W EE Power System 14 04 17 LS2 S
Serial : W_EE_Power System_140417_LS2
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CLASS TEST 2017-18
ELECTRICAL ENGINEERING
Subject : Power Systems Date of test : 14/04/2017 Answer Key 1.
(d)
7.
(a)
13.
(c)
19.
(b)
25.
(a)
2.
(c)
8.
(a)
14.
(a)
20.
(a)
26.
(a)
3.
(c)
9.
(b)
15.
(a)
21.
(c)
27.
(b)
4.
(c)
10.
(b)
16.
(c)
22.
(d)
28.
(d)
5.
(b)
11.
(d)
17.
(c)
23.
(a)
29.
(d)
6.
(b)
12.
(b)
18.
(b)
24.
(a)
30.
(c)
6
Electrical Engineering
Detailed Explanations 1.
(d) If the star neutral is grounded through reactor of impedance zn, an impedance 3 zn appears in series with z0 in the sequence network.
2.
(c) The pickup value of the relay is 8 A but since the relay setting is 130%, therefore, the operating current of the relay is 8 × 1.3 = 10.4 A The plug setting multiplier of the relay, PSM =
= 3.
Primary Current (fault current) Secondary Current = Relay Current Setting × CT ratio Relay Current Setting 5000 =6 10.4 × 80
(c) String efficiency, % η =
V × 100 nVn
Given, n = 5, V5 = 0.25 V = voltage across the bottom most unit. V × 100 5 × 0.25V %η = 80%
%η =
4.
(c) In LLG fault, Ipositive + Inegative + Izero = 0 Here, j1.653 – j 0.5 – j1.153 = 0
5.
(b) dP Synchronizing power coefficient = Pmax cos δ0 = e dδ Here, Pmax = 2, δ = 30° ∴
6.
Sp =
∂Pe ∂δ
δ = 30°
= 2 cos 30° = 2 ×
3 = 3 2
(b) We know that,
RRRV =
Vm LC
t sin LC
For short line, L and C decreases So, 7.
Vm increases LC
(a) For economic power operation, L
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dC = λ dP
(Where, L =
1 = Penalty factor) ∂PL 1 − ∂P © Copyright :
CTEE17 • Power Systems ∂C ∂P λ = ∂P 1 − loss ∂P
∴
or,
25 =
0.012P + 8 (1 − 0.2)
0.012 P = 20 – 8 P = 1000 MW
or, or, 8.
7
(a) X12 1
j 0.3
j 0.2
2 Infinite bus
Ef
j 0.15
j 0.15
F
X12 =
j (0.2 × 0.15 + 0.3 × 0.15 + 0.2 × 0.3) (Using star to delta transformation) 0.15
0.2 × 0.3 = j 0.2 + 0.3 + = j 0.9 p.u. 0.15
9.
(b) Insulation Resistance = Thus, ∴
10.
1 l 200 = 40MΩ Ri = 5
Ri α
(b)
VR(no-load) = 11.
ρ R ln 2πl r
VS 400 = = 500 kV 0.8 A
(d) For the given values of sending end and receiving end voltages, the power transfer will be maximum for δ = β. where δ is the phase angle between sending end and receiving end voltage. A = A∠α = 0.96∠1.0° B = B∠β = 100∠80° Ω The maximum power transmitted is given by
PR,max =
VS VR B
−
A VR B
2
cos(β − α) =
120 × 110 0.96 (110)2 cos(79°) − 100 100
PR,max = 109.83 MW
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8 12.
Electrical Engineering (b) According to the KVA triangle OAB. OA = kW1 = 120 kW (given)
OB = KVA1 = KVA = KVAR1 = KVA2 = KVAR2 =
kW1 120 = = 150 KVA 0. 8 0. 8
kW 2 + kVAR 2
O
φ1 φ
120 kW
13 3.3
1502 − 1202 = 90 KVAR
A
2
150 KVA
58 KVAR
KV A C
k.W. 120 = = 133. 3 KVA 0. 9 0. 9
90 KVAR
32 KVAR Capacitors
(133.3)2 − (120)2 = 58 KVAR
B
∴ The KVAR to be supplied by the shunt capacitor is KVAR1 – KVAR2 = 90 – 58 = 32 KVAR 13.
