Previous in Forum: Transformer Vector Symbols   Next in Forum: Centifugal Blower Motor
Close
Close
Close
16 comments
Rate Comments: Nested
Active Contributor
Engineering Fields - Power Engineering - New Member

Join Date: Jul 2010
Posts: 17

Short Circuit Current at Transformer Calculation

08/27/2010 11:12 AM

i want to know the short circuit current at transformer if fault happens at transformer?

following are the specifications of transformer -

1. 1MVA, secondary voltage=415V, %z=6.

2. 1MVA, secondary voltage=690V, %z=6.

3. 3.5MVA, secondary voltage=415V, %z=5.

4. 1MVA, secondary voltage=690V, %z=5.

as i am concerned about the low voltage systems. . .

Please tell me the answer. . .. thank a lot . .. in advance. . .

and what will be the effect of the system voltage on the system current obviously it will go down. . . but what will happen to the selection of protecting devices?

Register to Reply
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.

Good Answers:

These comments received enough positive votes to make them "good answers".

"Almost" Good Answers:

Check out these comments that don't yet have enough votes to be "official" good answers and, if you agree with them, vote them!
Guru

Join Date: Jul 2007
Location: Commissariat de Police, Nouvions, occupied France, 1942.
Posts: 2599
Good Answers: 77
#1

Re: calculation of short circuit current at transformer

08/27/2010 11:19 AM

It will be whatever the setting is of the circuit protective device, Uncle. That's what these things are installed for.

__________________
Good moaning!
Register to Reply
Guru
Technical Fields - Technical Writing - New Member Engineering Fields - Piping Design Engineering - New Member

Join Date: May 2009
Location: Richland, WA, USA
Posts: 21017
Good Answers: 795
#3
In reply to #1

Re: calculation of short circuit current at transformer

08/27/2010 1:04 PM

Short-circuit protective devices do not open instantly, and can pass very large currents for a few milliseconds until the fault clears. Thus a fuse or CB nominally rated to trip at 100a may need to interrupt a current of 28ka, to use kvsridhar's example.

__________________
In vino veritas; in cervisia carmen; in aqua E. coli.
Register to Reply
Guru
India - Member - New Member

Join Date: Sep 2009
Location: Temporarily at Ashburn, VA
Posts: 2744
Good Answers: 164
#2

Re: calculation of short circuit current at transformer

08/27/2010 11:40 AM

i am curious. You have given Voltages and Percent Impedances. And you say you don't know what the fault levels are? i am surprised. Anyway....

Giving in percent means an easy way to calculate the SC level. A 5% impedance means an SC level 20 times the rated value. Not in kA. but in kVA.

i will give a thumb rule .. a 5% impedance means that a 1000 kVA transformer will deliver a 28 kA fault at its terminals.

Can you calculate yourself now?

__________________
Nothing worthwhile can ever be taught, it can only be learnt.
Register to Reply
2
Power-User

Join Date: Jun 2009
Posts: 134
Good Answers: 11
#4

Re: Short Circuit Current at Transformer Calculation

08/27/2010 1:55 PM

Everyone seems to have a different answer, or no answer.

Here's my answer assuming
3 phase and no offset current

I=1000000/(415X3^1/2)=1391 amp rated

Isc=1391/0.06=23,190 amp

Register to Reply Good Answer (Score 2)
Associate

Join Date: Aug 2010
Location: Umdorman ,Sudan
Posts: 28
#5
In reply to #4

Re: Short Circuit Current at Transformer Calculation

08/27/2010 4:21 PM

That is right assuming short circuit impedance .06 ohm

Register to Reply
Guru
India - Member - New Member

Join Date: Sep 2009
Location: Temporarily at Ashburn, VA
Posts: 2744
Good Answers: 164
#7
In reply to #5

Re: Short Circuit Current at Transformer Calculation

08/28/2010 12:09 AM

It is not the impedance that is 0.06, it is the 6% that Carl Pugh is using as the multiple factor. i used 5% and got ≈ 28kA.

