Generator Stator Earth Fault Protection Relay Setting

The question asks us to determine the relay setting for Merz-Price differential protection used for a generator stator winding. This protection scheme detects internal faults by comparing the current entering and leaving the protected zone. For stator earth faults, the Merz-Price protection system is designed to detect faults within a specific percentage of the winding, usually from the line end towards the neutral point.

In this problem, 90% of the stator winding is protected. This means the unprotected zone is the 10% section of the winding closest to the neutral point. For the relay to provide 90% protection, it must be sensitive enough to detect an earth fault occurring right at the boundary between the protected and unprotected zones, which is at the point 10% away from the neutral end.

Calculating Rated Generator Current

First, we need to find the rated phase current of the generator. The formula for rated apparent power (MVA) is:

\(\text{Rated MVA} = \sqrt{3} \times V_{line} \times I_{rated}\)

Given:

• Rated MVA = 20 MVA = \(20 \times 10^6\) VA
• Rated line voltage \(V_{line}\) = 13.2 kV = \(13.2 \times 10^3\) V

We can calculate the rated current \(I_{rated}\):

\(I_{rated} = \frac{\text{Rated MVA}}{\sqrt{3} \times V_{line}}\)

\(I_{rated} = \frac{20 \times 10^6}{\sqrt{3} \times 13.2 \times 10^3}\)

\(I_{rated} = \frac{20 \times 10^3}{\sqrt{3} \times 13.2}\)

\(I_{rated} \approx \frac{20000}{1.732 \times 13.2} \approx \frac{20000}{22.8624}\)

\(I_{rated} \approx 874.88\) A

(Using more precise values: \(I_{rated} = \frac{20 \times 10^6}{\sqrt{3} \times 13200} \approx 874.77\) A)

Let's also calculate the phase voltage \(V_{phase}\):

\(V_{phase} = \frac{V_{line}}{\sqrt{3}} = \frac{13.2 \times 10^3}{\sqrt{3}} \approx \frac{13200}{1.732} \approx 7621\) V

(Using more precise value: \(V_{phase} = \frac{13200}{\sqrt{3}} \approx 7621.02\) V)

Calculating Minimum Earth Fault Current

The minimum earth fault current that must be detected occurs at the boundary of the protected zone, which is at 10% of the winding length from the neutral end. Let 'p' be the fraction of the winding length from the neutral end (p=0 at neutral, p=1 at line end). The fault is at p = 0.1.

The voltage driving the fault current at this point is approximately \(p \times V_{phase}\). The impedance in the fault loop consists of the impedance of the winding section up to the fault point and the neutral grounding resistance.

The impedance of the winding section up to point p is \(p \times X\), where \(X\) is the total winding reactance per phase (4 Ω). The neutral grounding resistance \(R_n\) is 8 Ω. Assuming the winding reactance is purely reactive and the neutral resistance is purely resistive, the total impedance in the fault loop is the vector sum:

\(Z_{fault}(p) = \sqrt{(p X)^2 + R_n^2}\)

The earth fault current at point p is:

\(I_f(p) = \frac{p V_{phase}}{Z_{fault}(p)} = \frac{p V_{phase}}{\sqrt{(p X)^2 + R_n^2}}\)

The minimum fault current within the protected zone occurs at p = 0.1 (10% from neutral):

\(I_{min} = I_f(0.1) = \frac{0.1 \times V_{phase}}{\sqrt{(0.1 \times X)^2 + R_n^2}}\)

Using \(V_{phase} \approx 7621.02\) V, \(X = 4 \, \Omega\), and \(R_n = 8 \, \Omega\):

\(I_{min} = \frac{0.1 \times 7621.02}{\sqrt{(0.1 \times 4)^2 + 8^2}}\)

\(I_{min} = \frac{762.102}{\sqrt{(0.4)^2 + 64}}\)

\(I_{min} = \frac{762.102}{\sqrt{0.16 + 64}}\)

\(I_{min} = \frac{762.102}{\sqrt{64.16}}\)

\(I_{min} \approx \frac{762.102}{8.01}\)

\(I_{min} \approx 95.14\) A

Determining Relay Setting Percentage

The Merz-Price relay setting is typically given as a minimum pickup current. To ensure 90% protection, the relay pickup current \(I_{pickup}\) must be set equal to or slightly less than the minimum fault current \(I_{min}\) calculated at the boundary (10% from neutral).

\(I_{pickup} = I_{min} \approx 95.14\) A

The relay setting is usually expressed as a percentage of the generator's rated phase current. Let the setting percentage be S.

\(S = \left(\frac{I_{pickup}}{I_{rated}}\right) \times 100\%\)

Using \(I_{pickup} \approx 95.14\) A and \(I_{rated} \approx 874.77\) A:

\(S = \left(\frac{95.14}{874.77}\right) \times 100\%\)

\(S \approx 0.10875 \times 100\%\)

\(S \approx 10.875\%\)

Comparing this calculated percentage to the given options, the closest value is 10.8%.

Verification

Let's verify if a relay setting of 10.8% of rated current indeed provides approximately 90% protection. A setting of 10.8% means the pickup current is \(I_{pickup} = 0.108 \times I_{rated} = 0.108 \times 874.77 \approx 94.475\) A. The unprotected zone is where the fault current \(I_f(p)\) is less than this pickup current. The boundary of the unprotected zone is at the point 'p' where \(I_f(p) = I_{pickup}\).

\(\frac{p V_{phase}}{\sqrt{(p X)^2 + R_n^2}} = I_{pickup}\)

\(\frac{p \times 7621.02}{\sqrt{(4p)^2 + 8^2}} = 94.475\)

\(\frac{7621.02p}{\sqrt{16p^2 + 64}} = 94.475\)

\(7621.02p = 94.475 \sqrt{16p^2 + 64}\)

Squaring both sides:

\((7621.02p)^2 = (94.475)^2 (16p^2 + 64)\)

\(58080000 p^2 \approx 8925.5 (16p^2 + 64)\)

\(58080000 p^2 \approx 142808 p^2 + 571232\)

\(57937192 p^2 \approx 571232\)

\(p^2 \approx \frac{571232}{57937192} \approx 0.00986\)

\(p \approx \sqrt{0.00986} \approx 0.0993\)

This shows that a 10.8% setting protects down to approximately 9.93% from the neutral end, leaving 9.93% unprotected. This is very close to the specified 10% unprotected zone (90% protected).

Therefore, the relay setting should be approximately 10.8%.