Short circuit current is calculated according to IEC 60909 Standard. However, in the standard there will be some statements related to short circuit current, such as prospective short circuit current, initial symmetrical short circuit current, peak short circuit current, breaking current, steady state current, dc component of the short circuit current. It could be more correct to say which current we mention.
Short circuit currents can be calculated on the faulted bus. IEC 60909 includes equations to calculate impedance of power system components, and short circuit analysis is carried out by impedance and currents.
Below is the straightforward example to calculate short circuit current. A generator is connected to transformer and there are faults on 10 kV and 0.38 kV buses. We will find the impedance of the generator and transformer in case of short circuited, and then using the impedances, initial symmetrical short circuit currents based on bus voltage level can be calculated. After the initial symmetrical short circuit currents, peak current, breaking current, steady state current and dc components will be obvious.
Given data for the equipments can be calculated from vendor, equipment datasheet or manufacturer. IEC 60909-2 gives some sample values for the components' impedance in complex number form.
IEC60909 refers the terms of fault currents which are calculated based on fault location on the single line:
- Initial Symmetrical Current = Maximum Short Circuit Current with c factor is 1.1
- Peak Short Circuit Current
- Breaking Current
- Steady State Current
- DC Component of the short circuit current
- Asymmetrical Breaking Current
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| Figure 1. Generator Circuit Short Circuit Currents |
F1 Fault at 10 kV Bus
Generator Reactance
Generator reactance (\(X_d^{''}\)) equation comes from IEC 60909 and generator reactance equals to subtransient reactance.
\[X_d^{''} = \frac{{x_d^{''}}}{{100\% }}.\frac{{{U_{rG}}^2}}{{{S_{rG}}}}\]
\[X_d^{''} = \frac{{0,16}}{{100\% }}.\frac{{{{10}^2}}}{{48}} = 0,3333\Omega\]
If the condition is; \({S_{rG}} < 100MVA\) (rated apperent power of generator) \({U_{rG}} > 1kV\) ( rated voltage of generator) then, fictitious resistance (\({R_{Gf}}\)) value will be obtained.
\({R_{Gf}} = 0,07.X_d^{''}\)
\({R_{Gf}} = 0,07.0,3333 = 0,02333\Omega\)
Correction Factor
Correction factor (\({K_G}\)) for generator has to be calculated.
\[{K_G} = \frac{{{U_n}}}{{{U_{rG}}}}.\frac{{{c_{\max }}}}{{1 + x_d^{''}.\sqrt {1 - {{\cos }^2}{\varphi _{rG}}} }}\]
\[{K_G} = \frac{{10}}{{10}}.\frac{{1,1}}{{1 + 0,16.\sqrt {1 - 0,{9^2}} }} = 1,02828\]
Armature Resistance
Armature Resistance of Generator
\({R_{Gf}} = {R_a}\)
Fictitious resistance equals to armature resistance
\[\frac{{X_d^{''}}}{{{R_a}}} = \frac{{0,3333}}{{0,0233}} = 14,2587\]
Generator Impedance and Corrected Impedance
Generator impedance is written down in complex number.
\({\underline Z _G} = 0,0233 + j0,3333\Omega\)
Corrected generator impedance can be calculated with correction factor.
\({\underline Z _{GK}} = {K_G}({R_a} + j{X_d}'')\)
\({\underline Z _{GK}} = 1,02828(0,0233 + jX0,3333)\Omega\)
Impedance at F1 fault location 10 kV bus
\({\underline Z _{GK}} = {\underline Z _{kF1}} = 0,023993 + j0,342762\Omega\)
Initial symmetrical short circuit current for F1 fault 10 kV (Un) bus
Since faulted bus is 10 kV, impedance at 10 kV of components (corrected impedance) is used.
\[I_k^{''} = \frac{{c{U_n}}}{{\sqrt 3 {Z_k}}} = \frac{{c{U_n}}}{{\sqrt 3 \sqrt {{R_k}^2 + {X_k}^2} }}\]
\(I_k^{''}\) is represented as initial symmeterical short circuit curent which is the maximum fault current since c value is 1.1 as per IEC60909.
