NCEES Question 530

[rmsg]

Can Anybody please explain me how to solve NCEES question 530. How did he get

Isc = 1.0 / (0.04 + 0.025)

I can understand 0.04 from 4% Transformer Impedance, but how and where did he get the term 0.025 ?

Please help I am confused ..

 

[Flyer_PE]

They solve the problem on a 1MVA base.  The impedance of the system is 100% (1pu) on a 40 MVA base.  When you convert that to a 1 MVA base you get (1pu*1MVA/40MVA) 0.025.

 

[rmsg]

They solve the problem on a 1MVA base.  The impedance of the system is 100% (1pu) on a 40 MVA base.  When you convert that to a 1 MVA base you get (1pu*1MVA/40MVA) 0.025.

that does makes some sense, so the total impedance is the impedance of system (with 40MVA = 0.025p.u.) + impedance of Transformer which is 4% impedance so z transf (p.u.) = 4% = 0.04

so total z(p.u.) = 0.04 + 0.025 = 0.065 right ?

if my interpretation is correct, than many thanks. I think I got it.

Cheers..

 

[Zach Stone P.E.]

4 hours ago, rmsg said:
that does makes some sense, so the total impedance is the impedance of system (with 40MVA = 0.025p.u.) + impedance of Transformer which is 4% impedance so z transf (p.u.) = 4% = 0.04

so total z(p.u.) = 0.04 + 0.025 = 0.065 right ?

if my interpretation is correct, than many thanks. I think I got it.

Cheers..


1

Correct. And since Vpu = 1, the per unit short circuit current is equal to the inverse of the equivalent per unit impedance, or: Ipu = 1/Zeq-pu. 

In my opinion, these type of simple fault problems are easier to solve with the MVA method, including this one.  We can use it to verify the solution. For example, the total apparent power at the fault is 15.38MVA:

• Sf = 40MVA//(1,000kVA/4%)
  
• Sf = 15.38MVA
  

The magnitude of the short-circuit current at the faulted 480V bus is 18.5kA:

• |Isc| = 15.38MVA/(√3·480V)
  
• |Isc| = 18.5kA
  

 

[rmsg]

Correct. And since Vpu = 1, the per unit short circuit current is equal to the inverse of the equivalent per unit impedance, or: Ipu = 1/Zeq-pu. 

In my opinion, these type of simple fault problems are easier to solve with the MVA method, including this one.  We can use it to verify the solution. For example, the total apparent power at the fault is 15.38MVA:

• Sf = 40MVA//(1,000kVA/4%)
  
• Sf = 15.38MVA
  

The magnitude of the short-circuit current at the faulted 480V bus is 18.5kA:

• |Isc| = 15.38MVA/(√3·480V)
  
• |Isc| = 18.5kA
  

Thanks Zach, and I agree one can check the work also.

 

[kris7o2]

What does 40MVA//(1,000kVA/4%) mean? I think I am confused as to what the // mean?

 

[Zach Stone P.E.]

What does 40MVA//(1,000kVA/4%) mean? I think I am confused as to what the // mean?

Hi Kris,

The double forward slash "//" is short hand for parallel, used to indicate adding two terms using the reciprocal of reciprocal sums, such as adding two resistors in parallel:

x//y = 1/[(1/x)+(1/y)]

Or:

x//y = (xy)/(x+y)

Even though the bus and transformer are in series, fault duty power quantities in volt-amps add in the reciprocal of reciprocal sums when in series, and they sum (x+y) when connected in parallel.

40MVA is the fault duty of the bus.

1,000kVA/4% is the fault duty of the transformer (power rating divided by percent impedance).