# Base Voltage Conversion from region to region- Kaplan 3.18 - 3.21



## trying2pass (Mar 4, 2010)

T1: 240MVA Delta/Y 45KV:345KV

T2: 210MVA Y/Y 345KV:138KV

T3: 180MVA Y/Delta 132KV:13.6KV

Region 1 includes the ground, voltage source Vg, and primary of T1.

Region 2 includes the secondary of T1 and primary of T2 with an impedance of Z12 between them.

Region 3 includes the secondary of T2 and primary of T3 with an impedance of Z12 between them.

Region 4 includes the secondary of T3 and a load of 8 ohms, and ground.

*3.18 states *=&gt; Choosing a base of 240MVA (three phase) and 345 KV (L-L) in Region 2. What is the per unit impedance of the load?

I understood how they solved it except one part. Getting the base voltage at Region 4. They used 345(138/345)(13.6/132) = 14.22KV. From there they found the base impedance and then the per unit value of the load impedance.

*3.19 states* =&gt;Choosing a base of 240MVA (three phase) and 138KV (L-L) in Region 3. Generator volt chosen as reference. Load current is .06 with an angle -10.35 pu. What is complex power delivered by generator?

On this one, they go the base voltage on Region1 as 138(345/138)(45/345) = 45KV.

Anyway, can someone explain to me how the base voltage conversion work? There is also problem 3.20 and 3.21 that does the voltage conversion, but I feel like it was done different each time.

Please help!! :sharkattack:

Thanks in advance.


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## Flyer_PE (Mar 4, 2010)

When you choose a base voltage for the system, your voltage is 1.0 pu at that point.

In 3.18, they picked 345 kV as the base. Now, the base voltage on the other parts of the system are dependent on the turns ratios of the transformers, not their voltage ratings.

On the 345 kV level, 345 kv = 1pu

Since the turns ratio for the transformer is 345kV:138kV, the base voltage for region 3 is 138 kV.

The place where you have to pay attention on these is where the turns ratio doesn't match the voltage level. The base voltage for region 4 will _not_ be 13.6 kV because the primary for T3 is given as 132 kV.

The base voltage for region 4 is then 138kV*(13.6kV/132kV) = 14.22 kV = 1 pu.

In 3.19, the voltages match the turns ratios so 138 kV in region 3 results in 45 kV being 1.0 pu in region 1.


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## trying2pass (Mar 5, 2010)

Flyer_PE said:


> When you choose a base voltage for the system, your voltage is 1.0 pu at that point.
> In 3.18, they picked 345 kV as the base. Now, the base voltage on the other parts of the system are dependent on the turns ratios of the transformers, not their voltage ratings.
> 
> On the 345 kV level, 345 kv = 1pu
> ...



I read your post more than a few times. It makes sense, but still confused, but not as confused after reading your post. I guess my next question would be how did you know which voltage or ratio goes on the numerator and which goes on the denominator?

When you say turns ratio I immediately think about a=Vpri/Vsec which is opposite on E3.18. On E3.19, it looks like they used the a=Vpri/Vsec. I'm obviously missing something here. Can you please explain? Thanks!


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## Flyer_PE (Mar 5, 2010)

It depends on which voltage you are trying to find.

If you have a 345 kV - 138 kV transformer, the ratio between primary and secondary voltage is then 345:138.

VPrimary/VSecondary=345/138

It's simple algebra from there. Just solve for the unknown.

VPrimary = VSecondary*(345/138)

VSecondary = VPrimary*(138/345)

A simple check on whether you have done it correctly is: Does the answer make any sense. If you have the turns ratio inverted, you're going to wind up with an answer that will be either MUCH smaller or MUCH larger than what you expect.


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## CLTEE49 (Mar 5, 2010)

This may not be a good place for this but I didnt want to start another topic.

When you are given a variable MVA transformer (30/40/50MVA) with a transformer impedance (say Xs = 10%), how do you go about calculations for the line? Do you look at it as a 50MVA xfmr?

You can change it to any new base you want, but what do you use as the old base, 30, 40, or 50 MVA? Zpu(new)=Zpu(old)*MVA(new)/Mva(old)

What MVA do you use for current calculations? I=S/(sqrt(3)*V)


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## Rei (Mar 5, 2010)

What about the base current for each region? Do we use the same base VA, in this case is 240MVA divided by the base voltage for each region? Each transformer has a different VA rating. I didnt get the same base current as the given answer.


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## Flyer_PE (Mar 5, 2010)

CLTEE49 said:


> This may not be a good place for this but I didnt want to start another topic.
> When you are given a variable MVA transformer (30/40/50MVA) with a transformer impedance (say Xs = 10%), how do you go about calculations for the line? Do you look at it as a 50MVA xfmr?
> 
> You can change it to any new base you want, but what do you use as the old base, 30, 40, or 50 MVA? Zpu(new)=Zpu(old)*MVA(new)/Mva(old)
> ...


When a transformer manufacturer gives multiple ratings, the percent impedance is based on the self-cooled rating of the transformer. In the case you describe, the 10% impedance is on a 30 MVA base. The MVA used to determine the base current will be the base MVA you have chosen for the system.


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## trying2pass (Mar 5, 2010)

Flyer_PE said:


> It depends on which voltage you are trying to find.
> If you have a 345 kV - 138 kV transformer, the ratio between primary and secondary voltage is then 345:138.
> 
> VPrimary/VSecondary=345/138
> ...



