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# Method C

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```NOOR MOHAMED.A
METHOD C
IEC-60909
NOOR MOHAMED
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Objective of Short Circuit:
•
The main purpose for calculating short circuit current is safety.
•
All equipment must be capable of carrying the very high currents that occur in the event
of a short circuit. All equipment must be capable of withstand thermal heating and
mechanical forces with the passage of fault current.
•
Circuit Breakers must be capable to close onto a fault and opening to isolate a fault.
•
Different standards exist to guide engineers for Short Circuit calculation.
•
Here we will focus on the problem of calculating the dc component of a short circuit
and look specifically at the IEC 60909 Method C calculation.
Single fault at feed
Figure 1 Simple RL circuit
πΈ = ππ sin(ο·π‘ + οͺ)
The Short Circuit current magnitude is limited by the impedance of the circuit,
ππ sin(ο·π‘ + οͺ) = ππ + πΏ
ππ
ππ‘
The general solution for this differential equation is
π(π‘) =
ππ
√π 2 − (ωL)2
sin [ππ‘ + π − πΌ tan (
ππΏ
) + π΄π −ππ‘/πΏ ]
π
At t=0, i(t)=0, and solving for A:
π΄=
ππ
2
√π −(ωL)2
Substituting
−ππ
√π 2 − (ωL)2
sin [π − πΌ tan (
ππΏ
)]
π
= √2 ∗ πΌπππ and π = ππΏ
π
π
π(π‘) = √2 ∗ πΌπππ sin(ο·π‘ + οͺ − πΌπ‘ππ π ) − √2 ∗ πΌπππ sin(οͺ − πΌπ‘ππ(π ) ∗ π
−ππ‘π
π
From this above equation,
π
•
The first term of equation √2 ∗ πΌπππ sin(ο·π‘ + οͺ − πΌπ‘ππ π ) is a constant sinusoidal ac
current,
•
The second term of equation −√2 ∗ πΌπππ sin(οͺ − πΌπ‘ππ(π ) ∗ π
exponential of (dc) current.
π
−ππ‘π
π
is a decaying
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π
The maximum initial value of the dc component will occur when sin (οͺ − πΌπ‘ππ π ) = 1.
Since X/R is typically higher, closing the switch when the voltage phase angle is close to π =
0 will produce the maximum dc current.
The total fault current is the sum of the ac and dc component of current.
When the maximum dc component of current is present, the total fault current (i.e.) the
sum of ac and dc components combine to produce a current waveform with maximum initial
asymmetric current.
Note: In practise the magnitude of the ac component will also decay with time, especially if
the fault occurs in apart of the network with a rotating machine.
As the source voltage is at 50 Hz, with π = 0 at t=10 ms the ac and dc components combine
to give the asymmetric peak current.
A circuit breaker does not break the peak current, as it will take time to detect a fault
and send a trip signal to the circuit breaker. The poles of the circuit breaker might start to open
at 50 to 90 ms, and with an X/R ratio of 10 dc component of current has not yet decayed to
zero.
Accurate calculation of the dc component of current is essential when determining both
the asymmetric peak current and the actual breaking current during the arcing period.
The initial rms value of the ac component of current I” is 0.995 and an empirical formula to
π
estimate the worst case peak current is: ππ = √2 ∗ πΌπππ sin(οͺ − πΌπ‘ππ(π ) ∗ π
−ππ‘π
π
ο ……
Table 1 RL circuit
Example data
X
1
X/R
10
R
0.1
Z
1.005
0
οͺ
V ac rms
1
I ac rms
0.995
What if multiple faults in feed?
When the circuit breaker closes, we can calculate the branch currents i1, i2, i3
separately as the equation A, will still apply to the loop formed by each branch, the fault current
and the source voltage. In this case when we use the equation IRMS is the steady state branch
and we use the X/R ratio of each branch. The total fault current is then the sum of i1, i2 and i3.
