KVA Example

[sdilucca]

check out this example for KVA calculation

THREE-PHASE

Given: We have a large EMC Symmetrix 3930-18/-36 storage system with 192 physical volumes. EMC's website shows a requirement for a 50-amp 208 VAC receptacle. For this calculation, we will use 21 amps. Do not calculate any value for the plug or receptacle.

KILOVOLT-AMPERES (kVA) = (VOLTS x AMPERES x 1.73)/1000
208 x 21 x 1.73 = 7,556.64
7,556.64 / 1000 = 7.556 kVA

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The way this is worded, it sounds like both voltage and current values are line to line. I am confused. What component is being multiplied by 1.73 and WHY? Are there any assumptions I should be making?

 

[charlie b]

Re: KVA Example

Voltage is "line-to-line." Current is "line." Current is not measured from one point in the system to another, but along a single line.

You are not multiplying voltage by the square root of three, and you are not multiplying current by the square root of three. What you are multiplying by the square root of three is the product of votlage and current. That 1.732 factor comes from the product, not from either component alone.

The "why" of that 1.732 factor is beyond what I can convey here. It is a trigonometric derivation, and it uses Greek letters that the Forum's word processor does not recognize. But if you want to see a copy (in Word format), send me a PM with your email address, and I will send you a copy of the derivation.

 

[rattus]

Re: KVA Example

J. C.,

Look at the problem this way:

Assume that you have a 120/208V wye--120V phase, 208V line. The line currents are equal to the phase currents, 21A.

Then Pa = Vphase x Iline x 3

= 120V x 21A x 3 = 7.560KVA

Now Vline = 2cos(30) x Vphase = 1.732 x Vphase

(this is the trig mentioned by Charlie B.)

Then,

Pa = (Vline/1.732) x Iline x 3

= Vline x Iline x 1.732

= 208V x 21A x 1.732 = 7.565KVA

You can perform a similar computation for a delta system in which Vphase = Vline, and Iline = Iphase x 2cos(30)

Results will be the same in either case.