{"id":5671,"date":"2018-03-18T13:43:58","date_gmt":"2018-03-18T20:43:58","guid":{"rendered":"https:\/\/brainfiller.com\/?p=5671"},"modified":"2025-11-19T09:40:32","modified_gmt":"2025-11-19T16:40:32","slug":"short-circuit-calculations-with-transformer-and-source-impedance","status":"publish","type":"post","link":"https:\/\/brainfiller.com\/es\/electrical-safety-codes-and-standards\/short-circuit-calculations-with-transformer-and-source-impedance\/","title":{"rendered":"C\u00e1lculos de cortocircuito con impedancia de transformador y fuente"},"content":{"rendered":"\t\t
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Short Circuit Calculations \u2013 Transformer and Source Impedance<\/strong><\/h2>

\"Short<\/a>An infinite bus transformer short circuit calculation can be used to determine the maximum short circuit current on the secondary side of a transformer circuit using only transformer nameplate data.\u00a0 This is a good (and simple) method for determining the worst case MAXIMUM short circuit current through the transformer since it ignores the source\/utility impedance.\u00a0 Ignoring the source impedance means it is assumed to be zero and voltage divided by zero is infinite, hence the often-used term \u201cinfinite bus\u201d or \u201cinfinite source\u201d.<\/p>

In my Infinite Bus Article<\/a><\/strong> at brainfiller.com, the infinite bus method is illustrated to calculated the maximum worst case short circuit current at the 480 Volt secondary of a 1500 kVA transformer.\u00a0 Using the \u201cinfinite bus\u201d or \u201cworst case\u201d approach indicated an available short circuit current of 31,374 Amps.<\/p>

\"impedance<\/a>However, what if you are evaluating the adequacy of a panel on the secondary that has a short circuit rating of 30,000 Amps?\u00a0 \u00a0With the infinite bus approach, it would mean the panel has an inadequate interrupting rating. But does it really? This could be an expensive conclusion based on assumed (infinite primary) data.<\/p>

Another concern about using the infinite bus approach is if the short circuit calculations are to be used for an arc flash study.\u00a0 This is something that I discuss in my Arc Flash Training class about using IEEE 1584 to perform arc flash calculations.<\/p>

With arc flash studies, the higher short circuit current may result in the worst-case incident energy but this is not always the case.\u00a0 It is possible that a lower short circuit current could result in a protective device taking longer to operate which would result in a longer arc flash duration and increase in the total incident energy.<\/p>

To provide more accurate short circuit calculations, the source impedance needs to be included.\u00a0 Let\u2019s see how to include the effect of the actual source short circuit current and the equivalent source impedance.\u00a0 To account for the source impedance, the same formula that was used for the infinite bus solution can be used but a few more steps need to be added.<\/p>

The infinite bus formula is based on transformer impedance as shown below.\u00a0 It ignores the source impedance:<\/p>

SCAsecondary<\/sub> =\u00a0 x ( FLAsecondary<\/sub> x 100 ) \/ (%Ztransformer<\/sub>)<\/p>

Source and Transformer Impedance<\/strong><\/h2>

\"transformer<\/a>The actual short circuit current available at a transformer\u2019s secondary bus is not just a function of transformer short circuit impedance, but it is also dependent on how strong the source is at the primary of the transformer.\u00a0 A transformer connected to a strong source such as close to a major utility substation, will have a greater secondary short circuit current than if the same transformer was connected to a weak source such as a long distribution line in a rural area.<\/p>

To factor in the strength \/ weakness of the source impedance we only need to add one extra variable, % Zsource<\/sub>\u00a0 to the previous equation.<\/p>

The new equation would be:<\/p>

SCAsecondary<\/sub> =\u00a0\u00a0 ( FLAsecondary<\/sub> x 100 ) \/ (%Z transformer<\/sub> + %Z source<\/sub>)<\/p>

By adding %Zsource<\/sub>\u00a0 to\u00a0 %Ztransformer<\/sub>\u00a0 the strength of the source is included.\u00a0 A stronger source will have a smaller value for\u00a0 %Zsource<\/sub>\u00a0 and a weaker source will have a larger value.<\/p>

