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The Per Unit System (Examples) Example 1 A three-phase, wye-connected system is rated at 50MVA and 120kV. Express 40MVA of three-phase apparent power as a per unit value referred to: a) The three-phase system as MVA base, and b) The per-phase system MVA as base. Solution a) For the three-phase system MVA as base, S B 3φ = 50 MVA S pu 3φ =
V B (line ) = 120 kV
S actual 40 = = 0.8 pu S B 3φ 50
b) For the per phase base S B 3φ
50 M = 16.67 MVA 3 3 VB ( line ) 120 k V B ( phase ) = = = 69.28kV 3 3 1 40 S pu 1φ = × = 0.8 pu 3 16.67 S B 1φ =
=
Example 2 Find the per unit value for XT1, XT2 and XT if the base values are 11kV and 60MVA. 60 MVA 11kV X G=0.5 EG=1.5pu
T1
T2 X T=20Ω
G 60 MVA 12/132kV X T1=10%
30 MVA 132/33kV X T2=10%
Solution Step 1: Draw the section of the network.
1
60 MVA 11kV X G=0.5 EG=1.5pu
T1
T2 XT=20Ω
G
30 MVA 132/33kV X T2=10%
60 MVA 12/132kV XT1=10%
Step 2: Find the base value of the voltage for each section. S B = 60 MVA V B1 = 11kV VB 2 =
V2 132 × V B1 = × 11kV = 121kV V1 12
VB 3 =
V3 33 × VB 2 = × 121kV = 30.25kV V2 132
Step 3: Find the per unit values of each component Transformer T1 and T2 Since both transformer voltage base are the same as their rated values, their p.u reactance on a 60MVA are: 2
V old S new × Bnew × Bold VB S B
Z
new pu
=Z
X
new T1
12 60 = 0.1 × × = 0.12 pu 11 60
old pu
2
:: Looking at the LV side of T1
OR 2
132 60 X Tnew × = 0.12 pu :: Looking at the HV side of T1 1 = 0.1 × 121 60 2
132 60 X Tnew × = 0.24 pu :: Looking at the HV side of T2 2 = 0.1 × 121 30 OR 2
X
new T1
33 60 = 0.1 × × = 0.24 pu :: Looking at the LV side of T2 30.25 30
NOTE: The per unit value for the transformer impedance is the same whether it is seen in the HV or LV side of the transformer- one of the advantages of per unit system.
2
Line Impedance Since the impedance given in actual value, we have to find the base value for the impedance.
(V B 2 ) 2
X T ( base ) = X T ( pu ) =
=
SB *
X T ( actual ) X T ( base )
(121k ) 2
=
60 M
= 244Ω
20 = 0.082 pu 244
Note that the line impedance has only the resistive value. Therefore the complex power conjugate value is the same since θ is equal to 0.
Example 3 The one-line diagram of three-phase power system is shown below. Select a common base of 100 MVA and 22 kV on the generator side. Draw an impedance diagram with all impedance including the load impedance marked in per-unit. 50 MVA 22/220 kV X = 10%
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T1
40 MVA 220/11 kV X = 6.0%
T2
2 Line 1 220 kV
66.5 MVA 10.45 kV X = 18.5%
X = 48.4 Ω
G 90 MVA 22 kV X = 18%
4
M T3
T4
3 Line 2 110 kV
40 MVA 22/110 kV X = 6.4%
X = 64.43 Ω
Load 40 MVA 110/11 kV X = 8.0%
57 MVA 0.6 pf lag 10.45 kV
Solution Step 1: Draw the zone in the circuit diagram
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50 MVA 22/220 kV X = 10%
40 MVA 220/11 kV X = 6.0%
T1
T2
2
1
4 Line 1 220 kV
66.5 MVA 10.45 kV X = 18.5%
X = 48.4 Ω
G
M
90 MVA 22 kV X = 18%
57 MVA 0.6 pf lag 10.45 kV
3 Line 2 110 kV T3
X = 64.43 Ω
40 MVA 22/110 kV X = 6.4%
Load T4
40 MVA 110/11 kV X = 8.0%
Step 2: Find the base voltage for each zone. Vb1 = 22kV Vb 2 =
V2 220 × Vb1 = × 22k = 220 kV V1 22
Vb 3 =
V3 110 × Vb1 = × 22k = 110kV V1 22
Vb 4 =
V4 11 × Vb 2 = × 220 k = 11kV V2 220
