ECE421HW03_ABM.pdf

Session 17Fall 2013ECE421:

Introduction to Power SystemsHomework # 03

Due Session 17 (October 4th)Arturo Barradas Munoz of 17

Homework #03

1. Problem 3.9 in Glover, Sarma and Overbye




5. Problem 3.42 in Glover, Sarma and Ovebye

General Definitions

≔ ≔ ≔ ≔

≔ ≔ ≔a ∠1 °120 ≔ 1 ≔Turns 1




Problem 01.


A single-phase transformer has 2000 turns on the primary winding and 500 turns on the secondary. Winding resistances are R1=2 and R2= 0.125 ; leakage reactances are X1=8 and X2=0.5 . The resistance load on the secondary is 12 .

(a) If the applied voltage at the terminals of the primary is 1000 V, determine V2 at the load terminals of the transformer, neglecting magnetizing current.

(b) If the voltage regulation is defined as the difference between the voltage magnitude at the load terminals of the transformer at full load and at no load in percent of full-load voltage with input voltage held constant, compute the percent voltage regulation.

≔TurnsHV_p1 2000 Turns ≔TurnsLV_p1 500 Turns ≔TRp1 ――――TurnsHV_p1

TurnsLV_p1

≔R1_HV_p1 2 ≔R2_LV_p1 0.125

≔X1_HV_p1 8j ≔X2_LV_p1 0.5j ≔RL_LV_p1 12.0

≔V1_HV_p1(( ∠1000 °0 )) ≔V1_LV_p1 ―――

V1_HV_p1

TRp1

=V1_LV_p1(( ∠250 °0 ))

≔Z1_HV_p1 +R1_HV_p1 X1_HV_p1 =Z1_HV_p1(( +2 8j))

≔Z1_LV_p1 ―――Z1_HV_p1

TRp12

=Z1_LV_p1(( +0.125 0.5j))

≔Z2_LV_p1 +R2_LV_p1 X2_LV_p1 =Z2_LV_p1(( +0.125 0.5j))




(a) If the applied voltage at the terminals of the primary is 1000 V, determine V2 at the load terminals of the transformer, neglecting magnetizing current.

≔ILV_p1 ―――――――――V1_LV_p1

++Z1_LV_p1 Z2_LV_p1 RL_LV_p1

=ILV_p1(( ∠20.3405 °−4.66686 ))

=⋅ILV_p1 RL_LV_p1(( ∠244.08603 °−4.66686 ))

≔V2_LV_p1 −V1_LV_p1 ⋅ILV_p1⎛⎝ +Z1_LV_p1 Z2_LV_p1

⎞⎠ =V2_LV_p1(( ∠244.08603 °−4.66686 ))

____________________________/




(b) If the voltage regulation is defined as the difference between the voltage magnitude at the load terminals of the transformer at full load and at no load in percent of full-load voltage with input voltage held constant, compute the percent voltage regulation.

⋅――――−VNL VFL

VFL

%100

=―――――――−||V1_LV_p1

|| ||V2_LV_p1||

||V2_LV_p1||

%2.4229




Problem 02.


A single-phase 50-kVA, 2400/240-volt, 60-Hz distribution transformer is used as a step-down transformer at the load end of a 2400-volt feeder whose series impedance is (1.0 + j 2.0) ohms. The equivalent series impedance of the transformer is (1.0 + j2.5) ohms referred to the high-voltage (primary) side. The transformer is delivering rated load at 0.8 power factor lagging and at rated secondary voltage. Neglecting the transformer exciting current, determine:(a) the voltage at the transformer primary terminals, (b) the voltage at the sending end of the feeder, and (c) the real and reactive power delivered to the sending end of the feeder.

