How to Calculate Effective and Low-impedance Grounding for Power Systems

The best method to control overvoltages is effective, or solid, grounding. It’s important to properly calculate the resistors required for this type of grounding. This article shows how to calculate effective and low-impedance grounding through a step-by-step example.

These methods supply phase-to-neutral connected loads without the risk of finding dangerous neutral-to-ground voltages when a ground-fault arises. An effectively grounded system also has the highest values of ground-fault currents, which may range from zero to three times the three-phase short-circuit current.

In the USA, low-resistance grounding is the most popular method utilized to limit ground-fault current. The value of resistance is much lower than that of the high-resistance method and ranges from 5% to 20% of the three-phase fault current. Some applications limit the ground current to around 50A to 600A.

A typical resistor of 400A will allow enough current flow to operate the protective relays for fast fault clearing. These resistors are also time-rated. A standard figure is 10s because, like in effective grounding, the branch will shut down after the first ground-fault.

 

Effective and Low-impedance Grounding Calculations

The example below will illustrate the impact of effective and low-impedance (reactance or resistance) grounding on the power system.

Note: 

Xd’’ = Direct-axis subtransient reactance (a synchronous generator parameter)

VLL = Voltage line-to-line

Zb = Base impedance  (This is part of the per-unit system used to analyze power systems)

A symmetrical synchronous generator operated at no load (i.e., with the main circuit breaker open), and rated 500 MVA, 22kV, is wye connected and solidly grounded. 

Generator constants:

X_d’’= X₁ = X₂ = 15% ; X₀ = 5%

 

1. Find the subtransient single line-to-ground fault current in the generator windings.

Base values kV_b = 22 and MVA_b = 500

Zb = (kV_b)²/MVA_b = (22)²/500 = 0.968 Ω

X_d’’ = X₁ = X₂ = 0.15 ∙ 0.968 = 0.1452 Ω

X₀ = 0.05 ∙ 0.968 = 0.0484 Ω

I₀ = V_LN/(X₁ + X₂ + X₀) = (V_LL/√3)/(X₁ + X₂ + X₀)

I₀ = (22/√3) kV/(0.1452 + 0.1452 + 0.0484) Ω = 12.7 kV/0.3388 Ω = 37.49 kA

I_f1ɸ = 3∙I₀ = 3 ∙ 37.49 = 112.47 kA

 

Figure 1. Unloaded, solidly grounded generator with a line-to-ground fault on phase a

 

2. Find the ratio Xₒ/X₁

Xₒ/X₁ = 5%/15% = 0.333

Note: Xₒ/X₁ < 3 —> system effectively grounded

 

3. Find the subtransient three-phase fault current in the generator windings.

I_f3ɸ = V_LN/X₁ = (V_LL/√3)/X₁ = (22/√3) kV/0.1452 Ω = 87.48 kA

 

Figure 2. Unloaded, solidly grounded generator with a three-phase short circuit at its terminals

 

4. Find the ratio I_f1ɸ/I_f3ɸ.

I_f1ɸ/I_f3ɸ = 112.47 kA/87.48 kA = 1.286

Note: The line-to-ground fault current is 28.6% higher than the three-phase fault current. It may damage the generator.

5. Find the reactance of a neutral grounding reactor to limit the line-to-ground fault current to 100% of the three-phase fault current.

I_f1ɸ = 3∙I₀ = I_f3ɸ = 87.48 kA = 3∙V_LN/(X₁ + X₂ + X₀ + 3∙X_N) = (√3∙√3∙ V_LL/√3)/(X₁ + X₂ + X₀ + 3∙X_N) = √3∙22 kV/(0.1452 + 0.1452 + 0.0484 + 3∙X_N) Ω = 38.106 kV/(0.3388 + 3∙X_N) Ω => X_N = 0.03226 Ω

Note: X₀ + 3∙X_N = X₁ = X₂ when I_f1ɸ = I_f3ɸ.

