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The fault current of a power transformer will depend on the following;
Transformer Rating (in KVA/MVA)
per unit impedence of the transformer (%p.u.)
line/phase Voltage (VL/VP)
the following formula can be used to find the fault current on the secondary side of a transformer
Fault Current = Transformer Rating /(per unit impedance x phase voltage)
The Values of Transformer Rating, per unit impedance & phase/line voltage will usually be mentioned on the transformer rating plate / data sheet
As an example a 500kVA, 11kV/400V/3-Phase/50Hz transformer with 5% p.u impendence will have the following fault levels on the secondary side
Fault level = 500/(5%)=10000kVA
S=1.732 * VPP * IP
Fault current = 500/(5% x 400 x 1.732) = 14.4 kA
Remember to use 3phase voltage!
Wiki User
∙ 13y ago
Wiki User
∙ 14y agoF = V / sqrt(P) / ((L x R / 1000) ^ 2 + (L x X / 1000 + V / (A x sqrt(P))) ^ 2) ^ 0.5 + 4 x M + 5 x N Where: {| |- | F = Available fault current at source (A) V = Circuit voltage (use line voltage if three-phase and phase voltage if single-phase) (V) P = Number of phases (1 or 3) L = Distance from source of feeder to point of interest (Ft.) R = Resistance of feeder per 1000 Ft. (Ω) X = Reactance of feeder per 1000 Ft. (Ω) A = Available fault current at the source of the feeder (A) M = Total fault current contribution of induction motors (A) N = Total fault current contribution of synchronous motors (A) |}
Wiki User
∙ 12y agoThe answer to this question is far too complicated to answer satisfactorily on WikiAnswers. However, in brief, the first step involves determining the equivalent impedance of all the lines and Transformers back to the source. There are various techniques for doing this, each having varying degrees of accuracy and involving various assumptions. This is a major topic area, and beyond a simple explanation here.
Wiki User
∙ 13y agoTake the VA of the transformer and divide it by the output voltage.
Wiki User
∙ 12y agoIt depends if the Transformer is a car or a fighting machine....
Wiki User
∙ 9y agofault level at Transformer terminal= 3xVphxIph
where voltage is expressed in kV and current in kA.
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What is the reason for using a neutral current transformer on a transformer?
A current transformer is primarily used at the neutral point of a transformer for earth fault protection. A neutral current transformer will measure any ground fault current which will essentially flow from the star point of the transformer. A fault-detection device other devices is connected to the current transformer and, if the fault current exceeds a certain trigger value, the fault-detection device will give a trip command to an earth-fault relay to disconnect the supply of electricity to the transformer.
Function of no-load in transformer in a open-circuit?
To calculate the no load current from transformer & core loss is also calculated.
Transformer shorted turn fault?
? Are you asking about a transformer turn to turn fault? Or if a transformer is shorted, does this turn into a fault? Please clarify the question.
How do you calculate MVA rating of power transformer?
The product of the secondary rated current and the secondary rated voltage will give you the rated V.A of the transformer.
How do you calculate internal fault loop impedance?
by calculating the loop current
How do you calculate maximum earth fault current?
transformer max earth fault current
What is the reason for using a neutral current transformer on a transformer?
A current transformer is primarily used at the neutral point of a transformer for earth fault protection. A neutral current transformer will measure any ground fault current which will essentially flow from the star point of the transformer. A fault-detection device other devices is connected to the current transformer and, if the fault current exceeds a certain trigger value, the fault-detection device will give a trip command to an earth-fault relay to disconnect the supply of electricity to the transformer.
How transformer conduct current through ground?
A transformer doesn't 'conduct current through the ground'. However, in the event of a ground fault in the circuit supplied by a grounded transformer, the resulting fault current will flow through the earth back to the grounded terminal of that transformer's secondary winding.
How do you calculate the current in a transformer?
By the volt metre.
Function of no-load in transformer in a open-circuit?
To calculate the no load current from transformer & core loss is also calculated.
What is meant by ground fault?
The ground ('earth'), because of its mass, is a reasonably-good conductor and is used as a 'reference' for a distribution transformer's neutral terminal. A ground fault ('earth fault') occurs when the line conductor from the distribution transformer makes accidental contact directly with the ground ('earth'). The resulting low-resistance earth path back to the transformer's neutral is such that the resulting ground-fault current('earth-fault current') will operate the transformer's overcurrent protection device (e.g. fuse).
Transformer shorted turn fault?
? Are you asking about a transformer turn to turn fault? Or if a transformer is shorted, does this turn into a fault? Please clarify the question.
Is the star-delta transformer commissioned without earthing transformer or NGR?
Yes, possible. NGR or earthing transformers are used when it is necessary to limit the fault current.
How do you calculate MVA rating of power transformer?
The product of the secondary rated current and the secondary rated voltage will give you the rated V.A of the transformer.
How do you calculate internal fault loop impedance?
by calculating the loop current
How do you calculate current on the primary side of a transformer only knowing the turns ratio and the voltage ratio?
The primary current on a loaded transformer depends on the secondary current, which is determined by the load. So, if you know the secondary load current, then you can use the turns ratio of the transformer to determine the primary current:Ip/Is = Ns/Np
Why earthing is use in transformer?
Earthing of a transformer is done to 1:reduce the sensitivity of protective relays. 2:reduce the damage due to fault current. 3:get rid of great loss of equipment and money in case of any earth fault.
