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# Activity 1

advertisement ```0.375∠78 Ω
=7.7479∠78 pu
(1100 V)2⁄
25 kVA
EE 506 - Power System Analysis & Design
Solution Manual
ZHV(pu) =
Problem No. 01
Base Voltage = 1.1 kV. Base KVA = 106. What is
base impedance?
ZHV(pu) =7.7479 ∠78 pu
Solution:
(1.1 kV)2
ZBase =
6
10 kVA
ZBase = 1.21 mΩ
Problem No. 02
If the resistance in ohms is 5 Ω, find the per unit
value. Given base kVA = 10 and base kV = 11.
Solution:
(11 kV)2
ZBase = 6
= 12.1 kΩ
10 kVA
5Ω
Rpu =
12.1 kΩ
Rpu = 413. 223 pu
Problem No. 03
A single phase two – winding transformer is rated
25 kVA, 1100/440 volts, 50 Hz. The equivalent
leakage impedance of the transformer referred to
the low voltage side is 0.06 ∠78° Ω. Using
transformer rating as base values, determine the
per – unit leakage impedance referred to low
voltage winding and referred to high voltage
winding.
Solution:
ZLV(pu) =
0.06 ∠78 Ω
=7.7479∠78 pu
(440 V)2⁄
25 kVA
ZLV(pu) =7.7479∠78 pu
1100 2
ZHV = (0.06 ∠78 Ω) (
) = 0.375∠78 Ω
440
Problem No. 04
A single-phase transformer is rated at 2.5 kVA,
11/0.4 kV. If the leakage reactance is 0.96 Ω when
referred to low – voltage side, then determine its
leakage reactance in per unit.
Solution:
Zpu =
0.96 Ω
(0.4 kV)2⁄
2.5 kVA
Zpu = 0.015 pu
Problem No. 05
For a 110/440 V, 25 kVA, single – phase
transformer, primary and secondary leakage
reactances are 0.04 Ω and 0.10 Ω respectively.
Show that net pu leakage reactance of the
transformer referred to LV side is same as referred
to HV side.
Solution:
440 V 2
Zeq(HV) = 0.10 Ω+0.04 Ω (
) = 0.74 Ω
110 V
0.74 Ω
Zeq(HV) =
(440 V)2⁄
pu
25 kVA
Zeq(HV) = 0.0956 pu
pu
110 2
Zeq(LV) = 0.04 Ω + 0.10 Ω ( ) = 0.04625 Ω
440
0.04625 Ω
Zeq(LV) =
pu
(110)2⁄
25 kVA
Zeq(LV) = 0.0956 pu
pu
Problem No. 06
An 11/0.4 kV, 200 kVA transformer has an
equivalent impedance of 2.4+j12.4 Ω referred to
HV side. Determine the base values for the pu
system, the per unit equivalent impedance and the
equivalent impedance rop at one – half rated
current.
Solution:
SBase = 200 kVA
VBase = 11 kV
(11 kV)2
ZBase(HV) =
200 kVA
ZBase(HV) = 605 Ω
(0.4 kV)2
ZBase(LV) =
200 kVA
ZBase(LV) = 0.8 Ω
IBase(HV) =
200 kVA
11 kV
IBase(HV) = 18.1818 A
IBase(LV) =
200 kVA
0.4 kV
IBase(LV) = 500 A
ZHV(pu) =
2.4+j12.4
ZBase(HV)
ZHV(pu) = 0.0209∠79.0459 pu
@ Half Load
ZHVpu =
2.4+j12.4
1210.0012
ZHVpu =0.0104 ∠79.0459 pu
Problem No. 07
Determine the per unit impedance of a
transmission line having an impedance of 30 +
j110 Ω on 100 MVA and 132 kV base voltage.
Solution:
Zpu =
30+j110
(132 kV)2⁄
100 MVA
Zpu = 0.6544∠74.7449 pu
Problem No. 08
A 30 MVA, 11 kV generator has a reactance of 0.2
pu referred to its ratings as bases. Determine the
per unit reactance when referred to base KVA of
50000 kVA and base kV of 33 kV.
Solution:
SNew VOld 2
XNew = XOld (
)(
)
SOld VNew
50 MVA 11 kV 2
Xg(New) =0.2 (
)(
)
30 MVA 33 kV
Xg(New) =0.0370 pu
Problem No. 09
The transformation ratio of the step – up
transformer is 11/220 kV. The base MVA = 100
and base kV = 11 on the generator side. What is
the base kV on the transmission side?