≔ST_mag_p2 50 ≔VT_HV_p2 2400 ≔VT_LV_p2 240

≔ZF_p2(( +1 2j)) ≔ZT_HV_p2

(( +1 2.5j)) ≔ZL_p2 50

≔fp2 60 ≔pfL_p2 0.8 Lagging

≔TRT_p2 ―――VT_HV_p2

VT_LV_p2

=TRT_p2 10

≔VL_LV_p2(( ∠240 °0 )) ≔SL_p2 ∠ST_mag_p2 acos ⎛⎝pfL_p2

⎞⎠

=SL_p2(( +40 30j))

≔IL_LV_p2 ―――‾‾‾‾SL_p2

‾‾‾‾‾‾‾VL_LV_p2

=IL_LV_p2(( −166.66667 125j))

=IL_LV_p2(( ∠208.33333 °−36.8699 ))

≔IL_HV_p2 ―――IL_LV_p2

TRT_p2

=IL_HV_p2(( −16.66667 12.5j))

=IL_HV_p2(( ∠20.83333 °−36.8699 ))




≔ZLoad_p2 =―――VT_LV_p2

IL_LV_p2

(( +0.9216 0.6912j))

≔VDrop_F_p2 ⋅IL_HV_p2 ZF_p2 =VDrop_F_p2(( ∠46.58475 °26.56505 ))

≔VDrop_T_p2 ⋅IL_HV_p2 ZT_HV_p2 =VDrop_T_p2(( ∠56.09547 °31.32869 ))

(a) the voltage at the transformer primary terminals,

≔VTer_HV_p2 +VT_HV_p2 VDrop_T_p2 =VTer_HV_p2(( ∠2448.09042 °0.68264 ))

__________________________________/

(b) the voltage at the sending end of the feeder, and

≔VSrc_HV_p2 +VTer_HV_p2 VDrop_F_p2 =VSrc_HV_p2(( ∠2490.08537 °1.15056 ))

__________________________________/

(c) the real and reactive power delivered to the sending end of the feeder.

≔SSrc_p2 ⋅VSrc_HV_p2‾‾‾‾‾‾IL_HV_p2 =SSrc_p2

(( +40.86806 31.95313i))

≔PSrc_p2 Re ⎛⎝SSrc_p2⎞⎠ =PSrc_p2 40.86806

_______________________/

≔QSrc_p2 Im ⎛⎝SSrc_p2⎞⎠ =QSrc_p2 31.95313

_______________________/




Problem 03.


Using the transformer ratings as base quantities, work Problem 3.14 in per-unit.

≔SB_p3 ST_mag_p2 =SB_p3 50

≔VB_HV_p3 VT_HV_p2 =VB_HV_p3 2400

≔VB_LV_p3 ―――VB_HV_p3

TRT_p2

=VB_LV_p3 240

≔ZB_HV_p3 ――――VB_HV_p3

2

SB_p3

=ZB_HV_p3 115.2

≔ZB_LV_p3 ――――VB_LV_p3

2

SB_p3

=ZB_LV_p3 1.152

≔IB_LV_p3 ―――SB_p3

VB_LV_p3

=IB_LV_p3 208.33333

≔IB_HV_p3 ―――SB_p3

VB_HV_p3

=IB_HV_p3 20.83333

≔VL_p3_pu ―――VL_LV_p2

VB_LV_p3

=VL_p3_pu(( ∠1 °0 ))

≔SL_p3_pu ――SL_p2

SB_p3

=SL_p3_pu(( ∠1 °36.8699 ))

≔IL_p3_pu ―――IL_LV_p2

IB_LV_p3

=IL_p3_pu(( ∠1 °−36.8699 ))

≔ZT_p3_pu ―――ZT_HV_p2

ZB_HV_p3

=ZT_p3_pu(( +0.00868 0.0217j))

≔ZF_p3_pu ―――ZF_p2

ZB_HV_p3

=ZF_p3_pu(( +0.00868 0.01736j))




(a) the voltage at the transformer primary terminals,

≔VTer_HV_p3_pu +VL_p3_pu ⋅IL_p3_pu ZT_p3_pu =VTer_HV_p3_pu ∠1.02004 °0.68264

≔VTer_HV_p3 ⋅VTer_HV_p3_pu VB_HV_p3 =VTer_HV_p3(( ∠2448.09042 °0.68264 ))

