# Thread: Short Circuit MVA

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## Short Circuit MVA

A question about short circuit MVA (MVAsc). If we have a generator of 2 MVA, then what will be its MVAsc ?

2. Originally Posted by timm333
A question about short circuit MVA (MVAsc). If we have a generator of 2 MVA, then what will be its MVAsc ?
Not enough information to answer...You need to know the generator's subtransient reactance.

Generator MVAsc = MVA/X''d

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Thanks. So if a 2 MVA generator has Xd'' = 0.2, then MVAsc = 2/0.2 = 10 MVAsc, this makes sense. I am just being curious that why MVA and MVAsc are not the same. In other words, if the generator is capable to produce 10 MVAsc, then why its useful output power is only 2 MVA (and not 10 MVA)?

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Originally Posted by timm333
Thanks. So if a 2 MVA generator has Xd'' = 0.2, then MVAsc = 2/0.2 = 10 MVAsc, this makes sense. I am just being curious that why MVA and MVAsc are not the same. In other words, if the generator is capable to produce 10 MVAsc, then why its useful output power is only 2 MVA (and not 10 MVA)?

the 2 is into a load
the 10 is limited only by its reactance, the windings will be destroyed if sustained

similar to a xfmr
rated mva 1
impedance pu is 5%
sc mva = 1/0.05 = 20

5. The subtransient (x") time period lasts from about one to three cycles, which is from 16.7 milliseconds to 50 milliseconds (assuming a power frequency of 60 Hz). So if you consider useful power output for about 20-50 milliseconds, then you are good to go.

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Originally Posted by timm333
Thanks. So if a 2 MVA generator has Xd'' = 0.2, then MVAsc = 2/0.2 = 10 MVAsc, this makes sense. I am just being curious that why MVA and MVAsc are not the same. In other words, if the generator is capable to produce 10 MVAsc, then why its useful output power is only 2 MVA (and not 10 MVA)?
Yet another reason is that the running MVA may well be limited by the size of the prime mover (engine or turbine for example), while the rotating mass of the system will allow the MVAsc to be limited only by the generator portion.
A vendor of cheap wind turbines once famously rated his wind generator at roughly twice the sustained production for the reference wind speed on the grounds that once it got up to speed and you closed the output circuit it would deliver the higher power "for awhile" as the system slowed down to a new equilibrium speed.

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generators produce heat and need cooled (air/fan, liquid, gas)

losses (cooling load) = i^2 R
assume rated i = 1 pu
therefore sc i = 10 pu
R is constant
so cooling load is 100 times greater
smoke lol

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Originally Posted by Ingenieur
generators produce heat and need cooled (air/fan, liquid, gas)

losses (cooling load) = i^2 R
assume rated i = 1 pu
therefore sc i = 10 pu
R is constant
so cooling load is 100 times greater
smoke lol
Also why large generators were once considered impractical because the misinterpretation of energy transfer conditions caused the designers to try to make source impedance and load impedance equal!

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The first page of the link above from the website of Cummins states that R for generators is negligible. So does it mean that the generators have no X/R ratio like transformers? When we say, for example, the X/R on the primary side of the transformer is 15, does it mean that all of this X/R is due to the cable between generate and transformer (without any contribution from generator)?

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