Jawaban Bab 5 (5.1) A 69-kV, three-phase short transmission line is 16 km long. The line has a per phase series impedance of 0.125 + j0.4375
per km. Determine the sending end voltage, voltage
regulation, the sending end power, and the transmission efficiency when the line delivers. (a) 70 MVA, 0.8 lagging power factor at 64 kV. (b) 120 MW, unity power factor at 64 kV. Use lineperf program to verify your results. (
The Line impedance is
)(
)
The receiving end voltage per phase is √ a. The complex power at the receiving end is (
)
The current per phase is given by (
)
The Sending end voltage is (
)(
)(
)
The sending end line-to-line voltage magnitude is |
(
)|
√ |
|
The Sending end power is (
)
(
)(
)
Transmission line efficiency is (
)
(
)
b. The complex power at the receiving end is (
)
The current per phase is given by (
)
The sending end voltage is (
)(
)(
)
The sending end line to line voltage magnitude is |
)|
(
√ |
|
The sending end power is (
(
)
Transmission line efficiency is (
)
(
)
)(
)
(5.4) Shunt capacitors are installed receiving end to improve the line performance of problem 5.3. the line deivers 200MVA, 0,8 lagging power factor at 220 kV a. determine the total Mvar and the capacitance per phase of the Y-connected capacitors when the sending end votage is 220 kV. Hint: use (5.85) and (5.86) to compute the power angel and the receiving end reactive power b. use lineperf to obtain the compensated line performance a. The complex load at the receiving end is (
)
For VR(LL) = 220 kV, from (5.85), we have (
)(
)
(
)
(
)(
)
(
Therefore (
)
Or
now from (5.86), we have ( (
)(
)
(
)
)
(
)
therefore, the required capacitor Mvar is |
|
(
⁄
)
The required shunt capacitance per phase is (
)(
)(
)
(
)(
)
SOAL 6.2
The values marked are impedances in per unit on a base of 100 MVA. The currents entering buses 1 and 2 are
I1 = 1:38 - j2:72 pu I2 = 0:69 - j1:36 pu
(a) Determine the bus admittance matrix by inspection. (b) Use the function Y = ybus(zdata) to obtain the bus admittance matrix. The
function argument zdata is a matrix containing the line bus numbers, resistance and reactance. (See Example 6.1.) Write the necessary MATLAB commands to obtain the bus voltages. Converting all impedances to admittances results in the admittance diagram shown in Figure 50
A. Determine the bus admittance matrix by insfection
I 1
j 0,5
j 1,0
1
I2
2
0,01 + j0,03
0,0125 + j0,025
3
0,4 + j0,2
y12 I1
-j2
-j 1
10 – j20
1 y23
2 y13
10 - j30
16 - j32
3 2 - j1
KCL I1
= y10V1 + y12(V1-V2) + y13 (V1-V3)
I2
= y20V2 + y21(V2-V1) + y23(V2-V3)
0
= y31(V3-V1) + y32(V3-V2)
Rearanging Equattion I1
= (y10+ y12+ y13)V1- y12V2- y13V3
I2
= -y13V1+(y20+y21+y23)V2- y23V3
0
= -y31V1 – y32V2 + (y31+y32+y33)V3
Admittances y11
= y10+ y12+ y13
= 20 - j52
y22
= y20+y21+y23
= 26 - j53
y33
= y31+y32+y33
= 28 - j63
y12
= y21 = - y12
= 10 - j20
y13
= y31 = - y13
= 16 - j32
y23
= y32 = - y23
= 10 - j30
I2
Equattion Reduces I1
= y11V1
-y12V2
-y13V3
I2
= -y13V1
y22V2
-y23V2
0
= -y31V1
-y32V2
y33V3
Matrix Form is I1 I2
=
0
y11
-y12
-y13
V1
-y13
y22
-y23
V2
-y31
-y32
y33
V3
Matrix Inspection
y bus
20 - j52
-10 + j20
-10 + j30
-10 + j20
26 - j53
-16 + j32
-10 + j30
-16 + j32
28 - j63
B. Fungtion y bus to MATLAB
Y = ybus (Z),
z
=
[0
1
0.0
0.5
0
2
0.0
1.0
0
3
0.4
0.2
1
2
0.02
0.04
1
3
0.01
0.03
2
3
0.0125
0.025];
Y
= ybus(z)
I
= [1.38-j*2.72; 0.69-j*1.36; 0];
V
= Y\I;
Vm
= abs (V)
phase = 180/pi*angle(V)
The result is Y= 20.0000
-52.000i
-10.0000
+20.000i
-10.0000
+30.000i
-10.0000
+20.000i
26.0000
-53.000i
-16.0000
+32.000i
-10.0000
+30.000i
-16.0000
+32.000i
28.0000
-63.000i
Vm = 1.0293 1.0217 1.0001 phase = 1.4596 0.9905 -0.0150
