Quick calculation of short-circuit current

Thread Content

Abstract: An “mnemonic” calculation method is introduced; by memorizing 7 mnemonics, one can master the method for calculating short-circuit currents. The original was published in \"Building Electricity\" many years ago. I can no longer remember the specific time and author. I have simply organized it a bit for reference by colleagues in need. Keywords: short-circuit current, calculation method, mnemonic phrases

I. Overview
When a short circuit occurs in a power supply network, the large amount of short-circuit current can cause electrical equipment to overheat or be damaged due to electromagnetic forces. It also leads to a drop in voltage within the network, thereby disrupting the normal operation of the equipment connected to it. To eliminate or mitigate the consequences of short circuits, it is necessary to calculate the short-circuit current, so as to properly select electrical equipment, design relay protection systems, and choose components capable of limiting short-circuit currents.

II. Calculation Conditions
1. Assume that the system has infinite capacity. After a short circuit occurs at a user’s location, the voltage at the system busbars can remain unchanged. In other words, the calculated impedance should be much larger than the actual system impedance. Specifically, for calculating short-circuit currents in 3–35 kV grids, it can be assumed that systems with voltages of 110 kV and above have infinite capacity. It is sufficient to calculate the impedance of components in grids with voltages of 35 kV and below.
2. When calculating short-circuit currents in high-voltage electrical equipment, only the reactance of generators, transformers, and reactors needs to be considered, while their resistance can be ignored. For overhead lines and cables, resistance needs to be taken into account only when it is greater than one-third of their reactance; generally, only the reactance is considered, with resistance being ignored.
3. The formulas or tables used for calculating short-circuit currents are based on three-phase short circuits, as the short-circuit currents in single-phase or two-phase short circuits are smaller than those in three-phase short circuits. Electrical equipment capable of breaking three-phase short-circuit currents will certainly also be able to break single-phase or two-phase short-circuit currents.