__________________________________/=VTer_HV_p2

(( ∠2448.09042 °0.68264 ))

(b) the voltage at the sending end of the feeder, and

≔VSrc_HV_p3_pu +VL_p3_pu ⋅IL_p3_pu⎛⎝ +ZT_p3_pu ZF_p3_pu

⎞⎠ =VSrc_HV_p3_pu ∠1.03754 °1.15056

≔VSrc_HV_p3 ⋅VSrc_HV_p3_pu VB_HV_p3 =VSrc_HV_p3(( ∠2490.08537 °1.15056 ))

__________________________________/=VSrc_HV_p2

(( ∠2490.08537 °1.15056 ))

(c) the real and reactive power delivered to the sending end of the feeder.

≔SSrc_p3_pu ⋅VSrc_HV_p3_pu‾‾‾‾‾‾IL_p3_pu =SSrc_p3_pu

(( +0.81736 0.63906i))

≔SSrc_p3 ⋅SSrc_p3_pu SB_p3 =SSrc_p3(( +40.86806 31.95313i))

≔PSrc_p3 Re ⎛⎝SSrc_p3⎞⎠ =PSrc_p3 40.86806

______________________/=PSrc_p2 40.86806

≔QSrc_p3 Im ⎛⎝SSrc_p2⎞⎠ =QSrc_p3 31.95313

_______________________/=QSrc_p2 31.95313




Problem 04.


Consider the single-line diagram of the power system shown in Figure 3.38. Equipmentratings are:

Generator 1: 1000 MVA, 18 kV, X"= 0.2 per unitGenerator 2: 1000 MVA, 18 kV, X"=0.2Synchronous motor 3: 1500 MVA, 20 kV, X"=0.2Three-phase –Y transformers 1000 MVA, 500 kV Y/20 kV , X=0.1T1, T2, T3, T4:Three-phase Y–Y transformer T5: 1500 MVA, 500 kV Y/20 kV Y, X=0.1

Neglecting resistance, transformer phase shift, and magnetizing reactance, (b)draw the equivalent reactance diagram. Use a base of 100 MVA and 500 kV for the 50-ohm line. (a) Determine the per-unit reactances.

≔SG1_p4 1000 ≔VG1_p4 18 ≔X''G1_rat_p4 0.2j

≔SG2_p4 1000 ≔VG2_p4 18 ≔X''G2_rat_p4 0.2j

≔SSM3_p4 1500 ≔VSM3_p4 20 ≔X''SM3_rat_p4 0.2j

≔ST1_p4 1000 ≔VT1_HV_p4 500 ≔XT1_rat_p4 0.1j

≔VT1_LV_p4 20





≔VT2_LV_p4 20


≔VT3_LV_p4 20


≔VT4_LV_p4 20


≔VT5_LV_p4 20

≔XL12_rat_p4 ⋅50j ≔XL13_rat_p4 ⋅25j ≔XL23_rat_p4 ⋅25j

≔SB_p4 100 ≔VB_HV_p4 500

≔ZB_HV_p4 ――――VB_HV_p4

2

SB_p4

=ZB_HV_p4 2500

≔VB_LV_p4 20

≔ZB_LV_p4 ――――VB_LV_p4

2

SB_p4

=ZB_LV_p4 4

＝⋅Zpu_old ZB_old Zohm ＝⋅Zpu_new ZB_new Zohm

＝⋅Zpu_old ―――ZB_old

ZB_new

Zpu_new ＝⋅⋅Zpu_old

⎛⎜⎝―――VB_old

VB_new

⎞⎟⎠

2

―――SB_new

SB_old

Zpu_new




(b) Determine the per-unit reactances.