If each branch has different R and X values (ex. Impedance), the steady state component
of the branch short circuit currents will differ, as will the initial dc component and the rate of
decay of the dc component in each branch.
In general, if the fault is supplied by N branches, the total dc current in the fault will be
the sum of N decaying exponentials with possibly N different time constants.
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Table 2 Three parallel RL circuit
Example data
#1
#2
#3
X
1
1
1
X/R
5
10
20
R
0.2
0.1
0.05
Z
1.02 1.005 1.001
0
0
0
οͺ
V ac rms
1
1
1
I ac rms 0.981 0.995 0.999
•
•
•
•
Each branch has the same reactance X, but different X/R ratios.
The total impedance of each branch is practically identical, so their Iac_rms values are
almost identical.
The dc component of current decays most quickly in the branch with the lowest X/R
ratio.
The total dc current at the fault is the sum of the three individual dc current.
Reducing the three parallel branches to a single equivalent branch results in a single
equivalent X/R ratio, which produces a dc component of current that decays faster than correct
solution.
IEC 60909 provides three methods to correct this error.
Figure 2 Equivalent circuit
Table 3 Equivalent circuit
Equivalent Branch
X
0.335
X/R
8.590
R
0.039
Z
0.337
0
οͺ
V ac rms
1
I ac rms 2.965
IEC 60909 METHODS FOR DC COMPONENT
METHOD A- Uniform X/R ratio:
For this method, determine the smallest ratio of R/X or the largest ratio of X/R of all
branches of the network is taken.
The Thevenin equivalent impedance is calculated for that short circuit location and
taking X/R value from that impedance assuming the whole network as uniform X/R ratio.
METHOD B- R/X ratio or X/R ratio at the short circuit location:
To reduce the complex impedance in the network, the factor k with the ratio Rk/Xk is
multiplied with 1.15.
As long as R/X remains smaller than 0.3 in all branches which carry a short circuit
current, it is not necessary to use the factor 1.15.
METHOD C- Equivalent frequency ππ
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An equivalent impedance ππ = ππ + πππ . of the positive sequence impedance of the
short circuit location is calculated assuming a frequency fc = 20 Hz (for a nominal frequency
of 50 Hz) or fc = 24 Hz (for a nominal frequency = 60 Hz). The X/R is then determined by the
below equation,
π
π
π
ππ
= ππ ∗ ππ
Where ππ is the equivalent frequency,
Table 4 Method C-Equivalent frequency
f*t
Fc/f
t(ms) 50Hz
fc (Hz) 50 Hz
t(ms) 60Hz
fc (Hz) 60 Hz
&lt;0.5
0.4
10
20
8.33
24
&lt;1
0.27
20
13.5
16.67
16.2
&lt;2.5
0.15
50
7.5
41.67
9
&lt;5
0.092
100
4.6
83.34
5.52
&lt;12.5
0.055
250
2.75
208.34
3.3
Methodology:
1. At require time, determine ππ
2. Calculate impedance of each branch at ππ
3. Combine all the parallel branches to get a single equivalent impedance ππ = ππ + πππ .
4. Determine Xc/Rc ratio of the single equivalent frequency ππ .
π
π
ππ
5. Scale back to obtain the X/R ratio π = ππ ∗ ππ
6. Calculate the impedance of each branch at nominal frequency f.
7. Combine all the parallel branches to get a single equivalent impedance at nominal
frequency f, π = π + ππ.
8. Calculate the steady state ac current IRMS at nominal frequency f from Z.
π
9. Calculate the dc current from √2 ∗ πΌπππ sin(οͺ − πΌπ‘ππ(π ) ∗ π
−ππ‘π
π
Conclusions:
Working through the method C methodology over a range of times or the dc current it
is apparent that method C does not underestimate the dc component as the simple Thevenin
equivalent did, and the results of method C are remarkably close to the accurately calculated
dc current.
Method C is clearly superior in Method A and B and should always be used when
calculating the DC component of the fault current.
Reference:
1.IEC-60909-0
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