DocumentD2

PRIMERA PRUEBA CURSO ELECTROSUR-353M

UNIVERSIDAD NACIONAL DE INGENIERIA

FACULTAD DE INGENIERIA ELECTRICA Y ELECTRONICA

UNIVERSIDAD NACIONAL DE INGENIERIA

FACULTAD DE INGENIERIA ELECTRICA Y ELECTRONICANmero de ordenNOTA

PRUEBA N P3 (D2)

CURSO: EE-353M

Entrega: 14/04/2015, 23:55hApellidos:PINARES SARMIENTONombres: FREDY

Cdigo: 20155013CPROBLEMA N 1 (3)

En el sistema 3 efectuar los clculos requeridos. Considerar grupo de conexin del transformador.

BASESIII

MVA100100

kV526.0563

2767.2939.69

A109.75916.43

T1

PosaS nSPos.nRbR

+563.00

+41.0462.40

+361.8

+2521.04+261.20

+11.02511.02+160.60

0501.00060.00

-1490.98-1

-2480.96-2

-3

-4

-5

T1a:1.02a/b:0.98

b:1.04

Calcular en la barra DCalcular en pu en la barra D

ND (MW+jMVAR)210.5 + j61.3ND (Cartesiano)2.105 + j0.6130

ND (Md,Ang)219.24 (( 16.24ND (Polar)2.1924 ( 16.2362

VD (kV, Ang)63.13 ( 0VD (Polar)1.0021 ( 0

ID (A, Ang)2005.04 ( -16.24ID (Polar)2.1879 ( -16.2362

Calcular en pu en la barra C

NC (Cartesiano)2.1052 + j0.7775

NC (Polar)2.2442 ( 20.2715

VC (Polar)1.0052 ( 4.0353

IC (Polar)2.2326 ( -16.2362

Calcular en la barra C

Vab(kV,Ang)528.79 ( 4.04Ia (A, Ang)245.03 ( -46.24

Vca(kV,Ang)528.79 ( -115.96Ib(A, Ang)245.03 ( -166.24

Vbc(kV,Ang)528.79 ( 124.04Ic (A, Ang)245.03 ( 73.76

Va(kV,Ang)305.30 ( -25.96NC (MW+jMVAR)210.52 + j77.75

Vb (kV,Ang)305.30 ( -145.96P(MW)0.02

Vc (kV,Ang)305.30 ( 94.04Q(MVAR)16.45

Calcular en la barra C considerando grupo de conexin del transformador

Vab(kV,Ang)528.79 ( -145.96Ia (A, Ang)528.79 ( -196.24

Vca(kV,Ang)528.79 ( -265.96Ib(A, Ang)528.79 ( -316.24

Vbc(kV,Ang)528.79 ( -25.96Ic (A, Ang)528.79 ( -76.24


Vb (kV,Ang)305.30 ( -295.96P(MW)0.02

Vc (kV,Ang)305.30 ( -55.96Q(MVAR)16.45

PROBLEMA N 2 (7)

Resolver el sistema 3. Considerar el grupo de conexin de los transformadores T1 y T3. (Usar mtodo pu)BASESIIIIII

MVA100100100

kV9.8023495.786960

A5889.95117.16962.25

0.96082428.389136

T1T3

PnSaSPnRbRPnSaSPnRaR

+5

+4

+362.831.03

+2+2+262.221.02

+1121.09+1564.061.02+1511.021.02+161.611.01

0111.000553.001.000501.001.00061.001.00

-110-1-1-1

-2-2-2

-3

-4

-5

T1a:1.09a/b:1.07T3a:1.02a/b:0.99

b:1.02b:1.03



ND (Md,Ang)219.24 ( 16.24ND (Polar)2.1924 ( 16.2362

VD (kV, Ang)63.13 ( 0VD (Polar)1.0522 ( 0

ID (A, Ang)2005.04 ( -16.24ID (Polar)2.0837 ( -16.2362

Calcular en pu en la barra A

NA (Cartesiano)2.1323 + j1.1675

NA (Polar)2.4310 ( 28.701

VA (Polar)1.2359 ( 12.4648

IA (Polar)1.9670 ( -16.2362

Calcular en la barra A sin grupo de conexinCalcular en la barra A con grupo de conexin

NA (MW+jMVAR)231.23 + j116.75NA (MW+jMVAR)231.23 + j116.75

Vab (kV, Ang)12.11 ( 12.46Vab (kV, Ang)12.11 ( -107.54

Vbc (kV, Ang)12.11 ( -107.54Vbc (kV, Ang)12.11 ( -227.54

Vca (kV, Ang)12.11 ( 132.46Vca (kV, Ang)12.11 ( 12.46

Va (kV, Ang)6.99 ( -17.54Va (kV, Ang)6.99 ( -137.54

Vb (kV, Ang)6.99 ( -137.54Vb (kV, Ang)6.99 ( -257.54

Vc (kV, Ang)6.99 ( 102.46Vc (kV, Ang)6.99 ( -17.54

Ia (A, Ang)11585.53 ( -46.24Ia (A, Ang)11585.53 ( -166.24

Ib (A, Ang)11585.53 ( -166.24Ib (A, Ang)11585.53 ( -286.24

Ic (A, Ang)11585.53 ( 73.76Ic (A, Ang)11585.53 ( -46.24

P(MW)2.73P(MW)2.73

Q(MVAR)55.45Q(MVAR)55.45

PROBLEMA N 3 (3)

En el sistema 3 efectuar los clculos requeridos. Considerar grupo de conexin del transformador.