(c) For a solid LG fault, 3E Fault current is: (IF)LG = (2X + X + 3 X ) 1 0 n Similarly, for a solid 3-φ fault (I F ) 3-φ =
(Here, X1 ≈ X2 for synchronous generator)
E X1
For LG fault current to be less than 3-φ fault current, 3E E 2X 1 + X 0 + 3 X n < X1
or, or,
2X1 + X0 + 3Xn > 3X1
Xn >
1 (X − X 0 ) 3 1
Hence, option (c) is correct. 14.
(a)
Zbase =
Now,
∴ 15.
Ybase =
Y (in pu) =
2 Vbase Vbase Vbase = = Ibase (VA)base (VA)base / Vbase
1 Z base Y Ybase
=
=
(VA)base 2 Vbase
2 YVbase pu (VA)base
(a) Since the generators are in parallel, they will operate at the same frequency at steady load. Let load on generator 1 (200 MW) = x MW load on generator 2 (400 MW) = (600 – x) MW and reduction in frequency = ∆f ∴
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0.04 × 50 ∆f = 200 x
... (i)
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CTEE17 • Power Systems
and Equating ∆f in (i) and (ii),
0.05 × 50 ∆f = 400 600 − x
... (ii)
x = 230.77 MW (load on generator-1) 600 – x = 369.23 MW (load on generator-2) ∴ 16.
System frequency = 50 −
QR =
Now, or, or, or, or,
VsVR AVR2 sin(β − δ) − sin(β − α) B B
275 × 275 0.85 sin(75° − δ) − × 2752 sin(75° − 5°) 200 200 378 sin(75 – δ) = 302 sin(75 – δ) = 0.798 δ = 22°
VsVR AVR2 cos( ) cos(β − α) β − δ − PR = B B =
275 × 275 0.85 × 2752 cos(75° − 22°) − cos70° 200 200
PR = 117.64 MW
or, (c)
C =
2π ∈0 ∈r 2π × 8.854 × 10 −12 × 2.4 F/m = F/m R 15 ln ln r 10
= 0.329 µF/km ∴ Total cable capacitance = 0.329 × 2.5 = 0.8225 µF
18.
α = 5°, β = 75°
0 =
Now,
17.
0.04 × 50 × 230.77 = 47.69 Hz. 200
(c) Given, Vs = VR = 275 kV, Since the power is received at unity power factor, therefore QR = 0
(Since l = 2.5 km)
So, dielectric loss,
2 ωC tan δ Pd = Vph
or,
= (11000)2 × (100 π) (0.8225 × 10–6 ) (0.031) Pd = 969 W
(b) no. of buses, n no. of Generator buses no. of Reactive power support buses no. of buses with fixed shunt capacitor no. of slack buses (ns) ∵ size of Jacobian matrix
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= = = = = = = =
9
300 20 25 15 20 + 25 – 15 = 30 2 (n – ns) × 2(n – ns) 2(300 – 30) × 2 (300 – 30) 540 × 540
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10 19.
Electrical Engineering (b) The network has two buses, Therefore, YBUS matrix is a 2 × 2 matrix Y11 = Sum of all admittances terminating at bus 1 = –j 0.45 + (–j 0.75) = –j 1.2 Y22 = Sum of all admittances terminating at bus 2 = –j 0.6 + (–j 0.75) = –j 1.35 Y12 = Y21 = –y12 = –(–j 0.75) = j 0.75
[YBUS ] 20.
Y11 Y12 − j1. 2 j 0.75 = = Y Y 21 22 j 0. 75 − j1. 35
(a) Since the LT side is delta connected, the CTs on that side will be star connected. Therefore, if 400 A is line current, the CT secondary current is 5 A. The line current on the star side of the power transformer will be 6. 6 = 80 A 33 The CTs on the star side are delta connected and the current required on the relay side of the CT is 5 A. 400 ×
Therefore, the current in the CT secondary (phase current) is The CT ratio on the HT side will be 80: 21.
5 . 3
(c) The swing equatin is M
22.
5 . 3
d 2δ dt 2
= PS – PE
(d) Receiving-end phase voltage,
VR =
110 × 103 = 63508 V 3
Magnitude of receiving-end current, IR =
Load in MW × 106 3 VRL cos φR
=
30 × 106 3 × 110 × 103 × 0.8
= 196.82 A
Taking receiving-end phase voltage as reference phasor, we have VR = 63508 ∠0° V Receiving-end current, IR = 196.82 ∠–36.87° A Sending-end phase voltage, Vs = AVR + BIR = 0.96 ∠1.0° × 63508 ∠0° + 100 ∠80° × 196.82 ∠–36.87° = 76709 ∠10.91° V Magnitude of sending-end line voltage, VSL =
3 × 76709 = 132863 V or 132.863 kV
Voltage regulation =
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VSL−VRL VRL
× 100 =
132.863 − 110 × 100 = 20.78% 110
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CTEE17 • Power Systems 23.
