tarea electricas

Ejemplo de calculo de caídas de tensión:

 Linea de transporte trifásica a 50 Hz de 20km, con conductores de cobre de

50mm2 y diámetro efectivo 9,2mm colocados en triángulo equilátero de 1m.

 Suministra 5000kw con fdp 0.8 inductivo a 20kV.  Calcular el porcentaje de caída de tensión y el rendimiento de la línea. VL ≔ 20

long ≔ 20

d ