# Spin-up questions version 2nd edition 2014



## supra33202 (Sep 12, 2017)

I hope you can help me with the following questions.

1) Q1-64, can you show me the reference regarding equivalent circuit fault?

2) Q2-11, can you show me the reference regarding equivalent circuit fault?

3) Q3-04, how do you get 6 MVA for the load? Should that information (load MVA) given in the question?

4) Q3-47, can you show me how to get reactance?

5) Q3-68, can you show me the reference regarding equivalent circuit fault?

6) Q3-78, how do you get 12 MVA for the load? Should that information (load MVA) given in the question?

7) Q4-48, how do you get 6MVA for the load? Should that information (load MVA) given in the question?

8) Q5-61, how do you get 6MVA for the load? Should that information (load MVA) given in the question?

Thanks!


----------



## rg1 (Sep 12, 2017)

supra33202 said:


> I hope you can help me with the following questions.
> 
> 1) Q1-64, can you show me the reference regarding equivalent circuit fault?
> 
> ...


Can you post the questions please


----------



## supra33202 (Sep 13, 2017)

1) Q1-64:






3) Q3-04:





4) Q3-47:










Thanks!


----------



## rg1 (Sep 13, 2017)

supra33202 said:


> 1) Q1-64:Explained in Glover, very nicely.
> 
> 
> 
> ...


Hope is has made sense. Thanks


----------



## supra33202 (Sep 14, 2017)

rg1 said:


> Hope is has made sense. Thanks


1) I understand Q3-47 now. Thanks!

2) I still need help with Q3-04. Here is the solution.






I still don't know how to get 6 MVA for the load.

Now, I am thinking maybe because generator A has max capacity of 3MVA and to match generator A, generator B should use 3 MVA out of 4 MVA capacity only. So the total MVA load should max out at 6MVA?

Actually, I still don't understand the principle. Generator B can contribute more than generator A. Why limit generator B's capacity?

Please advise.

Thanks!


----------



## rg1 (Sep 15, 2017)

supra33202 said:


> 1) I understand Q3-47 now. Thanks!
> 
> 2) I still need help with Q3-04. Here is the solution.
> 
> ...


To supply power while operation in parallel by any two sources, the terminal Voltage has to be equal because they are connected to same terminals. My advise will be instead of using the formula, understand the problem, go from basics of sharing load by two resistances in parallel.The condition of I1*Z1=I1Z2 has to be met while operation in parallel. where Z is actual Z, not per unit. Because Actual Zs of both generators are same in our case; when put in parallel they share same current and hence same MVA. I hope it is okay now. If you still have a doubt we can discuss a little more.


----------



## supra33202 (Sep 15, 2017)

rg1 said:


> To supply power while operation in parallel by any two sources, the terminal Voltage has to be equal because they are connected to same terminals. My advise will be instead of using the formula, understand the problem, go from basics of sharing load by two resistances in parallel.The condition of I1*Z1=I1Z2 has to be met while operation in parallel. where Z is actual Z, not per unit. Because Actual Zs of both generators are same in our case; when put in parallel they share same current and hence same MVA. I hope it is okay now. If you still have a doubt we can discuss a little more.


I got it now.

Thanks!


----------



## rg1 (Sep 15, 2017)

supra33202 said:


> I got it now.
> 
> Thanks!


It is my pleasure.


----------



## rg1 (Sep 16, 2017)

rg1 said:


> It is my pleasure.


There is a word of caution here. Generally synchronous Gen do not share power like this because in that case other conditions of Frequency and Excitation Voltage etc comes into play. This simple method of power sharing ican be applied to Xmers, Batteries etc. but is not to be applied when other info pointing to Synchronous Gen is given in the question.  I mean we have to see the question to decide on the approach.

This question  itself can be twisted by including different no load voltages of the sources. In that case there will be circulating currents among the sources and Load current will be less than the current supplied by sum of currents  from individual sources. The equation I1Z1=I2Z2 will become V1-I1Z1=V2-I2Z2=load Voltage.

However these are all applications of basic Electrical Engg laws.


----------



## Stephen2awesome (Sep 16, 2017)

So, MVAload would just be the lowest MVA rated xfmr *2?


----------



## rg1 (Sep 16, 2017)

Stephen2awesome said:


> So, MVAload would just be the lowest MVA rated xfmr *2?


If I have understood you correctly and if the question is pointing to my posts regarding the question in question, you are absolutely right. In this particular question the answer should be 6MVA (3*2 or 3+3).


----------



## Stephen2awesome (Sep 17, 2017)

rg1 said:


> If I have understood you correctly and if the question is pointing to my posts regarding the question in question, you are absolutely right. In this particular question the answer should be 6MVA (3*2 or 3+3).


Yes I was referring to the question. Thanks. I've found quite a few errors in Spin Up exams so far.


----------



## supra33202 (Sep 19, 2017)

NCEES practice exam question 125 has a similar problem as spin-up Q3-04. However, they use different method to solve it. It changes the impedance base.

I used Spin-up method, and I got 2000 KVA.

NCESS answer is 2500 KVA.

Could you talk about it?


----------



## rg1 (Sep 19, 2017)

supra33202 said:


> NCEES practice exam question 125 has a similar problem as spin-up Q3-04. However, they use different method to solve it. It changes the impedance base.
> 
> I used Spin-up method, and I got 2000 KVA.
> 
> ...


Both methods give same results of 2500KVA. Can you share your calculations to locate the error.


----------



## supra33202 (Sep 20, 2017)

rg1 said:


> Both methods give same results of 2500KVA. Can you share your calculations to locate the error.




*Problem:*

Transformer#1: 1000 KVA, Z1= 4.5%

Transformer#2 : 2000 KVA, Z2= 6.0%

What is the max total load (KVA) without overloading?

