Transformers Q

[cruzy]

Anyone can answer this, but for anyone who's got the Chelapati book, this is on page 3-48, why do they divide the 208 by sqrt 3 to transfer the l-n voltage on the low side to the l-n voltage on the high side?? The high side is delta, so the single phase equivalent would still just be the 2400V. Am I wrong here?? I assume that they say that "the stated voltage require the three single-phase transformers to be connected Delta-Wye..." because the generator nominal voltage (which means is L-L) is the same as the given high side rating of the transformer, and the load is 208V but the rating of the single phase low side of the transformer is 120V. Otherwise, how can you tell just by the voltages what the configuration of the transformer (without assuming standard ratings).

Thanks,

Cruz

 

[cableguy]

I think they showed the 208/sqrt(3) so that you understand that you're now working with a 3-phase, line to neutral voltage for the voltage drop equation, instead of the single phase 120 volt transformer that you started with.

This ends up being a neat little voltage drop plus three single phase transformer delta/wye connection problem.

Combine NCEES Practice Exam 501 and 506 and that's very similar to what you're working with here.

From 501: you calculate the source voltage at the transformer secondary. Here it's the generator source voltage.

From 506: you calculate the turns ratio of 3 single phase transformers connected in a delta-wye format

 

[kulalv]

If at all possible, could you post the question with the solution? Trying to work as many good sample questions I can.

thanks!

 

[cableguy]

I don't have time to post the solution, maybe someone else can.

Three single phase 2400-120V transformers are used to make a 3 phase transformer. This is fed by a (nominal) 2400V generator that is separated from the transformer by a line impedance of 2+3j ohms. The load (on the low side of the transformer) is 200kW at 208V, 3 phase.

Find the voltage at which the generator must operate.

I'll make it multiple choice. Most nearly:

(_a_) 2400

(_b_) 2490

(_c_) 2590

(_d_) 2650

Answer later.

 

[kulalv]

I don't have time to post the solution, maybe someone else can.
Three single phase 2400-120V transformers are used to make a 3 phase transformer. This is fed by a (nominal) 2400V generator that is separated from the transformer by a line impedance of 2+3j ohms. The load (on the low side of the transformer) is 200kW at 208V, 3 phase.

Find the voltage at which the generator must operate.

I'll make it multiple choice. Most nearly:

(_a_) 2400

(_b_) 2490

(_c_) 2590

(_d_) 2650

Answer later.

Need clarification on problem:

Is the "2400-120V" stated in the problem? Or, is the "208V, 3 phase" stated? Just want to make sure we have the correct problem statement.

Took a stab at it:

Isecondary = 200kW/sqrt(3)/208V = 555.2A

a=2400/120 = 20

Iprimary = Isecondary/a = 27.8A

Vd,cable = IZ = (27.8)(2+j3) = 100V

Not sure here, I'm thinking convert back to three phase for generator? Which would be Vd, cable*sqrt(3)? Not sure since the "answers" given are around the 2400V range.

Any help?

 

[cableguy]

No answers yet!

The 2400-120 is stated in the problem. You have 3 single phase transformers that are 2400-120. You have to hook them up in a manner to get 208 volts 3 phase (that is also stated) on the equipment side, and 2400 on the source side. You are NOT given the information as to how to hook them up. That's up to you...

I will say, you need to think some more about NCEES problem 506. In that problem, you're given a delta-wye 115kV / 24kV bank of 3 single phase transformers, and they want to know the actual turns ratio. Is it 115:24?

 

[DK PE]

Took a stab at it:Isecondary = 200kW/sqrt(3)/208V = 555.2A

a=2400/120 = 20

Iprimary = Isecondary/a = 27.8A

Vd,cable = IZ = (27.8)(2+j3) = 100V

Not sure here, I'm thinking convert back to three phase for generator? Which would be Vd, cable*sqrt(3)? Not sure since the "answers" given are around the 2400V range.

Any help?

Be careful with your use of a = turns ratio of the INDIVIDUAL single phase transformers...

One thought process that may help when under the pressure of a test in a case like this is to think of the delta-wye Txfmer as a black box. You know the power on the output side (low side) of the transformer is 200kW in this case --> Therefore the input to a lossless transformer is 200kW. What is the line current on the primary side of a black box with 200kW dissipation with a Vline = 2400 V? The answer should be 48.1 A. You can also find this as the ratio of the line voltages on each side of the transformer i.e. 555 *( 208/2400) = 48.1A.

 

[knight1fox3]

The answer I would "most nearly" select is C. I came up with a VgenL-L of 2579V.

 

[cableguy]

C is correct. Chelapati says 2592.

 

[knight1fox3]

C is correct. Chelapati says 2592.

Interesting. Perhaps I have a newer/older version of the Chelapati book. Attached is what I have for the problem that was referenced. Answer 2579.0V. Still "most nearly"

 

[DK PE]

The answer I would "most nearly" select is C. I came up with a VgenL-L of 2579V.

I agree and got 2578V. I also quickly did drop using IEEE-141 approx and got close enough to get "most nearly" but was 10 - 12 volts low. I think 2592 is slightly high.

 

[cableguy]

D'OH! I read it wrong. My bad. My Chelapati book says 2579 as well. No idea where I got 2592.

I'm going to shut up now and finish a few last things to get ready.