BASESIII

MVA100100

kV10500

12500

A5773.5115.47

T1

PosnSaSPos.nRbR

+2

+1121.2+1564.061.13

0110553.00

-110-1

-2

T1a:1.2a/b:1.0619

b:1.13



ND (Md,Ang)219.24 ( 16.24ND (Polar)2.1924 ( 16.2362

VD (kV, Ang)505.13 ( 0VD (Polar)1.0103 ( 0

ID (A, Ang)250.59 ( -16.24ID (Polar)2.1700 ( -16.2362

Calcular en pu en la barra C

NC (Cartesiano)2.1049 + j0.8635

NC (Polar)2.2751 ( 22.3052

VC (Polar)1.1134 ( 6.069

IC (Polar)2.0434 ( -16.2362

Calcular en la barra C

Vab(kV,Ang)11.13 ( 6.07Ia (A, Ang)11797.57 ( -46.24

Vca(kV,Ang)11.13 ( -113.93Ib(A, Ang)11797.57 ( -166.24

Vbc(kV,Ang)11.13 ( 126.07Ic (A, Ang)11797.57 ( 73.76


Vb (kV,Ang)6.43 ( -143.93P(MW)0

Vc (kV,Ang)6.43 ( 96.07Q(MVAR)25.05


Vab(kV,Ang)11.13 ( -143.93Ia (A, Ang)11797.57 ( -194.46

Vca(kV,Ang)11.13 ( -263.93Ib(A, Ang)11797.57 ( -314.46

Vbc(kV,Ang)11.13 ( -23.93Ic (A, Ang)11797.57 ( -74.46


Vb (kV,Ang)6.43 ( -293.93P(MW)0

Vc (kV,Ang)6.43 ( -53.93Q(MVAR)25.05

PROBLEMA N 4 (7)

Resolver el sistema 3. Considerar el grupo de conexin de los transformadores T1 y T3. (Usar mtodo pu)

BASESIIIIII

MVA100100100

kV1050060

A5773.50115.47962.25

1250036

T1T3

PnSaSPnRbRPnSaSPnRaR

+5

+4

+363.441.06

+2+2523.12+262.83

+1121.2+1562.021.12+1513.061.03+161.61

0110551.000503.00061.00

-110-1-1-1

-2-2-2

-3

-4

-5

T1a:1.20a/b:1.07T3a:1.03a/b:0.97

b:1.12b:1.06



ND (Md,Ang)219.24 ( 16.24ND (Polar)2.1924 ( 16.2362

VD (kV, Ang)63.13 ( 0VD (Polar)1.0522 ( 0

ID (A, Ang)2005.04 ( -16.24ID (Polar)2.0836 ( -16.2362

Calcular en pu en la barra CCalcular en pu en la barra B

NC (Cartesiano)NB (Cartesiano)

NC (Polar)NB (Polar)

VC (Polar)VB (Polar)

IC (Polar)IB (Polar)

Calcular en pu en la barra A

NA (Cartesiano)2.1233 + j1.1829

NA (Polar)2.4306 ( 29.12

VA (Polar)1.2108 ( 12.88

IA (Polar)2.0075 ( -16.24

Calcular en la barra A sin grupo de conexinCalcular en la barra A con grupo de conexin

NA (MW+jMVAR)212.33 + j118.29NA (MW+jMVAR)212.33 + j118.29

Vab (kV, Ang)12.11 ( 12.88Vab (kV, Ang)12.11 ( -107.12

Vbc (kV, Ang)12.11 ( -107.12Vbc (kV, Ang)12.11 ( -227.12

Vca (kV, Ang)12.11 ( 132.88Vca (kV, Ang)12.11 ( 12.88

Va (kV, Ang)6.99 ( -17.12Va (kV, Ang)6.99 ( --137.12

Vb (kV, Ang)6.99 ( -137.20Vb (kV, Ang)6.99 ( -257.12

Vc (kV, Ang)6.99 ( -102.88Vc (kV, Ang)6.99 ( -17.12

Ia (A, Ang)11590.30 ( -46.24Ia (A, Ang)11590.30 ( -166.24

Ib (A, Ang)11590.30 ( -166.24Ib (A, Ang)11590.30 ( -286.24

Ic (A, Ang)11590.30 ( -73.76Ic (A, Ang)11590.30 ( -46.24

P(MW)1.83P(MW)1.83

Q(MVAR)56.99Q(MVAR)56.99

PRUEBA N P3 (D2)

