# Homework Help: Delta Power System

1. Dec 9, 2009

### camino

1. The problem statement, all variables and given/known data
http://img42.imageshack.us/img42/3537/powersyst.jpg [Broken]

2. Relevant equations

P = I^2 R

3. The attempt at a solution

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!!

Last edited by a moderator: May 4, 2017
2. Dec 10, 2009

### rootX

I believe the power lost in the line is the loss hence providing you efficiency of less than 100% (power consumed by the delta load)

3. Dec 25, 2009

### Mr.Green

I don't check whether the numerical values are true or not, but there are errors in calculating the real power in the resistors.

On assumption that the currents values are true.

PA12 is OK.

PB12 = I1 - I2 = (|(49.68A<-68.6 - 49.68A<-128.6)|^2) * 12 = something
By "|" I mean the magnitude.

Here we are dealing with phasors, so first obtain the current through the resistor (don't forget the angle of the current) and then calculate the power.

The same should be done for Pab400 and Pbc400.

By considering lines "Aa", "Bb", and "Cc" as transmssion lines, the impedance (12+j16) is the line impedance and there are power loss in the line. The loss is the power dissipated in the 12 Ohm resistors. So, the efficiency can be calculated.