\[I_{kF1}^{''} = \frac{{1,1.10}}{{\sqrt 3 .\sqrt {0,{{023993}^2} + 0,{{342762}^2}} }} = 18,4832kA\]
For angle, conjugate math operations,
\[I_{kF1}^{''} = \frac{{1,1.10}}{{\sqrt 3 .(0,023993 + j0,342762)}} = \frac{{6,350853.(0,023993 - j0,342762)}}{{(0,{{023993}^2} + 0,{{342762}^2})}}\]
\[I_{kF1}^{''} = 53,79288.(0,023993 - j0,342762) = 1,290669 - j18,43813kA\]
And the angle \(\theta\) can be calculated based on imaginary and real side of complex number.
\[\theta = \arctan \left( {\frac{{ - 18,43813}}{{1,290669}}} \right) = - 85,9958^\circ\]
Peak short circuit current for F1 fault 10 kV bus
The equation is given in standard for \(\kappa\) value as:
\[\kappa = 1,02 + 0,98{e^{ - 3R/X}}\]
And peak short circuit current is represented as \({i_p}\).
\[{i_p} = \kappa \sqrt 2 {I_k}''\]
The equations are from IEC 60909. R/X value resistance by reactance is needed for peak current.
\[\frac{{{R_{kF1}}}}{{{X_{kF1}}}} = \frac{{0,023993}}{{0,342762}} = 0,07\]
Please notice that already found from fictitious resistance equation for 0.07.
\[\kappa = 1,02 + 0,98.{e^{ - 3.(0,07)}} = 1,8143\]
Peak short circuit current can be calculated which is 47,426 kA and initial symmetrical current was 18.48kA
\[{i_p} = 1,8143.\sqrt 2 .18,483 = 47,426kA\]
Breaking Current for F1 fault 10 kV bus
Breaking current \({I_b}\) is given as for circuit breaker.
\[{I_b} = \mu .I_{k\max }^{''}\]
Time value is choosen. If the time value is choosen as 0,1 s,
\[t = 0,10s\]
\[\mu = 0,62 + 0,72{e^{ - 0,32I_{kG}^{''}/{I_{rG}}}}\]
However, as seen in \(\mu\) formula, generator currents are required to find the value.
Rated generator current can be calculated easily from traditional power formula.
\[{I_{rG}} = \frac{{{S_{rG}}}}{{\sqrt 3 .{U_{rG}}}}\]
\[{I_{rG}} = \frac{{48}}{{\sqrt 3 .10}} = 2,77128kA\]
So we see the generator rated current 2,77 kA.
Initial symmetrical current on 10 kV will be rated generator current. As there are no other elements contribution to fault current on the 10 kV bus.
\[I_{kF1}^{''} = I_{kG}^{''} = 18,483kA\]
\[\frac{{I_{kG}^{''}}}{{{I_{rG}}}} = \frac{{18,483}}{{2,7712}} = 6,6695\]
So defined mu value is obvious.
\[\mu = 0,62 + 0,72{e^{ - 0,32.6,6695}} = 0,70519\]
Breaking current for breaker is now clear.
\[{I_b}_{F1} = 0,70519.18,483 = 13,0343kA\]
Steady State Current (Ik) for F1 Fault
Maximum steady state current is \({I_{k\max }}\)
\[{I_{k\max }} = {\lambda _{\max }}.{I_{rG}}\]
IEC 60909-1 can be used for equation based calculation. You can find the below equation in the IEC 60909-1.
\[\lambda = \lambda \max = \frac{{{u_{f\max }}.\sqrt {1 + 2.{x_{dsat}}.\sin {\varphi _{rG}} + x_{dsat}^2} }}{{{x_{dsat}} - x_d^{''} + (1 + x_d^{''}.\sin {\varphi _{rG}}).{I_{rG}}/I_{kG}^{''}}}\]
\[\sin {\varphi _{rG}} = 0,4358\]
Remind that generator cosinus value is given as 0.9 in the single line
Now \(\lambda \max\) value is obvious.