CURSO: EE-353M

Entrega: 14/04/2015, 23:55hApellidos: PINARES SARMIENTO Nombres: FREDY

Cdigo: 20155013CHOJA DE RESULTADOS D2PROBLEMA N 1 (3)


Vbc(kV,Ang)528.79 ( -25.96Ic (A, Ang)141.47 ( -76.24

Vc(kV,Ang)305.30 ( -55.96NC (MW+jMVAR)210.52 + j77.75

Q(MVAR)16.45P(MW)0.02

PROBLEMA N 2 (7)

Calcular en la barra A considerando grupo de conexin del transformador

NA (MW+jMVAR)213.23 + j116.75Va (kV, Ang)6.99 ( -137.54

Vab (kV, Ang)12.11 ( -107.54Vb (kV, Ang)6.99 ( -257.54

Vca (kV, Ang)12.11 ( -12.46Ia (A, Ang)11585.53 ( -166.24

Ic (A, Ang)11585.53 ( -17.54P(MW)/Q(MVAR)2.73 / 55.45

T1: a/b1.07T2: a/b0.99

PROBLEMA N 3 (3)


Vab(kV,Ang)11.13 ( -143.93Ia (A, Ang)11797.57 ( -194.46

Vc(kV,Ang)6.43 ( -53.93NC (MW+jMVAR)210.5 + j86.35

Q(MVAR)25.05P(MW)0

PROBLEMA N 4 (7)

Calcular en la barra A con grupo de conexin del transformador

NA (MW+jMVAR)212.33 + j118.29Va (kV, Ang)6.99 ( -137.12

Vab (kV, Ang)12.11 ( -107.12Vb (kV, Ang)6.99 ( -257.12

Vca (kV, Ang)12.11 ( 12.88Ia (A, Ang)11590.30 ( -166.24

Ic (A, Ang)11590.30 ( -46.24P(MW)/Q(MVAR)1.83 / 56.99

T1: a/b1.07T2: a/b0.97

El Profesor del Curso

VBaseD = 6Z kV

Zona II: barra D

T1: 300MVA, (50Y2x2%)/(6X5x1%)kV, VCC=10.5%, Yd5

Ubicacin actual de los taps:

Lado de AT: (2da. Posicin ms alta)

Lado de BT: (2da. Posicin ms alta)

Para el clculo de bases utilizar las posiciones centrales de T1

ND = 2YX.W+ j6Y.Z MVA

VD = 63.YZ kV

T1

Y

D

C

L: 1W+8Xj

VBaseD = 6X kV

Zona III: barra D

T1: 300MVA, (55Z2x2%)/(10-11-12)kV, VCC=12.5%, Yd1

T3: 300MVA, (50Y2x2%)/(6Y5x1%)kV, VCC=10.5%, Yd5

Ubicacin actual de los taps:

T1: Lado de AT 500kV (2da. Posicin ms alta +1)/Lado de BT 10kV (En 12kV)

T3: Lado de AT 500kV (2da. Posicin ms alta +1)/Lado de BT 60kV (En la posicin +3)

Para el clculo de bases utilizar as posiciones centrales de T1 y T3

A

B

C

D

T1

L


VD = 63.YZ kV

T3

Y

Y

NC

IC

NB

IB

NA

IA

T1

VBII = 50X kV

VBI = 10.X kV


VD = 50W.YZ kV

Ubicacin de los taps:

Lado de AT: 2da. Posicin ms alta +1

Lado de BT: 12kV

Y

C

D

T1: 300MVA, (55Z2x2%)/(10-11-12)kV, VCC=12.5%, Yd5

L: 1X+8Wj

VBIII = 6X kV

VBII = 50X kV

VBI = 10.X kV

T1: 300MVA, (55Y2x2%)/(10-11-12)kV, VCC=12.5%, Yd1

T3: 300MVA, (50Z2x2%)/(6Y5x1%)kV, VCC=10.5%, Yd5

Ubicacin de los taps:

T1: Lado de AT 500kV (2da. Posicin ms alta +1)/Lado de BT (En 12kV)

T3: Lado de AT 500kV (2da. Posicin ms alta +1)/Lado de BT (En la posicin +3)

A

B

C

D

T1

L


VD = 63.YZ kV

T3

Y

Y

NC

IC

NB

IB

NA

IA

Curso: ANALISIS DE SISTEMAS ELECTRICOS DE POTENCIA IProfesor: Moiss Ventosilla Zevallos1

Curso: ANALISIS DE SISTEMAS ELECTRICOS DE POTENCIA IProfesor: Moiss Ventosilla Zevallos9

Documents

DocumentD2