# Problem 530 (NCEES)



## jeanbj2000

A 1,000 KVA, 12.47KV- 480Y/277-V transformer with 4% impedance feeds a 480-V bus. The fault duty of the 12.47-KV system is 40MVA. Assuming the transformer and the 12.47KV system have the same X/R ratio, the 3-phase short circuit current(Amperes) avaliable at the 480-V bus is mostlynearly:

I am having difficulties visualizing this problem. Can you anyone tackle this problem on a step by step process?.

-What so they mean by same "X/R ratio"?

Thanks!!


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## willsee

Look up the MVA Method for solving this problem.


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## sam314159

jeanbj2000 said:


> A 1,000 KVA, 12.47KV- 480Y/277-V transformer with 4% impedance feeds a 480-V bus. The fault duty of the 12.47-KV system is 40MVA. Assuming the transformer and the 12.47KV system have the same X/R ratio, the 3-phase short circuit current(Amperes) avaliable at the 480-V bus is mostlynearly:
> 
> I am having difficulties visualizing this problem. Can you anyone tackle this problem on a step by step process?.
> 
> -What so they mean by same "X/R ratio"?
> 
> Thanks!!


This is a Per Unit problem for me.

1. Choose a system 3-phase power base to be 1000 KVA (I chose the transformer base just for simplicty)

2. Your I_fault in per unit will be V/Z_total

V in per unit is 1 at an angle of zero degrees for our voltage source

Z_total = Z_transformer per unit + Z_transmission line per unit

Z_transformer per unit is given on a 1 MVA base which matches our chosen MVA base so we don't have to convert it, Z_transformer = j0.04

Z_transmission per unit is equal to MVA_Base_3_Phase/Fault_Duty_3_Phase = 1 MVA/40MVA = j0.025 (I remember this formula from grad school)

Z_total = j0.025 + j0.04 = j0.065 per unit (transformer in series with transmission line)

I_Fault = ( per unit voltage)/(0.065) = 15.38 per unit

Now we need to find our I_Base at the 480V bus to obtain I_actual:

I_Base_480V = (S_Base_Single_Phase)/(V_Base_Phase) = (333,333 VA/277 V) = 1203.37 A

I_Actual = I_Fault_Per_Unit * I_Base_480V = 15.38 * 1203.37 A = 18,507.82 A

Not sure if this matches how they solved it exactly, but that's how I would solve it.

Hope this helps.


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## Dolphin P.E.

jeanbj2000 said:


> A 1,000 KVA, 12.47KV- 480Y/277-V transformer with 4% impedance feeds a 480-V bus. The fault duty of the 12.47-KV system is 40MVA. Assuming the transformer and the 12.47KV system have the same X/R ratio, the 3-phase short circuit current(Amperes) avaliable at the 480-V bus is mostlynearly:
> 
> I am having difficulties visualizing this problem. Can you anyone tackle this problem on a step by step process?.
> 
> -What so they mean by same "X/R ratio"?
> 
> Thanks!!


Go thru this for "X/R ratio"

http://www.brainfiller.com/documents/XRRatio.pdf


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## jeanbj2000

Very Neat!!!!!

I didn't know about the following formula: ":Transmission per unit is equal to MVA_Base_3_Phase/Fault_Duty_3_Phase " but your approach makes perfectly sense to me. I would be curious to know how they derived this formula. Can you email, post or refer me to a book talking about Fault_Duty_3_Phase? Not much time left for April exam.

Thanks!!!!


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## jeanbj2000

sam314159 said:


> jeanbj2000 said:
> 
> 
> 
> A 1,000 KVA, 12.47KV- 480Y/277-V transformer with 4% impedance feeds a 480-V bus. The fault duty of the 12.47-KV system is 40MVA. Assuming the transformer and the 12.47KV system have the same X/R ratio, the 3-phase short circuit current(Amperes) avaliable at the 480-V bus is mostlynearly:
> 
> I am having difficulties visualizing this problem. Can you anyone tackle this problem on a step by step process?.
> 
> -What so they mean by same "X/R ratio"?
> 
> Thanks!!
> 
> 
> 
> This is a Per Unit problem for me.
> 
> 1. Choose a system 3-phase power base to be 1000 KVA (I chose the transformer base just for simplicty)
> 
> 2. Your I_fault in per unit will be V/Z_total
> 
> V in per unit is 1 at an angle of zero degrees for our voltage source
> 
> Z_total = Z_transformer per unit + Z_transmission line per unit
> 
> Z_transformer per unit is given on a 1 MVA base which matches our chosen MVA base so we don't have to convert it, Z_transformer = j0.04
> 
> Z_transmission per unit is equal to MVA_Base_3_Phase/Fault_Duty_3_Phase = 1 MVA/40MVA = j0.025 (I remember this formula from grad school)
> 
> Z_total = j0.025 + j0.04 = j0.065 per unit (transformer in series with transmission line)
> 
> I_Fault = ( per unit voltage)/(0.065) = 15.38 per unit
> 
> Now we need to find our I_Base at the 480V bus to obtain I_actual:
> 
> I_Base_480V = (S_Base_Single_Phase)/(V_Base_Phase) = (333,333 VA/277 V) = 1203.37 A
> 
> I_Actual = I_Fault_Per_Unit * I_Base_480V = 15.38 * 1203.37 A = 18,507.82 A
> 
> Not sure if this matches how they solved it exactly, but that's how I would solve it.
> 
> Hope this helps.
Click to expand...

