Ece421hw04 Abm

Session 21Fall 2013ECE421:

Introduction to Power SystemsHomework # 04

Due Session 21 (October 14th)Arturo Barradas Munoz of 23

Homework #04

1. Problem 3.48 in Glover, Sarma and Overbye




5. Problem 3.59 in Glover, Sarma and Ovebye



General Definitions

≔ ≔ ≔ ≔

≔ ≔ ≔a ∠1 °120 ≔ 1 ≔Turns 1




Problem 01.


With the same transformer banks as in Problem 3.47, Figure 3.41 shows the one-line diagram of a generator, a step-up transformer bank, a transmission line, a step-down transformer bank, and an impedance load. The generator terminal voltage is 15 kV (line-to-line).

(a) Draw the per-phase equivalent circuit, accounting for phase shifts for positive sequence operation.

(b) By choosing the line-to-neutral generator terminal voltage as the reference, determine the magnitudes of the generator current, transmission-line current, load current, and line-to-line load voltage. Also, find the three-phase complex power delivered to the load.




≔ZLoad(( +5 1j))

≔ZLine((100j))

≔XT1_Sph_LV 0.24j

≔XT1_LV ――――XT1_Sph_LV

3=XT1_LV 0.08i

≔XT2_Sph_LV 0.24j

≔XT2_LV ――――XT2_Sph_LV

3=XT2_LV 0.08i




(a) Draw the per-phase equivalent circuit, accounting for phase shifts for positive sequence operation.

(b) By choosing the line-to-neutral generator terminal voltage as the reference, determine the magnitudes of the generator current, transmission-line current, load current, and line-to-line load voltage. Also, find the three-phase complex power delivered to the load.

≔Vg_T

⎛⎜⎜⎝

∠――15

‾‾3

°0⎞⎟⎟⎠

=Vg_T(( ∠8.66025 °0 ))

≔ZRight_LV +ZLoad XT2_LV =ZRight_LV(( +5 1.08j))

≔ZRight_HV ⋅ZRight_LV⎛⎝ ⋅10 ‾‾3

⎞⎠

2

=ZRight_HV(( +1500 324j))

≔ZMiddle_HV +ZLine ZRight_HV =ZMiddle_HV(( +1500 424j))

≔ZMiddle_LV ⋅ZMiddle_HV

⎛⎜⎜⎝―――

1

⋅10 ‾‾3

⎞⎟⎟⎠

2

=ZMiddle_LV(( +5 1.41333j))

≔ZLeft_LV +ZMiddle_LV XT1_LV =ZLeft_LV(( +5 1.49333i))

≔Ig_T ―――Vg_T

ZLeft_LV

=Ig_T(( ∠1659.61158 °−16.62913 ))

=||Ig_T|| ⎛⎝ ⋅1.65961 103 ⎞⎠

_______________________/




≔ILine ⋅⋅Ig_T ―――1

⋅10 ‾‾3

(( ∠1 °30 )) =ILine(( ∠95.81772 °13.37087 ))

=||ILine|| 95.81772

_______________________/

≔ILoad ⋅⋅⋅ILine 10 ‾‾3 (( ∠1 − °30 )) =ILoad⎛⎝ ∠⋅1.65961 103 °−16.62913 ⎞⎠

=||ILoad|| 1659.61158

_______________________/

≔VLoad ⋅ILoad ZLoad =VLoad⎛⎝ ∠⋅8.46239 103 °−5.3192 ⎞⎠

≔VLoad_LL ⋅VLoad‾‾3

=||VLoad_LL|| 14.65729

_______________________/

≔SLoad ⋅⋅3 VLoad‾‾‾‾ILoad =SLoad

(( +41.31466 8.26293i))

=SLoad(( ∠42.13285 °11.30993 ))

________________________________/




Problem 02.