III. Simplified Calculation Methods
Even with certain assumptions made, it remains difficult to accurately calculate short-circuit currents, and this is not necessary for ordinary users. Some design manuals provide simplified calculation tables, which make the calculation process easier. But what if no such manual is available? Here, a “mnemonic phrase”-based calculation method is introduced. By memorizing 7 mnemonic phrases, one can master the method of calculating short-circuit currents. Before introducing the simplified methods, it is necessary to understand some basic concepts.
1. Key Parameters
Sd: Three-phase short-circuit capacity (in MVA); also known as short-circuit capacity, used to verify the breaking capacity of switches.
Id: RMS value of the periodic component of the three-phase short-circuit current (in KA); also known as short-circuit current, used to verify the breaking current and thermal stability of switches.
IC: RMS value of the total current during the first cycle of a three-phase short circuit (in KA); also known as the RMS value of the inrush current, used to verify dynamic stability.
ic: Peak value of the total current during the first cycle of a three-phase short circuit (in KA); also known as the peak value of the inrush current, used to verify dynamic stability.
x: Reactance (in Ω). Among these, the system’s short-circuit capacity Sd and the reactance x at the point of calculation are crucial.
2. Per-unit Values
During calculations, a reference capacity (Sjz) and a reference voltage (Ujz) are selected. All parameters involved in short-circuit calculations are converted into ratios relative to these reference values; these ratios are known as per-unit values. This is a key feature of short-circuit current calculations, as it simplifies the calculations.
(1) References
Reference capacity: Sjz = 100 MVA
Reference voltages UJZ are defined in 8 levels: 230, 115, 37, 10.5, 6.3, 3.15, 0.4, 0.23 KV. With these two values, the reference currents for each voltage level can be calculated. For example:
UJZ (KV): 37, 10.5, 6.3, 0.4
Since S = 1.73 × U × I, then IJZ (KA) = 1.56, 5.59, 9.16, 144.
(2) Calculation of Per-unit Values
Per-unit capacity: S* = S/Sjz. For example, if the short-circuit capacity on a 10 kV busbar is 200 MVA, then its per-unit capacity is S* = 200/100 = 2.
Per-unit voltage: U* = U/UJZ
Per-unit current: I* = I/IJZ
3. Formula for Calculating Three-Phase Short-Circuit Currents in Systems with Infinite Capacity
Per-unit value of short-circuit current: I*d = 1/x* (the reciprocal of the total per-unit reactance).
RMS value of short-circuit current: Id = IJZ × I*d = IJZ/x* (in KA)
RMS value of inrush current: IC = Id × √[(1 + KC – 1)/2] (in KA), where KC is the impact factor, typically taken as 1.8. Therefore, IC = 1.52 × Id.
Peak value of inrush current: ic = 1.41 × Id × KC = 2.55 × Id (in KA)
When there is a short circuit on the secondary side of a transformer with a capacity of 1000 kVA or less, the impact factor KC is taken as 1.3. In this case, the RMS value of the inrush current is IC = 1.09 × Id (in KA), and the peak value is ic = 1.84 × Id (in KA).
With this knowledge, one can calculate short-circuit currents. The formulas are simple and few in number. However, the problem lies in determining the total reactance at the point of short circuit. This includes the reactance of transformers in substation areas, transmission lines, and industrial substations, among others. One method is to consult design manuals, which provide per-unit values for the reactance of common transformers, transmission lines, and reactors. Once the total reactance is determined, the short-circuit current can be calculated using the formulas above. Design manuals also contain tables that allow direct calculation of short-circuit currents.
Here, another “mnemonic phrase”-based calculation method is introduced. By memorizing 7 mnemonic phrases, one can master the method of calculating short-circuit currents.
4. Simplified Algorithms
[1] Calculation of System Reactance
System reactance is expressed in units of hundred megohms. As capacity increases or decreases, reactance decreases in proportion. 100 divided by system capacity. Example: Base capacity is 100 MVA. When the system capacity is 100 MVA, the system reactance is XS* = 100/100 = 1. When the system capacity is 200 MVA, the system reactance is XS* = 100/200 = 0.5. When the system capacity is infinite, the system reactance is XS* = 100/∞ = 0. The unit for system capacity is MVA; this value should be provided by the local power supply authority. When it is not available, the breaking capacity of the power supply outlet switch can be used as the system capacity. It is known that the outgoing switch of the power supply department is of type W-VAC 12KV 2000A, with a rated breaking current of 40KA. Then the system capacity S can be considered to be 1.73*40*10000 V = 692 MVA, and the system reactance XS* is 100/692 = 0.144. 