How can I get a Kv base? I don't know why I get the kv base as (5x10^3/10x10^3)x9600 7. Prob. 7 below:For the system shown in Figure 3.34, draw an impedance diagram in per unit,by choosing 100 kVA to be the base kVA and 2400 V as the base voltage for thegenerators.10 kVAG1Z= j0.2pu2400 VT1Transmission LineT2MZ= (50 + j200)N20 kVA40 kVA80 kVA25 kVA2400 V(G2,2400/9600 V10/5 kV4 kVZ= j0.2puZ= j0.1puZ= j0.1pu The per unit value of the voltage will bekV actualkV basekVpukVbase = ( Sx10³(9600) p.u.kVbase = 4800VkVbase = 4.8VTherefore,4kVpu =s p.u.4.8 P.u.kVpu = 0.833p.u.

7. Prob. 7 below: For the system shown in Figure 3.34, draw an impedance diagram in per unit, by choosing 100 kVA to be the base kVA and 2400 V as the base voltage for the generators. 10 kVA 2400 V G Z = j0.2pu T, Transmission Line T2 M Z= (50 + j200)N 20 kVA 2400 V 40 kVA 80 kVA 25 kVA (G2 Z = j0.2pu 2400/9600 V 10/5 kV 4 kV Z= j0.1pu Z = j0.1pu

7. Prob. 7 below: For the system shown in Figure 3.34, draw an impedance diagram in per unit, by choosing 100 kVA to be the base kVA and 2400 V as the base voltage for the generators. 10 kVA 2400 V G Z = j0.2pu T, Transmission Line T2 M Z= (50 + j200)N 20 kVA 2400 V 40 kVA 80 kVA 25 kVA (G2 Z = j0.2pu 2400/9600 V 10/5 kV 4 kV Z= j0.1pu Z = j0.1pu

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.33P: Consider the three single-phase two-winding transformers shown in Figure 3.37. The high-voltage...

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