voltage spikes can breakdown winding insulation,
damage rectifiers, and damage other power supply components
When switching power supplies are first turned on, they
present high initial currents as a result of filter
capacitor impedance. These large filter capacitors act like
a short circuit, producing an immediate inrush surge
current with a fast rise time. The peak inrush current is
several orders of magnitude greater than the circuit's
steady state current. This power surge can seriously damage
other components
Calculate the Steady State Current:
(Available data from ur mail)
Transformer power: 3 KVA
Primary Voltage : 11 KV
Secondary Voltage : 415 V
Efficiency : 85 % (assumed)
Impedance : 4 % (assumed)
Steady State Current: (KVA of transformer) / (efficiency
of transformer) X (Minimum Input voltage)
: 3 KVA / (0.70) X (415)
: 10.32 Amps.
Short Circuit Calculation: 3 KVA / 415
: 7.2 A
: 7.2 * 25 (impedance)
: 180.7 A (0.18 KA)
Short circuit current is 187.5 Amps.
| Is This Answer Correct ? | 5 Yes | 1 No |
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