__________________
Nothing worthwhile can ever be taught, it can only be learnt.
Register to Reply
2
Guru
India - Member - New Member

Join Date: Sep 2009
Location: Temporarily at Ashburn, VA
Posts: 2744
Good Answers: 164
#6

Re: Short Circuit Current at Transformer Calculation

08/27/2010 10:10 PM

What i learnt is school about '5% impedance' means that: with the secondary shorted, one requires 5% of the rated voltage at the primary to pass rated current through the secondary.

Here is the calculation :

__________________
Nothing worthwhile can ever be taught, it can only be learnt.
Register to Reply Good Answer (Score 2)
Associate

Join Date: Aug 2010
Location: Umdorman ,Sudan
Posts: 28
#12
In reply to #6

Re: Short Circuit Current at Transformer Calculation

08/28/2010 7:19 PM

Is this method of calculating short circuit current value is valid for low power/voltage transformer such as 1KW 220V/24V transformer ?

Register to Reply
Guru
India - Member - New Member

Join Date: Sep 2009
Location: Temporarily at Ashburn, VA
Posts: 2744
Good Answers: 164
#14
In reply to #12

Re: Short Circuit Current at Transformer Calculation

08/28/2010 10:12 PM

i dare say that the practice of giving "Percent Impedance" is more in Power transformers. Yours would be a control transformer, and i doubt if the manufacturer has given a %Impedance figure on the nameplate. However, if he has, it is the same method.Quick and simple.

(Full Load Current X 100) ÷ %Impedance.

__________________
Nothing worthwhile can ever be taught, it can only be learnt.
Register to Reply
Anonymous Poster
#8

Re: Short Circuit Current at Transformer Calculation

08/28/2010 8:30 AM

But why the transformer short circuit current must be calculated ? any specific reason?

Register to Reply
Guru
Technical Fields - Technical Writing - New Member Engineering Fields - Piping Design Engineering - New Member

Join Date: May 2009
Location: Richland, WA, USA
Posts: 21017
Good Answers: 795
#11
In reply to #8

Re: Short Circuit Current at Transformer Calculation

08/28/2010 2:08 PM

So that the short circuit protective devices are correctly selected to be able to interrupt the maximum instantaneous current to which they may be subjected. Then the devices do deal with it by opening successfully (without blowing apart and showering sparks everywhere).

__________________
In vino veritas; in cervisia carmen; in aqua E. coli.
Register to Reply
Guru

Join Date: Oct 2009
Posts: 2061
Good Answers: 169
#9

Re: Short Circuit Current at Transformer Calculation

08/28/2010 9:40 AM

If the fault is at the transformer secondary (LV Side) terminals, the follwing will be fault currents (assuming three phase symmetrical short circuit):

i) 1 MVA, 415V, 6% = 23.19kA

ii) 1 MVA, 690V, 6% = 13.95kA

iii) 3.5 MVA, 415V, 5% = 97.39kA

iv) 1 MVA, 690V, 5% = 16.74kA

You need not worry about the low voltage systems, as the fault is at the transformer terminals and currennt would not flow beyond this point.

And, also, the secondary side voltage would sink to zero for a three phase dead short circuit. Obviously, the protective devices need to be current operatyed only. That too, for a fault at the transformer LV terminals, only HV relays would operate, as LV CTs would be mounted after the LV terminals.

Register to Reply Score 1 for Good Answer
Anonymous Poster
#10

Re: Short Circuit Current at Transformer Calculation

08/28/2010 12:14 PM

I still asking my question:

Why the transformer short circuit current must be calculated ? any specific reason?

That is very high current and no wire or device can deal with it.

Register to Reply
Guru

Join Date: Oct 2009
Posts: 2061
Good Answers: 169
#13
In reply to #10

Re: Short Circuit Current at Transformer Calculation

08/28/2010 7:35 PM

Why? Because of the good old "Murphy's Law" that "if something can go wring, it will". Yes! It is very high current and that is why it is cause for concern. And, it is not true that no wire or device can deal with it. It is a matter of Current & Time. Every wire or Device has got a certain Thermal Withstand Characterstics. If if the high current flow exceeds the permissible time, the device or wire would not be able to handle it. If detection & isolation is effected within the permissible time, no danger will happen to any wire or device. After all, that is the job of an electrical engineer - to ensure that in the case of such abnormality, no wire or device is damaged.