≔X''G1_p4 ⋅X''G1_rat_p4

⎛⎜⎝―――

VG1_p4

VB_LV_p4

⎞⎟⎠

2 ⎛⎜⎝―――SB_p4

SG1_p4

⎞⎟⎠

=X''G1_p4 0.0162j

_____________________/

≔X''G2_p4 ⋅X''G2_rat_p4

⎛⎜⎝―――

VG2_p4

VB_LV_p4

⎞⎟⎠

2 ⎛⎜⎝―――SB_p4

SG2_p4

⎞⎟⎠

=X''G2_p4 0.0162j

_____________________/

≔X''SM3_p4 ⋅X''SM3_rat_p4

⎛⎜⎝―――VSM3_p4

VB_LV_p4

⎞⎟⎠

2 ⎛⎜⎝―――

SB_p4

SSM3_p4

⎞⎟⎠

=X''SM3_p4 0.01333j

_____________________/

≔XT1_p4 ⋅XT1_rat_p4

⎛⎜⎝――――VT1_HV_p4

VB_HV_p4

⎞⎟⎠

2 ⎛⎜⎝――SB_p4

ST1_p4

⎞⎟⎠

=XT1_p4 0.01j

_____________________/


⎛⎜⎝――――VT2_HV_p4

VB_HV_p4

⎞⎟⎠

2 ⎛⎜⎝――SB_p4

ST2_p4

⎞⎟⎠

=XT2_p4 0.01j

_____________________/


⎛⎜⎝――――VT3_HV_p4

VB_HV_p4

⎞⎟⎠

2 ⎛⎜⎝――SB_p4

ST3_p4

⎞⎟⎠

=XT3_p4 0.01j

_____________________/


⎛⎜⎝――――VT4_HV_p4

VB_HV_p4

⎞⎟⎠

2 ⎛⎜⎝――SB_p4

ST4_p4

⎞⎟⎠

=XT4_p4 0.01j

_____________________/


⎛⎜⎝――――VT5_HV_p4

VB_HV_p4

⎞⎟⎠

2 ⎛⎜⎝――SB_p4

ST5_p4

⎞⎟⎠

=XT5_p4 0.00667j

_____________________/

≔XL12_p4 ――――XL12_rat_p4

ZB_HV_p4

=XL12_p4 0.02j

_____________________/

≔XL14_p4 ――――XL13_rat_p4

ZB_HV_p4

=XL14_p4 0.01j

_____________________/

≔XL24_p4 ――――XL23_rat_p4

ZB_HV_p4

=XL24_p4 0.01j

_____________________/




(b)draw the equivalent reactance diagram.




Problem 05.


For the power system in Problem 3.41, the synchronous motor absorbs 1500 MW at 0.8 power factor leading with the bus 3 voltage at 18 kV. Determine the bus 1 and bus 2 voltages in kV. Assume that generators 1 and 2 deliver equal real powers and equal reactive powers. Also assume a balanced three-phase system with positive-sequence sources.

≔SSM3_op_p5(( ∠1500 −acos ((0.8)))) =SSM3_op_p5

(( −1200 900j))

≔SSM3_p5 ――――SSM3_op_p5

SB_p4

=SSM3_p5(( ∠15 °−36.8699 ))

≔VB3_op_p5 18 ≔VB3_p5 ―――VB3_op_p5

VB_LV_p4

=VB3_p5 0.9

≔ISM3_p5 ――――‾‾‾‾‾‾SSM3_p5

⋅‾‾3 ‾‾‾‾‾VB3_p5

=ISM3_p5(( ∠9.6225 °36.8699 ))

≔ZSM3_p5 ―――VB3_p5

ISM3_p5

=ZSM3_p5(( −0.07482 0.05612j))