Consider the single-line diagram of a power system shown in Figure 3.42 with equipment ratings given below:

Generator G1: 50 MVA, 13.2kV, x=0.15 puGenerator G2: 20 MVA, 13.8kV, x=0.15 pu3Ph -Y T1: 80 MVA, 13.2 /165 Y kV, X=0.1 pu3Ph Y- T2: 40 MVA, 165 Y/13.8 kV, X=0.1 puLoad: 40 MVA, 0.8 PF lagging, operating at 150kV

Choose a base of 100 MVA for the system and 132-kV base in the transmission-line circuit. Let the load be modeled as a parallel combination of resistance and inductance. Neglect transformer phase shifts. Draw a per-phase equivalent circuit of the system showing all impedances in per unit.

≔SG1_rat 50 ≔VG1_rat 13.2 ≔xG1_rat 0.15j

≔SG2_rat 50 ≔VG2_rat 13.2 ≔xG2_rat 0.15j

≔ST1_rat 80 ≔VT1_HV_rat 165 ≔xT1_rat 0.1j

≔VT1_LV_rat 13.2

≔ST2_rat 40 ≔VT2_HV_rat 165 ≔xT2_rat 0.1j

≔VT2_LV_rat 13.8




≔SLoad_rat(( ∠40 acos ((0.8)))) =SLoad_rat

(( +32 24j)) ≔VLoad_rat 150

≔ZLine1_Ω(( +50 200j)) ≔ZLine2_Ω

(( +25 100j)) ≔ZLine3_Ω(( +25 100j))

≔SB 100 ≔VB_HV 132 ≔ZB_HV ―――VB_HV

2

SB

=ZB_HV 174.24

≔VB_G1 ⋅VB_HV ―――13.2

165=VB_G1 10.56 ≔ZB_G1 ―――

VB_G12

SB

=ZB_G1 1.11514

≔VB_G2 ⋅VB_HV ―――13.8

165=VB_G2 11.04 ≔ZB_G2 ―――

VB_G22

SB

=ZB_G2 1.21882

Change of base

≔xG1 ⋅⋅xG1_rat

⎛⎜⎝―――VG1_rat

VB_G1

⎞⎟⎠

2 ⎛⎜⎝―――

SB

SG1_rat

⎞⎟⎠

=xG1 0.46875j

___________________/

≔xG2 ⋅⋅xG2_rat

⎛⎜⎝―――VG2_rat

VB_G2

⎞⎟⎠

2 ⎛⎜⎝―――

SB

SG2_rat

⎞⎟⎠

=xG2 0.42888j

___________________/

≔xT1 ⋅⋅xT1_rat

⎛⎜⎝――――VT1_HV_rat

VB_HV

⎞⎟⎠

2 ⎛⎜⎝―――

SB

ST1_rat

⎞⎟⎠

=xT1 0.19531i

___________________/

≔xT2 ⋅⋅xT2_rat

⎛⎜⎝――――VT2_HV_rat

VB_HV

⎞⎟⎠

2 ⎛⎜⎝―――

SB

ST2_rat

⎞⎟⎠

=xT2 0.39063i

___________________/

≔ZLine1 ―――ZLine1_Ω

ZB_HV

=ZLine1(( +0.28696 1.14784i))

____________________________/


ZB_HV

=ZLine2(( +0.14348 0.57392i))

____________________________/


ZB_HV

=ZLine3(( +0.14348 0.57392i))

____________________________/




≔SLoad ―――SLoad_rat

SB

=SLoad(( +0.32 0.24i))

≔VLoad ―――VLoad_rat

VB_HV

=VLoad 1.13636

≔RLoad_Parallel ――――||VLoad

||2

Re ⎛⎝SLoad⎞⎠

=RLoad_Parallel 4.03538

____________________________/

≔XLoad_Parallel ――――⋅||VLoad

||2

1j

Im ⎛⎝SLoad⎞⎠

=XLoad_Parallel 5.38051j

____________________________/

Draw a per-phase equivalent circuit of the system showing all impedances in per unit.




