Transformer short circuit current

You have not given the required information. Please provide adequate info to answer your question.

data given is inadequate to calculate s.c current

Ok, I'll bite. How do you know that your answer is 8.21KA? Was that the peak current reading before copper vaporized? There's always a fuse somewhere in a circuit.

Give us the complete details of your transformer such as single phase with impedance ohmic value at nameplate?... 4% of what..KVA rating?, the primary/secondary voltages.So I have a hint for Short circuit test refer to primary side at shorted secondary winding.

Note that the rated primary transformer current is equivalent to short circuit current refered to primary at shorted secondary.

thus = KVA x'former rating/KV(primary), assuming that no wattmeter and voltmeter readings that put on the line and across the line during the test.

An 200 KVA transformer at single phase is not logic

Sorry people I also got the question like that without any more information from a exam so didnt know if more details were required hmmm thanks anyways

Short circuit current is always indicated in a % value of nominal voltage by it's manufacturer, on the transformer's label.

This means when you shorted the secondary (ex. low voltage side of the transformer) and gradually increase the primary voltage to reach a certain value, of which you can read the nominal current (Amp.) value at the shorted secondary side.

This point is the short current voltage at the primary side. And it is indicated %value of the nominal primary voltage at the label by it's manufacturer.

This is the situation, the transformer is shorted under control, without no harm.

This % value can be used later for calculation of many purposes. Such as the copper windings losses etc. But also for one other purpose, the breaker's (KA) releasing setpoint.

Now that you know the transformation ratio of the transformer, you can calculate the real short circuit current value at the primary or secondary side under nominal voltage conditions. You cannot test the short current, under nominal voltage conditions in any way without giving damage to the transformer, but you can calculate the valu by his way. and set the magnetic breaker of the switch to feed the transformer accordingly

Hope this explanation is useful.

Kindest Regards

In addtion to determining the short circuit current, from this result you can also compute the equivalent resistance(Re) and reactance (Xe). Other option for short test is to use wattmeter and voltmeter mehtods can be useful to determine equivalent impedance(Ze)

Useful Formulas:

Ip = KVA x1000 / Ep = short circuit current in amps...refer to shorted secondary winding.

Equilavalent impedance(Ze) and resistance(Re) refer to primary;

Ze = Voltmeter reading / short circuit current

Re = wattmetter reading / (short circuit current)^2

Gen Equation:

Ze^2 = Re^2 + Xe^2

"Ze = Voltmeter reading / short circuit current" Do you mean when shorted and you measure the voltage at the shorted side?

Voltmeter reading on what side?. Voltage is "0" at the shorted side terminals, because it is shorted.

However in fact there is a voltage at the shorted side causing to pass the short current. you can not measure it, but you are able calculate since you know the primary voltage and the transformation ratio.

Possibly I could not get your message correctly, and if yes I apologise

It is an isolation type transformer even if shorted secondary still there is a voltage reading at primary side.. this is the frequent question given during the P.E.E state board exam.. the commissioner always insists that there is..to justify the issue we made an experiment here in factory G.E. dry type isolation transformers so the result it is not "0" volts at primary side. Yes the voltmeter reads small amount of voltage at primary side.

Sorry for misunderstanding. This is my fault.

You meant Voltage measurement at the primary side.

It is O.K. Of course this is your test voltage.

Ze = Voltmeter reading / short circuit current

Since the Voltmeter reading at the primary, then you mean also the current at the primary side, (of which is the rational conversion of the current, shorted at secondary side (Rational conversion of the Nominal secondary current in other words))

Thanks for explanation

Kindest Regards

short ckt level; fault level I= Transformer MVA / %impedance,

Many complex responses to this question, but i think the mpedance is given in percentage to enable quick SC calculation.

Here is extract from a webpage -

The impedance (or resistance to current flow) is important and used to calculate the maximum short circuit current which is needed for sizing, circuit breakers and fuses. Impedance is expressed as a percent. This percentage represents the amount of normal rated primary voltage which must be applied to the transformer to produce full rated load current when the secondary winding is short circuited. The maximum short circuit current that can be obtained from the output of the transformer is limited by the impedance of the transformer and is determined by the multiplying the reciprocal of the impedance times the full load current . Thus, if a transformer has 5% impedance, the reciprocal of .05 is 20 and maximum short circuit current is 20 times the full load current

I believe there is a confussion to understand the meaning of the web page.

Impedance is a constant value for a certain winding, whether it is primary or secondary. Impedance is neither dependent with the voltage applied, nor with the passing current.

Therefore it will not be correct to speak on the % values of the impedance. This is the ratio in % of Test voltage to Nominal working voltage

This % ratio, is essentially named as "Short Circuit Ratio".

