NCEES #509

Where do they get the -30 and +30 degrees in the answer? And do the * dots in the wattmeters have anything to do with how they are read?

cruzy said:

Where do they get the -30 and +30 degrees in the answer? And do the * dots in the wattmeters have anything to do with how they are read?

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I'm going to dig in to this one later today, but in short, the 30 degrees come from the phase shift between the reference voltage (typically Van at 0 degrees) and the line-to-line voltages. Pull out your phasor diagram and look at Van vs Vab - you'll see the 30 degree phase shift.

And yes, the * is the polarity of the inductive coil, just like a transformer. If they reverse it, things get really really weird.

OK, I ran the numbers on 509.

They put their voltage reference at Vcn = 0 degrees - associated with W1. This is not given to us, but I ran with the equations I've been using:

W = V I cos (voltage angle - current angle)

If Vcn = 0, then Vca = 30 degrees. Ic = -36.86 degrees. W1 = 432 watts

With Vcn = 0, Vba = 90 degrees. Ib = 83.14 degrees (120 minus 36.86). Therefore W2 = 1092 watts

If you put voltage reference at Van with this problem, you get crazy numbers. The above numbers match what their formulas give.

If they throw a wattmeter question at us on the exam, I'll first run with the assumption that whatever phase W1 is on, will be the reference angle.

I posted a message for a thread similar to this, see this link.

When the load power factor is unity, corresponding to a purely resistive load, both wattmeters will indicate the same wattage.

Click to expand...

That quote alone answers this problem.

Additionally,

Setting the two equal to one another, and the use of a trig identity gives:

WA = WC

VIcos(30º - Φ) = VIcos(30º + Φ)

cos(30º - Φ) = cos(30º + Φ)

cos(30º)cos(Φ) + sin(30º)sin(Φ) = cos(30º)cos(Φ) - sin(30º)sin(Φ)

2sin(30º)sin(Φ) = 0

sin(Φ) = 0

Φ = 0 -> pf=cos(Φ) = 1.0