Problem 03.


The ratings of a three-phase, three-winding transformer are:

Primary: Y connected, 66 kV, 15 MVASecondary: Y connected, 13.2 kV, 10 MVATertiary: connected, 2.3 kV, 5 MVA

Neglecting resistances and exciting current, the leakage reactances are:

XPS= 0.07 per unit on a 15-MVA; 66-kV baseXPT= 0.09 per unit on a 15-MVA; 66-kV baseXST= 0.08 per unit on a 10-MVA; 13.2-kV base

Determine the per-unit reactances of the per-phase equivalent circuit using a base of 15 MVA and 66 kV for the primary.

≔SB 15 ≔VB_P 66 ≔VB_S 13.2 ≔VB_S 2.3

≔XPS_rat 0.07 ≔XPT_rat 0.09 ≔XST_rat 0.08

≔XPS ⋅XPS_rat

⎛⎜⎝―――

SB

15

⎞⎟⎠

=XPS 0.07

≔XPT ⋅XPT_rat

⎛⎜⎝―――

SB

15

⎞⎟⎠

=XPT 0.09

≔XST ⋅XST_rat

⎛⎜⎝―――

SB

10

⎞⎟⎠

=XST 0.12

≔XP ―1

2⎛⎝ −+XPS XPT XST

⎞⎠ ≔XS ―1

2⎛⎝ +−XPS XPT XST

⎞⎠ ≔XT ―1

2⎛⎝ ++−XPS XPT XST

⎞⎠

=XP 0.02 =XS 0.05 =XT 0.07

_____________/ _____________/ _____________/





A single-phase two-winding transformer rated 90 MVA, 80/120 kV is to be connected as an autotransformer rated 80/200 kV. Assume that the transformer is ideal.

(a) Draw a schematic diagram of the ideal transformer connected as an autotransformer, showing the voltages, currents, and dot notation for polarity.

(b) Determine the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the rated values as a two-winding transformer.

How much of the kVA rating is transferred by magnetic induction?

≔IPrim ―――90

80≔ISec ―――

90

120≔ITot +IPrim ISec

=IPrim 1125 =ISec 750 =ITot 1875

≔VPrim 80 ≔VSec 120 ≔VTot =+VPrim VSec 200

(a) Draw a schematic diagram of the ideal transformer connected as an autotransformer, showing the voltages, currents, and dot notation for polarity.




(b) Determine the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the rated values as a two-winding transformer.

≔SAuto ⋅VPrim ITot =SAuto 150000

=⋅VTot ISec 150000

______________________/

How much of the kVA rating is transferred by magnetic induction?

≔STrans_Ind ⋅VPrim IPrim =STrans_Ind 90000

______________________/




Problem 05.


The two parallel lines in Example 3.13 supply a balanced load with a load current of 1.0 /30 degrees per unit. Determine the real and reactive power supplied to the load bus from each parallel line with (a) no regulating transformer, (b) the voltage-magnitude regulating transformer in Example 3.13(a), and (c) the phase-angle-regulating transformer in Example 3.13(b). Assume that the voltage at bus abc is adjusted so that the voltage at bus a'b'c' remains constant at 1.0/ 0 per unit. Also assume positive sequence.

Comment on the e¤ects of the regulating transformers.

≔ILoad(( ∠1 − °30 )) ≔VLoad

(( ∠1 °0 ))

≔SLoad ⋅VLoad‾‾‾‾ILoad =SLoad

(( +0.86603 0.5j))

≔PLoad Re ⎛⎝SLoad⎞⎠ =PLoad 0.86603

≔QLoad Im ⎛⎝SLoad⎞⎠ =QLoad 0.5

≔ZLine_0.20 ⋅0.20j ≔ZLine_0.25 ⋅0.25j




Determine the real and reactive power supplied to the load bus from each parallel line with (a) no regulating transformer. Assume that the voltage at bus abc is adjusted so that the voltage at bus a'b'c' remains constant at 1.0/ 0 per unit. Also assume positive sequence.

