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Download PUB‐NLH‐470 Island Interconnected System Supply Issues and Power Outages Page 1 of 5 Q.
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PUB‐NLH‐470 Island Interconnected System Supply Issues and Power Outages Page 1 of 5 1 Q. 2 Further to the response to PUB‐NLH‐238, explain how the pre‐insertion resistor is used to discharge the dc cable prior to connection of the cable to the other pole. 3 4 5 A. For discussion purposes, assume that the spare cable had been connected in 6 parallel to the positive pole cable and is operating at +350 kV when there is a 7 requirement to switch the spare cable to the negative pole due to a loss of the 8 negative pole cable. The first step of the cable‐switching scheme requires that the 9 spare cable be disconnected from the positive pole. When the switches connecting 10 the spare cable to the positive pole are opened and the spare cable is isolated, the 11 effect of the cable capacitance is to result in a trapped charge being maintained on 12 the spare cable, similar to a shunt capacitor that has just been switched out of a 13 circuit. For discussion purposes, the spare cable can be assumed to have a +350 kV 14 voltage on it despite being isolated from the positive pole. Before the spare cable 15 can be connected to the negative pole operating at ‐350 kV, the trapped charge on 16 the spare cable must be dissipated such that the spare cable voltage is brought to 17 zero voltage (i.e., discharge the cable). This is accomplished by connecting the 18 spare cable to ground through a resistor as shown in Figure 1. PUB‐NLH‐470 Island Interconnected System Supply Issues and Power Outages Page 2 of 5 1 POLE 1 CABLE + SPARE CABLE ‐ Spare Cable parallel with Pole 1 Cable Fault occurs on Pole 2 Cable 2 POLE 2 CABLE FAULT POLE 1 CABLE + SPARE CABLE ‐ POLE 2 CABLE Isolate faulted Pole 2 Cable Remove Spare Cable from Pole 1 circuit 3 POLE 1 CABLE + SPARE CABLE ‐ POLE 2 CABLE Discharge Spare Cable through Resistor 4 Resistor, R Spare Cable Capacitance, C + Vc ‐ Cable Discharge Circuit Figure 1 – Simplified Spare Cable Discharge Scheme 1 2 PUB‐NLH‐470 Island Interconnected System Supply Issues and Power Outages Page 3 of 5 The connection provides an RC circuit in which the voltage on the cable is calculated by the equation: 3 ∗ ∗ 4 Where, 5 Vc is the cable voltage in kV 6 Vco is the initial cable voltage at the time the ground switches are closed 7 t is the time in seconds 8 R is the pre‐insertion resistor resistance in ohms, Ω 9 C is the cable capacitance in Farads, F 10 11 Given a total spare cable capacitance of 16.72 µF (16.72 x 10‐6 F), one can calculate 12 the cable voltage versus time for varying pre‐insertion resistor ohmic values during 13 the spare cable discharge sequence. 14 15 Figure 2 provides a plot of the spare cable voltage versus time for both a 100 Ω and 16 a 400 Ω pre‐insertion resistor. One notes that the lower pre‐insertion resistor value 17 results in a faster discharge time for the spare cable. PUB‐NLH‐470 Island Interconnected System Supply Issues and Power Outages Page 4 of 5 Spare Cable Discharge Through Resistor Voltage versus Time 400 350 Cable Voltage ‐ kV 300 250 200 400 Ohm 150 100 Ohm 100 50 0 0 0.01 0.02 0.03 0.04 0.05 Time ‐ seconds Figure 2 – Spare Cable Voltage versus Time for Discharge Through a 100Ω and a 400Ω Pre‐insertion Resistor 1 2 Similarly, the spare cable discharging current can be calculated for the RC circuit 3 using the following equation: 4 Where 5 ic is the cable current in kA ∗ ∗ 6 Vco is the initial cable voltage at the time the ground switches are closed in kV 7 t is the time in seconds 8 R is the pre‐insertion resistor resistance in ohms, Ω 9 C is the cable capacitance in Farads, F 1 PUB‐NLH‐470 Island Interconnected System Supply Issues and Power Outages Page 5 of 5 Figure 3 provides a plot of the spare cable discharge current versus time for both a 2 100 Ω and a 400 Ω pre‐insertion resistor1. One notes that the lower pre‐insertion 3 resistor value results in a greater discharge current for the spare cable. For the 4 100Ω resistor value the peak discharge current equals 3.5 kA versus a peak 5 discharge current of 0.9 kA for the 400Ω resistor. 6 7 Consequently, the selection of the pre‐insertion resistor size is a balance between 8 the discharge time and the peak discharge current. The pre‐insertion resistor size 9 will be selected during the final design stage. 10 Spare Cable Discharge Through Resistor Current versus Time 5 Cable Current ‐ kA 4 3 400 Ohm 2 100 Ohm 1 0 0 0.01 0.02 0.03 0.04 0.05 Time ‐ seconds Figure 3 – Spare Cable Discharge Current versus Time for Discharge Through a 100Ω and a 400Ω Pre‐insertion Resistor 1 The cable current is depicted in Figure 3 as a positive current for clarity despite the current flow being out of the cable, and therefore by convention, considered negative in value.