When you test a transformer to find the SC ratio, a smaller test voltage is applied to the primary side of the transformer.

This test voltage is such a value that you can obtain the nominal output current at the shorted secondary side.

The rational % value of this test voltage against the full load grid voltage is the value that named as Short Circuit ratio. This is also a constant value,

Of course the current at the shorted secondary side is dependent with the sum of primary and secondary impedances, of which are constant. Those unchanging impedances are determining the current amount flowing a the secondary side, during the SC tests.

But the given % value only indicates the ratio of the test voltage to full load voltage. The impedances are not variable.

As mentioned in your message, the main goal of SC tests is, to find the exact Short circuit KA value when output side is accidentally shorted under normal voltage working conditions.

The way of calculation you described, to find exact Short circuit current in working conditions, is correct.

By knowing the exact short current value, you can adjust the release settings of the breakers, of which are protecting the transformer.

I am correcting my previous message of #14

Impedance is variable. This is my fault.

Dear Guest,

There are so many mathematical approach for computing short circuit fault.

such as MVA base, Kva base and Impedance (Zbase) especially in determining a point-to-point circuit fault in main substation up to sub-distribution board and at to where the fault occured.

The question is.... it 200kva transformer not 200MVA transformer. Different approaches but it will results into the same expectation because it is derived from the same theorem.

For example algebraic equation with 3-unknowns.

First;

4x + 2y + z = 4.... eq 1

2x - 2y + 3z= 2 ....eq 2

5x + 3y - 2z = 1 ...eq 3

These three equations you can detrmine the values of x , z, y in different approaches.

a. Determinant method

b. Determinant pivotal theorem

c. Substitution of equation method.

d. Elemination of equation method

c. and many more

Second;

(xty)^9 this can be solved thru expansion method but consume much papers for the solution but it can save paper and ink if you gonna expand it by using Pascal's triangle theorem... could be only one line solution in the paper.

I am assuming that by percentage impedance you mean the percentage of the primary voltage that has to be applied such that the secondary shirt-circuit current shall be the same as the normal(rated) secondary current and that by tolerance impedance you mean that it can vary with a total of 10% ( that is 0.04*0.95 minimum and 0.04 * 1.05 max) ( +/- 5%).

The short circuit current was 8210A. That means that the secondary rated current is 8210 * 0.042 = 344.82A (max) respectively 8210 * 0.038 = 311.98A (min).

Given that the rated power of the transformer is 200 kVA, the secondary voltage should be between 335V ( for I2=344.82A) and 370V (for I2=311.98A).

The primary voltage does not matter in this case.

If the 10% tolerance is meant as +/- 10%, the secondary currents are between 295.56 and 361.24 A and consequently the possible secondary voltage between 319.64 and 390.68V.

If the secondary voltage (3-phase line-to-line) is not between 320 and 391V, the numbers do not fit.

If by percentage impedance and tolerance impedance you mean something else, the rationing has to be redone.

without knowing primary/ sec vtg it is impossible to calculate s.c of transformer.The data is inadequate.

As far as my knowledge goes, the SC kVA = 200 * 100/4 = 5 MVA of SC power. It appears that your system voltage is approx 350V (rather non-standard?) .. so that 5000/sqrt(3)/350 = 8.25 kA. Am i missing something here?

SC Current = FLC / PI (PI= Percentage impedance)

= KVA/√3*V*PI

with the tolerance on impedance T the effective PI = PI * (100+T%)

SCC = 8.21 KA = 8210 = 200,000/√3*V*(0.04/1.10)

or V = 200,000/(√3*8210*0.04/1.10) = 386.77 V

Again non standard Value.

Alright Guys,

Herewith is my little knowledge to impart on this Forum. This probably out of topic.

SYSTEM EQUIVALENT IMPEDANCE CALCULATIONS

MDB %Z---fault impedances at MDB

1. MV System Impedance contribution

System voltage

Utility Power Short Circuit MVA, MVA_USC

System Fault

System X/R ratio, X/R

Per Unit Impedance, Resistance & Reactance at 100 MVA base

Impedance p.u., Z_{NT p.u.}=Base MVA / MVA_USC

Resistance p.u., R_{NT p.u.}= √[(Z_{NT p.u.})²/(1+X/R²)]

Reactance p.u., X_{NT p.u.}= R_{NT p.u.} x (X/R)

Utility Per Unit impedance:

Per Unit Impedance, Resistance & Reactance at 100 MVA base

Ohm's Law

VA = I/V ------- eqn. 1

VA = V² / Zu ------- eqn. 2

Multiply eqn. 2 by(1000)²

MVAusc = Kv² /Zu ------- eqn. 3

Solving for Utility impedance Zu

Zu = Kv² / MVAusc ------- eqn. 4

For Zbase

Zbase = Kv² / MVA base ------- eqn.5

For Zpu(per unit)

