2-wattmeter power factor question: Shorebrook PE Exam Question 74

[akyip]

Hey guys,

In this Shorebrook Exam question 74 about two-wattmeter power factor, my initial attempt to solve for the power factor was using:

tan(theta) = sqrt(3) * (Phigh - Plow) / (Phigh + Plow) = 1.732 * (12 KW - 8.7 KW) / (12 KW + 8.7 KW) = 0.276

power factor angle: theta = tan^-1 (0.276) = 15.4295 degrees

power factor: p.f. = cos(15.4295 degrees) = 0.964

However, the given solution calculates pf = P/S, with:

P = P WA + P WB = 8.7 KW + 12 KW = 20.7  KW

S = sqrt(3) * V LL * I L = 1.732 * 300 V * 50 A = 25.98 KVA

P/S = 20.7 KW / 25.98 KVA = 0.8

What's the best explanation as to why these 2 power factor equations are yielding 2 different power factor answers for this problem?

Thanks for any input on this!

 

[Orchid PE]

Let's look at the base equation for the 2-wattmeter method:



Rearranging we get:



P1 + P2 is his "effective" (real) power, and sqrt3*V*I is his apparent power, cos(theta) is power factor.

The equation you used is correct as well:



As for why the two methods are different? My best guess is that Shorebrook randomly came up with the V and I values. Because if the correct power factor is 0.9639, at a line-to-line voltage of 300 the line current would be (P1+P2) / (sqrt3 * VL * cos theta) = 41 Amps. I wouldn't be surprised, since he also incorrectly labeled his apparent power as kW instead of kVA.

 

[akyip]

Let's look at the base equation for the 2-wattmeter method:

View attachment 18857

Rearranging we get:

View attachment 18858

P1 + P2 is his "effective" (real) power, and sqrt3*V*I is his apparent power, cos(theta) is power factor.

The equation you used is correct as well:

View attachment 18859

As for why the two methods are different? My best guess is that Shorebrook randomly came up with the V and I values. Because if the correct power factor is 0.9639, at a line-to-line voltage of 300 the line current would be (P1+P2) / (sqrt3 * VL * cos theta) = 41 Amps. I wouldn't be surprised, since he also incorrectly labeled his apparent power as kW instead of kVA.

Chattaneer, revisiting this. Thanks for your input. I usually use the P1 + P2 = P 3-ph = sqrt(3) x V LL x I L x cos(theta) and tan(theta) = ( sqrt(3) x (P high - P low) / (P high + Plow) ) approach for 2-wattmeter problems. Seems like that should be the standard way to go for 2-wattmeter problems.

I do see some mistakes here and there on the Shorebrook PE exam...