Three Phase WYE - DELTA Problem

[Marcin H]

Homework Statement

Homework Equations

Power Triangle and 3-phase equations (listed in my solutions)

The Attempt at a Solution

Here are the instructor solutions:

The instructor solutions just go straight into using equations, but I like to use the power triangle because it is more intuitive for me. The problem is I am not sure what I am doing wrong in my work... The first power triangle for load one looks to match the first chunk ( 100(0.9 +jsin(25.84)) of work in the instructor solutions, but then I get stuck at the second one. I don't see what I am doing wrong and I don't get what the instructor solutions are doing...

 

[magoo]

Your second load is a leading one instead of a lagging one. Your power triangle looks OK except that the leading load results in a negative value for the reactive load (vertical component).

 

[Marcin H]

Your second load is a leading one instead of a lagging one. Your power triangle looks OK except that the leading load results in a negative value for the reactive load (vertical component).

Right I caught that mistake so I made my Q for load 2 negative, but I didn't redraw my power triangle. I still get the wrong answer though. My solution above uses a negative Q for load 2.

 

[magoo]

I think I see the problem.

On L3, its a delta load. So if you just know the line current and the phase to phase voltage, the single phase kVA is given by kVph-ph * Iline. To get the three phase kVA, just multiply by 3, or 3*kVph-ph*Iline.

 

[Marcin H]

I think I see the problem.

On L3, its a delta load. So if you just know the line current and the phase to phase voltage, the single phase kVA is given by kVph-ph * Iline. To get the three phase kVA, just multiply by 3, or 3*kVph-ph*Iline.

Ya I figured out my mistake and got that same equation like you have. The voltage we we're given was line to line, but the current for L3 was a phase current so I had to change it to a line current by using iL=sqrt(3)i_phasor. so that and the original equation for S gave me the right answer. Thanks!

 

[Marcin H]

I think I see the problem.

On L3, its a delta load. So if you just know the line current and the phase to phase voltage, the single phase kVA is given by kVph-ph * Iline. To get the three phase kVA, just multiply by 3, or 3*kVph-ph*Iline.

I am now stuck on part C... Can you explain how we can find the Qcap for this problem? I am not sure how this works with a PF=1 because that gives us an angle of 0 meaning there is no Q? That doesn't make much sense. I don't understand the solutions either. They don't use Qcap = |Qnew - Qold| which is what I wanted to use.

 

[magoo]

Yes, a power factor of 1.0 means that there is no Q. So you want to size a capacitor bank equal to the net lagging Q - which you got from part a.

 

[Marcin H]

Yes, a power factor of 1.0 means that there is no Q. So you want to size a capacitor bank equal to the net lagging Q - which you got from part a.

Ohhhhhh, so in part A we have the total 3 phase complex power and to get the perphase power here in part C we just divide by 3. Got it!

 

[Marcin H]

Yes, a power factor of 1.0 means that there is no Q. So you want to size a capacitor bank equal to the net lagging Q - which you got from part a.

I have a question about a different problem:

Here are the solutions for that problem:

In my solution I had the exact same start. Trying to set up 2 equations and then use substitution and etc. But what I don't understand is what happens in the solutions when they are squaring everything. Firstly, the reason we want to square is to get rid of the j correct? but j^2 = -1. Why don't they have a negative in their answer in the second line?

Edit* Also, when the substitute for Qnew... They write Qnew = Qold - Qcap, but that does not seem right... The equation for caps we have been using this whole time is Qcap = Qnew - Qold which means Qnew is Qcap+Qold. There seems to be a lot of weird sign things happening here and I don't understand why

 

[Marcin H]

Ohhhhhh, so in part A we have the total 3 phase complex power and to get the perphase power here in part C we just divide by 3. Got it!

Ok I am sorry for all these messages, but this is driving me crazy. I tried a different approach (check picture below) but I think I am making a small sign error somewhere and I can't catch it. If that number at the very bottom on the left was negative and i subtract 8000^2 from it and then divide by 16,000 I would get the correct magnitude. My answer would just have a negative. But I can't figure out what I did wrong here... Can you please check?