Register to Reply
3
Guru

Join Date: Jul 2006
Location: Eastern Kansas USA
Posts: 1503
Good Answers: 128
#15
In reply to #10

Re: Short Circuit Current at Transformer Calculation

08/29/2010 9:28 AM

Guest,

Your question is fair, and understandable for when a person is first learning about electricity. The answer you have received is correct. A transformer is capable of delivering its nameplate current forever. In doing its work, it is storing and transforming the incoming power in a magnetic field to convert the voltage from the primary to the secondary voltage. If you short out the secondary, either at the terminals of the transformer or further downstream, the impedance (AC resistance) of the load circuit is greatly reduced. This stored energy now is being output very quickly into the resulting short, and it results in a much higher current being possible.

Every electrical wire has a current rating for how much it can carry continuously in a safe manner. As the current is raised above this level, the wire will overheat--initially destroying the insulation (as almost every electrician has seen at one time or another), and eventually so hot it can melt. At very high current levels, this melting can occur in milliseconds or less. Then, as the current is flowing where the wire was, vaporized particles of the metal form a conductive pathway, called plasma. The temperature of this plasma can be as high as 30,000 °C. Because of the high current, this arc also has a very strong magnetic field. The vaporization of the metal and the strong magnetic field combine into a violent explosion. This has injured and killed many unfortunate electricians and maintenance workers, and is the reason for use of protective clothing and safety distances.

If the equipment is designed so it includes devices that can safely interrupt this high current quickly, the energy liberated in this arc can be reduced to a level where it does little or even no harm to the surrounding equipment. If not, it can blow heavy sheet metal enclosures apart, lift people into the air as they are blown back as much as 3-10 meters, and light fires.

Take a simple circuit breaker, and look at the back end of it--you will see some slots. When the circuit breaker is switched off, the arc between the contacts is forced out through these slots, which serve to separate the arc until it can quench itself (by being cooled and lengthened, as well as the time lapse until the AC voltage crosses through zero). This design is capable of interrupting a current much higher than the breaker's labeled "size". In the USA, a 20A (or 15A, 40A, etc.) breaker must be able to safely interrupt a short circuit of 5000A. If it is exposed to a short circuit current that is higher, it will probably be unable to quench this arc, and will fly apart--if the current is much higher, this will be very violent as described above (an arc that has not been safely interrupted becomes a ground-fault arc and/or a phase-phase fault, and can vaporize multiple sections of switchgear in seconds--I have seen the results of this!). For more money, you can buy and install breakers capable of interrupting 10,000A, or 22,000A, and more; you can buy fuses capable of interrupting as much as 300,000A.

Engineers, code enforcement personnel, electricians, and many others need to know that the electrical equipment is chosen to have components that can safely interrupt the available fault current--no explosion, no fire, no equipment damage. Therefore, calculations are done to determine what the available short circuit fault current is, at the main equipment, and further downstream (if it is over 5000A or sometimes 10,000A). After these calculations are done, then the proper equipment can be selected for the wires, wireways, breakers, fuses, motor controllers, and everything else.

Its just not nice when things go boom.

--JMM

Register to Reply Good Answer (Score 3)
Anonymous Poster
#16

Re: Short Circuit Current at Transformer Calculation

09/17/2010 6:01 AM

You have provided sufficient datas which are needed to calculated short ckt. current of a transformer.

For 1 MVA, 415V with 6%-

Falut current in MVA= base MVA / per unit % impedance

= 1/0.06=16.66MVA

Fault current= 16.66/1.732/0.415

=23kA

so, the Maximum short ckt. current will be 23kA.

By following above steps you can calculate for other configurations.

Register to Reply
Register to Reply 16 comments

Good Answers:

These comments received enough positive votes to make them "good answers".

"Almost" Good Answers:

Check out these comments that don't yet have enough votes to be "official" good answers and, if you agree with them, vote them!
Copy to Clipboard

Users who posted comments:

AfricanGiraffe (2); Anonymous Poster (3); Carl Pugh (1); Crabtree (1); electricalexpert65 (2); jmueller (1); kvsridhar (4); Tornado (2)

Previous in Forum: Transformer Vector Symbols   Next in Forum: Centifugal Blower Motor

Advertisement