Thanks sam314159. Neat job!!!


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## jeanbj2000

dolphin said:


> jeanbj2000 said:
> 
> 
> 
> A 1,000 KVA, 12.47KV- 480Y/277-V transformer with 4% impedance feeds a 480-V bus. The fault duty of the 12.47-KV system is 40MVA. Assuming the transformer and the 12.47KV system have the same X/R ratio, the 3-phase short circuit current(Amperes) avaliable at the 480-V bus is mostlynearly:
> 
> I am having difficulties visualizing this problem. Can you anyone tackle this problem on a step by step process?.
> 
> -What so they mean by same "X/R ratio"?
> 
> Thanks!!
> 
> 
> 
> Go thru this for "X/R ratio"
> 
> http://www.brainfiller.com/documents/XRRatio.pdf
Click to expand...

I will double check the site. Thanks!!


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## willsee

jeanbj2000 said:


> A 1,000 KVA, 12.47KV- 480Y/277-V transformer with 4% impedance feeds a 480-V bus. The fault duty of the 12.47-KV system is 40MVA. Assuming the transformer and the 12.47KV system have the same X/R ratio, the 3-phase short circuit current(Amperes) avaliable at the 480-V bus is mostlynearly:
> 
> I am having difficulties visualizing this problem. Can you anyone tackle this problem on a step by step process?.
> 
> -What so they mean by same "X/R ratio"?
> 
> Thanks!!


MVA Method

40MVA in series with 25 MVA (1MVA/.04)

40 X 25 / (40+25) = 15.38 MVA

Fault is on the 480v bus

15.38MVA/(sqrt(3) * 480e-6MV) = 18,500A


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## jeanbj2000

willsee said:


> jeanbj2000 said:
> 
> 
> 
> A 1,000 KVA, 12.47KV- 480Y/277-V transformer with 4% impedance feeds a 480-V bus. The fault duty of the 12.47-KV system is 40MVA. Assuming the transformer and the 12.47KV system have the same X/R ratio, the 3-phase short circuit current(Amperes) avaliable at the 480-V bus is mostlynearly:
> 
> I am having difficulties visualizing this problem. Can you anyone tackle this problem on a step by step process?.
> 
> -What so they mean by same "X/R ratio"?
> 
> Thanks!!
> 
> 
> 
> MVA Method
> 
> 40MVA in series with 25 MVA (1MVA/.04)
> 
> 40 X 25 / (40+25) = 15.38 MVA
> 
> Fault is on the 480v bus
> 
> 15.38MVA/(sqrt(3) * 480e-6MV) = 18,500A
Click to expand...

The MVA method seems to be shorter but it feels good to know more than one method.

Thanks for your quick response. Great forum!!


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## cableguy

http://www.arcadvisor.com/pdf/ShortCircuitABC.pdf

MVA method. VERY useful to learn. Doesn't take much to "master". It's very good for double-checking your work on the exam. IIRC, there were a couple problems on the exam, I worked them different ways (one way was MVA method) and got the same answer, so I was confident in my answer on those.


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## jeanbj2000

cableguy said:


> http://www.arcadvisor.com/pdf/ShortCircuitABC.pdf
> MVA method. VERY useful to learn. Doesn't take much to "master". It's very good for double-checking your work on the exam. IIRC, there were a couple problems on the exam, I worked them different ways (one way was MVA method) and got the same answer, so I was confident in my answer on those.


Thanks a lot cableguy. Very useful!! I am getting very confident for the April exam. 2 more weeks to go.


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## ros

cableguy said:


> http://www.arcadvisor.com/pdf/ShortCircuitABC.pdf
> MVA method. VERY useful to learn. Doesn't take much to "master". It's very good for double-checking your work on the exam. IIRC, there were a couple problems on the exam, I worked them different ways (one way was MVA method) and got the same answer, so I was confident in my answer on those.


good material,,,,

i have a question in the phase - ground claculation shown on page 4.7.6 in the document u attached... i dont know how they have shown three branches (fig 18) .

shouldn't there be only two branches... where the third branch came from... any help thanks....


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## jayache80

I searched for "branches" in this forum for the exact question you have, Ros. I do not understand this MVA method paper's calculation of SLG fault. Does anyone know why there are three branches?