≔ILine_0.20 ―――――――⋅ILoad ZLine_0.25

+ZLine_0.20 ZLine_0.25

=ILine_0.20(( −0.48113 0.27778i))

≔ILine_0.25 ―――――――⋅ILoad ZLine_0.20

+ZLine_0.20 ZLine_0.25

=ILine_0.25(( −0.3849 0.22222i))

≔SLine_0.20 ⋅VLoad‾‾‾‾‾‾ILine_0.20 =SLine_0.20

(( +0.48113 0.27778i))

=SLine_0.20(( ∠0.55556 °30 ))

≔PLine_0.20 Re ⎛⎝SLine_0.20⎞⎠ =PLine_0.20 0.48113

≔QLine_0.20 Im ⎛⎝SLine_0.20⎞⎠ =QLine_0.20 0.27778

≔SLine_0.25 ⋅VLoad‾‾‾‾‾‾ILine_0.25 =SLine_0.25

(( +0.3849 0.22222i))

=SLine_0.25(( ∠0.44444 °30 ))

≔PLine_0.25 Re ⎛⎝SLine_0.25⎞⎠ =PLine_0.25 0.3849

______________________/

≔QLine_0.25 Im ⎛⎝SLine_0.25⎞⎠ =QLine_0.25 0.22222

______________________/




Determine the real and reactive power supplied to the load bus from each parallel line with (b) the voltage-magnitude regulating transformer in Example 3.13(a). Assume that the voltage at bus abc is adjusted so that the voltage at bus a'b'c' remains constant at 1.0/ 0 per unit. Also assume positive sequence.

Comment on the effects of the regulating transformers.

From page 133 in Duncan Glover's Book the admittance parameters of the equation shown below are,

＝I1

−I2

⎡⎢⎣

⎤⎥⎦

⋅Y11 Y12

Y21 Y22

⎡⎢⎣

⎤⎥⎦

V1

V2

⎡⎢⎣

⎤⎥⎦

＝Y11 YEq ＝Y12 ⋅−c YEq

＝Y21 ⋅−c YEq ＝Y22 ⋅||c||2 YEq

From page 138 in Duncan Glover's Book the admittance parameters are computed as follows:

≔c 1.05−1 =c 0.95238

The admitance parameters of the regulating transformer in series with line of 0.25j pu are:

≔Y11Line_0.25 ―――1

ZLine_0.25

=Y11Line_0.25 −4j

≔Y22Line_0.25 ⋅||c||2 Y11Line_0.25 =Y22Line_0.25 −3.62812j

≔Y12Line_0.25 ⋅−c Y11Line_0.25 =Y12Line_0.25 3.80952j

≔Y21Line_0.25 ⋅−‾c Y11Line_0.25 =Y21Line_0.25 3.80952j




The admitance parameters of the line of 0.20j pu are:

≔Y11Line_0.20 ―――1

ZLine_0.20

=Y11Line_0.20 −5j

≔Y22Line_0.20 Y11Line_0.20 =Y22Line_0.20 −5j

≔Y12Line_0.20 −Y11Line_0.20 =Y12Line_0.20 5j


Combining both admitances in parallel,

≔Y11 +Y11Line_0.25 Y11Line_0.20 ≔Y12 +Y12Line_0.25 Y12Line_0.20


=Y11 −9j =Y12 8.80952j

=Y21 8.80952j =Y22 −8.62812j

＝I1

−ILoad

⎡⎢⎣

⎤⎥⎦

⋅Y11 Y12

Y21 Y22

⎡⎢⎣

⎤⎥⎦

V1

VLoad

⎡⎢⎣

⎤⎥⎦

Solving for V1

＝−ILoad +⋅Y21 V1 ⋅Y22 VLoad

≔V1 ――――――−−ILoad ⋅Y22 VLoad

Y21

=V1(( ∠1.04082 °5.41968 ))