\[\lambda \max = \frac{{1,3.\sqrt {1 + 2.1,5.0,4358 + 1,{5^2}} }}{{1,5 - 0,16 + (1 + 0,16.0,4358).2,7712/18,483}} = 1,8496\]
\[{I_{kF1}} = 1,8496.2,7712 = 5,1259kA\]
Then steady state fault current \({I_{kF1}}\) at F1 location is 5,12 kA.
DC Component of the short circuit current for F1 Fault
Frequency ( 50 Hz), time 0.1 sn and R/X is now given data.
\[{i_{DC}} = \sqrt 2 .I_k^{''}.{e^{ - 2\pi .f.t.R/X}}\]
\[{i_{DC}} = \sqrt 2 .18,483.{e^{ - 2\pi .50.0,1.0,07}} = 2,898kA\]
And for the values, dc component of the fault current \({i_{DC}}\) is 2,89kA
Asymmetrical Breaking Current for F1 Fault
It is easy to calculate with dc and breaking current which we have found before and putting into IEC formula.
\[{I_{basyn}} = \sqrt {I_b^2 + i_{DC}^2}\]
\[{I_{basyn}} = \sqrt {13,{{0343}^2} + 2,{{898}^2}} = 13,352kA\]
Asymmetrical breaking current \({I_{basyn}}\) is 13,35 kA and near to breaking current aside DC component.
F2 Fault (0.38 kV)
Transformer Impedance
Transformer Impedance based on 0.4 kV ( Note: Bus voltage is 0,38 kV)
\[{Z_T} = \frac{{{u_{kr}}}}{{100\% }}.\frac{{{U_{rT}}^2}}{{{S_{rT}}}}\]
\[{Z_T} = \frac{{0,0715}}{{100\% }}.\frac{{0,{4^2}}}{5} = 0,002288\Omega\]
Transformer Resistance :
Transformer Resistance based on 0.4 kV and the power loss as 41,75 kW and so 0,04175 MW. Apparent power was 5 kA in the single line.
\[{R_T} = \frac{{{P_{krT}}}}{{\frac{{{S_{rT}}^2}}{{{U_{rT}}^2}}}}\]
\[{R_T} = \frac{{0,04175}}{{\frac{{{5^2}}}{{0,{4^2}}}}} = 0,0002672\Omega\]
Relative Resistive Component :
Relative resistive component of the relative impedance \({u_{Rr}}\%\) or \({{u_{kr}}}\%\)
\[{u_{Rr}} = \frac{{{P_{krT}}}}{{{S_{rT}}}}.100\%\]
\[{u_{Rr}} = \frac{{0,04175}}{5} = 0,00835 = 0,835\%\]
Power loss and apparent power value is sufficient to find the relative resistive component of the transformer. It is the component of z% given as 7.15%.
Transformer reactance :
Transformer reactance is not a relative value so transformer impedance ( 0,002288 ohm and 0,0002672 resistance) is valid.
\[{X_T} = \sqrt {{Z_T}^2 - {R_T}^2}\]
\[{X_T} = \sqrt {0,{{002288}^2} - 0,{{0002672}^2}} = 0,002272344\Omega\]
and please notice that the reactance of transformer is 0,0022 ohm which is more than resistance and roughly equals to impedance.
Relative Reactance:
Relative reactance component of the relative impedance(z%) \({{u_{kr}}}\%\) is defined in the IEC60909.
\[{x_T} = \frac{{{X_T}}}{{\left( {\frac{{U_{rT}^2}}{{{S_{rT}}}}} \right)}}\]
\[{x_T} = \frac{{0,002272344}}{{\left( {\frac{{0,{4^2}}}{5}} \right)}} = 0,0710\]
As we know the transformer reactance, nominal voltage as 0,4 kV and apparent power 5 MVA, the relative reactance is 0,0710.