That makes so much more sense!! I tried it on 3.18 and 3.19 and it worked! However, I did try this very same concept to the next two problems (3.20 and 3.21) and got one of the conversions wrong. Here are the ratings:

(One line connected from left to right)

Generator G: 200MVA, 22kv, x=.2pu

T1: 300MVA, 220/25kv delta/Y, x=.1pu

T2: three single phase trans. each rated at 100MVA, 130/13KV, x=.1pu

Motor M: 300MVA, 13kv, x=.1pu

They asked what is the p.u. impedance of motor M with a base of 200MVA(three phase) and 22kv(L-L) in the generator section.

Using the concept of VSecondary = VPrimary*(138/345) with VPrimary = 22kv, I get Vsecondary=(22)*(25/220)(13/130/sqrt(3)).

Their answer to convert this is (22)*(220/25)(13/130/sqrt(3)). I hope that this is one of those mistakes that they have in this book, but then again maybe not.

How come the second part of the conversion is (220/25) and not (25/220)?

Thanks in advance!!


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## Flyer_PE (Mar 5, 2010)

trying2pass said:


> (One line connected from left to right)Generator G: 200MVA, 22kv, x=.2pu
> 
> T1: 300MVA, 220/25kv delta/Y, x=.1pu
> 
> ...


The generator voltage is 22 kV, the transformer ratio is 25kv:220 kV. The expectation is that the voltage will be slightly lower than 220 kV. The voltage on the high-side of T1 will be 22kV*(220kV/25kV) = 193.6 kV.

The beauty of pu analysis is that T1 and T2 are not even part of the problem. All you need to do is convert the motor impedance to a new base.

There are two conversions at work here. The first is a change from the 300 MVA base of the motor to the 200 MVA base they are looking for.

ZpuNew=ZpuOld*(MVAnew/MVAOld)

The second conversion is for voltage. Impedance changes with the inverse square of voltage:

ZpuNew=ZpuOld*(kVOld/kVNew)^2

ZpuNew = 0.1*(200/300)*(13/22)^2 = 0.023 pu


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## trying2pass (Mar 6, 2010)

The generator voltage is 22 kV, the transformer ratio is 25kv:220 kV. The expectation is that the voltage will be slightly lower than 220 kV. The voltage on the high-side of T1 will be 22kV*(220kV/25kV) = 193.6 kV.

The beauty of pu analysis is that T1 and T2 are not even part of the problem. All you need to do is convert the motor impedance to a new base.

There are two conversions at work here. The first is a change from the 300 MVA base of the motor to the 200 MVA base they are looking for.

ZpuNew=ZpuOld*(MVAnew/MVAOld)

The second conversion is for voltage. Impedance changes with the inverse square of voltage:

ZpuNew=ZpuOld*(kVOld/kVNew)^2

ZpuNew = 0.1*(200/300)*(13/22)^2 = 0.023 pu


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## Flyer_PE (Mar 6, 2010)

trying2pass said:


> *correction*Reading the post, I forgot to mention T2 for the three single phase transformers is a y/delta configuration.


No problem. It had to be connected that way or the motor wouldn't be supplied anywhere near the correct voltage.



trying2pass said:


> How did you get the transformer ratio to be 25:220 when the given ratio is 220:25 (primary:secondary) for T1?


The which coil is considered primary and which is secondary is purely arbitrary. Transformers don't much care which way the power is going. The same transformer can be used for either a step-up or step-down. The 22 kV generator had to be connected to the 25 kV winding.



trying2pass said:


> When I first answered this question, I did exactly the same as you just did with ZpuNew = 0.1*(200/300)*(13/22)^2 = 0.023 pu and thought it was correct cuz that's how I learned it.
> The answer is .0992pu. What they did is they found the base voltage on the motor side to be 11.1775KV. Used that value instead of the 13KV on the motors side. I definitely would have gotten this wrong becuase I would have gotton .023pu and moved on.


You were and are correct. Their answer is wrong. The motor impedance is on the motor nameplate base voltage, NOT the voltage actually supplied to it. I've never seen a nameplate impedance on a 11.18 kV base.


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## glnn01 (Mar 10, 2010)

For Problem 3.19, how did they get the actual voltage for the generator to be 25kV?


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## Bluekayak (Mar 10, 2010)

glnn01 said:


> For Problem 3.19, how did they get the actual voltage for the generator to be 25kV?


I'm thinking that's a mistake. The problem states that the generator voltage is 45kV (line-to-line). So I used 1 per unit for the generator voltage and got S=(14.2+j2.59)MVA for the complex power delivered by the generator. Does that sound reasonable?

I don't agree with the solution to #10 either. Section 310.4 of the 2008 NEC clearly states that conductors of size 1/0 and larger shall be permitted to be connected in parallel. The problem statement does not meet the criteria for either one of the two exceptions. Therefore, I think the answer to #10 should be 1/0.


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## glnn01 (Mar 11, 2010)

OK, thanks. Yeah, I thought I wasn't sure if I was missing something here. Generally, they would have to provide the actual voltage parameter for the generator to arrive at the S (actual) but I think they forgot to add that to circuit diagram. On 3.18, at least they provide the actual impedance for the load to get the correct per-unit impedance.


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