≔ILine_0.20 ――――−V1 VLoad

ZLine_0.20

=ILine_0.20(( ∠0.52373 °−20.1976 ))

≔SLine_0.20_b ⋅VLoad‾‾‾‾‾‾ILine_0.20 =SLine_0.20_b

(( +0.49153 0.18082j))

≔PLine_0.20_b Re ⎛⎝SLine_0.20_b⎞⎠ =PLine_0.20_b 0.49153

______________________/

≔QLine_0.20_b Im ⎛⎝SLine_0.20_b⎞⎠ =QLine_0.20_b 0.18082

______________________/

≔ILine_0.25 ――――−V1 ⋅VLoad c

ZLine_0.25

=ILine_0.25(( ∠0.51666 °−40.44026 ))

≔SLine_0.25_b ⋅VLoad‾‾‾‾‾‾‾‾⋅ILine_0.25 c =SLine_0.25_b

(( +0.3745 0.31918j))


______________________/


______________________/

or in a simpler way

≔SLine_0.25_b −SLoad SLine_0.20_b =SLine_0.25_b +0.3745 0.31918j


______________________/


______________________/





=SLine_0.20_b(( +0.49153 0.18082j)) =SLine_0.20

(( +0.48113 0.27778j))

=SLine_0.25_b(( +0.3745 0.31918j)) =SLine_0.25

(( +0.3849 0.22222j))

The real power flow in line of 0.20j pu increases a little when comparing to the previous case (from 0.48 to 0.29pu). In contrast the reactive power flow reduces significantly from 0.27 pu to 0.18 pu.

The real power flow in line of 0.25j pu changes a little too when comparing to the previous case (from 0.38 to 0.37 pu) whereas the reactive power flow increases from 0.22 to 0.31 pu, that is to say an approximately increase of 41%.

Therefore the regulating transformer with an increased voltage magnitude will affect mostly the reactive power flow.

Determine the real and reactive power supplied to the load bus from each parallel line with (c) the phase-angle-regulating transformer in Example 3.13(b). Assume that the voltage at bus abc is adjusted so that the voltage at bus a'b'c' remains constant at 1.0/ 0 per unit. Also assume positive sequence.


≔c (( ∠1.0 − °3 ))−1

=c (( ∠1 °3 ))

The admitance parameters of the regulating transformer in series with line of 0.25j pu are:

≔Y11Line_0.25 ―――1

ZLine_0.25

=Y11Line_0.25 −4j

≔Y22Line_0.25 ⋅||c||2 Y11Line_0.25 =Y22Line_0.25 −4j

≔Y12Line_0.25 ⋅−c Y11Line_0.25 =Y12Line_0.25(( +−0.20934 3.99452j))

≔Y21Line_0.25 ⋅−c Y11Line_0.25 =Y21Line_0.25(( +−0.20934 3.99452j))




The admitance parameters of the line of 0.20j pu are:

≔Y11Line_0.20 ―――1

ZLine_0.20

=Y11Line_0.20 −5j

≔Y22Line_0.20 Y11Line_0.20 =Y22Line_0.20 −5j



Combining both admitances in parallel,



=Y11 −9j =Y12 +−0.20934 8.99452j

=Y21 +−0.20934 8.99452j =Y22 −9j

＝I1

−ILoad

⎡⎢⎣

⎤⎥⎦

⋅Y11 Y12

Y21 Y22

⎡⎢⎣

⎤⎥⎦

V1

VLoad

⎡⎢⎣

⎤⎥⎦

Solving for V1

＝−ILoad +⋅Y21 V1 ⋅Y22 VLoad

≔V1 ――――――−−ILoad ⋅Y22 VLoad

Y21

=V1(( ∠1.06029 °3.87542 ))




