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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 1 of 13
Homework #02
1. Problem 2.40 in Glover, Sarma and Overbye
2. Problem 2.43 in Glover, Sarma and Overbye
3. Problem 2.44 in Glover, Sarma and Overbye
4. Problem 2.52 in Glover, Sarma and Overbye
5. Problem 2.38 in Glover, Sarma and Ovebye
a 1 120
Solutions
Problem 01.
1. Problem 2.40 in Glover, Sarma and Overbye
A balanced three-phase 208-V source supplies a balanced three-phase load. If the line current IA is measured to be 10 A and is in phase with the line-to-line voltage VBC, find the per-phase load impedance if the load is (a) Y-connected, (b) -connected.
VLL_Mag 208 IA_Mag 10
VAG VLL_Mag
3
0 =VAG (( 120.08886 0 ))
VBG VAG a2
=VBG (( 120.08886 120 ))
VCG VAG a =VCG (( 120.08886 120 ))
VAB VAG VBG =VAB (( 208 30 ))
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 2 of 13
VBC VBG VCG =VBC (( 208 90 ))
VCA VCG VAG =VCA (( 208 150 ))
IA IA_Mag argVBC IB IA_Mag argVCA IC IA_Mag argVAB
=IA (( 10 90 )) =IB (( 10 150 )) =IC (( 10 30 ))
Find the per-phase load impedance if the load is (a) Y-connected, (b) -connected.
ZY_Connected VAG
IA=ZY_Connected (( 12.00889 90 ))
=ZY_Connected 12.00889j
_______________________________/
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 3 of 13
Find the per-phase load impedance if the load is (b) -connected.
IA IAB ICA IA
VAB
Z_Connected
VCA
Z_Connected
Z_Connected VAB VCA
IA=Z_Connected (( 36.02666 90 ))
=Z_Connected 36.02666j
_______________________________/
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 4 of 13
Problem 02.
2. Problem 2.43 in Glover, Sarma and Overbye
A three-phase line, which has an impedance of (2 +j4) Ohms per phase, feeds two balanced three-phase loads that are connected in parallel. One of the loads is Y connected with an impedance of (30 + j40) Ohms per phase, and the other is -connected with an impedance of (60 - j45) Ohms per phase. The line is energized at the sending end from a 60-Hz, three-phase, balanced voltage source of 120 sqr(3) V (rms, line-to-line).
Determine (a) the current, real power, and reactive power delivered by the sending end source; (b) the line-to-line voltage at the load; (c) the current per phase in each load; and (d) the total three-phase real and reactive powers absorbed by each load and by the line. (e) Check that the total three-phase complex power delivered by the source equals the total three-phase power absorbed by the line and loads.
VLL 120 3 VSrc_A VLL
30 ZLine (( +2 1j 4))
ZY_Load (( +30 1j 40)) Z_Load (( 60 1j 45))
ZYEq__Load Z_Load
3=ZYEq__Load (( 20 15j))
ZLoad_Eq +ZYEq__Load
1ZY_Load
11
=ZLoad_Eq (( 22 4j))
ZTot +ZLine ZLoad_Eq =ZTot (( 24 0 ))
Determine (a) the current, real power, and reactive power delivered by the sending end source
ISrc_A VSrc_A
ZTot=ISrc_A (( 5 0 ))
_____________________/
S, P and Q Delivered by the sendiung-end source
SSrc 3 VSrc_AISrc_A =SSrc (( 1800 0 ))
_____________________/
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 5 of 13
PSrc ReSSrc =PSrc 1800_____________________/
QSrc ImSSrc =QSrc 0_____________________/
(b) the line-to-line voltage at the load;
VLoad_A VSrc_A ISrc_A ZLine =VLoad_A (( 111.8034 10.30485 ))_________________________________/
(c) the current per phase in each load, c.1 Y-connected load
ILoad_Y_A VLoad_A
ZY_Load=ILoad_Y_A (( 2.23607 63.43495 ))
_________________________________/
(c) the current per phase in each load, c.2 -connected load
ILoad_YEq__A VLoad_A
ZYEq__Load=ILoad_YEq__A (( 4.47214 26.56505 ))
_________________________________/
ILoad__AB
VAB
Z_Connected
ILoad__AB VLoad_A 1 a2
Z_Load=ILoad__AB (( 2.58199 56.56505 ))
_________________________________/
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 6 of 13
(d) the total three-phase real and reactive powers absorbed by each load and by the line.
