# AC power analysis

Discussion in 'Homework Help' started by righteous818, Nov 9, 2013.

1. ### righteous818 Thread Starter New Member

Nov 9, 2013
3
0

I have worded out the power factor correction capcaitor to be 580 uF and i have converted the combined load to a resistor and inductor pf value 3.84 ohms and 0.01 henries from using the apparent and real power equation. i am unsure of how to interprert this graph and dervive and expression for the load voltage. any assitance would be grateful thank you. i dont want answers i want to know how to figure it out. thank u

2. ### jegues Well-Known Member

Sep 13, 2010
735
44
Why don't you start by showing your work and results for parts 2a and 2b?

3. ### righteous818 Thread Starter New Member

Nov 9, 2013
3
0
i have figured out parts 2 a and 2 b , i have worked out the powre factor correction part buts it really long to post here. I dont need the ans for the question 3 but an idea on how to go about working it

thank you

4. ### WBahn Moderator

Mar 31, 2012
20,232
5,755
I'm not convinced your answers are correct to the first parts. If the resistance is 3.84Ω, then even if there is no inductance at all, the total real power would only be 60kW. Yet your two loads total to an average power of 70kW even with the reactive component.

Did you note that the loads were given as average power, not apparent power?

The impedance of a 0.01H inductor at 60Hz (377r/s) is going to be j3.77Ω. So the magnitude of your reactance is almost the same as your resistance, which means that you will have a power factor of right at 0.7. Does that make sense given your loads?

Did you check if your answer was correct? One of the nice things about engineering is that it is usually the case that we can check the validity of our answers from the answers themselves. You need to get in the hibit of doing that.

If you would show your work, we could help you find where you've gone wrong (if you have). But, no work no partial credit.

As for the third part, the graph is give you the current that is injected into the line by the lightning strike. Just think of it as a current source that has that profile.

The diagram is admittedly confusing. The text says that it is a 1.2/50 μs waveform. But the diagram shows the rising part going from zero to the full 20kA in 1.2μs but the second part only dropping 50% in 50μs. Thus the two would seem to be contradictory. Unless the waveform is assumed to be a linear ramp followed by an exponential decay, but the diagram sure looks like a linear decay, too.

5. ### jegues Well-Known Member

Sep 13, 2010
735
44
Just to clarify this, lightning impulse voltages are often referred to as a T1/T2 impulse, where T1 is the front time, and T2 is the time to half value.

The front time is defined as 1.67 times the interval T between the instants when the impulse is 30 percent and 90 percent of the peak value for full or chopped lightning impulses, while the time to half value is the time it takes from inception of the impulse, to when it reaches half its peak value. The time to half point is obviously measured after the impulses peak has occured. (i.e. there are two points on the curve that satisfy this criteria, one before the peak value and one after, we use the one after)

The 1.2/50 impulse is the accepted standard lighting impulse voltage today.

Last edited: Nov 10, 2013
6. ### WBahn Moderator

Mar 31, 2012
20,232
5,755
Thanks.

So, if it is being modeled for simulation purposes what is the shape of the front and back waveforms. The diagram shows them as linear but the description you talk about is more consistent with exponential. When inductors are involved, I would think that the difference could be significant. Of course, whether the modeled circuit is adequate at those ramp rates is another matter.