problemas transformadores y PU 2.pdf

1. Datos: ≔ MVAG 90 ≔ kVG 22 ≔ XG 0.18j ≔ MVAT1 50 ≔ kVLT1 22 ≔ kVHT1 220 ≔ XT1 0.10j ≔ MVAT2 40 ≔ kVLT2 11 ≔ kVHT2 220 ≔ XT2 0.06j ≔ MVAT3 40 ≔ kVLT3 22 ≔ kVHT3 110 ≔ XT3 0.064j ≔ MVAT1 50 ≔ kVLT1 22 ≔ kVHT1 220 ≔ XT1 0.10j ≔ MVAT4 40 ≔ kVLT4 11 ≔ kVHT4 110 ≔ XT4 0.08j ≔ MVAM 66.5 ≔ kVM 10.45 ≔ XM 0.185j ≔ FPmotor 0.8 ≔ SignoFPmotor 1

2. ≔ MVAM 66.5 ≔ kVM 10.45 ≔ XM 0.185j ≔ FPmotor 0.8 ≔ SignoFPmotor 1 ≔ MVAcarga 57 ≔ kVcarga 10.45 ≔ FPcarga 0.6 ≔ SignoFPcarga -1 ≔ MVAb 100 ≔ kVbG 22 ≔ kVbL1 220 ≔ kVbL2 110 ≔ kVbM 11 ≔ XL1 48.4j ≔ XL2 64.43j Solución Se convertirán todas las reactancias a la base de 100 MVA ≔ XG ⋅ XG ――― MVAb MVAG = XG 0.2j ≔ XT1 ⋅ XT1 ――― MVAb MVAT1 = XT1 0.2j ≔ XT2 ⋅ XT2 ――― MVAb MVAT2 = XT2 0.15j ≔ XT3 ⋅ XT3 ――― MVAb MVAT3 = XT3 0.16j ≔ XT4 ⋅ XT4 ――― MVAb MVAT4 = XT4 0.2j ≔ VTM ―― kVM kVbM = VTM 0.95

3. ≔ VTM ―― kVM kVbM = VTM 0.95 ≔ θmotor ⋅ SignoFPmotor acos⎛ ⎝FPmotor ⎞ ⎠ = θmotor 36.87 deg ≔ Ibase_motor ―――― ⋅ MVAb 103 ⋅ ‾‾ 3 kVbM = Ibase_motor 5248.639 ≔ Imotor ∠ ―――― ⋅ MVAM 103 ⋅ ‾‾ 3 kVM θmotor = Imotor ∠ 3674.047 ° 36.87 ≔ Imotor_pu ―――― Imotor Ibase_motor = Imotor_pu ∠ 0.7 ° 36.87 ≔ θcarga ⋅ SignoFPcarga acos⎛ ⎝FPcarga ⎞ ⎠ = θcarga -53.13 deg ≔ Icarga ∠ ――――― ⋅ MVAcarga 103 ⋅ ‾‾ 3 kVcarga θcarga = Icarga ∠ 3149.183 - ° 53.13 ≔ Icarga_pu ―――― Icarga Ibase_motor = Icarga_pu ∠ 0.6 - ° 53.13 ≔ Zb_linea1 ――― kVbL1 2 MVAb = Zb_linea1 484 ≔ XL1pu ――― XL1 Zb_linea1 = XL1pu 0.1j ≔ Zb_linea2 ――― kVbL2 2 MVAb = Zb_linea2 121 ≔ XL2pu ――― XL2 Zb_linea2 = XL2pu 0.532j ≔ Xs + + XT1 XL1pu XT2 = Xs 0.45j ≔ Xi + + XT3 XL2pu XT4 = Xi 0.892j ≔ Itotal + Imotor_pu Icarga_pu = Itotal ∠ 0.922 - ° 3.731 ≔ Is ⋅ Itotal ――― Xi + Xs Xi = Is ∠ 0.613 - ° 3.731 ≔ VN1 + VTM ⋅ ⎛ ⎝Is ⎞ ⎠ ⎛ ⎝ + + XT1 XL1pu XT2 ⎞ ⎠ = VN1 ∠ 1.006 ° 15.873