1
IBaseHV =18.1818 A ( )
2
Solution:
IBaseHV =9.0909 A
VBaseTL =11 kV (
ZBaseHV =
11 kV
=1210.0012 Ω
9.0909 A
ZActual = Zpu (1210.0012 Ω)
ZActual =25.2890∠79.0459 Ω
VBaseTL =220 kV
220 kV
)
11 kV
Problem No. 10
Three generators are rated as follows:
Generator 1: 100 MVA, 33 kV, reactance = 10%
Generator 2: 150 MVA, 32 kV, reactance = 8%
Generator 3: 110 MVA, 30 kV, reactance = 12%
Choosing 200 MVA and 35 kV as base quantities,
compute per unit reactances of the three
generators referred to these base quantities. Draw
reactance diagram and mark per unit reactances.
The three generators are connected to common
bus – bars.
Problem No. 11
Figure 17 shows single line diagram of a single –
phase circuit. Using the base values of 3 kVA and
230 V, draw the per – unit circuit diagram and
determine the per – unit impedances and per –
unit source voltage. Also, calculate the load
current both in per unit and in Amperes.
Solution:
Reactance Diagram
Solution:
Per- Unit Diagram
Xg1
XT1
Xg3
Xg2
XL
XT2
Xg
Vg1
Ipu
Vg3
Vg2
Vg
SBase = 20 MVA
VBase = 35 kV
Vg1 = 33⁄35 = 0.9429 pu
Vg2 = 32⁄35 = 0.9143 pu
Vg3 = 30⁄35 = 0.8571 pu
10 200 33 2
Xg1 = ( ) ( ) ( )
100 100 35
@Section 1
SBase = 3 kVA
VBase1 = 230 V
Vg = 220 V⁄230 V
8
200 32 2
Xg2 = ( ) ( ) ( )
100 150 35
Vg = 0.9565 pu
Xg = 0 pu
3 kVA 230 V 2
XT1 =0.1 (
)(
)
3 kVA 230 V
Xg2 = 0.0892 pu
XT1 =0.1 pu
Xg1 = 0.1778 pu
Xg3 = (
2
12 200 30
)( )( )
100 110 35
Xg1 = 0.1603 pu
@Section 2
433 V
VBase2 =230 V (
) = 433 V
230 V
Zload
XT1 = 0.1 (
3 kVA 433 V
)(
) =0.1 pu
3 kVA 433 V
20% reactance each. Draw the per – unit circuit
diagram.
(433)2
ZBase2 =
=62.4963 Ω
3 kVA
XL = 3⁄Z
Base2
XL = 0.0480 pu
2
XT2 =0.1 (
3 kVA 440 V
)(
)
2 kVA 433 V
Solution:
Per-Unit Diagram
@Section 3
120 V
VBase3 = 433 V (
) = 118.0909 V
440 V
2
3 kVA
120 V
XT2 = 0.1 (
)(
) = 0.1549 pu
2 kVA 118.0909 V
XL
XT1
XT2 = 0.1549 pu
Xm1 Xm2 Xm3
Xg
M
Vg
@Region 1
SBase = 110 MVA
VBase1 = 33 kV
Vg = 33 kV⁄33 kV
ZLoad = 0.1838∠20.5560 pu
Vg = 1 pu
By Ohm’s Law:
Ipu = 2.3574∠-64.9029 pu
3 kVA
ILoad = (Ipu )(IBase3 ) = (Ipu ) (
)
√3 (118.0909)
ILoad = 14.6675 ∠ -64.9029 A
Problem No. 12
A 100 MVA, 33 kV, three phase generator has a
reactance of 15%. The generator is connected to
the motors through a transmission line and
transformers as shown. Motors have rated inputs
of 40 MVA, 30 MVA and 20 MVA at 30 kV with
M
M
Vm1 Vm2 Vm3
(118.0909 V)2
ZBase3 =
= 4.6485 Ω
3 kVA
0.8+j0.3
ZLoad =
ZBase3
0.9565
Ipu =
0+j0.1+j0.048+j0.1549+0.1838∠20.5560
XT2
Xg = 0.15 (
110 MVA 33 kV 2
)(
)
100 MVA 33 kV
Xg = 0.1650 pu
110 MVA 32 kV 2
XT1 = 0.08 (
)(
)
110 MVA 33 kV
XT1 = 0.0752 pu
@Region 2
110 kV
VBase2 =33 kV (
) =113.4375 kV
32 kV
ZBase2 =
(113.4375 kV)2
=116.9824 Ω
110 MVA
XL =
60
Solution:
ZBase2
15 MVA
66 kV
XL = 0.5129 pu
Zp
2
110 MVA
110 kV
XT2 =0.08 (
)(
)
110 MVA 113.4375 kV
XT2 = 0.0752
Zp =
110 MVA 30 kV 2
Xm1 =0.2 (
)(
)
40 MVA 33 kV
Xm1 = 0.4545 pu
ZP = j0.0873 pu
Zp =
2
110 MVA 30 kV
Xm3 =0.2 (
)(
)
20 MVA 33 kV
Xm3 = 0.9091 pu
Problem No. 13
The three – phase ratings of a three – winding
transformer are:
Primary – Y – connected, 66 kV, 15 MVA
Secondary – Y – connected, 13.2 kV, 10 MVA
Tertiary – Δ – connected, 2.3 kV, 5 MVA
Neglecting resistances, the leakage impedances
are:
ZPS = 0.08 pu on 15 MVA, 66 kV base
ZPT = 0.10 pu on 15 MVA, 66 kV base
ZST = 0.09 pu on 10 MVA, 13.2 kV base
Find the per unit impedances of the star –
connected equivalent circuit for a base of 15 MVA,
66 kV base in the primary circuit.
