Σ = 10 100 mVA V=20kV Zs=al+jo.2 SG=120 MVA 2012 DY ++ 20/138k | Z₁₁ = 2 +5 j 2 126 = 1381² 150 STI = 150 MVA 2b= 120m = 3.33 0.1 70.2 Pu =0.03+0.06 3.33 2 TI Pu Pu = 25 Pa b Pu now Zmew= Zold Sb old -=126.96 =0.015+j 0039 Pu = ZTLold TI-new ZLine 0.04 +01 Yin = 74x100 5xm ( Zline =0.01 +10.026 Y 6 pu → 25 new = (0.03 +0.06j) (1000) n² =0.025+1005 Pu_0.04+10.1 126.96 -6 YY m 138/13.8V 100+407 ZT2=1.8Hj STOMYA pu_18443 = -0.0140031 Y pu=14x100 - 14×106126.96 Pu Line Yb = b Pu=0000315+0.000787 2Line Pu old YP = 0.000507 new Line pu_new Line 126.96 pu-new (272 =0.003+10.021 = 0.00021+10.00053 Pu-new Y line = 0.00076 Generator G: 120 MVA, 20 kV, synchronous impedance Zs = 0.2 +j 1.2 / phase Y. Transformers: T1: 150 MVA, 20A / 138Y kV, per phase winding resistance and leakage reactance of the high voltage side: R = 22, X = 5. T2: 150 MVA, 138Y/ 13.8 YkV, per phase winding resistance and leakage reactance of the high voltage side: R = 1.82, X = 4. Transmission line: z = 0.04 +j 0.1 2/km, y=j 4×10-6 S/km, length = 100 km. Load: 100 MW +j 40 Mvar (constant power load). G T1 T2 transmission line B AY YY FIGURE E4-1 Load A system base of 100 MVA, 20 kV is chosen on the low voltage side of transformer T1. Use the nominal- circuit model for the transmission line. Solve the power flow problem for this system and find the following 1) The load (bus 4) voltage. 2) The transmission line currents at the sending and receiving ends. 3) The real and reactive power supplied by the generator at bus 1. for these cases: starting from 20% of the nominal value given above, load power is increased in steps of 20% up to the nominal value.
Σ = 10 100 mVA V=20kV Zs=al+jo.2 SG=120 MVA 2012 DY ++ 20/138k | Z₁₁ = 2 +5 j 2 126 = 1381² 150 STI = 150 MVA 2b= 120m = 3.33 0.1 70.2 Pu =0.03+0.06 3.33 2 TI Pu Pu = 25 Pa b Pu now Zmew= Zold Sb old -=126.96 =0.015+j 0039 Pu = ZTLold TI-new ZLine 0.04 +01 Yin = 74x100 5xm ( Zline =0.01 +10.026 Y 6 pu → 25 new = (0.03 +0.06j) (1000) n² =0.025+1005 Pu_0.04+10.1 126.96 -6 YY m 138/13.8V 100+407 ZT2=1.8Hj STOMYA pu_18443 = -0.0140031 Y pu=14x100 - 14×106126.96 Pu Line Yb = b Pu=0000315+0.000787 2Line Pu old YP = 0.000507 new Line pu_new Line 126.96 pu-new (272 =0.003+10.021 = 0.00021+10.00053 Pu-new Y line = 0.00076 Generator G: 120 MVA, 20 kV, synchronous impedance Zs = 0.2 +j 1.2 / phase Y. Transformers: T1: 150 MVA, 20A / 138Y kV, per phase winding resistance and leakage reactance of the high voltage side: R = 22, X = 5. T2: 150 MVA, 138Y/ 13.8 YkV, per phase winding resistance and leakage reactance of the high voltage side: R = 1.82, X = 4. Transmission line: z = 0.04 +j 0.1 2/km, y=j 4×10-6 S/km, length = 100 km. Load: 100 MW +j 40 Mvar (constant power load). G T1 T2 transmission line B AY YY FIGURE E4-1 Load A system base of 100 MVA, 20 kV is chosen on the low voltage side of transformer T1. Use the nominal- circuit model for the transmission line. Solve the power flow problem for this system and find the following 1) The load (bus 4) voltage. 2) The transmission line currents at the sending and receiving ends. 3) The real and reactive power supplied by the generator at bus 1. for these cases: starting from 20% of the nominal value given above, load power is increased in steps of 20% up to the nominal value.
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 3.37P: Three single-phase two-winding transformers, each rated 25MVA,54.2/5.42kV, are connected to form a...
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Please do theoretical calculation q1 q2 and q3 step by step no ai solution thank you
Transcribed Image Text:Σ = 10
100 mVA
V=20kV
Zs=al+jo.2
SG=120 MVA
2012
DY
++
20/138k
| Z₁₁ = 2 +5 j
2
126 = 1381²
150
STI = 150 MVA
2b=
120m = 3.33
0.1 70.2
Pu
=0.03+0.06
3.33
2 TI
Pu
Pu
=
25
Pa
b
Pu now
Zmew= Zold
Sb
old
-=126.96
=0.015+j 0039
Pu
= ZTLold
TI-new
ZLine 0.04 +01
Yin = 74x100 5xm
(
Zline
=0.01 +10.026
Y
6
pu
→ 25 new = (0.03 +0.06j) (1000) n²
=0.025+1005
Pu_0.04+10.1
126.96
-6
YY
m
138/13.8V
100+407
ZT2=1.8Hj
STOMYA
pu_18443
=
-0.0140031
Y pu=14x100 - 14×106126.96
Pu
Line
Yb = b
Pu=0000315+0.000787
2Line
Pu
old YP = 0.000507
new
Line
pu_new
Line
126.96
pu-new
(272 =0.003+10.021
= 0.00021+10.00053
Pu-new
Y line = 0.00076
Transcribed Image Text:Generator G: 120 MVA, 20 kV, synchronous impedance Zs = 0.2 +j 1.2 / phase Y.
Transformers:
T1: 150 MVA, 20A / 138Y kV, per phase winding resistance and leakage
reactance of the high voltage side: R = 22, X = 5.
T2: 150 MVA, 138Y/ 13.8 YkV, per phase winding resistance and leakage
reactance of the high voltage side: R = 1.82, X = 4.
Transmission line:
z = 0.04 +j 0.1 2/km, y=j 4×10-6 S/km, length = 100 km.
Load: 100 MW +j 40 Mvar (constant power load).
G
T1
T2
transmission line
B
AY
YY
FIGURE E4-1
Load
A system base of 100 MVA, 20 kV is chosen on the low voltage side of transformer
T1. Use the nominal- circuit model for the transmission line. Solve the power flow
problem for this system and find the following
1) The load (bus 4) voltage.
2) The transmission line currents at the sending and receiving ends.
3) The real and reactive power supplied by the generator at bus 1.
for these cases: starting from 20% of the nominal value given above, load power is
increased in steps of 20% up to the nominal value.
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