H A N D S O N R

H A N D S O N R

H A N D S O N R E L A Y S C H O O L 2 0 1 Ground-Distance Concepts for Relay Technicians Steve Laslo System Protection and Control Specialist Bonneville Power Administration 9 H A

N D S O N R E L A Y S C H O O L 2 Special Thanks Quintin Jun Verzosa Jr. Doble Engineering Company Schweitzer Engineering Laboratories, Inc. 0 1 9 H A N D

S O N R E L A Y S C H O O L 2 0 1 9 Presentation Information Our Objectives for this presentation are to enhance attendee knowledge by: Reviewing Ground-Distance relay fundamentals that affect relay settings and relay decision-making. Review calculations for test quantities for testing basic Ground-Mho and Ground-Quad characteristics. Explore testing of basic Ground-Mho and Ground-Quad characteristics. H A N D

S O N R E L A Y S C H O O L 2 0 1 9 Some Background Historically, ground-distance relays were not commonly used in the era of electro-mechanical relaying. In that era protection was commonly: Phase-distance Directional Ground-overcurrent With the advent of solid-state and now microprocessor relays, the ground-distance functions are more easily accomplished and have become more commonly applied. H A

N D S O N R E L A Y S C H O O L 2 0 1 9 Some Background Training Programs for Relay Technicians have historically educated new technicians well in how to test and work on phase-distance relays because they have been around for so long. Not all training programs have done as well when it comes to ground-distance concepts and practical application. This presentation will attempt to fill in some of the conceptual gaps that might be present between phase-distance and ground-distance relaying. H

A N D S O N R E L A Y S C H O O L 2 0 1 9 Protection Key Concept Relays make impedance calculations based on the values of voltage and current at the relay location. H A N D

S O N R E L A Y S C H O O L 2 0 1 Phase-Distance Decisions Fault current is limited by the line impedance. Relays see the correct impedance to the fault location. 9 H A N D S O N

R E L A Y S C H O O L 2 0 Phase-Distance Decisions Test values are relatively easy to calculate: Assume test voltage = 40.4V I = E / Z = 40.4V / 9.36 = 4.32A = 4.32A Test Quantities: VA = 40.40V, VV, VB = 40.4-120V, VV, VC = 40.4120V, VV IA = 4.32-83.97V, VA, IB = 4.32-203.97V, VA, IC = 4.3236 = 4.32A.03V, VA 1 9 H A N D S O N

R E L A Y S C H O O L 2 0 1 9 Relay Settings for our example - Positive and Zero-Sequence impedance of the protected line (magnitude and angle) in Secondary Ohms. - Reach of Zone 1, Zone 2, and Zone 3 for Phase Mho Characteristic in Secondary Ohms. - Overcurrent Supervision settings; ignored for our discussion. - Reach of Zone 1, Zone 2, and Zone 3 for Ground Mho Characteristic in Secondary Ohms. - Reactive* Reach of Zone 1, Zone 2, and Zone 3 for Ground-Quadrilateral Characteristic in Secondary Ohms. - Resistive Reach of Zone 1, Zone 2, and Zone 3 for Ground-Quadrilateral Characteristic in Secondary Ohms. - Zero-Sequence Compensation Factor Settings; magnitude and angle. T is a correction factor we will ignore in our discussion. Note that all settings above (for this manufacturer) are Per-Phase Values H A N D

S O N R E L A Y S C H O O L 2 0 1 9 H A N D S O N R E

L A Y S C H O O L 2 0 1 9 H A N D S Associated with Normal conditions where we only have positive sequence. This diminishes when things go wrong. O N R E L A

Y S C Associated with unbalance, in any form. Does not exist when we are under ideal normal conditions. H O O L 2 0 1 9 Associated with ground, as in ground-faults. Does not exist when we are under ideal normal conditions. H A N D S O N R E L

A Y S C H O O L 2 Zone 2 Settings If the impedance to the fault is 9.36 = 4.32A83.97V, V Ohms secondary, and the reach of our relay based on our setting is the same value, then our relay is reaching right up to the fault and should operate. 0 1 9 H A N D S O N R E L A Y

S C H O O L 2 0 Per-Phase Values Are just what they sound like: In this case, they are the values of impedance per-phase, as shown in the picture below. 1 9 H A N D S O N R E L A Y S C

H O O L 2 Ground-Distance Decisions Fault current is limited by conductor impedance + ground impedance. Relays see an incorrect* loop impedance to the fault location. 0 1 9 H A N D S O N R E L A Y S C H O

