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

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akyip

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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!

Shorebrook PE Exam Question 74.jpg

Shorebrook PE Exam Solution 74.jpg

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

image.png

Rearranging we get:

image.png

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:

image.png

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.

 
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...

 
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