4. ≔ VN1 + VTM ⋅ ⎛ ⎝Is ⎞ ⎠ ⎛ ⎝ + + XT1 XL1pu XT2 ⎞ ⎠ = VN1 ∠ 1.006 ° 15.873 ≔ kVN1 ⋅ kVbG VN1 = kVN1 ∠ 22.139 ° 15.873 ≔ EM - VTM ⋅ Imotor_pu XM = EM ∠ 1.033 - ° 5.756 ≔ kVMotor ⋅ kVbM EM = kVMotor ∠ 11.362 - ° 5.756 ≔ EG + VN1 ⎛ ⎝ ⋅ Itotal XG ⎞ ⎠ = EG ∠ 1.082 ° 25.109 ≔ kVgenerador ⋅ kVbG EG = kVgenerador ∠ 23.809 ° 25.109 Datos: ≔ MVA1ϕ 9 ≔ kVHT1ϕ 7.2 ≔ kVLT1ϕ 4.16 ≔ ZTH + 0.12 0.82j ≔ MVAcarga 18 ≔ FPcarga 0.8 ≔ SignoFPcarga -1 ≔ kVcarga 4.16 Solución ≔ MVAb 27 ≔ kVHT ⋅ ‾‾ 3 kVHT1ϕ = kVHT 12.471 ≔ ZT_pu ⋅ ZTH ――― MVAb kVHT 2 = ZT_pu + 0.021 0.142j

5. ≔ θcarga ⋅ SignoFPcarga acos⎛ ⎝FPcarga ⎞ ⎠ = θcarga -36.87 deg ≔ Icarga_pu ∠ ―――― MVAcarga MVAb θcarga = Icarga_pu - 0.533 0.4j = Icarga_pu ∠ 0.667 - ° 36.87 Forma alterna de la corriente de carga ≔ Icarga ∠ ――――― ⋅ MVAcarga 103 ⋅ ‾‾ 3 kVcarga θcarga = Icarga - 1998.52 1498.89j = Icarga ∠ 2498.15 - ° 36.87 ≔ Ibase ―――― ⋅ MVAb 103 ⋅ ‾‾ 3 kVcarga = Ibase 3747.225 ≔ Icarga_pu ―― Icarga Ibase = Icarga_pu - 0.533 0.4j = Icarga_pu ∠ 0.667 - ° 36.87 ≔ Vcarga_pu ――― kVcarga 4.16 = Vcarga_pu 1 ≔ VH + Vcarga_pu ⋅ Icarga_pu ZT_pu = VH + 1.068 0.068j = VH ∠ 1.07 ° 3.621 ≔ kVH ⋅ kVHT VH = kVH + 13.319 0.843j = kVH ∠ 13.346 ° 3.621 2.10

6. 2.10 El esquema unifilar de un sistema de energ��a ele�ctrica trifa�sico se muestra en la Figura 2.16. Las impedancias esta�n indicadas en tanto por unidad sobre una base de 100 MVA y 400 kV. La carga en el nudo 2 es S2 = 15.93 MW − j33.4 MVAR y en el nodo 3 es S3 = 77 MW + 14j MVAR. Se quiere mantener la tensio�n en el Nodo 3 en 400 kV. Trabajando en el sistema por unidad, determine la tensio�n en los nodos 2 y 1. Datos: ≔ MVAb 100 ≔ kVb 400 ≔ S2 - 15.93 33.4j ≔ S3 + 77 14j ≔ X12 0.5j ≔ X23 0.4j ≔ I2_pu ――― S2 MVAb = I2_pu - 0.159 0.334j = I2_pu ∠ 0.37 - ° 64.501 ≔ I3_pu ――― S3 MVAb = I3_pu + 0.77 0.14j = I3_pu ∠ 0.783 ° 10.305 ≔ kV3 400 Solución ≔ V3pu ―― kV3 kVb = V3pu 1 ≔ V2pu + V3pu ⋅ I3_pu X23 = V2pu + 0.944 0.308j = V2pu ∠ 0.993 ° 18.07 ≔ kV2 ⋅ kVb V2pu = kV2 ∠ 397.19 ° 18.07 ≔ V1pu + V2pu ⋅ ⎛ ⎝ + I2_pu I3_pu ⎞ ⎠ X12 = V1pu ∠ 1.296 ° 36.584 ≔ kV1pu = ⋅ kVb V1pu ∠ 518.562 ° 36.584