O L 2 0 Ground-Distance Decisions If the relay sees an impedance of 16 = 4.32A.1582.42V, V, then the fault appears to be much farther away than it actually is on the transmission line. How does the relay properly locate the fault? 1 9 H A N D S O N R E L A Y S C H O O L

2 0 1 9 Zero Sequence Compensation We factor out the ground impedance using a compensation factor: KN What does KN do for us? First remember that for a ground fault a relay sees a combination of line impedance + ground impedance. Compensation factors allow a relay to factor out the portion of impedance seen at the relay location that is the ground impedance. This allows the relay to estimate the transmission line impedance to the fault location. H A N D S O N R E L A Y S C H O

O L 2 Ground-Distance Decisions How does the relay properly locate the fault? Relay sees an impedance of 16 = 4.32A.1582.42V, V Relay uses KN to factor out 6 = 4.32A.8080.28V, V of ground impedance. Relay knows impedance to fault is 9.36 = 4.32A83.97V, V Relay takes proper logical actions 0 1 9 H A N D S O N R E L A Y S C H O O

L 2 0 1 9 H A N D S O N R E L A Y S C H O O L 2 0 1 9

How does this relay properly locate the fault? This relay uses the settings highlighted on the right to make compensation calculations. The first two are used for Zone 1 while the second two are used for other Zones. From the SEL-321 Manual: H A N D S O N R E L A Y S C H O O L 2 0 Ground-Mho Where did this test current come from? Since relay manufacturers vary in their exact form of compensation we need the form specifically used by this relay:

ZAG = (VA / IA) / (1+k0) ZPer-Phase = ZLoop / (1+k0) 1 9 H A N D S O N R E L A Y S C H O O L 2 0 Ground-Mho How do we use this? Using our knowledge that the loop impedance is the perphase impedance times the compensation factor we can take the Zone 2 settings, multiply the compensation factor and our result is the relay reach for Zone 2. Lets do it:

Zone 2 per-phase setting/reach is Z2MGZ1ANG Z=9.36 = 4.32A83.97 The compensation factor=1+k0 where k0=k0mk0AV, V KN = 1 + 0.726 = 4.32A-3.6 = 4.32A9V, V = 1.725-1.552V, V ZLoop= 9.36 = 4.32A83.97 * 1.725-1.552V, V = 16 = 4.32A.1582.42 Look familiar? 1 9 H A N D S O N R E L A Y S C H O O L 2 Ground-Distance Decisions How does the relay properly locate the fault? Relay sees an impedance of 16 = 4.32A.1582.42V, V

to factor out 6.8080.28 of zero of zero - Relay Zloop / uses KN =KZN per-phase sequence impedance 16 = 4.32A.1582.42V, V / (1+ 0.726 = 4.32A-3.6 = 4.32A9V, V ) = 9.36 = 4.32A83.97V, V Relay knows impedance to fault is 9.36 = 4.32A83.97V, V Relay takes proper logical actions 0 1 9 H A N D S O N R E L A Y S C H O O L 2

0 1 9 Ground-Distance Graphical Analysis Here we see: ZPh = the per-phase impedance: Z=9.36 = 4.32A83.97 ZKN = the compensated impedance: Z=6 = 4.32A.8080.28 KN = 1 + 0.726 = 4.32A-3.6 = 4.32A9V, V = 1.7251.552V, V ZLp = the loop impedance: Z=16 = 4.32A.1582.42 Note that the relay response is defined by the Loop impedance H A N D S O N R E L A Y S C H

O O L 2 0 1 9 Ground-Distance Graphical Analysis Any point on the per-phase characteristic can be translated to an equivalent point on the loop characteristic. Any per-phase impedance multiplied by the compensation factor gives the equivalent loop impedance. H A N D S O N R E L A Y S C

H O O L Ground Quadrilateral Key Ground-Quadrilateral Settings: XG2 = Reactive* reach of the quadrilateral element. Z=9.36 = 4.32A83.97 RG2 = Resistive reach of the quadrilateral element. Z=5.000 2 0 1 9 H A N D S O N R E L A Y S C

H Ground Quadrilateral XG2 RG2 O O L 2 0 1 9 H A N D S O N R E L A Y S C H Ground Quadrilateral

XG2 RG2 O O L 2 0 1 9 H A N D S O N R E L A Y S C H Ground Quadrilateral Z2MG XG2

RG2 O O L 2 0 1 9 H A N D S O N R E L A Y S C H O O L 2

0 1 9 Testing Ground-Distance Relays Review: Relays must make decisions based on the voltage and current at the relay location. Raw relay response to ground-impedance functions happens in the loop-impedance plane. The relay has to factor out the zero-sequence impedance to calculate a fault position on the protected line. Testing: Essentially happens in the loop-impedance plane. Once a test voltage is determined, one divides the loop impedance to calculate the test current for the impedance point being tested. H A N D S O N R E L A Y S C H O Calculating Test Quantities