Correction factor :
Transformer Correction Factor is required to find the actual impedance but note that we are in low votlage level for secondary side of transformer so the "c" value will change from 1.1 to 1.05. Correction factor \({K_T}\) is below.
\[{K_T} = 0,95.\frac{{{c_{\max }}}}{{1 + 0,6{x_T}}}\]
\[ c = 1,05\]for low voltage level
\[{K_T} = 0,95.\frac{{1,05}}{{1 + 0,6.0,0710}} = 0,9567\]
\[{\underline Z _{TK}} = {K_T}({R_T} + j{X_T})\]
We need to multiply the correction factor value with the impedance value of transformer in complex number.
\[{\underline Z _{TK}} = 0,9567(0,0002672 + j0,002272344)\]
\[{\underline Z _{TK}} = 0,000256 + j0,002174\Omega\]
Reduction to 0,38 kV
For F2 Fault, fault at F1 location \({Z_{kF1}}\) is transferred to 0.38 kV by using transformer turn ratio square \({\textstyle{1 \over {t_r^2}}}\). In converting from high voltage to low voltage, the inverse of the transformer conversion ratio is used.
\[t_r^2 = \frac{{{{10}^2}}}{{0,{4^2}}} = {25^2}\]
\[{\underline Z _{kF1@0,38kV}} = \left( {0,023993 + j0,342762} \right).\frac{1}{{{{25}^2}}} = 3,{83.10^{ - 5}} + j0,00055\Omega\]
Total Impedance at Fault Location
Total impedance at F2 fault location \({\underline Z _{kF2}}\) can be calculated as total of impedances of F1 and F2 fault location.
\[{\underline Z _{kF2}} = {\underline Z _{TK}} + {\underline Z _{kF1@0,38kV}} = 0,0002940 + j0,00272\Omega\]
Initial symmetrical short circuit current for F2 0,38 kV
Similar to initial symmetrical short circuit current at F1 fault location, \(I_{kF2}^{''}\) which is the initial symmetrical short circuit current for F2 and note that it is 0,38 kV and for low voltage c factor is 1,05.
\[I_k^{''} = \frac{{c{U_n}}}{{\sqrt 3 {Z_k}}}\]
\[I_{kF2}^{''} = \frac{{1,05.0,38}}{{\sqrt 3 .\sqrt {0,{{0002940}^2} + 0,{{00272}^2}} }} = 84,1266kA\]
Fault current for F2 location is 84,1266 kA and higher than fault current 18,48 kA at F1 location.
Peak short circuit current for F2 fault 0,38 kV bus
\[\frac{{{R_{kF2}}}}{{{X_{kF2}}}} = \frac{{0,0002940}}{{0,00272\Omega }} = 0,1080\]
Impedance components at F2 location is used for \({R_{kF2}\) and \({X_{kF2}\).
\(\kappa\) can be calculated now.
\[\kappa = 1,02 + 0,98{e^{ - 3.0,1080}} = 1,728\]
\[{i_p}_{F2} = 1,728.\sqrt 2 .82,1266 = 205,678kA\]
Peak current at F2 \({i_p}_{F2}\) is 205,678 kA
Breaking Current for F2 fault
\[{I_b}_{F2} = I_{kF2}^{''} = 84,1266kA\]
Breaking current \({I_b}_{F2}\) will be 84,126 kA.
Steady State Current (Ik) for F2 Fault
\[{I_k}_{F2} = I_{kF2}^{''} = 84,1266kA\]
Steady state current \({I_k}_{F2}\) at F2 fault location will be 84,12 kA.
DC Component of the short circuit current for F2 Fault
\[{i_{DC}} = \sqrt 2 .84,1266.{e^{ - 2\pi .50.0,1.0,1080}} = 3,998kA\]
Asymmetrical Breaking Current for F2 Fault
\[{I_{basyn}} = 84,2216kA\]
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