NCEES # 111 Power

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Can someone please explain why 12.5KV is divided by under-root 3 in NCEES solution?

This is how I solved this problem;

Vab=12.5 angle 0 + voltage drop across single phase line

= 12.5 angle 0 + 70 angle -20 * (5+j10)

= 13.08 KV angle 2.35

Since the load is Delta, phase voltage would be equal to line voltage.

Am I missing something?

 
What you are missing is that there are two segments of the cable in play. One between points 'a' and 'A' and the other between points 'b' and 'B'. The currents running through these two segments are 120 degrees out of phase.

The reason for converting the voltage to a phase equivalent is to make the math easier. It reduces the problem to a single loop with only one segment of the cable. If you do a loop analysis accounting for the phase angles of the currents through line aA and bB, you end up with the same reduction but it's a lot easier to miss that way.

 
I have a similar question. In the NCEES solution I follow the per phase conversion to calculate Van but what I'm missing is the conversion from 12.5/1.73 angle 0 to 12.5/1.73 and -30.

The -30 is confusing me because I worked it with a +30.(obviously got it wrong)

Can anyone explain why it was -30 instead of +30, was I suppose to pick that up from the ABC sequence or because it would stand to reason with the line current being at angle -20?

 
jvg2010 said:
I have a similar question. In the NCEES solution I follow the per phase conversion to calculate Van but what I'm missing is the conversion from 12.5/1.73 angle 0 to 12.5/1.73 and -30.
The -30 is confusing me because I worked it with a +30.(obviously got it wrong)

Can anyone explain why it was -30 instead of +30, was I suppose to pick that up from the ABC sequence or because it would stand to reason with the line current being at angle -20?
Click to expand...
Remember, the Voltage P-N always lags the Voltage P-P by 30 degrees. Hence Voltage P-N angle is (-30)

 
bacchi said:
jvg2010 said:
I have a similar question. In the NCEES solution I follow the per phase conversion to calculate Van but what I'm missing is the conversion from 12.5/1.73 angle 0 to 12.5/1.73 and -30.
The -30 is confusing me because I worked it with a +30.(obviously got it wrong)

Can anyone explain why it was -30 instead of +30, was I suppose to pick that up from the ABC sequence or because it would stand to reason with the line current being at angle -20?
Click to expand...
Remember, the Voltage P-N always lags the Voltage P-P by 30 degrees. Hence Voltage P-N angle is (-30)
Click to expand...
But isn't that for a Wye configuration? I thought the equation for a Wye is Vab = SQRT(3)Vp angle 30. In a Delta isn't is Vab = Vp angle -120?

 
Last edited by a moderator:
^For a delta connection, there is no phase-neutral voltage since there is no neutral connection. However, mathematically, it is often convenient to reduce a delta system to a single-phase equivalent circuit. Mathematically, the "phase" voltage (VL-n) voltage will still be lagging the "line" voltage (VL-L) by 30 degrees.

 
Flyer_PE said:
^For a delta connection, there is no phase-neutral voltage since there is no neutral connection. However, mathematically, it is often convenient to reduce a delta system to a single-phase equivalent circuit. Mathematically, the "phase" voltage (VL-n) voltage will still be lagging the "line" voltage (VL-L) by 30 degrees.
Click to expand...
Ah. Makes sense when you talk about in mathmatical terms.

 
Been working on this problem and I got the answer correct the first time, but mainly due to rather large difference between answers. I used a bit different approach, actually two and was wondering if it's just plain luck or correct thinking:

1) Find voltage drop in line Vd = IaA/sqrt(3) x Z = 452V , add it to load voltage , get the answer. Basically I am going for a very basic V=IR principle (since both I & R are given), but also paying attention to fact that IaA is different from IAB.

2) Second approach is formula from Red book for Voltage Drop Vd= IRcos (theta) + IXsin(theta) where theta is 20 (difference between load voltage and load current). This solution gives me 568V answer, which adds up to 13068V, which is close enough to answer C.

So in short, I understand it's a voltage drop problem and found 2 ways to solve it, am I just plain lucky or both approaches make sense?

Thank you,

 
I was working on this problem yesterday and I understand the way the solution manual works in theory but I didn't like the method. Van does not exist. I used a Kirchoff's Voltage loop accounting with the understanding that the problem states it is a balanced ABC system. The equation I had was this:

Vab = (5+j10) IaA + VAB - (5+j10) IbB

where is IbB = (70ang-140)

The result is the exact same answer they had in the book. 12.95kV

If you don't take into account the voltage drop in B phase then you don't get the correct answer for Vab.

 
BatteryPowered said:
Been working on this problem and I got the answer correct the first time, but mainly due to rather large difference between answers. I used a bit different approach, actually two and was wondering if it's just plain luck or correct thinking:

1) Find voltage drop in line Vd = IaA/sqrt(3) x Z = 452V , add it to load voltage , get the answer. Basically I am going for a very basic V=IR principle (since both I & R are given), but also paying attention to fact that IaA is different from IAB.

2) Second approach is formula from Red book for Voltage Drop Vd= IRcos (theta) + IXsin(theta) where theta is 20 (difference between load voltage and load current). This solution gives me 568V answer, which adds up to 13068V, which is close enough to answer C.

So in short, I understand it's a voltage drop problem and found 2 ways to solve it, am I just plain lucky or both approaches make sense?

Thank you,
Click to expand...
In your first method... how did you know to use IAB rather than IaA?

 
BatteryPowered said:
Been working on this problem and I got the answer correct the first time, but mainly due to rather large difference between answers. I used a bit different approach, actually two and was wondering if it's just plain luck or correct thinking:

1) Find voltage drop in line Vd = IaA/sqrt(3) x Z = 452V , add it to load voltage , get the answer. Basically I am going for a very basic V=IR principle (since both I & R are given), but also paying attention to fact that IaA is different from IAB.

2) Second approach is formula from Red book for Voltage Drop Vd= IRcos (theta) + IXsin(theta) where theta is 20 (difference between load voltage and load current). This solution gives me 568V answer, which adds up to 13068V, which is close enough to answer C.

So in short, I understand it's a voltage drop problem and found 2 ways to solve it, am I just plain lucky or both approaches make sense?

Thank you,
Click to expand...


I'm missing something very basic and crucial here. I'm assuming that IaA/sqrt(3) is used to get IAB? Why isn't the drop found by simply using the current through Zline (IaA)? Any help would be appreciated.

 
Strebe

I was trying to take the same approach as you on this problem. It wasn't readily apparent to me to find Van, etc. I was looking at is as a simple loop equation like you. I just got a little confused on the signs. If I back track your first equation the loop equation would be:

Vab - (5+j10)IaA - VAB + (5+j10)IbB = 0

I understand that the sign of the second and fourth term are opposite becasue IaA is in the opposite direction of IbB, but how did you determine that VAB was negative?

 
I assigned polarity in the classical/conventional method. for a proper voltage loop the equation would be

0 = -Vab + IaA(ZL) + VAB - IbB (ZL)

 

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