7. Datos ≔ kVLT1 23 ≔ kVHT1 230 ≔ MVAT1 150 ≔ XT1 0.1j ≔ kVLT2 23 ≔ kVHT2 230 ≔ MVAT2 150 ≔ XT2 0.1j ≔ kVLT3 69 ≔ kVHT3 230 ≔ MVAT3 150 ≔ XT3 0.1j ≔ S2 + 150 60j ≔ S4 + 120 60j ≔ Xlinea 60j ≔ kV4 69 Solución ≔ kVb1 23 ≔ kVb2 230 ≔ MVAb 150 ≔ S2pu ――― S2 MVAb = S2pu + 1 0.4j ≔ S4pu ――― S4 MVAb = S4pu + 0.8 0.4j

8. ≔ S4pu ――― S4 MVAb = S4pu + 0.8 0.4j ≔ kVN4_deseado 69 ≔ VN4_deseado_pu ―――― kVN4_deseado kVLT3 = VN4_deseado_pu 1 ≔ Xlinea_pu = ⋅ Xlinea ――― MVAb kVb2 2 0.17j ≔ I2_4 ‾‾‾‾‾‾‾‾‾‾‾‾ ⎛ ⎜ ⎝ ――――― S4pu VN4_deseado_pu ⎞ ⎟ ⎠ = I2_4 - 0.8 0.4j = I2_4 ∠ 0.894 - ° 26.565 ≔ VN2 + VN4_deseado_pu ⋅ ⎛ ⎝I2_4 ⎞ ⎠ ⎛ ⎝ + Xlinea_pu XT3 ⎞ ⎠ = VN2 + 1.108 0.216j = VN2 ∠ 1.129 ° 11.036 ≔ I2n ‾‾‾‾‾ ⎛ ⎜ ⎝ ―― S2pu VN2 ⎞ ⎟ ⎠ = I2n - 0.937 0.178j = I2n ∠ 0.954 - ° 10.765 ≔ IT1 + I2_4 I2n = IT1 - 1.737 0.578j = IT1 ∠ 1.831 - ° 18.409 ≔ Xeq12 ―――― ⋅ XT1 XT2 + XT1 XT2 = Xeq12 0.05j ≔ VN1 + VN2 ⋅ ⎛ ⎝IT1 ⎞ ⎠ ⎛ ⎝Xeq12 ⎞ ⎠ = VN1 + 1.137 0.303j = VN1 ∠ 1.177 ° 14.921 ≔ kVN1 ⋅ kVb1 VN1 = kVN1 + 26.15 6.968j = kVN1 ∠ 27.063 ° 14.921 ≔ Z2 ―― 2302 S2 = Z2 - 304.023 121.609j ≔ Zbase ―― 2302 100 = Zbase 529 ≔ Z2pu ―― Z2 Zbase = Z2pu - 0.575 0.23j

9. ≔ V ⋅ 69 103 ≔ I4 ‾‾‾‾‾‾‾‾ ⎛ ⎜ ⎜ ⎝ ――― ⋅ S4 103 ⋅ ‾‾ 3 69 ⎞ ⎟ ⎟ ⎠ = I4 - 1004.087 502.044j ≔ Ibase4 ――― ⋅ 150 103 ⋅ ‾‾ 3 69 = Ibase4 1255.109 ≔ I4pu ―― I4 Ibase4 = I4pu - 0.8 0.4j

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