≔ILine_0.20 ――――−V1 VLoad

ZLine_0.20

=ILine_0.20(( ∠0.46054 °−38.92064 ))

≔SLine_0.20_b ⋅VLoad‾‾‾‾‾‾ILine_0.20 =SLine_0.20_b

(( +0.35831 0.28933j))


______________________/


______________________/

≔SLine_0.25_b −SLoad SLine_0.20_b =SLine_0.25_b +0.50771 0.21067j


______________________/≔QLine_0.25_b Im ⎛⎝SLine_0.25_b

⎞⎠ =QLine_0.25_b 0.21067

______________________/


=SLine_0.20_b(( +0.35831 0.28933j)) =SLine_0.20

(( +0.48113 0.27778j))

=SLine_0.25_b(( +0.50771 0.21067j)) =SLine_0.25

(( +0.3849 0.22222j))

The reactive power flow in line of 0.20j pu increases a little when comparing to the previous case (from 0.27 to 0.28pu). In contrast the real power flow reduces significantly from 0.48 pu to 0.35 pu.

The reactive power flow in line of 0.25j pu changes a little too when comparing to the previous case (from 0.22 to 0.21 pu) whereas the real power flow increases from 0.38 to 0.50 pu.

Therefore the regulating transformer with an increased phase angle will affect mostly the real power flow.




Problem 06.


PowerWorld Simulator case Problem 3.60 duplicates Example 3.13 except that a resistance term of 0.06 per unit has been added to the transformer and 0.05 per unit to the transmission line. Since the system is no longer lossless, a field showing the real power losses has also been added to the one-line. With the LTC tap fixed at 1.05, plot the real power losses as the phase shift angle is varied from -10 to +10 degrees. What value of phase shift minimizes the system losses?.

≔θ

− °10

− °8

− °6

− °4

− °2

− °1

− °0.2

°0

°0.2

°1

°2

°4

°6

°8

°10

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

≔PLosses ⋅

24.917

22.002

19.711

18.049

17.016

16.736

16.626

16.614

16.609

16.65

16.843

17.703

19.192

21.309

24.051

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

≔QLosses ⋅

100.288

88.67

79.611

73.123

69.211

68.223

67.897

67.881

67.890

68.184

69.133

72.968

79.378

88.358

99.897

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

The system losses are minimized at near 0 degrees, P=16.609 MW and Q=67.89 MVAR




18.3

19.15

20

20.85

21.7

22.55

23.4

24.25

16.6

17.45

25.1

-6 -4 -2 0 2 4 6 8-10 -8 10

θ (( ))

PLosses(( ))

73.5

76.5

79.5

82.5

85.5

88.5

91.5

94.5

97.5

67.5

70.5

100.5

-6 -4 -2 0 2 4 6 8-10 -8 10

θ (( ))

QLosses(( ))




Problem 07.


Repeat Problem 3.60, except keep the phase-shift angle fixed at 3.0 degrees, while varying the LTC tap between 0.9 and 1.1. What tap value minimizes the real power losses?

≔Tap

0.900

0.925

0.950

0.975

0.9875

1.000

1.0125

1.025

1.050

1.075

1.100

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

≔PLosses ⋅

19.578

18.241

17.320

16.785

16.652

16.604

16.638

16.750

17.194

17.914

18.885

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

≔QLosses ⋅

79.794

74.457

70.814

68.738

68.249

68.106

68.296

68.805

70.728

73.773

77.846

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

The system losses are minimized at TAP=1, P=16.604 MW and Q=68.106 MVAR




The system losses are minimized at near 0 degrees, P=16.609 MW and Q=67.89 MVAR

17.2

17.5

17.8

18.1

18.4

18.7

19

19.3

16.6

16.9

19.6

0.94 0.96 0.98 1 1.02 1.04 1.06 1.08 1.10.9 0.92 1.12

Tap

PLosses(( ))

70

71

72

73

74

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