Pilih 100 Mva Base Gabungkan Saluran Paralel Menjadi Satu Z = (0,0338 + J 0,227) Pu Y’ = J0,212 Pu

200 MVA

XT = 20 %

A

B

Z = (0,0676 + j 0,454) pu y’ = j0,106 pu

per circuit Base 100 MVA

Hitung tegangan terminal generator bila VB = 1,0 pu

Pilih 100 MVA Base Gabungkan saluran paralel menjadi satu Z = (0,0338 + j 0,227) pu y’ = j0,212 pu

P = 100 MW Q = 40 MVAR

J 0,1

IT

Daya Reaktif oleh y’

A P4, Q4

P3, Q3

IL

P = 1,0 pu Q = 0,4 pu

P1, Q1

(0,0338 + J 0,227)

Q2

Q5

B

Q2 = j 0,106 (1,0)2 = j 0,106

P1 + jQ1 = (1,0 + j0,4) – j0,106 = (1,0 + j0,294) pu (IL

)*

(1,0 + j0,294) = (1,0 + j0)

IL = (1,0 - j0,294)

|IL|2 = 1,086

ΔV = IL Z = (1,0 - j0,294)(0,038 + j 0,227) = (0,1 + j 0,215) VA = VB + ILZ = (1,0 + j0) + (0,1 + j0,215) = (1,1 + j0,215) pu

|VA| = 1,12 pu

|VA|2 = 1,256

Losses (Line) |IL|2Z = 1,086 (0,0338 + j0,227) = (0,0367 + j0,246)pu P3 + jQ3 = P1 + jQ1 + |IL|2Z = (1,0 + j0,294) + (0,0367 + j0,246)pu = (1,0367 + j0,540) pu Daya Reaktif y’ Q5 = j 0,106 (1,256) = j 0,133 P4 + jQ4 = P3 + jQ3 - j Q5 = = (1,0367 + j0,540) - (j0,133) = (1,0367 + j0,407) pu (IT )* =

(1,0367 + j0,407) (1,1 + j0,215)

=

(1,0367 + j0,407) (1,1 + j0,215)

𝑥

(1,1− j0,215) (1,1− j0,215)

=

(1,228 + j0,225) 1,256

(IT ) =

(1,228 − j0,225) 1,256

ΔV di Trafo (1,228 − j0,225) IT XT = 1,256

x j0,1 =

(0,0225 − j0,1228) 1,256

= (0,0179 +j0,098) pu VG = VZ + IT XT = (1,1 + j0,215) + (0,0179 + j0,098) = (1,1179 + j0,313) pu

A

ZAB =0,06 + j 0,14 y’ = j0,3

B

200 MVA X = 20%

C G1

G2

P = 100 MW Q= 50 MVAR

P = 230 MW Q= 110 MVAR VB =1,0 pu

Bila Daya dari trafo ke Busbar B sebesar 160 MW + J 100 MVAR, hitung output G1 dan G2 serta tegangan VA dan VC

BB

ZBA =0,02 + j 0,06 y’ = j0,01

G2

SB =1 + j 0,25

A ZAC =0,01 + j 0,03 y’ = j0,02

SA =0,5 + j 0,2

Hitung output G2 dan tegangan VB

C

SG1 = 1 + j 0,3 G1

VC =1,0

B

ZBA =0,02 + j 0,06 y’ = j0,01

G2

SB = 0,3 + j 0,1

A ZAC =0,01 + j 0,03 y’ = j0,02

SA = 0,4 + j 0,1

Hitung output G2 dan tegangan VB

VC =1,0 C SG1 = 0,4 + j 0,15 G1 SC = 0,5 + j 0,2