 

[magoo]

Let me set up the problem a little differently than the way your solution does.

You start with

S = P + jQ

which corresponds to a line current of 23 A at 480 V

So the magnitude of S equals 480*1.732*23 or 19121.3 VA or 19.12 kVA
As you guessed, when you deal with magnitudes, you can get rid of the “j”, so S2 = P2 + Q2 (this is meant to be superscripts, like S^2)

REMEMBER: this equation is based on magnitudes! You can write the magnitude of S like |S|, but this takes a lot longer for me, so just follow my warning note.
You then add 8 kvar of capacitors to the load. This is 8000 vars.

You wind up with

S’ = P + jQ’

which corresponds to a line current of 18 A.
So now we’re working with the magnitude equation for S’, namely S’2 = P2 + Q’2 (this is meant to be superscripts, like S^2)

The P term did not change from the initial case. The Q term did change, which is why I called it simply Q’, where Q’ = Q – 8, if we’re working in kvar.

So the magnitude of S’ equals 480*1.732*18 or 14964.5 VA or 14.96 kVA.
The result is two equations (which involve magnitudes):

S2 = P2 + Q2

S’2 = P2 + Q’2

If we subtract the second from the first, we get:

S2 - S’2 = Q2 - Q’2
Substituting, it becomes

(19.121)2 – (14.965)2 = Q2 – (Q – 8)2 (this is meant to be superscripts, like S^2)

16 Q = 205.688

Q = 12.8555
The other question you had was about the sign of Q’. Again, since I dealt with magnitudes, it is always positive.

 

[Marcin H]

As you guessed, when you deal with magnitudes, you can get rid of the “j”, so S2 = P2 + Q2 (this is meant to be superscripts, like S^2)

REMEMBER: this equation is based on magnitudes! You can write the magnitude of S like |S|, but this takes a lot longer for me, so just follow my warning note.

I still don't understand why we can just get rid of the j here. I thought we should square everything to get rid of the j to make it nicer/easier to deal with. I'm not sure why we are dealing with magnitudes like you said.

Here i tried it again, except I changed the minuses to pluses to see if I would get a different answer:

I still wasn't sure why I have to do it like this, but it gave me a closer answer, just the wrong sign this time. I left out the minus in my final answer but if you look at my previous step you can see that my final answer would come out to be negative. The only difference between my substitution and your substitution [(19.121)2 – (14.965)2 = Q2 – (Q – 8)2] is the Q-8 part. In my work I have (Q+8)^2... Can you explain why you have Q-8? I thought the equation was Qcap=Qnew-Qold. So Qnew=Qcap+Qold = Qold +8... My book has the same equation for Qnew. That small difference seems to be the problem where I get a negative final answer.

So in short I am confused about why we just ignore the (-) when doing j^2 giving us different equations to then substitute and the (Q-8)^2 part...

 

[magoo]

My notation was a little different than your printed solution.

I used Q to represent the original load Q with no capacitors. I then used Q' to be the resulting Q after adding 8 kvar of capacitors.

So, my Q' = Q - 8

In any complex quantity, like z = a + jb, you can save a lot of grief by working with magnitudes. z-squared = a-squared + b-squared.

If you deal with magnitudes, the j disappears. Consider a 3 - 4 - 5 right triangle, where the base is 4 and the vertical length is 3. The largest side is 5 whether you make the vertical side 3 or -3. Do you understand that?

 

[Marcin H]

I used Q to represent the original load Q with no capacitors. I then used Q' to be the resulting Q after adding 8 kvar of capacitors.

So, my Q' = Q - 8

It looks like your Q' is the same thing as my Qnew, but the only difference is the sign inside. I have Qnew = Q+0 and you have Q'=Q-8. I am not sure how or why you get the minus 8 part. I am getting kinda lost with both our notations tho, lol...

And for the right triangle, i understand what you are saying, but i still don't understand why we just care about magnitude here. What in the original problem gave it away that we are dealing with magnitudes exactly? I feel like the direction of Q is important no? It will change the angle on our power triangle which changes whether current/voltage are leading or lagging no?