SLine 3 ||ISrc_A||2ZLine =SLine (( +150 300j))
SY_Load 3 VLoad_AILoad_Y_A =SY_Load (( +450 600j))
S_Load 3 VLoad_AILoad_YEq__A =S_Load (( 1200 900j))
(e) Check that the total three-phase complex power delivered by the source equals the total three-phase power absorbed by the line and loads.
SAbs ++SLine SY_Load S_Load =SAbs 1800
_____________________/
=SSrc (( 1800 0 ))_____________________/
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 7 of 13
Problem 03.
3. Problem 2.44 in Glover, Sarma and Overbye
Two balanced three-phase loads that are connected in parallel are fed by a three-phase line having a series impedance of (0.4 + j 2.7) Ohms per phase. One of the loads absorbs 560 kVA at 0.707 power factor lagging, and the other 132 kW at unity power factor. The line-to-line voltage at the load end of the line is 2200 sqr(3) V. Compute (a) the line-to-line voltage at the source end of the line, (b) the total real and reactive power losses in the three-phase line, and (c) the total three-phase real and reactive power supplied at the sending end of the line. Check that the total three-phase complex power delivered by the source equals the total three-phase complex power absorbed by the line and loads.
VLoad_LL 2200 3 VLoad_A VLoad_LL
30 =VLoad_A (( 2200 0 ))
ZLine (( +0.4 1j 2.7))
SLoad1 (( 560 acos ((0.707)))) =SLoad1 (( +395.92 396.03959j))
SLoad2 (( 132 acos ((1)))) =SLoad2 132
STotal +SLoad1 SLoad2 =STotal (( 659.95976 36.87681 ))
IA
STotal3
VLoad_A=IA (( 99.9939 36.87681 ))
Compute (a) the line-to-line voltage at the source end of the line
VSrc_A +IA ZLine VLoad_A =VSrc_A (( 2401.69541 4.58448 ))
VSrc_LL VSrc_A 3 30
=VSrc_LL (( 4159.85847 34.58448 ))
_________________________________/
=||VSrc_LL|| 4159.85847
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 8 of 13
(b) the total real and reactive power losses in the three-phase line
SLine 3 ||IA||2ZLine =SLine (( +11998.53664 80990.12231j))
PLine ReSLine =PLine 11.99854_____________________/
QLine ImSLine =QLine 80.99012_____________________/
(c) the total three-phase real and reactive power supplied at the sending end of the line
SSrc_3ph 3 VSrc_AIA =SSrc_3ph (( +539.91854 477.02971j))
PSrc_3ph ReSSrc_3ph =PSrc_3ph 539.91854_____________________/
QSrc_3ph ImSSrc_3ph =QSrc_3ph 477.02971_________________________/
Check that the total three-phase complex power delivered by the source equals the total three-phase complex power absorbed by the line and loads.
=+STotal SLine (( +539.91854 477.02971j))__________________________________________/
=SSrc_3ph (( +539.91854 477.02971j))__________________________________________/
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 9 of 13
Problem 04.
4. Problem 2.52 in Glover, Sarma and Overbye
A balanced three-phase load is connected to a 4.16-kV, three-phase, four-wire, grounded-wye dedicated distribution feeder. The load can be modeled by an impedance of ZL =(4.7 + j 9) Ohms/phase, wye-connected. The impedance of the phase conductors (0.3 + j1) Ohms. Determine the following by using the phase A to neutral voltage as a reference and assume positive phase sequence:
(a) Line currents for phases A, B, and C.(b) Line-to-neutral voltages for all three phases at the load.(c) Apparent, active, and reactive power dissipated per phase, and for all three phases in the load.(d) Active power losses per phase and for all three phases in the phase conductors.
VSrc_a =
4.16
30
2.40178 ZLine_a (( +0.3 1j)) ZLoad_a (( +4.7 9j))
(a) Line currents for phases A, B, and C.