1.0 Network reduction at HV side

≔XB12_p5 ++XT1_p4 XL12_p4 XT2_p4

=XB12_p5 0.04j

≔XB14_p5 +XT3_p4 XL14_p4

=XB14_p5 0.02j

≔XB24_p5 +XT4_p4 XL24_p4

=XB24_p5 0.02j

2.0 Delta-Star Conversion of the HV side

≔XB1_p5 ―――――――――⋅XB12_p5 XB14_p5

++XB12_p5 XB14_p5 XB24_p5

=XB1_p5 0.01j

≔XB2_p5 ―――――――――⋅XB12_p5 XB24_p5

++XB12_p5 XB14_p5 XB24_p5

=XB2_p5 0.01j

≔XB4_p5 ―――――――――⋅XB14_p5 XB24_p5

++XB12_p5 XB14_p5 XB24_p5

=XB4_p5 0.005j




3.0 Second Network reduction

≔XLeft_p5 +X''G1_p4 XB1_p5

=XLeft_p5 0.0262j

≔XRight_p5 +X''G2_p4 XB2_p5

=XRight_p5 0.0262j

≔ZBottom_p5 ++XB4_p5 XT5_p4 ZSM3_p5

=ZBottom_p5(( −0.07482 0.04445j))

≔ZThev_p5

⎛⎜⎝

++―――1

XLeft_p5

―――1

XRight_p5

――――1

ZBottom_p5

⎞⎟⎠

−1

=ZThev_p5(( +0.00195 0.01392i))

4.0 Voltage on Bus "4" at T5 HV Side

≔VT5_HV_p5 +VB3_p5 ⋅ISM3_p5 XT5_p4 =VT5_HV_p5 ∠0.86304 °3.40907

5.0 Voltage on Center Bus at the Star Equivalent Circuit

≔VCB_p5 +VB3_p5 ⋅ISM3_p5⎛⎝ +XT5_p4 XB4_p5

⎞⎠ =VCB_p5 ∠0.83747 °6.15621

≔IBottom_p5 ――――VCB_p5

ZBottom_p5

≔ILeft_p5 ―――VCB_p5

XLeft_p5

=IBottom_p5(( ∠9.6225 °36.8699 )) =ILeft_p5

(( ∠31.96458 °−83.84379 ))

≔IRight_p5 ―――VCB_p5

XRight_p5

≔IThev_p5 ―――VCB_p5

ZThev_p5

=IRight_p5(( ∠31.96458 °−83.84379 )) =IThev_p5

(( ∠59.5915 °−75.86397 ))




≔ISrc1_p5 +ILeft_p5 ―――IBottom_p5

2≔ISrc2_p5 +IRight_p5 ―――

IBottom_p5

2

≔VSrc1_p5 ⋅ISrc1_p5 XLeft_p5 ≔VSrc2_p5 ⋅ISrc2_p5 XRight_p5

≔ISrc1_p5 ――――――−VSrc1_p5 VCB_p5

XLeft_p5

≔ISrc2_p5 ――――――−VSrc2_p5 VCB_p5

XRight_p5

≔SG1 ⋅VSrc1_p5‾‾‾‾‾‾ISrc1_p5 ≔SG2 ⋅VSrc2_p5

‾‾‾‾‾‾ISrc2_p5

=SG1 ∠3.7559 − °22.73386 =SG2 ∠3.7559 − °22.73386

≔VB1_p5 −VSrc1_p5 ⋅ISrc1_p5 X''G1_p4 ≔VB2_p5 −VSrc2_p5 ⋅ISrc2_p5 X''G2_p4

≔VB1_p5_kV ⋅VB1_p5 VB_LV_p4 ≔VB2_p5_kV ⋅VB2_p5 VB_LV_p4

=ISM3_p5(( ∠9.6225 °36.8699 ))

=VB3_p5(( ∠0.9 °0 ))

=VSrc1_p5(( ∠0.7806 °14.136 ))

=ISrc1_p5(( ∠4.8113 °36.8699 ))

=VSrc2_p5(( ∠0.7806 °14.136 ))

=ISrc2_p5(( ∠4.8113 °36.8699 ))

=VB1_p5(( ∠0.814 °9.0692 ))

=VB2_p5(( ∠0.814 °9.0692 ))

Determine the bus 1 and bus 2 voltages in kV.

=VB1_p5_kV(( ∠16.279 °9.069 )) =VB2_p5_kV

(( ∠16.279 °9.069 ))_____________________________/ _____________________________/

ECE421HW03_ABM.pdf

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