When a short circuit occurs in an electrical power system, it can result in significant damage to equipment due to excessive current and voltage drops. This can lead to overheating, insulation failure, or even complete system collapse. To prevent such scenarios, it's crucial to calculate the short-circuit current accurately. This helps in selecting appropriate electrical equipment, designing effective relay protection systems, and choosing components that limit fault currents.
### Key Calculation Conditions
1. **Assume Infinite System Capacity**
In most cases, the power system is considered to have infinite capacity, meaning the voltage remains constant during a short circuit. The impedance of the network is much smaller than the system impedance. For 3–35 kV grids, the system above 110 kV is typically treated as infinite, while only the impedance of lower-voltage components is calculated.
2. **Reactance vs Resistance**
When calculating short-circuit current for high-voltage devices, only reactance (X) is considered, not resistance (R). For overhead lines and cables, if R is less than one-third of X, only X is used in calculations.
3. **Three-Phase Short-Circuit Assumption**
Short-circuit current calculations are generally based on three-phase faults, as they produce the highest current. Single-phase or two-phase faults result in lower current values, so equipment designed for three-phase faults will automatically handle these cases.
### Simplified Calculation Method
Even with some assumptions, precise calculation of short-circuit current can be complex. A simplified "mouth-style" method allows users to remember seven key rules to estimate the values quickly. This method avoids the need for detailed charts or design manuals.
### Basic Parameters
- **Sd**: Three-phase short-circuit capacity (in MVA), used to check switch breaking capacity.
- **Id**: RMS value of the periodic component of the short-circuit current (in KA), used to check switch breaking and thermal stability.
- **Ic**: RMS value of the first-cycle full current (in KA), used to check dynamic stability.
- **ic**: Peak value of the first-cycle full current (in KA), also used for dynamic stability checks.
- **x**: Reactance (in Ω).
The key parameters are the system short-circuit capacity (Sd) and the total reactance (x) at the point of the fault.
### Reference Values
To simplify calculations, reference values are used:
- **Reference Capacity (Sjz)**: Typically 100 MVA.
- **Reference Voltage (Ujz)**: Varies by system level (e.g., 10.5 kV, 6.3 kV, 0.4 kV).
- **Reference Current (Ijz)**: Calculated using $ I = \frac{S}{\sqrt{3} \times U} $.
For example:
- At 10.5 kV: $ Ijz = \frac{100}{\sqrt{3} \times 10.5} \approx 5.59 \, \text{kA} $
- At 0.4 kV: $ Ijz = \frac{100}{\sqrt{3} \times 0.4} \approx 144 \, \text{kA} $
### Short-Circuit Current Formulas
- **Short-Circuit Current (Id)**: $ Id = \frac{Ijz}{x^*} $
- **Inrush Current (Ic)**: $ Ic = 1.52 \times Id $ (for transformers over 1000 kVA)
- **Peak Impulse Current (ic)**: $ ic = 2.55 \times Id $
For transformers under 1000 kVA, the formulas adjust slightly:
- $ Ic = 1.09 \times Id $
- $ ic = 1.84 \times Id $
### Simplified Rules for Quick Estimation
Here are seven simple rules to estimate short-circuit current without detailed calculations:
1. **System Reactance**:
- If the system capacity is 100 MVA, its reactance is 1 per unit.
- If the system is larger, the reactance decreases inversely.
- Example: If the system capacity is 200 MVA, the reactance is 0.5 per unit.
2. **Transformer Reactance**:
- 110 kV: Divide the transformer capacity (in MVA) by 10.5
- 35 kV: Divide by 7
- 10 kV: Divide by 4.5
- Example: 3200 kVA at 35 kV → $ x^* = 7 / 3.2 = 2.1875 $
3. **Reactor Reactance**:
- Reactor reactance is calculated as $ x^* = \frac{\% \text{reactance}}{\text{rated MVA}} \times 0.9 $
- Example: 6 kV, 0.3 kA reactor with 4% reactance → $ x^* = \frac{4}{3.12} \times 0.9 = 1.15 $
4. **Overhead Line and Cable Reactance**:
- Overhead line: 6 kV = km; 10 kV = 1/3 km; 35 kV = 3%
- Cable: Multiply by 0.2 of the overhead value
- Example: 10 kV, 6 km overhead line → $ x^* = 6 / 3 = 2 $
5. **Short-Circuit Capacity**:
- $ Sd = \frac{100}{x^*} $
- Example: $ x^* = 2 $ → $ Sd = 50 \, \text{MVA} $
6. **Short-Circuit Current**:
- 6 kV: $ Id = 9.2 / x^* $
- 10 kV: $ Id = 5.5 / x^* $
- 35 kV: $ Id = 1.6 / x^* $
- 110 kV: $ Id = 0.5 / x^* $
- Example: $ x^* = 2 $ at 6 kV → $ Id = 4.6 \, \text{kA} $
7. **Inrush Current**:
- For transformers < 1000 kVA: $ Ic = Id $, $ ic = 1.8 \times Id $
- For transformers > 1000 kVA: $ Ic = 1.5 \times Id $, $ ic = 2.5 \times Id $
- Example: $ Id = 4.6 \, \text{kA} $ → $ Ic = 7.36 \, \text{kA}, \, ic = 11.5 \, \text{kA} $
By following these rules, you can quickly estimate short-circuit current and ensure proper selection of protective devices and equipment. The key is understanding how to calculate the total reactance before the fault point, which includes the system, transformers, lines, and reactors.
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