A The Series Impedance Per Phase Is Z R Jwl L 0 15 J2m X 60 1 3263 10 40 6 J20 146 5 Line Model And Pe 1 (39.98 KiB) Viewed 24 times
Can someone explain how they did the math for sending volage andwhy they used 10^-3
(a) The series impedance per phase is Z= (r+ jwL)l= (0.15+j2m x 60 × 1.3263 × 10-³)40 = 6 + j20 146 5. LINE MODEL AND PERFORMANCE The receiving end voltage per phase is 220/0° √3 VR The apparent power is SR(34) = 381/cos ¹0.8= 381/36.87° 304.8 +j228.6 MVA The current per phase is given by SR(30) 3 VR From (5.3) the sending end voltage is IR=- = 127/0⁰ kV 381-36.87° × 10³ 3 x 127/0⁰ = 1000/- 36.87° A Vs =VR+ZIR=127/0° +(6+j20) (1000/-36.87°) (10-³) = 144.33/4.93⁰ kV
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!