Figure 3. Unloaded, low-impedance grounded generator with a line-to-ground fault on phase a

 

6. Find the ratio Xₒ/X₁ with the neutral grounding reactor of question 5.

3∙X_N = (3∙0.03226 Ω/0.968 Ω) ∙100 = 10%

With the neutral grounding reactor inserted in the zero-sequence circuit X0 = 5% + 10% = 15% = X1

Xₒ/X₁ = 15%/15% = 1

Note 1: The fault current contribution of one winding of this generator to a line-to-ground fault, outside the machine, will not be higher than the three-phase fault current of the generator.

Note 2: Xₒ/X₁ < 3 —> By definition, the system is not low-reactance grounded, despite having a reactance inserted in the neutral.

    

7. Find the reactance of a neutral grounding reactor to limit the line-to-ground fault current to 60% of the three-phase fault current.

I_f1ɸ = 3∙I₀ = 0.60∙I_f3ɸ = 0.60∙87.48 kA = 52.49 kA = 38.106 kV/(0.3388 + 3∙X_N) Ω => X_N = 0.1291 Ω

    

8. Find the ratio Xₒ/X₁ with the neutral grounding reactor of question 7.

3∙X_N = (3∙0.1291 Ω/0.968 Ω) ∙100 = 40%

X₀ = 5% + 40% = 45%

Xₒ/X₁ = 45%/15% = 3

Note: 60% of I_f3ɸ is the maximum ground-fault current for the system to be low-reactance grounded.

     

9. Find the reactance of a neutral grounding reactor to limit the line-to-ground fault current to 25% of the three-phase fault current.

I_f1ɸ = 3∙I0 = 0.25∙I_f3ɸ = 0.25∙87.48 kA = 21.87 kA = 38.106 kV/(0.3388 + 3∙X_N) Ω => X_N = 0.4679 Ω

 

10. Find the ratio Xₒ/X₁ with the neutral grounding reactor of question 9.

3∙X_N = (3∙0.4679 Ω/0.968 Ω) ∙100 = 145%

X₀ = 5% + 145% = 150%

Xₒ/X₁ = 150%/15% = 10

Note: To control transient overvoltages, the ground-fault current with a reactor should not be less than 25% of the three-phase fault current

 

11. Find the resistance of a neutral grounding resistor to limit the line-to-ground fault current to 100% of the three-phase fault current.

I_f1ɸ = 3∙I₀ = I_f3ɸ = 87.48∠ kA = 3∙V_LN/j(X₁ + X₂ + X₀) + 3∙R_N = 38.106 kV/j(0.3388) + 3∙R_N Ω

Ɩ87.48∠Ɩ kA = Ɩ38.106 kV/j(0.3388) + 3∙R_N ΩƖ

Ɩj(0.3388) + 3∙R_NƖ = √(0.3388)² + 9∙(R_N)² = 38.106 kV/87.48 kA = 0.436 Ω

(R_N)² = 0.00836 Ω²

R_N = +/- √0.00836 = +/- 0.0914 Ω

Take the positive root R_N = 0.0914 Ω

 

12. Find the resistance of a neutral grounding resistor to limit the line-to-ground fault current to 400 A.

I_f1ɸ = 3∙I₀ = 0.4∠ kA

Ɩ0.4∠Ɩ kA = Ɩ38.106 kV/j(0.3388) + 3∙R_N ΩƖ

Ɩj(0.3388) + 3∙R_NƖ = √(0.3388)² + 9∙(R_N)² = 38.106 kV/0.4 kA = 95.265 Ω

(R_N)² = 1 008.32 Ω²

R_N = +/- √1 008.32 = +/- 31.75 Ω

Take the positive root R_N = 31.75 Ω

 

13. Repeat question 12 applying the expression R_N = V_LN/I_f1ɸ.

R_N = V_LN/I_f1ɸ = 22/√3 kV/0.4 kA = 31.75 Ω

Note: when the resistance value is significant compared with the system impedance, its value can be computed directly as R_N = V_LN/I_f1ɸ.