Question for electrical experts / Lightman1

Serendipity

A 3-phase 500kvA 7.2kV load has a Power Factor of 0.8 lagging. For the rated line voltage, what would be the following:
(a) The Line Current
(b) 3-phase Real Power
(c) 3-phase Apparent Power
(d) 3-phase Reactive Power

I only have knowledge of residential power but haven't dealt much with 3-phase. As a result, any help would be greatly appreciated.

Thanks!

disneyjoe7

What are you building, that's a lot of power for an amp!

I-SIG

Serendipity wrote: »

A 3-phase 500kvA 7.2kV load has a Power Factor of 0.8 lagging. For the rated line voltage, what would be the following:
(a) The Line Current = 40.09A (assuming 7.2kV is phase-phase voltage) 23.15A if 12.47kV is the 3-phase voltage.
(b) 3-phase Real Power = 400 kW
(c) 3-phase Apparent Power = 500 kVA
(d) 3-phase Reactive Power = 100 kVAR

I only have knowledge of residential power but haven't dealt much with 3-phase. As a result, any help would be greatly appreciated.

Thanks!

See the red. 7.2kV is a typical single phase primary voltage. The 3-phase voltage would be 12.47kV. Not sure about Russ, but most of my system that I design is 14.4/25kV.

Remember: If the current lags, the voltage sags! (I made that one up all by myself!)

I'm sure if I'm wrong, someone will beat me down.

Wes

Serendipity

I-SIG wrote: »

See the red. 7.2kV is a typical single phase primary voltage. The 3-phase voltage would be 12.47kV. Not sure about Russ, but most of my system that I design is 14.4/25kV.

Thanks so much!

I was wondering if you could explain how you got those numbers, as I'm more interested in knowing how to do it in principle than just the solution.

lightman1

Nailed it, Wes!
Most of the 3 phase around here runs between 12,470 and 14,400, depending on load and demand factors for the grid. i.e. light commercial vs. industrial.
There are a few resi hoods that have 13k to them only because the the friggin houses are HUGE and on the end of ten mile long penninsula on the lake. No way to back loop the circuits to the sub station. Brown outs were occuring at 7.2K and the rich folks didn't like that.
What are you building, Dippity-doo?

disneyjoe7

Ok that's what I was going to say....



























NOT

I-SIG

Serendipity wrote: »

Thanks so much!

I was wondering if you could explain how you got those numbers, as I'm more interested in knowing how to do it in principle than just the solution.

I was lazy for the current and just used a little wallet chart I have from ABB. 500kVA is really common 3-phase transformer size so it was on the chart. Just had to find your voltage(s).

Think of kW, kVA, and PF as a right triangle. KVA is the hypotenuse, and kW is the base along the X-axis. (Don't forget SOH CAH TOA)

If PF is 1, 100% or unity, kW, or real power, will be equal to kVA, or apparent power. and the rise on your Y-axis is 0 also. The origin is 0,0.

In your problem, the PF and kVA were given, so all we needed was some simple multiplication to kind the kW and kVAR, or kilo-VoltAmp Reactive, or imaginary power. (this is where AC gets really complicated.)

To find the kW (real power, same as DC), simply multiply the kVA by the PF (500 * 80% = 400kW) What's left is your reactive power, or 100 kVAR.

Wes

Serendipity

Thanks for your help!