The calculation procedure is similar to the infinite bus calculation, but we now have to add the additional step of calculating the source impedance:<\/p>

Step 1 \u2013 <\/strong>The source impedance formula used to calculate the equivalent source impedance in percent:<\/p>

%Zsource<\/sub> = (kVAtransformer<\/sub> \/ kVAshort circuit<\/sub>) x 100<\/p>

where:<\/p>

kVAshort circuit<\/sub>\u00a0 = kVL-L<\/sub> x Sqrt (3) x SCAprimary<\/sub><\/p>

\"transformer<\/a>This seems simple enough but where do you obtain the SCAprimary<\/sub>?\u00a0 Great question!\u00a0 If the transformer is going to be connected to the utility system, the utility company is usually the source of this information.\u00a0 It is best to start out by determining who is the utility account representative and ask if they can either provide you with the information or direct you to someone that might have the information.<\/p>

If the transformer is not directly connected to the utility but is further downstream in a power distribution system, you will need to obtain short circuit calculations for the upstream part of the system.\u00a0 This means someone (perhaps you) will have to perform short circuit calculations from the utility down through the power distribution system.<\/p>

If you are unable to determine any of this information, and you are concerned about worst case highest magnitude short circuits, you can always default to the simpler and generally more conservative infinite bus calculation.<\/p>

Care must be taken!\u00a0 Infinite bus calculations are good for the evaluation of the maximum worst case short circuit current through the transformer (excluding motor contribution and impedance tolerances for transformers not yet delivered\/tested).\u00a0 However, if you are interested in minimum short circuit currents for analysis such as arc flash, voltage flicker or harmonic resonance, an infinite bus calculation is not<\/em> the way to go.<\/p>

Derivation of Step 1 <\/strong><\/h3>

\"transformer<\/a>The source impedance calculation may seem a little odd the first time you see it.\u00a0 Dividing two different kVA\u2019s magically becomes an impedance? However, this method has its origin with the Per Unit System.\u00a0 The %Zsource<\/sub> is actually the true primary source impedance of the source in ohms, divided by the transformer base impedance in ohms.\u00a0 Her is how the derivation of step 1 works:<\/p>

%Zsource<\/sub> = (Zsource ohms<\/sub> \/ Ztransformer base<\/sub> ) x 100<\/p>

\u00a0%Zsource<\/sub> = [(kVL-L2<\/sub>)^2\/MVAshort circuit<\/sub>\u00a0] \/ [(kVL-L2<\/sub>)^2\/MVAtransformer<\/sub>] \u00a0x 100<\/p>

where:<\/p>

\u00a0Zsource ohms\u00a0 <\/sub>=\u00a0(kVL-L2<\/sub>)^2\/MVAshort circuit<\/sub><\/p>

Ztransformer base<\/sub>\u00a0 = kVL-L2<\/sub>)^2\/MVAtransformer<\/sub><\/p>

The kVL-L2<\/sub> in the numerator and denominator cancel each other and you are left with:<\/p>

%Zsource<\/sub> = [ (1 \/MVAshort circuit <\/sub>) \/ (1 \/MVAtransformer<\/sub> ) ]\u00a0 x 100<\/p>

which becomes:<\/p>

%Zsource<\/sub> = ( MVAtransformer<\/sub> \/ MVAshort circuit <\/sub>) x 100<\/p>

or in our case we use Kilo instead of Mega so our numbers are scaled by 1000:<\/p>

\u00a0%Zsource<\/sub> = ( kVAtransformer<\/sub> \/ kVAshort circuit<\/sub> ) x 100<\/p>

\u00a0<\/strong>Step 2 \u2013 <\/strong>Calculate the secondary full load current rating of the transformer:<\/p>

FLAsecondary <\/sub>=\u00a0 kVA3phase<\/sub>\u00a0 \/ ( kVL-L<\/sub> x Sqrt (3))<\/p>

