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MVA Method Short Circuit Calculation

A Short Circuit Study is an important tool in determining the ratings of electrical equipment to be installed in a project. It is also used as a basis in setting protection devices. Computer software simplifies this process however, in cases where it is not available, alternative methods should be used. The per-unit and ohmic method are very tedious manual calculation. These hand calculations are very prone to errors due to so many conversion required. In per unit, base conversion is a normal part of the calculation method while in ohmic method, complex entities conversion.The easy way to do hand calculation is the MVA method.

In this example, we shall be presenting a short circuit study of a power system. Motors are are already lumped with ratings 37kW and below assigned an impedance value of 25% while larger motors are 17%. A 4MVA generator is also included into the system to augment the utility.

sld2_a

Figure 1

Utility: 33KV, 250 MVAsc
Transformer 1: 10 MVA, 33/11KV, 9% Z

11KV Bus
Generator: 3MVA, X"d = 0.113
Transformer 2: 5 MVA, 11/6.6KV, 7% Z
Motor 1: 5MVA (Lumped), 17% Z

6.6KV Bus
Transformer 3: 2 MVA, 6.6KV/400V, 6% Z
Motor 3: 6.8 MVA (Lumped), 17% Z

400V Bus
Motor 4: 300 KVA (Lumped), 17% Z
Motor 5: 596 KVA (Lumped), 25% Z

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13 Responses

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  1. paulwilkie says

    this is very good example of short circuit calc. by MVA method.i like this.thanks a lot!

  2. davidcris castro says

    Very good article!……I presume that the single phase fault mentioned in page 5 is a line to ground fault. How about a line to line fault? a line to line to ground fault ? Mabuhay ka kabayan!…

  3. ABHIK CHATTERJEE says

    In the MVA method of short circuit calculation
    The generator MVAsc = 35.4 + 133.39 = 168.79 MVA
    From the calculation it is not clear that where from the value 133.39 is derived. Is it possible to explain that where from the value derived.

    The motor 1 MVAsc = 139.38 + 29.41 = 168.79 MVA
    From the calculation it is not clear that where from the value 139.38 is derived. Is it possible to explain that where from the value derived.

  4. Ver says

    Both values are calculated:
    133.39 = 76.87 + 27.11 + 29.41

    139.38 = 76.87 + 27.11 + 35.4

  5. AB says

    What will be the Single Phase Fault Calculation at the 11 KV Bus ? Can you please explain.

  6. Joseph says

    Your example is an interessant one for the beginning of the power system analysis.
    The two books where i find are:
    Electric power system Protection and coordination from Michael A.Anthony 1995 Mac Graw Hill
    The second one is a german Book
    Elektrische Krafwerke und Netze from Happoldt Oeding 1978 or Oeding Oswald 2004 Springer Verlag
    I search also in the Brown Book from IEEE, but don’t find it.

    How can we bridge the MVA Method with the %or per unit method?
    See on a research motor the document “Short circuit ABC-Learn it in one Hour Use it Anywhere, Memorize no Formula” from Moon H.Yuen Senior Member IEEE.
    You will find there the comparaison between MVA method and ohmic and/or per unit method

    I don’t know if any softwares use it.

    Anyway it is a good tool during commissioning.

    How can we connect the transient to the MVA method is another question.
    Joseph

  7. Ver says

    The relationship between % or per unit method is just OHM’s Law as I have explained in this topic.

  8. James says

    To determine the Faults Current at any bus on the power system, add the MVA values above and below the arrows. The sum should be the same on any branch.
    Example:
    11 KV Bus:
    From Transformer 1: MVAsc = 76.87 + 91.92 = 168.79 MVA
    From Generator : MVAsc = 35.4 + 133.39 = 168.79 MVA
    From Transformer 2: MVAsc = 141.68 + 27.11 = 168.79MVA
    From Motor 1: MVAsc = 139.38 + 29.41 = 168.79 MVA

    Hi, I just need a clarification. From you answer, are you telling that Generator and Motor contributed up to 168MVA short circuit? Let’s say we us your example, and I wanted to know how much Motor 1 contribute to the fault on the 11kV bus. You’re saying it’s 168MVA? I know how you got the answer, but it’s hard to believe that a small motor can contribute that much fault current. Help me understand here. Thanks!

  9. Ver says

    The total fault MVA (168MVA) in the 11kv branch are conbtributed by all sources, the normal power, the generator and the motor.

  10. Pramod Chavan says

    Dear Sir

    This is a good example. Can you suggest me some text books which can cover entire system fault calculation, relay co-ordination part?

    Thanks

    Pramod Chavan

  11. Ver says

    Here are some books I could recommend:
    1. Protective Relaying Theory and Application – Walter Elmore
    2. Practical Power Systems Protection – Les Hewitson
    3. Protection of Electrical Networks – Christophe Prévé

  12. jhunsg says

    The switchboard (600VAC) is powered by diesel generator and supplied power to 1000HP, 750VDC motor through 3 phase rectifier (SCR). How to compute DC motor’s contribution if fault occurs in switchboard side.

  13. Ver says

    Theoretically, there will be no motor contribution if fault occurs at the switchboard. The SCRs will block any back flow of current.



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