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## jayache80

I think I figured out Moon's MVA Method for SLG fault (single phase to ground fault). He has three branches, as if it is a 3PH fault. And then he calculates the current as if it were a 3 phase fault, meaning he uses the formula Isc_SLG = MVAsc / (rad(3) * V_LL). In this case, MVAsc = 3*MVAsc_SLG.

To me, it is more intuitive to just use one branch (all three branches are the same anyway, that's why he could simply multiply by 3). The key here is to remember to use the Line-to-neutral, or line-to-ground voltage. After all, this is a single-LINE-To-GROUD fault, right??

Isc_SLG = MVAsc_SLG / (V_LN) = MVAsc_SLG / ( V_LL / rad(3) )

You'll get the same answer either way. It's more intuitive to me to use one branch because that's what the sequence network looks like in my power book!


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## jayache80

http://www.jmpangseah.com/wp-content/uploads/2003/01/chapter-5.pdf

This is what cleared it up for me (SLG fault with MVA method)


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## Wheretostart

this is a very helpful thread


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## StinkyTofu

Sorry for bringing an old thread back to life, but what is the significance of the problem statement "Assuming the transformer and the 12.47KV system have the same X/R ratio...
"? What if this were not the case?


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## Flyer_PE

They're just telling you the angles for the impedances of the system and the transformer are the same. It makes the vector math a little easier. It usually isn't a factor though since one impedance will typically be several orders of magnitude away from the other making the delta in the angles irrelevant.


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## JB66money

This problem can also be solved using the per-unit method.


When using the per-unit method there is an impedance associated with the fault duty.
This obtained first by getting the per-unit apparent power which is found by using the 1000kVA rating of the transformer the system's base apparent power and converting it into MVA, which is 1MVA. Next the fault duty's per-unit value is S=40MVA / 1MVA = 40p.u.
Next the per-unit impedancnce associated with the fault duty is z =vpu2 / spu= 12 / 40 = .025pu.
Next we find the per-unit fault current is
i.fault_pu = 1 / (.04+.025) = 15.38pu.Next the base current on the secondary side of the transformer where the fault occurs is Ib = (1000*103) / (sqrt(3) * 480 ) = 1202.81Amps
Finally we determine the acctual fault current by multiplying the per-unit fault current by the base current on the secondary side of the transformer.
IFAULT = i.fault_pu * Ib = (15.38)*(1202.81) = 18499 Amps.


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## Kalika PE

willsee said:


> MVA Method
> 
> 40MVA in series with 25 MVA (1MVA/.04)
> 
> 40 X 25 / (40+25) = 15.38 MVA
> 
> Fault is on the 480v bus
> 
> 15.38MVA/(sqrt(3) * 480e-6MV) = 18,500A


Thank you for this.  I started the PU method but it looked too lengthy.  The MVA method is much simpler and easier to solve with.


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## LJEngineer

willsee said:


> MVA Method
> 
> 40MVA in series with 25 MVA (1MVA/.04)
> 
> 40 X 25 / (40+25) = 15.38 MVA
> 
> Fault is on the 480v bus
> 
> 15.38MVA/(sqrt(3) * 480e-6MV) = 18,500A


Wow, thank you so much! This was definitely easy to solve.


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## yaoyaodes

what is 12.47-kv-480Y/277-V mean?

Does it mean that the primary is in the range 12.47-480 kv, Y connection? And the secondary is 277-V?


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## no_concentrate

It means Primary is 12.47kV ( Delta) and secondary is 480V (Y connection) 480V Line voltage and 277 is Line to Neutral Voltage.


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## David_EE

yaoyaodes said:


> what is 12.47-kv-480Y/277-V mean?
> 
> Does it mean that the primary is in the range 12.47-480 kv, Y connection? And the secondary is 277-V?


12.47 kV is the distribution line voltage (High side voltage). It could be 3 wire Delta 12.47 kV (no ground wire connection) or 4 wire WYE 7.2 kV (12.47/sqrt(3)=7.2) line-ground, 12.47 kV line-line). It depends on how you connect the transformer.

The low side winding of each single phase transformer is 277V line to ground. WYE configuration is three phases line to ground. If you measure between two phases [email protected] 0 deg and 277V @120 deg, you will read 480 V line to line. Or 277*sqrt(3) = 480V.

So each single phase transformer could be: 12.47kV/277V (Delta HS/ WYE LS) or 7.2kV/277V (WYE HS/WYE LS).
There is no 480V winding, it is a product of the phase angles.

Note, it could be a 3 phase transformer, which includes all three single phase sets of windings inside one transformer. In practice an overhead transformer bank (on the pole) is usually 3 individual transformers. And underground pad mount (on the ground) transformers are generally 3 phase transformers.


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## salmanshu322

we can use MVA Method, 40MVA in series with 25 MVA (1MVA/.04)
and 40 X 25 / (40+25) = 15.38 MVA
we find out that Fault is on the 480v bus
15.38MVA/(sqrt(3) * 480e-6MV) = 18,500A


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