11
(a) The mutual GMD between sides A and B is
DM =
6
(D14 .D15 )(D24 .D25 )(D34 .D35 )
From the figure it is obvious that, D14 = D24 = D25 = D35 =
82 + 22 =
68 m
D15 = D34 = 82 + 62 = 64 + 36 =10 m DM = 24.
6 (68)2
× 100 = 8.793 m ≈ 8.8 m
(a) ∴ Fault current (If)LLG depends on all the three impedance Z1, Z2 and Z0 . Z1
Z2
E
26.
Z0
(a) Ia1 =
=
Ea Z1 + Z 2 + Z 0
1 + j 0.0 j 0.25 + j 0.35 + j 0.1
= –j 1.428 pu (If)LG = p.u. fault current = 3Ia1 = –j 4.285 pu Base current =
25 × 1000 3 × 13.2
= 1093.46 A
fault current (in amps) = 1093.46 × 4.285 = 4685.47 A ≈ 4685 A 28.
(d) Let C1 be the capacitance between a conductor and sheath while C2 be the capacitance between two conductors. C1
C2
C1
C2
C2
C1
Now, equivalent capacitance between two bunched conductors with shealth and 3rd conductor is 2C2 + C1 = 0.8 µF www.madeeasy.in
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12
Electrical Engineering Also, equivalent capacitance between three bunched conductors & shealth is 3C1 = 0.9 µF Thus, C1 = 0.3 µF C2 = 0.25 µF C/Phase = 3C2 + C1 = 3 × 0.25 + 0.3 = 1.05 µF
29.
(d) 30 units = 30 kWh 30 kWh/(day = 24 hours) 1 hour =
30.
30 = 1.25 kW 24
∴
% load factor =
Average load × 100 Max. demand
∴
% load factor =
1.25 × 100 = 62.5% 2
(c) Voltage transmitted in the cable =
=
2 Z2 Z1 + Z 2
2 × 40 2 = = 0.18 V 400 + 40 11
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Delhi | Noida | Bhopal | Hyderabad | Jaipur | Lucknow | Indore | Pune | Bhubaneswar | Kolkata | Patna Web: www.madeeasy.in | E-mail: [email protected] | Ph: 011-45124612
CLASS TEST 2017-18
ELECTRICAL ENGINEERING
Subject : Power Systems Date of test : 14/04/2017 Answer Key 1.
(d)
7.
(a)
13.
(c)
19.
(b)
25.
(a)
2.
(c)
8.
(a)
14.
(a)
20.
(a)
26.
(a)
3.
(c)
9.
(b)
15.
(a)
21.
(c)
27.
(b)
4.
(c)
10.
(b)
16.
(c)
22.
(d)
28.
(d)
5.
(b)
11.
(d)
17.
(c)
23.
(a)
29.
(d)
6.
(b)
12.
(b)
18.
(b)
24.
(a)
30.
(c)
6
Electrical Engineering
Detailed Explanations 1.
(d) If the star neutral is grounded through reactor of impedance zn, an impedance 3 zn appears in series with z0 in the sequence network.
2.
(c) The pickup value of the relay is 8 A but since the relay setting is 130%, therefore, the operating current of the relay is 8 × 1.3 = 10.4 A The plug setting multiplier of the relay, PSM =
= 3.
Primary Current (fault current) Secondary Current = Relay Current Setting × CT ratio Relay Current Setting 5000 =6 10.4 × 80
(c) String efficiency, % η =
V × 100 nVn
Given, n = 5, V5 = 0.25 V = voltage across the bottom most unit. V × 100 5 × 0.25V %η = 80%
%η =
4.
(c) In LLG fault, Ipositive + Inegative + Izero = 0 Here, j1.653 – j 0.5 – j1.153 = 0
5.
(b) dP Synchronizing power coefficient = Pmax cos δ0 = e dδ Here, Pmax = 2, δ = 30° ∴
6.
Sp =
∂Pe ∂δ
δ = 30°
= 2 cos 30° = 2 ×
3 = 3 2
(b) We know that,
RRRV =
Vm LC
t sin LC
For short line, L and C decreases So, 7.