*Solution:*

Since the max capacity that transformer#1 can contribute without overloading is 1000 KVA, the load should be 1000 * 2 = 2000 KVA.\

Transformer#1: (1000KVA/4.5) / [ (1000KVA/4.5)+(2000KVA/6) ]   *   2000KVA = 800 KVA

Transformer#1: (2000KVA/6) / [ (1000KVA/4.5)+(2000KVA/6) ]   *   2000KVA = 1200 KVA

Max total load = 800 + 1200 = 2000 KVA

Answer shows 2500 KVA. Please advise.


----------



## rg1 (Sep 20, 2017)

supra33202 said:


> *Problem:*
> 
> Transformer#1: 1000 KVA, Z1= 4.5%
> 
> ...


I have explained in red. Hope it will make sense. I have not fully solved it for you to make you understand the concept and use of formula. If it does not make sense, we can discuss further. Think about using the basic formula of equalizing I1Z1 and I2Z2 where Z are actual Zs not the per unit and do the maths after that.


----------



## supra33202 (Sep 20, 2017)

rg1 said:


> I have explained in red. Hope it will make sense. I have not fully solved it for you to make you understand the concept and use of formula. If it does not make sense, we can discuss further. Think about using the basic formula of equalizing I1Z1 and I2Z2 where Z are actual Zs not the per unit and do the maths after that.


1) Q3-04 has the percent impedance not per unit impedance?

2) NCEES 125 has per unit impedance not percent impedance?

How do we know that?


----------



## rg1 (Sep 20, 2017)

supra33202 said:


> 1) Q3-04 has the percent impedance not per unit impedance?
> 
> 2) NCEES 125 has per unit impedance not percent impedance?
> 
> How do we know that?


Percent and per unit are same. 5% means 5/100=0.05 per unit. Now their relation with actual quantity in case of Z (Impedance) is Zpu=Zactual/Zbase where Z base is KV**2/MVA so Zpu=Za*MVA/KV**2 or Za=Zpu*KV**2/MVA.

This per unit or percent was not making any difference in the answer here because they were appearing in both Numerator as well as denominator.


----------



## supra33202 (Sep 20, 2017)

rg1 said:


> Percent and per unit are same. 5% means 5/100=0.05 per unit. Now their relation with actual quantity in case of Z (Impedance) is Zpu=Zactual/Zbase where Z base is KV**2/MVA so Zpu=Za*MVA/KV**2 or Za=Zpu*KV**2/MVA.
> 
> This per unit or percent was not making any difference in the answer here because they were appearing in both Numerator as well as denominator.


I am still confused.

Why can't I use the same method (spin-up Q3-04) to solve NCEES 125 problem?

Based on my understanding, they are asking the same thing - maximum total load (KVA) without overloading either transformers.

" ( In the earlier question both generators had equal Zs ( Not pu Zs), which resulted in equal sharing of loads, that may not be the case always, so better use I1Z1=I2Z2)  "

I know both problem's transformers have different per unit impedance.

Q3-04 has equal actual impedance? How do you know that?

How do we know NCEES doesn't have equal actual impedance?

"Percent and per unit are same. 5% means 5/100=0.05 per unit. Now their relation with actual quantity in case of Z (Impedance) is Zpu=Zactual/Zbase where Z base is KV**2/MVA so Zpu=Za*MVA/KV**2 or Za=Zpu*KV**2/MVA."

I agree.

Do you think you can spend some time to solve NCEES 125 problem?

Thanks!


----------



## rg1 (Sep 20, 2017)

supra33202 said:


> I am still confused. Okay let us do it.
> 
> Spin Up Formula Sa=(MVAa/puZa)*Stotal/(MVAa/puZa+MVAb/puZb) So 1000=(1000/4.5)*Stotal/(1000/4.5+2000/6) Solve this for Stotal; it gives Stotal =2500kVA Is it okay now. Now you can ask why not use the same formula for T2 taking 2000KVA. You will get a different result for Stotal and  you will find that T1 will be overloaded in that case. Just do it and see for yourself for better understanding of the case.
> 
> ...


----------



## rg1 (Sep 20, 2017)

rg1 said:


> A small addition- My diagnosis is you should improve on pu system. Read some good stuff on pu and do some questions on pu. Of course our discussion must have benefited you in that direction. I have given basic formula for Zpu in my posts. Just have a deep look into that. Try to derive those formula yourself.


----------



## supra33202 (Sep 20, 2017)

I think I got it now.

NCEES 125)

Since transformer 1 has the least capacity (1000kVA) and we don't want to overload it, then Sa=(MVAa/puZa)*Stotal/(MVAa/puZa+MVAb/puZb) So 1000=(1000/4.5)*Stotal/(1000/4.5+2000/6) Solve this for Stotal; it gives Stotal =2500kVA 

The maximum total load (KVA) is 2500KVA. The NCEES answer is correct.

Just to confirm:

Transformer 1 will contribute 1000 KVA of the 2500 KVA total load.

For transformer 2:

Sb=(MVAb/puZa)*Stotal/(MVAa/puZa+MVAb/puZb) So Sb=(2000/6)*Stotal/(1000/4.5+2000/6) Solve this for Sb; it gives Sb =1500kVA 

Transformer 2 will contribute 1500 KVA of the 2500 KVA total load.

And it makes sense that Sa + Sb = 1000 KVA + 1500 KVA = 2500 KVA total load.

Thanks!


----------



## rg1 (Sep 20, 2017)

supra33202 said:


> I think I got it now.
> 
> NCEES 125)
> 
> ...


----------



## rg1 (Sep 20, 2017)

rg1 said:


> Please do not always think that lower capacity source will be fully loaded, it will depend on actual Z of the source, you can have it otherwise also if Zs are like that. So be careful.


----------