ISrc_a VSrc_a
+ZLine_a ZLoad_a=ISrc_a (( 214.82148 63.43495 ))
________________________________/
ISrc_b ISrc_a a2
=ISrc_b (( 214.82148 176.56505 ))________________________________/
ISrc_c ISrc_a a =ISrc_c (( 214.82148 56.56505 ))________________________________/
(b) Line-to-neutral voltages for all three phases at the load.
VLoad_a VSrc_a ISrc_a ZLine_a =VLoad_a (( 2.18115 1.00951 ))________________________________/
VLoad_b VLoad_a a2
=VLoad_b (( 2.18115 121.00951 ))________________________________/
VLoad_c VLoad_a a =VLoad_c (( 2.18115 118.99049 ))________________________________/
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 10 of 13
(c) Apparent, active, and reactive power dissipated per phase, and for all three phases in the load.
SSrc 3 VSrc_aISrc_a =SSrc (( +692.224 1384.448j))
SLoad_Phase ||ISrc_a||2ZLoad_a =SLoad_Phase (( +216.89685 415.3344j))
_____________________________________/
PLoad_Phase ReSLoad_Phase =PLoad_Phase 216.89685_____________________________________/
QLoad_Phase ImSLoad_Phase =QLoad_Phase 415.3344_____________________________________/
SLoad_3Ph SLoad_Phase 3 =SLoad_3Ph (( +650.69056 1246.0032j))_____________________________________/
PLoad_3Ph ReSLoad_3Ph =PLoad_3Ph 650.69056_____________________________________/
QLoad_3Ph ImSLoad_3Ph =QLoad_3Ph 1246.0032_____________________________________/
(d) Active power losses per phase and for all three phases in the phase conductors.
SLine_Phase ||ISrc_a||2ZLine_a =SLine_Phase (( +13.84448 46.14827j))
PLine_Phase ReSLine_Phase =PLine_Phase 13.84448__________________________/
QLine_Phase ImSLine_Phase =QLine_Phase 46.14827
SLine_3Ph 3 SLine_Phase =SLine_3Ph (( +41.53344 138.4448j))
PLine_3Ph ReSLine_3Ph =PLine_3Ph 41.53344__________________________/
QLine_3Ph ImSLine_3Ph =QLine_3Ph 138.4448
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 11 of 13
Problem 05.
5. Problem 2.38 in Glover, Sarma and Overbye
Given the impedance diagram of a simple system as shown in Figure 2.31, draw the admittance diagram for the system and develop the 4 x 4 bus admittance matrix Ybus by inspection.
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 12 of 13
Draw the admittance diagram for the system
__________________________________________________________________________________/
Develop the 4 x 4 bus admittance matrix Ybus by inspection.
YBus
++1
1j1
0.4j1
0.2j1
0.4j1
0.2j0
1
0.4j++
1
0.8j1
0.4j1
0.2j1
0.2j0
1
0.2j1
0.2j++
1
0.08j1
0.2j1
0.2j1
0.08j
0 0 1
0.08j1
0.08j
=YBus
8.5j 2.5j 5j 0
2.5j 8.75j 5j 0
5j 5j 22.5j 12.5j
0 0 12.5j 12.5j
_______________________________________/
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Session 1Fall 2013ECE421:
Introduction to Power SystemsHomework # 02
Due Session 10 (September 18)Arturo Barradas Munoz Page 13 of 13
Verifying against circuit reduction - hand calculations
=YBus1
0.5j 0.4j 0.45j 0.45j
0.4j 0.48j 0.44j 0.44j
0.45j 0.44j 0.545j 0.545j
0.45j 0.44j 0.545j 0.625j
Za 0.2j 0.4j
++0.4j 0.2j 0.2jZb
0.4j 0.2j
++0.4j 0.2j 0.2jZc
0.2j 0.2j
++0.4j 0.2j 0.2j
Z44 =++ + +1j Za
1
+0.8j Zb1
1
Zc 0.08j 0.625i
Z33 =+ + +1j Za
1
+0.8j Zb1
1
Zc 0.545i
Z22 = + ++1j Za Zb
1
0.8j1
1
0.48i
Z11 = + ++0.8j Zb Za
1
1j1
1
0.5i