Vm increases LC
(a) For economic power operation, L
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dC = λ dP
(Where, L =
1 = Penalty factor) ∂PL 1 − ∂P © Copyright :
CTEE17 • Power Systems ∂C ∂P λ = ∂P 1 − loss ∂P
∴
or,
25 =
0.012P + 8 (1 − 0.2)
0.012 P = 20 – 8 P = 1000 MW
or, or, 8.
7
(a) X12 1
j 0.3
j 0.2
2 Infinite bus
Ef
j 0.15
j 0.15
F
X12 =
j (0.2 × 0.15 + 0.3 × 0.15 + 0.2 × 0.3) (Using star to delta transformation) 0.15
0.2 × 0.3 = j 0.2 + 0.3 + = j 0.9 p.u. 0.15
9.
(b) Insulation Resistance = Thus, ∴
10.
1 l 200 = 40MΩ Ri = 5
Ri α
(b)
VR(no-load) = 11.
ρ R ln 2πl r
VS 400 = = 500 kV 0.8 A
(d) For the given values of sending end and receiving end voltages, the power transfer will be maximum for δ = β. where δ is the phase angle between sending end and receiving end voltage. A = A∠α = 0.96∠1.0° B = B∠β = 100∠80° Ω The maximum power transmitted is given by
PR,max =
VS VR B
−
A VR B
2
cos(β − α) =
120 × 110 0.96 (110)2 cos(79°) − 100 100
PR,max = 109.83 MW
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8 12.
Electrical Engineering (b) According to the KVA triangle OAB. OA = kW1 = 120 kW (given)
OB = KVA1 = KVA = KVAR1 = KVA2 = KVAR2 =
kW1 120 = = 150 KVA 0. 8 0. 8
kW 2 + kVAR 2
O
φ1 φ
120 kW
13 3.3
1502 − 1202 = 90 KVAR
A
2
150 KVA
58 KVAR
KV A C
k.W. 120 = = 133. 3 KVA 0. 9 0. 9
90 KVAR
32 KVAR Capacitors
(133.3)2 − (120)2 = 58 KVAR
B
∴ The KVAR to be supplied by the shunt capacitor is KVAR1 – KVAR2 = 90 – 58 = 32 KVAR 13.
(c) For a solid LG fault, 3E Fault current is: (IF)LG = (2X + X + 3 X ) 1 0 n Similarly, for a solid 3-φ fault (I F ) 3-φ =
(Here, X1 ≈ X2 for synchronous generator)
E X1
For LG fault current to be less than 3-φ fault current, 3E E 2X 1 + X 0 + 3 X n < X1
or, or,
2X1 + X0 + 3Xn > 3X1
Xn >
1 (X − X 0 ) 3 1
Hence, option (c) is correct. 14.
(a)
Zbase =
Now,
∴ 15.
Ybase =
Y (in pu) =
2 Vbase Vbase Vbase = = Ibase (VA)base (VA)base / Vbase
1 Z base Y Ybase
=
=
(VA)base 2 Vbase
2 YVbase pu (VA)base
(a) Since the generators are in parallel, they will operate at the same frequency at steady load. Let load on generator 1 (200 MW) = x MW load on generator 2 (400 MW) = (600 – x) MW and reduction in frequency = ∆f ∴
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0.04 × 50 ∆f = 200 x
... (i)
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CTEE17 • Power Systems
and Equating ∆f in (i) and (ii),
0.05 × 50 ∆f = 400 600 − x
... (ii)
x = 230.77 MW (load on generator-1) 600 – x = 369.23 MW (load on generator-2) ∴ 16.
System frequency = 50 −
QR =
Now, or, or, or, or,
VsVR AVR2 sin(β − δ) − sin(β − α) B B
275 × 275 0.85 sin(75° − δ) − × 2752 sin(75° − 5°) 200 200 378 sin(75 – δ) = 302 sin(75 – δ) = 0.798 δ = 22°
VsVR AVR2 cos( ) cos(β − α) β − δ − PR = B B =
275 × 275 0.85 × 2752 cos(75° − 22°) − cos70° 200 200
PR = 117.64 MW
or, (c)
C =
2π ∈0 ∈r 2π × 8.854 × 10 −12 × 2.4 F/m = F/m R 15 ln ln r 10
= 0.329 µF/km ∴ Total cable capacitance = 0.329 × 2.5 = 0.8225 µF
18.
α = 5°, β = 75°
0 =
Now,
17.
