# Delta Power System

Discussion in 'Homework Help' started by camino, Dec 9, 2009.

1. ### camino Thread Starter New Member

Sep 17, 2009
8
0

Here's what I have so far:

I1 = 49.68A<-68.6 degrees (top loop)
I2 = 49.68A<-128.6 degrees (bottom loop)
I3 = 28.68<-98.6 degrees (triangle loop)

a)
PA12 = (49.68A)^2(12) = 29,617.23 W
PB12 = (49.68A-49.68A)^2(12) = 0 W
PC12 = (49.68A)^2(12) = 29,617.23 W
Ptotal12 = 29,617.23W + 0W + 29,617.23W = [59,234 W] answer

b)
Pab400 = (49.68A-28.68A)^2(400) = 176,400 W
Pbc400 = (49.68A-28.68A)^2(400) = 176,400 W
Pca400 = (28.68A)^2(400) = 329,017 W
Ptotal400 = 176,400W + 176,400W + 329,017W = [681,817 W] answer

Now c) and d) I am unsure of. I was thinking power factor would be 59,234 W/681,817W which would give me a p.f. of 0.087 but that doesn't seem right, and efficiency I have no clue how to calculate.

I'm not 100% sure I even did parts a) and b) right. Any help/explanations would be greatly appreciated!!

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
All phase loads and line impedances are balanced [or equal] so the individual line current and load current magnitudes will be equal. In that case the power in each 12 Ω line resistor and 400 Ω load resistor will be the same - which is not what your solution indicates.