Paralelling Gear Size

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Alwayslearningelec

Senior Member
Location
NJ
Occupation
Estimator
Is is possible to estimate approximately the size of paralalling needed for a 2.5MW genset with only having the genset size? THanks.
 
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rcwilson

Senior Member
Location
Redmond, WA
If you know the voltage you can calculate the generator full load amps.

Example 480V generator, 2.5 MVA = 2500kVA

FLA = (2500 kVA x 1000) / (480V x 1.732) = 3007 amps.

Switchgear should be sized 115% minimum = 1.15 x 3007 A = 3458 A,
usually sized for 125% = 1.25 x 3007 = 3,759 A

If it is 480V use 4,000 A gear.

I just noted you said this is a 2.5 MW generator which would make it a 2.5/.85 pf = 2941 MVA.

FLA would be 3538A. +15% would be 4,068A at 480V. That's pushing it for 4,000A gear, need 5,000A.
 
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jdsmith

Senior Member
Location
Ohio
rcwilson said:
That's pushing it for 4,000A gear, need 5,000A.
Click to expand...

A slight tangent....

To optimize safety and reliability a few features are preferred in switchgear: fully insulated insulated bus, minimal or no ventilation openings in the front panels, and no active ventilation (fans). With these and other practical restrictions, 4000A is the largest fully featured, safe, reliable switchgear available at 480V, and 3000A is the largest available at 5kV.

2.5 MVA/.85 pf = 2941 kVA.
FLA would be 3538A. +15% would be 4,068A at 480V.
Click to expand...

As a client I would prefer you put a 13% margin on the switchgear instead of 15% so you can stay under the 4000A limit and not end up with goofy gear. Switchgear LV power circuit breakers are 100% rated breakers, and the load will rarely be at the generator's nameplate - it's not necessary to stack up a bunch of design margins on top of one another.
 
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