Digital Energy Multilin GE Multilin technical note eutral voltage unbalance function at grounded wye capacitor banks GE publication number: GET-8544 Copyright 00 GE Multilin eutral voltage unbalance function at grounded wye capacitor banks Introduction The C70 neutral overvoltage unbalance function (ASI 59U) compensates for both the system unbalance ( 0 ), and the bank inherent unbalance using balancing factors k AB and k AC. The operating quantity equation for a perfectly balanced bank (k AB = k AC = ) takes the following familiar simplified form. op = X 0 (Eq ) In this equation, X is a bank neutral point voltage and 0 is a bus (system voltage) which can either calculated from phase voltages or supplied externally from broken delta T. It is preferable to have the bus voltage supplied from a three-phase voltage T, since the C70 can compensate for bank inherent unbalance by applying balancing coefficients. The auto-setting procedure is available to automatically calculate these coefficients. For greater sensitivity and security, the relay applies the differential principle with the restraint quantity rest calculated as follows. rest = X + 0 The neutral overvoltage unbalance element operates if: (Eq ) op > S rest (Eq 3) Where S is a user-programmable slope factor. The neutral overvoltage unbalance element can be applied to both ungrounded and grounded capacitor banks. Applying neutral overvoltage unbalance to ungrounded banks appears logical and simple. However, it may not be so obvious for grounded banks. In this case, the banks are grounded through a neutral capacitor with a single-phase T across the capacitor or grounded through a CT to measure the neutral current or equivalent neutral voltage. This application note attempts to clarify the C70 application for these type of grounded bank configurations. GE MULTILI TECHICAL OTE EUTAL OLTAGE UBALACE FUCTIO AT GOUDED WYE CAPACITO BAKS
eutral point voltage equivalent The following figure depicts two possible grounding arrangements for the capacitor bank. Grounded through neutral capacitor. Grounded through neutral CT. In any configuration the value of the resistor has to be chosen so that the voltage which appears at the neutral point (one phase voltage at the bus equals zero) across resistor through the CT/auxiliary T input matches the bus 0 value. Figure : eutral arrangement for grounded wye banks Alternative: measured bus 30 Phase CT bank T Phase T bank C70 Capacitor Bank Protection and Control System T Alternative CT In the case of neutral CT configuration, it can be achieved by calculating value of the resistor in neutral CT secondary as follows: 834769A.CD BLL ( ) CT = ------------------------------ 3 S B T (Eq 4) In the above equation: B(LL) is bus phase-to-phase bus voltage (in k). S B is the bank reactive power rating (in MA). T is the auxiliary T input ratio setting, which may be same as bus T ratio or used to achieve balance if an exact resistor value is not possible. CT is the neutral CT ratio. The voltage derived from the current in the neutral secondary resistor is shifted 90 from the bus 0 voltage. The relay automatically compensates for this if the eutral oltage Unbalance Ground setting is programmed as CTx (grnd). GE MULTILI TECHICAL OTE EUTAL OLTAGE UBALACE FUCTIO AT GOUDED WYE CAPACITO BAKS
GEEAL CALCULATIOS Application example Consider a capacitor bank with the following ratings. Bus phase-to-phase bus voltage B(LL) =69k. Bank reactive power rating S B =6.MA. 900 ka capacitor can rated 9.9 k. Phase T ratio is 600:. T secondary = 67. eutral CT is 75/5A. Each capacitor can consists from = 0 capacitor sections. Two capacitors cans are connected in series (P = ) and there are three parallel strings (S = 3) in each phase. Figure : Sample bank configuration 69 k Measured bus 30 000/5A Phase CT bank 6. MA total T C70 Capacitor Bank Protection and Control System 4000/67 75/5A T 834770A.CD General calculations The nominal current of the capacitor bank is calculated as follows. S I B 0 3 6. MA 0 3 nom = --------------------------- = -------------------------------------- = 33.55 A 3 BLL ( ) 3 69 The impedance of the healthy phase capacitors stack is calculated as follows. (Eq 5) Z P -------------- BLL ( ) ( 69 ) = = ----------------------- = 93.89 Ω S B 6. MA (Eq 6) The impedance of the capacitor can is calculated as follows. Z Z P S 93.89 Ω 3 C = -------------- = -------------------------------- = 440.84 Ω P (Eq 7) GE MULTILI TECHICAL OTE EUTAL OLTAGE UBALACE FUCTIO AT GOUDED WYE CAPACITO BAKS 3