Zpu = Zu / Zbase ------- eqn.6

Divide eqn. 4 by eqn. 5 Then

Zpu = (MVAbase/MVAusc) x (Kv² / kVbase²)

For kVu = Kv base, then

Zpu = (MVA base / MVA usc)

Or

MVA usc = MVAbase / Zpu

1. Station Service Transformer Impedance contribution

KVA Rating, KVA_T

% Impedance at rated kVA, %Z_T1

Secondary voltage, KV_s

Copper loss, P_L

Impedance, Z_T1=10 x %Z x KV_S² / KVA_T

% Resistance, %R_T1=(P_L x 100) / KVA_T

Reactance, %X_T1=√(%Z_T1)² - (%R_T1)²%

Transformer per unit impedance:

Per Unit Impedance, Resistance & Reactance at 100 MVA base

%Impedance Formula

%Zt = I x Zt x 100 /V ------- eqn.1

Multiply eqn. 1 by (V/V), then

%Zt =Zt x I x V x 100/V² ------- eqn.2

Simplify;

Multiply eqn. 2 by(1000/1000)² , then

%Zt = MVAt x 100/Kv² ------- eqn.3

Base impedance

Zpu = Zt x MVA base / Kv² base ------- eqn.4

Divide eqq. 4 by eqn. 3 and solve for Zpu, then

Zpu = (%Zt /100) x (MVA base / MVAt) x (Kv² / kVVbase²)

If kVt = kVbase, then

Zpu = ((%Zt /100) x ( MVA base / MVAt)

Or

Zpu = (%Zt /100) x (MVAbase x 1000/kVAt)

Generator Impedance contribution

MVA Base

Data

MVA Base

KVA Rating

% Sub-transient Impedance at rated kVA, %"Z_G

Secondary voltage, KV_s

Generator X/R, X_G/R_G

Per Unit Impedance, Resistance & Reactance at 100 MVA base

Impedance p.u., Z_{G P.U}.= (%Z_G x Base MVA x 1000)/(100 x Gen.KVA)

Resistance p.u., R_{G p.u.}= √[(Z_{G p.u.})²/(1+X_G/R_G²)]

Reactance p.u., X_{G p.u.}= R_{G p.u.} x (X_G/R_G)

Generator per unit impedance

%Impedance Formula

%Zg = I x Zg x 100 /V ------- eqn.1

Multiply eqn. 1 by (V/V), then

%Zg =Zg x I x V x 100/V² ------- eqn.2

Simplify;

Multiply eqn. 2 by(1000/1000)² , then

%Zg = MVAg x 100/Kv² ------- eqn.3

Base impedance

Zpu = Zg x MVA base / Kv² base ------- eqn.4

Divide eqq. 4 by eqn. 3 and solve for Zpu, then

Zpu = (%Zg /100) x (MVA base / MVAg) x (Kv² / kVVbase²)

If kVg = kVbase, then

Zpu = ((%Zg /100) x ( MVA base / MVAg)

Or

Zpu = (%Zg /100) x (MVAbase x 1000/kVAg)

Cable Data from Generator to MDB

SINGLE CORE CABLE, CU/XLPE/PVC, 0.6/1kV

Resistance, R_G

Re Length, L_G

Total Resistance, R_TG=R_G x L_G /1000

Total Resistance, R_TG=R_G x L_G /1000

Total Impedance, Z_TG=√R_GT² + X_GT²

No. of conductor run(s) per phase, N_GC

Final Resistance, R_FG=R_TG/N_GC

Per Unit Impedance, Resistance & Reactance at 100 MVA base

Resistance per unit, R_{CG P.U.}=(R_FG x Base MVA)/(KVs²)

Reactance per unit, X_{CG P.U.}=(X_FG x Base MVA)/(KVs²)

Impedance per unit, Z_{CG P.U.}=(Z_FG x Base MVA)/(KVs²)

Cables/Line Per Unit impedance

Ohm's Law

VA = I/V ------- eqn. 1

VA = V^2 / Zc (cable) ------- eqn. 2

Multiply eqn. 2 by(1000)²

MVAusc = Kv² /Zc ------- eqn. 3

Solving for Zc

Zc = Kv² / MVA ------- eqn. 4

For Zbase

Zpu = Kv² / MVA base ------- eqn.5

For Zpu (per unit)

Zpu = Zc / Zbase ------- eqn.6

Substutute the value of Zbase from eqn. 5 in eqn.6. Then

Zpu = (Zc x MVAbase) / kVbase²)

Notes:

u--- unit

pu --- per unit

usc ---utility power short circuit

Z ---- impedance

t -----transformer