ZS =
5 MVA
2.3 kV
Zt
ZPS + ZPT - ZST
2
@Region 3
VBase3 = VBase2 = 33 kV
30kV
Vm1 = Vm2 = Vm3 =
33kV
Vm1 = Vm2 = Vm3 = 0.9091 pu
110 MVA 30 kV 2
Xm2 =0.2 (
)(
)
30 MVA 33 kV
Xm2 = 0.6061 pu
Zs 10 MVA
13.2 kV
15 13.2 2
0.08+0.01- [0.09 ( ) (
)]
10 66
2
ZST + ZSP - ZPT
2
2
ZS =
2
10 66
10 66
0.09+ [0.08 ( ) (
) ] - [0.1 ( ) (
)]
15 13.2
15 13.2
2
ZS = - j0.1217 pu
ZT =
ZTS + ZTP - ZPS
2
5 66 2
5 13.2 2
[0.10 ( ) ( ) ] + [0.09 ( ) (
) ]−
15 2.3
10 2.3
2
Zp =
5 66
[0.08 ( ) ( ) ]
15 2.3
2
ZP = j3.4839 pu
Problem No. 14
Draw the per – unit impedance diagram of the
system shown in Figure 19. Assumed base values
are 100 MVA and 100 kV.
G1: 50 MVA, 12.2 kV, Xg1 = 0.10 pu
G2: 20 MVA, 13.8 kV, Xg2 = 0.10 pu
T1: 80 MVA, 12.2/132 kV, XT1 = 0.10 pu
T2: 40 MVA, 13.8/132 kV, XT2 = 0.10 pu
Load: 50 MVA, 0.8 pf lagging operating at 124 kV.
Problem No. 17
Figure 22 shows single – line diagram of a power
system.
Solution:
Problem No. 15
Figure 20 shows a sample power system network.
Find the current supplied by the generator, the
transmission line current, the load current, the
load voltage and the power consumed by the load.
Solution:
Problem No. 16
The single line diagram of a three – phase system
is shown in Figure 21. Select a common base of
100 MVA and 13.8 kV on the generator side. Draw
per – unit impedance diagram.
G: 90 MVA, 13.8 kV, Xg = 18%
T1: 50 MVA, 13.8/220 kV, XT1 = 10%
T2: 50 MVA, 220/11 kV, XT2 = 10%
T3: 50 MVA, 13.8/132 kV, XT3 = 10%
T4: 50 MVA, 132/11 kV, XT4 = 10%
M: 80 MVA, 10.45 kV, Xm = 20%
Load: 57 MVA, 0.8 pf (lagging) at 10.45 kV.
Xline1 = 50 Ω
Xline2 = 70 Ω
Solution:
The ratings of the generators and transformers are
given below:
G1: 25 MVA, 6.6 kV, Xg1 = 0.2 pu
G2: 15 MVA, 6.6 kV, Xg2 = 0.15 pu
G3: 30 MVA, 13.2 kV, Xg3 = 0.15 pu
T1: 30 MVA, 6.6 ∆ - 115 Y kV, XT1 = 0.1 pu
T2: 15 MVA, 6.6 ∆ - 115 Y kV, XT2 = 0.1 pu
T3: 1 – phase unit each rated 10 MVA, 6.9/69 kV,
XT3 = 0.1 pu
Draw the per – unit circuit diagram using base
values of 30 MVA and 6.6 kV in the circuit of
generator 1
Solution:
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