UBALACE CALCULATIOS The impedance of the capacitor section is calculated as follows. Z Z C S = ----- = 440.84 ----------------------- Ω = 44.084 Ω 0 (Eq 8) From the formula above, the value of resistor in the neutral CT secondary is estimated as follows. BLL ( ) CT ( 69 ) 5 = ------------------------------ = 3 S B T 3 ------------------------------------------------ = 6. MA 600.449 Ω (Eq 9) In the above equation, the voltage auxiliary input ratio is assumed same as the phase voltage input ratio. The calculations of and 0(bus) provide a quick check that the resistor value for one system phase voltage is equal to zero. The value is calculated as follows. I nom 33.55 A.449 Ω = ------------------- = ------------------------------------------------ =.3 CT 5 (Eq 0) The 0(bus) value is calculated as follows. ( BLL ( ) 0 3 ) ( 3) 69000 3 0( bus) = -------------------------------------------------- = 3 T ------------------------------- = BUS 3 600.3 (Eq ) The and 0(bus) values are practically the same. Unbalance calculations For a fuseless sample bank as shown above with S C = P = 0 = 0 series capacitor sections in one string and S = 3 parallel strings, the general equation to calculate ground current for n failed sections is shown below. n I G = I ( -------------------------------------- S C n) S + n nom ( S C n) S (Eq ) The neutral point secondary voltage with one failed section (n = ) is calculated as follows. ( n = ) n = I G -------- 33.55 A (Eq 3) CT ( -------------------------------------- 0 ) 3 + ( 0 ) 3.449 Ω = = -------------------- = 0.385 5 ( n = ) This can also be expressed in per-unit values as ( pu) = 0.385 / 67 = 0.0057 pu. The neutral point secondary voltage with two failed sections (n = ) is calculated as follows. ( n = ) ( ) ( n = ) I G -------- 33.55 A CT ( -------------------------------------- 0 ) 3 + ( 0 ) 3.449 Ω = = -------------------- = 0.8076 5 (Eq 4) ( n = ) This can also be expressed in per-unit values as ( pu) = 0.8076 / 67 = 0.0 pu. The neutral point secondary voltage with three failed sections (n = 3) is calculated as follows. ( n = 3) ( n = 3) I G -------- 33.55 A CT ( -------------------------------------- 0 3) 3 + 3 ( 0 3) 3.449 Ω = = -------------------- =.86 5 (Eq 5) ( n = 3) This can also be expressed in per-unit values as ( pu) =.86 / 67 = 0.09 pu. From the calculations above, the alarm pickup setting is chosen for the failure of one section as follows. ( n = ) 59U alarm = 0.9 = 0.9 0.0057 pu = 0.005 pu (Eq 6) 4 GE MULTILI TECHICAL OTE EUTAL OLTAGE UBALACE FUCTIO AT GOUDED WYE CAPACITO BAKS
ELAY COFIGUATIO The trip pickup setting is chosen as midway between the second and third capacitor section failures as follows. ( n = ) ( n = 3) 59U + trip = ---------------------------------------- = 0.0 -------------------------------------------------------- pu + 0.09 pu = 0.056 pu (Eq 7) elay configuration If the external 3 0 voltage is brought into the relay, then the C70 model with two T banks must be ordered to protect capacitor bank according to the application example. One T bank is sufficient for calculated 3 0. For externally brought 3 0 and two type L CT/T modules in the F and M slots. Program the following settings in the System Setup > AC Inputs > Current menu. Program the following settings in the System Setup > AC Inputs > oltage menu. Program the following settings in the System Setup > Signal Sources menu. GE MULTILI TECHICAL OTE EUTAL OLTAGE UBALACE FUCTIO AT GOUDED WYE CAPACITO BAKS 5
ELAY COFIGUATIO Program the following settings in the Grouped Elements > Group > oltage Elements > eutral oltage Unbalance menu. Appendix As was mentioned above in the case of neutral CT connections, the value of resistor must be chosen so that the voltage drop across at the neutral point through the CT/auxiliary T input transformation matches the bus 0 value for a single phase to ground fault. This value is derived as follows. If the fault occurs on phase A, then 0 is calculated as follows. 0 = -- ( 0 + 3 B + C ) = -- ( 3 B + C ) The impedance of each leg of bank is calculated as follows. (Eq 8) Z = -------------- BLL ( ) S B (Eq 9) The neutral current is calculated as follows. I B + = ----------------------- C BLL ( ) S B The secondary current in neutral resistor is calculated as follows. I sec ( ) -------- B + C = ----------------------- CT BLL ( ) S B The voltage across secondary resistor is calculated as follows. (Eq 0) (Eq ) sec ( ) -------- B + C = ----------------------- CT BLL ( ) S B (Eq ) Assume that the voltage setting for the auxiliary T ratio, T, is set the same as the bus T ratio. It is desired that following magnitudes are equivalent. T -------- B + C ----------------------- = CT 0 = -- ( BLL ( ) S 3 B + C ) B (Eq 3) 6 GE MULTILI TECHICAL OTE EUTAL OLTAGE UBALACE FUCTIO AT GOUDED WYE CAPACITO BAKS
ELAY COFIGUATIO We therefore have: CT ----------------------- BLL ( ) S B = -- 3 CT -------- BLL ( ) = -------------- T 3 S B (Eq 4) GE MULTILI TECHICAL OTE EUTAL OLTAGE UBALACE FUCTIO AT GOUDED WYE CAPACITO BAKS 7