Step 3<\/strong> \u2013 Calculate the short circuit current on the transformer secondary bus, but this time we use the transformer impedance AND the source impedance.<\/p>

\u00a0SCAsecondary<\/sub> = ( FLAsecondary<\/sub> x 100 ) \/ (% Ztransformer<\/sub> + %Zsource<\/sub>)<\/p>

Here is an example of the calculation<\/strong>:<\/h2>

Let\u2019s say we have a transformer rated 1500 kVA with a secondary voltage of 480Y\/277V, a primary voltage of 13.2 kVL-L<\/sub> and an impedance of 5.75%.\u00a0 \u00a0Suppose the utility informs us that their maximum short circuit current available at the transformer\u2019s primary is 6,740 Amps at 13.2 kV.<\/p>

Step 1 <\/strong>\u2013 Calculate the source impedance:<\/p>

\u00a0kVAshort circuit<\/sub>\u00a0 = 6,740 Amps x 13.2 kVL-L<\/sub> x\u00a0 sqrt(3)<\/p>

\u00a0kVAshort circuit<\/sub>\u00a0 = 154,097 kVA<\/p>

(some utility companies might refer to this as 154 MVA )<\/p>

% Zsource<\/sub> =\u00a0 ( 1500 kVA \/ 154,097 kVA ) x 100<\/p>

% Zsource<\/sub> =\u00a0 0.97%<\/p>

\u00a0<\/strong>Step 2<\/strong> \u2013 Just like last month, calculate the secondary full load current rating of the transformer.<\/p>

\u00a0FLAsecondary<\/sub> = 1500 kVA \/ (0.48 kVL-L<\/sub> x Sqrt (3))<\/p>

\u00a0FLAsecondary<\/sub> = 1804 Amps<\/p>

\u00a0<\/strong>Step 3<\/strong> \u2013 Calculate the short circuit current on the transformer secondary bus.<\/p>

SCAsecondary<\/sub> = 1804 Amps x 100 \/ (5.75% + 0.97%)<\/p>

SCAsecondary = 26,845 Amps<\/p>

If this calculation ignored the source and assumed it was infinite, the short circuit current at the secondary would be<\/p>

SCAsecondary <\/sub>= 31,374 Amps<\/p>

You can see that factoring in the source impedance (source strength) has a significant effect on the magnitude of short circuit current at the transformer secondary terminals.<\/p>

All of the variables listed above are:<\/p>

FLAsecondary<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = Secondary Full Load Amps<\/p>

kVL-L2<\/sub> \u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = Secondary voltage in kV<\/p>

kVA3phase <\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = Transformer three phase kVAself cooled <\/sub><\/p>

Sqrt (3)\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = Square root of three (1.73)<\/p>

% Ztransformer<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = Transformer percent impedance<\/p>

% Zsource<\/sub> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = Source percent impedance referenced to the transformer base<\/p>

kVAshort circuit<\/sub>\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0 = Short circuit power<\/p>

SCAsecondary<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 3 Phase Short Circuit Amps at the secondary bus<\/p>

SCAprimary<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 3 Phase Short Circuit Amps at the primary bus<\/p>

A few more words of caution!\u00a0 The impedance of a transformer must be the actual nameplate and not an assumed value.\u00a0 Impedances of transformers that have not yet been built or tested can vary by + \/ – 7.5% of the specified impedance.\u00a0 The above calculation does not include motor contribution which will also need to be considered.<\/p>

Adding the source and transformer impedances like we just did is good for a close approximation but it is not perfect.\u00a0 The impedances should be added using vector addition which means divide each impedance into its respective X and R value and combine the individual terms to determine the true magnitude of the total impedance.\u00a0 The X\/R ratio is discussed in the X\/R article<\/a><\/u><\/strong>.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<\/div>\n\t\t","protected":false},"excerpt":{"rendered":"

Short Circuit Calculations \u2013 Transformer and Source Impedance An infinite bus transformer short circuit calculation can be used to determine the maximum short circuit current on the secondary side of a transformer circuit using only transformer nameplate data. 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