Question

Eb1 = 32kV Sb = 11kVA

2. Per Unit (50 pts) (SHOW ALL WORK: Using the base values found on the last page for Esi and Sa, fill in the Base Values for the 3 zones of the circuit in Table 1. Using the parame Table 2, fill in Table 3 with the per unit impedances of the elements. Using Eo -34.5 kV, find the values for Table 4 ters of the transformers and the values of the impedances found in (ZONE 2 LOAD Table 1: Base Values Table 2: Element Values Element 6kj 200j 200j 1.0j 2.0 Zone 1 Zone 2 Zone 3 units units VA 22 22.7+8.3 z, l 20.2, 17Oj | 4.5k/480 VIV VA x,q= 0.0151 pu 10k VIV VA X0.015 pu 40.0k/800 T: 20k

Answer 1

S_B = Vertical-bar Vertical-bar KVA E_b = 32 KV Pu_value = Actual value in same units /Base (or) surface value in some until For Zone S_B = 11 KVA E_b1 = 32 KV Z_B = Z_Base Z_B = (E_b)^2/S_B = (32 times 103)^2/11 alpha 10^3 (Because 1 KV = 1000) (Therefore 32 KV = 32000 V 32 times 10^3 V) = 1024 times 10^6/11 times 10^3 = 1024 times 10^3/11 = 93.090.90 V Z_B = 93.090 K Ohm I_B = E_b1/Z_B = 32 KV/93.090 KV = 32 times 10^3/93.090 times 10^3 (Because 1 KV = 1000 V = 10^3 V Therefore 32 K = 32000 V = 32 times 10^3 V) = 0.34375 Amps I_n a network containing transformer base value must be selected as per the following rules rightarrow Two different voltage are selected as base voltage in such a way that there ratio must be equal to transformation ratio original transformer Tran saturation ratio (or) curve ratio = 34.5/480 KV a = 0.071875 times 10^3 a = 71.875 In zone transformer 2 is in high voltage side in zone transformer is in low voltage