【2】Calculation of transformer reactance: 110 KV, 10.5 divided by transformer capacity ； 35KV, 7 divided by transformer capacity ； 10KV{6KV}, 4.5 divided by transformer capacity. Example: The reactance X* of a 35KV 3200KVA transformer is 7/3.2 = 2.1875, while the reactance X* of a 10KV 1600KVA transformer is 4.5/1.6 = 2.813. The unit for transformer capacity is MVA. The coefficients 10.5, 7, and 4.5 actually represent the percentage of the transformer’s short-circuit reactance. Different voltage levels have different values. 【3】Calculation of the reactor’s reactance: The rated reactance of the reactor is calculated by taking 90% of its rated capacity. Example: There is a reactor with U=6 KV, I=0.3 KA, and a rated reactance of X=4%. Rated capacity S=1.73*6*0.3=3.12 MVA. Reactance of the reactor X*={4/3.12}*0.9=1.15; unit of reactor capacity: MVA. 【4】Calculation of reactance for overhead lines and cables: For overhead lines at 6 KV, it is equal to the length in kilometers ； 10KV, take 1/3 ； For 35KV, use 3% for cables: multiply by 0.2 based on overhead lines. Example: 10KV 6KM overhead line. The reactance of the overhead line is X* = 6/3 = 2 for 10 KV 0.2 KM cables. Cable reactance X*={0.2/3}*0.2=0.013. Simplifications have been made here; in reality, the reactance of overhead lines and cables is related to their cross-sectional area, with a larger area resulting in lower reactance. 【5】Calculation of short-circuit capacity: Reactance is fixed, and 100 is removed. Example: Given that the sum of the per-unit reactance values of the components before the short-circuit point is X*∑=2, then the short-circuit capacity at the short-circuit point is Sd=100/2=50 MVA. Short-circuit capacity unit: MVA 【6】Calculation of short-circuit current: 6KV, 9.2 divided by reactance ； 10KV, 5.5 reactance division; 35KV, 1.6 reactance division; 110KV, 0.5 reactance division. 0.4KV, 150 per reactance. Example: Given that the sum of the per-unit reactance values of all components before the short-circuit point is X*∑=2, and the voltage level at the short-circuit point is 6KV, then the short-circuit current at that point is Id=9.2/2=4.6KA. Unit of short-circuit current: KA 【7】 Calculation of short-circuit inrush current: When there is a short circuit on the secondary side of transformers with a capacity of 1000 KVA or less, the effective value of the inrush current Ic = Id, and the peak value of the inrush current ic = 1.8Id. For transformers with a capacity greater than 1000 KVA, the effective value of the inrush current Ic = 1.5Id, and the peak value of the inrush current ic = 2.5Id. Example: Given that the short-circuit current Id at the short-circuit point {secondary side of a 1600 KVA transformer} is 4.6 KA, then the effective value of the inrush current at this point is Ic = 1.5Id = 1.5*4.6 = 7.36 KA, while the peak value of the inrush current is ic = 2.5Id = 2.5*4.6 = 11.5 KA. It can be seen that the key to calculating short-circuit current is to determine the total reactance {per unit value} before the short-circuit point. However, the system reactance must also be included. As an example, the system is shown in the figure below. A 10 KV overhead line is sent from the regional substation operated by the power utility, and after traveling 10 KM, it reaches the enterprise substation. There is a 200 M cable section before reaching the substation. The substation is equipped with a 1600 KVA transformer. We need to determine the short-circuit parameters at points K1 and K2. The reactance values for the system can be easily calculated using the formula above; there are 4 such values in total. System reactance: X0 = 100/573 = 0.175. Reactance of the 10 KM, 10 KV overhead line: X1 = 10/3 = 3.333. Reactance of the 200 M, 10 KV cable line: X2 = (0.2/3)*0.2 = 0.133. Reactance of the 1600 KVA transformer: X3 = 4.5/1.6 = 2.81. Note that all these reactance values are per unit values (X*). By adding up the reactances of each section, we get the total reactance at point K1 as X0 + X1 = 3.51. The total reactance at point K2 is X0 + X1 + X2 + X3 = 6.45 (not 2.94!). Using the formula, we can then calculate the short-circuit current: U (KV) × X* × I_d (KA) / I_c (KA) = S_d (MVA). Formulas: 5.5/X* × 1.52 × I_d² = 2.55I_d; 100/X* = K1 = 10.5/3.51 = 1.56; 2.37/4.02 = 8.5. Another formula: 150/X* × 1.52 × I_d² = 2.55I_d; 100/X* = K2 = 0.4/6.45 = 233559/15.5. What’s the difference between using these formulas and the method described in Section 3? The formulas yield values in actual units such as KA and MVA, while the method using the formulas gives per unit values. Those who pay close attention will notice that the coefficients 150, 9.2, 5.5, and 1.6 in the formulas actually represent the voltage levels, simplified versions of them. The accurate values should be 144, 9.16, 5.5, and 1.56. What’s the use of these short-circuit parameters? They are used to verify the performance of switches. For example, the low-voltage main switch of this 1600 KVA transformer is of type M25, N1, with a rated current of 2500 A and a rated breaking current of 55 KA. Verification: The rated current of the transformer is 2253 A, and the rated current of the switch is greater than that of the transformer; moreover, the rated breaking current of the switch is greater than the short-circuit current I_d. Thus, the verification is successful.