Ground-Mho Characteristic: Test voltage = 300V What is our test current at: The Line Angle? [email protected] Line Angle = 16 = 4.32A.1582.42V, V I = E / Z = 300V, VV / 16 = 4.32A.1582.42V, V I = 1.86 = 4.32A-82.42V, VA At 45V, V? [email protected], V = 16 = 4.32A.15 * COS(82.42V, V-45V, V) = 12.83 I = E/Z = 300V, VV / 12.8345V, V I = 2.34-45V, VA At 90V, V? [email protected], V = 16 = 4.32A.15 * COS(82.42V, V-90V, V) = 16 = 4.32A.01 I = E/Z = 300V, VV / 16 = 4.32A.0190V, V I = 1.87-90V, VA Hint: ZR = ZMAX * COS(MTA-)) O L 2 0 1 9 H A N D S O N R E L A

Y S C H O O Interpreting Test Results Examine how your results are reported: L 2 0 1 9 H A N D S O N R E L A Y S C

Interpreting Test Results H O O L 2 0 1 9 H A N D S O N R E L A Y S C H O O Calculating Test Quantities Ground-Quad Characteristic:

Test voltage = 300V What is our test current at: The Line Angle? [email protected] Line Angle = 16 = 4.32A.1582.42V, V I = E / Z = 300V, VV / 16 = 4.32A.1582.42V, V I = 1.86 = 4.32A-82.42V, VA At 90V, V? [email protected], V = y component of: 16 = 4.32A.1582.42V, V 16 = 4.32A.1582.42V, V = (2.13 + j16 = 4.32A.01); y = 16 = 4.32A.01 I = E/Z = 300V, VV / 16 = 4.32A.0190V, V I = 1.87-90V, VA L 2 0 1 9 H A N D S O N R E L A Y S C H

O O L Calculating Test Quantities Ground-Quad Characteristic: Test voltage = 300V What is our test current at: At 0V, V? [email protected], V = 5 I = E/Z = 300V, VV / 50V, V I = 6 = 4.32A0V, VA At Top Right Corner of Quad? [email protected] = 16 = 4.32A.1582.42V, V + 50V, V = 17.526 = 4.32A5.99V, V I = E/Z = 300V, VV / 17.526 = 4.32A5.99V, V I = 1.71-6 = 4.32A5.99V, VA 2 0 1 9 H A N D S O N R E L A Y

S C H O O L 2 0 1 Calculating Test Quantities Ground-Quad Characteristic: Test voltage = 300V What is our test current at: A point in the per-phase plane: ZP1 = 4.6 = 4.32A26 = 4.32A0V, V? ZP1 = 4.6 = 4.32A26 = 4.32A0V, V = 2.31 + j4.00 Z1X = Tan(90-83.97)V, V*4.00 = 0.422 Z1 = 0.422 + j4.00 = 4.02283.97V, V Z1 = Z1*KN = 4.02283.97V, V * 1.725-1.552V, V KN = 1 + 0.726 = 4.32A-3.6 = 4.32A9V, V = 1.725-1.552V, V Z1 = 6 = 4.32A.9482.42V, V = 0.915 + j6 = 4.32A.88 ZL1 = Z1 + (1.89 + j0) = 2.80 + j6 = 4.32A.88 = 7.436 = 4.32A7.82V, V I = E/Z = 300V, VV / 7.436 = 4.32A7.82V, V I = 4.04-6 = 4.32A7.82V, VA 9 H A N D S

O N R E L A Y S C H Some Test Considerations O O L 2 0 1 9 H A N D S O N R E

L A Y S C H Some Test Considerations O O L 2 0 1 9 H A N D S O N R E L A Y S

C H O O L 2 0 1 9 Synopsis - Fundamentals 1. Relays make decisions based on voltage and current at the relay location. 2. Relays are generally set in terms of per-phase impedance. 3. For ground faults relays respond to loop impedance. Relays are tested in the loop impedance plane. 4. A Compensation Factor (KN) is used to factor out zero sequence impedance to allow a relay to make a fault location decision based on the per-phase settings. Relay manufacturers use a variety of forms of compensation but their fundamental application is the same. 5. Relay test software may need interpretation to resolve discrepancies between raw values of voltage and current and per-phase settings. H A N D S O N R

E L A Y S C H O O L 2 0 Thanks for your time! Please provide feedback to the school on this presentation and its content. 1 9

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