Complex Imaginary Test 4 Problem 41

[applepieordie]

Can someone please explain the solution to me?

Question:

A wye-wye, 230 kV, 3-ph transmission line is being stepped down to 200 kV using three autotransformers connected between the neutral and the line. For each phase, if the autotransformer is rated for 500 MVA, and the turns ratio is 5:2 (common:series) what is the apparent power in the common coil?

Their Solution:

(Nc/Ns) = (5/2), S = 500 MVA

(S/Sc) = (Ns+Nc)/Ns = 1+(Nc/Ns)

Sc = S/ (1+(Nc/Ns)) = 500MVA/ (1+(5/2)) = 143 MVA

To me it seems like they solved for the apparent power in the series coil. Or am I thinking about this wrong? please help.

 

[Stephen2awesome]

I just did this problem. I solved it using the turns ratio. I found the current going to the low side first. Turns ratio for the system is (5+2):5. I2 going out onto the 200kv/1.732 line is 4330A. I2 coming from the 230kv/1.732 is 3090A. The Ic is I2 - I1 = 1240A. Use Ic to find S in the common(shared) winding). Try that out and see if you it makes sense.

 

[rg1]

In this question the Voltage Ratio and turns ratio do not match. So take turns Ratio (5:2) and MVA to get the answer of the solution. Do not mix Voltage with turns ratio. 

 

[rg1]

Can someone please explain the solution to me?

Question:

A wye-wye, 230 kV, 3-ph transmission line is being stepped down to 200 kV using three autotransformers connected between the neutral and the line. For each phase, if the autotransformer is rated for 500 MVA, and the turns ratio is 5:2 (common:series) what is the apparent power in the common coil?

Their Solution:

(Nc/Ns) = (5/2), S = 500 MVA

(S/Sc) = (Ns+Nc)/Ns = 1+(Nc/Ns)

Sc = S/ (1+(Nc/Ns)) = 500MVA/ (1+(5/2)) = 143 MVA

To me it seems like they solved for the apparent power in the series coil. Or am I thinking about this wrong? please help. The apparent power of series coil and common coil are same in Auto Xmer so you can solve for anything. May be you have confusion with KVA of common wdg with output KVA. Let me put it like this. There 4 KVAs here. Input KVA=Output KVA and Series wdg KVA=Common wdg KVA. Try to locate all the four, you will feel good!!!!

 

[applepieordie]

I was able to solve the problem by solving for low side current, I2, first. But I was unable to solve the problem when I try to solve the problem by finding the high side current first. I am doing: I1 = (500MVA)/(200/sqrt(3) KV) = 3765A. What current did I just solve for? when I solve for this current I assume it is the current from the high side into the series winding.

@rg1 im sorry i will have to ask you to further explain the 4 KVAs you mentioned. I agree with input KVA = output KVA.  i havent fully understood why the series and common wdg KVA is equal. 

 

[rg1]

I was able to solve the problem by solving for low side current, I2, first. But I was unable to solve the problem when I try to solve the problem by finding the high side current first. I am doing: I1 = (500MVA)/(200/sqrt(3) KV) = 3765A. What current did I just solve for? when I solve for this current I assume it is the current from the high side into the series winding.

@rg1 im sorry i will have to ask you to further explain the 4 KVAs you mentioned. I agree with input KVA = output KVA.  i havent fully understood why the series and common wdg KVA is equal. 

Let me put is like this. Imagine an Auto Xmer made from a single phase Xmer having two wingdings. Now we know from our 1phase Xmer knowledge that both these windings always have same KVA ( V1I1=V2I2). While using the same windings as series/common winding in auto Xmer mode we do not change anything, then they should have same kVA . Hope you can now it make out now. If you need additional input, we can discuss it further. 

 

[rg1]

I was able to solve the problem by solving for low side current, I2, first. But I was unable to solve the problem when I try to solve the problem by finding the high side current first. I am doing: I1 = (500MVA)/(200/sqrt(3) KV) = 3765A. What current did I just solve for? when I solve for this current I assume it is the current from the high side into the series winding.

@rg1 im sorry i will have to ask you to further explain the 4 KVAs you mentioned. I agree with input KVA = output KVA.  i havent fully understood why the series and common wdg KVA is equal. 

As I mentioned earlier, the  Voltage  Ratio and turns ratios do not match in this question. Please do not solve this Q by Voltages. Solve it by Turns Ratio. Assume any Voltage for primary and the answer will be same. 

 

[applepieordie]

@Stephen2awesome @rg1

thanks for your explanations. I think i understand now.

I initially tried to solve for Ipri in the Sauto equation under fig. 17 then I tried to solve for Icomb by plugging in Ipri and Isec into the equation above it - it is from a reference i purchased from a website. can you tell me is the equation wrong? 

here is a pic of the the reference:

 

[rg1]

@Stephen2awesome @rg1

thanks for your explanations. I think i understand now.

I initially tried to solve for Ipri in the Sauto equation under fig. 17 then I tried to solve for Icomb by plugging in Ipri and Isec into the equation above it - it is from a reference i purchased from a website. can you tell me is the equation wrong? 

here is a pic of the the reference:

The equations given here (fig17) are good. The problem is in the question  of CI itself. The Voltage Ratios are not matching with the given turns ratio. Please do not use the Secondary Voltage given by CI Question, make your own secondary Voltage by turns ratios given. The Q can be done without even taking into account Voltages. Assume Primary Voltage as 1 or V , the secondary voltage you will get from turns ratio, and try to do it from your equations you will get it.

One good thing of a bad/wrong question is, it makes you learn more than a right question. 

 

[rg1]

The equations given here (fig17) are good. The problem is in the question  of CI itself. The Voltage Ratios are not matching with the given turns ratio. Please do not use the Secondary Voltage given by CI Question, make your own secondary Voltage by turns ratios given. The Q can be done without even taking into account Voltages. Assume Primary Voltage as 1 or V , the secondary voltage you will get from turns ratio, and try to do it from your equations you will get it.

One good thing of a bad/wrong question is, it makes you learn more than a right question. 

The equations given  at figure 17 are exactly what I was mentioning  about 4 KVAs in my previous reply. 

 

[applepieordie]

@rg1 OH I see now. this question really had me confused but I understand why i had so much trouble solving it now. Thank you for your help, really appreciate it!

 

[rg1]

@rg1 OH I see now. this question really had me confused but I understand why i had so much trouble solving it now. Thank you for your help, really appreciate it!

It is my pleasure!