0.04 × 50 × 230.77 = 47.69 Hz. 200
(c) Given, Vs = VR = 275 kV, Since the power is received at unity power factor, therefore QR = 0
(Since l = 2.5 km)
So, dielectric loss,
2 ωC tan δ Pd = Vph
or,
= (11000)2 × (100 π) (0.8225 × 10–6 ) (0.031) Pd = 969 W
(b) no. of buses, n no. of Generator buses no. of Reactive power support buses no. of buses with fixed shunt capacitor no. of slack buses (ns) ∵ size of Jacobian matrix
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= = = = = = = =
9
300 20 25 15 20 + 25 – 15 = 30 2 (n – ns) × 2(n – ns) 2(300 – 30) × 2 (300 – 30) 540 × 540
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10 19.
Electrical Engineering (b) The network has two buses, Therefore, YBUS matrix is a 2 × 2 matrix Y11 = Sum of all admittances terminating at bus 1 = –j 0.45 + (–j 0.75) = –j 1.2 Y22 = Sum of all admittances terminating at bus 2 = –j 0.6 + (–j 0.75) = –j 1.35 Y12 = Y21 = –y12 = –(–j 0.75) = j 0.75
[YBUS ] 20.
Y11 Y12 − j1. 2 j 0.75 = = Y Y 21 22 j 0. 75 − j1. 35
(a) Since the LT side is delta connected, the CTs on that side will be star connected. Therefore, if 400 A is line current, the CT secondary current is 5 A. The line current on the star side of the power transformer will be 6. 6 = 80 A 33 The CTs on the star side are delta connected and the current required on the relay side of the CT is 5 A. 400 ×
Therefore, the current in the CT secondary (phase current) is The CT ratio on the HT side will be 80: 21.
5 . 3
(c) The swing equatin is M
22.
5 . 3
d 2δ dt 2
= PS – PE
(d) Receiving-end phase voltage,
VR =
110 × 103 = 63508 V 3
Magnitude of receiving-end current, IR =
Load in MW × 106 3 VRL cos φR
=
30 × 106 3 × 110 × 103 × 0.8
= 196.82 A
Taking receiving-end phase voltage as reference phasor, we have VR = 63508 ∠0° V Receiving-end current, IR = 196.82 ∠–36.87° A Sending-end phase voltage, Vs = AVR + BIR = 0.96 ∠1.0° × 63508 ∠0° + 100 ∠80° × 196.82 ∠–36.87° = 76709 ∠10.91° V Magnitude of sending-end line voltage, VSL =
3 × 76709 = 132863 V or 132.863 kV
Voltage regulation =
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VSL−VRL VRL
× 100 =
132.863 − 110 × 100 = 20.78% 110
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CTEE17 • Power Systems 23.
11
(a) The mutual GMD between sides A and B is
DM =
6
(D14 .D15 )(D24 .D25 )(D34 .D35 )
From the figure it is obvious that, D14 = D24 = D25 = D35 =
82 + 22 =
68 m
D15 = D34 = 82 + 62 = 64 + 36 =10 m DM = 24.
6 (68)2
× 100 = 8.793 m ≈ 8.8 m
(a) ∴ Fault current (If)LLG depends on all the three impedance Z1, Z2 and Z0 . Z1
Z2
E
26.
Z0
(a) Ia1 =
=
Ea Z1 + Z 2 + Z 0
1 + j 0.0 j 0.25 + j 0.35 + j 0.1
= –j 1.428 pu (If)LG = p.u. fault current = 3Ia1 = –j 4.285 pu Base current =
25 × 1000 3 × 13.2
= 1093.46 A
fault current (in amps) = 1093.46 × 4.285 = 4685.47 A ≈ 4685 A 28.
(d) Let C1 be the capacitance between a conductor and sheath while C2 be the capacitance between two conductors. C1
C2
C1
C2
C2
C1
Now, equivalent capacitance between two bunched conductors with shealth and 3rd conductor is 2C2 + C1 = 0.8 µF www.madeeasy.in
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12
Electrical Engineering Also, equivalent capacitance between three bunched conductors & shealth is 3C1 = 0.9 µF Thus, C1 = 0.3 µF C2 = 0.25 µF C/Phase = 3C2 + C1 = 3 × 0.25 + 0.3 = 1.05 µF
29.
(d) 30 units = 30 kWh 30 kWh/(day = 24 hours) 1 hour =
30.
30 = 1.25 kW 24
∴
% load factor =
Average load × 100 Max. demand
∴
% load factor =
1.25 × 100 = 62.5% 2
(c) Voltage transmitted in the cable =
=
2 Z2 Z1 + Z 2
2 × 40 2 = = 0.18 V 400 + 40 11
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