Step 3: Find the per unit value Generator and Transformer Since generator & transformer voltage base are the same as their rated values, their p.u reactance on a 100 MVA 2
V Bold S Bnew Z = Z new old VB S B new new old S B old Z pu = Z pu SB 100 Z Gnew = 0.18 = 0.2 pu 90 100 Z Tnew = 0.2 pu 1 = 0.10 50 100 Z Tnew = 0.15 pu 2 = 0.06 40 100 Z Tnew = 0.16 pu 3 = 0.064 40 Motor 100 Z Tnew = 0.2 pu 4 = 0.08 40 new pu
old pu
4
S Bnew = 100 MVA
S Bold = 66.5MVA
V Bnew = 11kV
V Bold = 10.45kV
old Z motor = 0.18 2
Z
new motor
=Z
old motor
V old S new × Bnew × Bold SB VB 2
10.45 100 = 0.18 × × 11 66.5 = 0.25 pu
Line Impedance Line 1 V B 2 = 220 kV
Line 2
S B = 100 MVA Z B line1 =
(V )
2
B2
=
SB
Z line1( pu ) =
( 220k )
V B 3 = 110 kV
2
100 M
S B = 100 MVA
= 484Ω
Z B line 2 =
Z actual 48.4 = = 0.1 pu Z B line1 484
(V B 3 ) 2
Z line 2 ( pu ) =
SB
=
(110k ) 2 100 M
= 121Ω
Z actual 65.43 = = 0.54 pu Z B line 2 121
Load
S 3φ = 57 MVA VL = 10.45kV pf = 0.6 lagging
θ = cos −1 0.6 = 53.13° S 3φ load = 57 ∠53.13°MVA Z load ( act ) = Z load (base )
VL
(10.45k ) 2
2
S 3φ load *
=
57 M∠53.13°
2 2 ( ( VB 4 ) 11k ) = =
Z load ( pu ) =
SB
Z load ( act ) Z load ( base )
100 M
=
= 1.1495 + j1.5327 Ω
= 1.21Ω
1.1495 + j1.5327 Ω = 0.95 + j1.267 pu 1.21Ω
Step 4: Draw the per unit impedance diagram. 5
XG= j0.2p.u
XT1= j0.2p.u
ZL1= j0.10p.u
XT3= j0.16p.u
ZL1= j0.54p.u
XT2= j0.15p.u
XT4= j0.20p.u
Zmotor= j0.25p.u
M
G ZLoad= 0.95+j1.267
Example 4 Using base values of 30kVA and 240V in the generator side, draw the per unit circuit, and determine the per unit impedances and the per unit source voltage. Then calculate the load current both in per unit and in amperes. Transformer winding resistance and shunt admittance branches are neglected. T1 30 kVA VG=220∟0ºV
Xline=2Ω
G 30 kVA 240/480V X T1=0.1pu
T2
20 kVA 460/115V XT2=0.1pu
Zload=0.9+j0.2Ω
Solution Step 1: Draw the zone in the network. T2
T1 30 kVA VG=220∟0ºV
Xline=2Ω
G 30 kVA 240/480V X T1=0.1pu
20 kVA 460/115V XT2=0.1pu
Zload=0.9+j0.2Ω
Step 2: Draw the per unit circuit
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IS(pu)
VS(pu)
XT1(pu)
Zline(pu)
XT2(pu)
Iload(pu)
Zload(pu)
G
Step 3: Find the base values SB for the entire network is 30kVA. Find the base voltages and impedances of each zone. 2
ZB =
VB SB
V B1 = 240V
240 2 = 1.92Ω 30k 480 2 = = 7.68Ω 30k
Z B1 =
VB 2 =
V2 480 × V B1 = × 240 = 480V V1 240
Z B2
VB 3 =
115 × 480 = 120V 460
Z B3 =
120 2 = 0.48Ω 30k
Find base current in zone 3 [!:to calculate the load current later] S 30k I B3 = B = = 250 A V B 3 120 Step 4: Find the per unit values
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Vs ( pu ) =
Vs 220∠0° = = 0.9167 ∠0° pu V B1 240
old X Tnew 1 = X T 1 = 0.1 pu
X line ( pu ) =
X line 2 = = 0.2604 pu Z B2 7.68 2
X
new T2
=X
old T2
V Bold S Bnew × new × old VB S B 2
460 30 = 0.1 × × = 0.1378 pu 480 20 or
:: calculation using base value
2
X
new T2
=X
old T2
V old S new × Bnew × Bold VB S B 2
Z load(pu)
115 30 = 0.1 × × = 0.1378 pu :: calculation using voltage ratio 120 20 Zload 0.9 + j 0.2 = = = 1.875 + j 0.4167 pu ZB3 0.48
Step 5: Find the load current I load ( pu ) = I s ( pu ) = =
Vs ( pu ) Z total ( pu ) Vs ( pu )
j ( X T 1 + X line + X T 2 ) + Z load ( pu )
0.9167 ∠0° j (0.1 + 0.2604 + 0.1378) + (1.875 + j 0.4167 ) 0.9167 ∠0° = 2.086∠26.01° = 0.4395∠ − 26.01° pu = I load ( pu ) × I B 3 =
I load
= 0.4395∠ − 26.01° × 250 = 109.9∠ − 26.01° A
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