Question

18. The primary conductor in Figure 10.2 is one phase of a three-phase trans- mission line operating at 345 kV, 700 MVA, 0.95 power factor lagging. The CT ratio is 1200:5, and the VT ratio is 3000:1. Determine the CT second- ary current and the VT secondary voltage V. Assume zero CT error. A CO-8 relay with a current tap setting of 5 amperes is used with the 100:5 CT in Example 10.1. The CT secondary current I' is the input to the relay operating coil. The CO-8 relay burden is shown in the following table for various relay input currents CO-8 relay input current l'. A 5 8 10 13 15 CO-8 relay burden Z, 2 05 08 1.0 13 1.5 Primary current and CT error are computed in Example 10.1 for the 5-,8-, and 15-A relay input currents. Compute the primary current and CT error for (a) I'=1° A and ZB-1.0Ω, and for (b) I' =13Aand Zs-1.3Ω. (c) Plot I' versus I for the above five values of I'. (d) For reliable relay operation, the fault-to-pickup current ratio with minimum fault current should be greater than two. Determine the minimum fault current for application of this CT and relay with 5-A tap setting.

Answer 1

ne 1 8 en by one phase of tranwninon lire Current txaraformer(c) eltoge tvonsformer (vi) ry Jineo-line voltage ofgnsemmex Volb LL345 kV Cos. o.95 Cuvrient TYansfomer Ratio 1200: 5 to 120o:5 10 こ3000 : 1 Now, we calaulah the oP voltage t

rarwrorm 3000 115 V voltage , v' oF VT is 115 V C co in Mow, we calculate the cwvient,1 which oP cwient trans formex 700 × 10 3)(34-5 X l0 59 557,52 ⑩owwe draw the current transforme I: n

no 3 at us calealste the secondory cuvent of CT, we knew that 5 ( 1200 11子1-43 4-881 A circuit, the secondovy from the above CT equivaloné c components, 1 and 1" nt cwrent of CT, we corite To determine tha secendouy eden eCaciting Cuwrent 0 e.

"Thin from e uation。冫we got 1e 나,881-140 T: 488 A decondory curent of CT:() 4881A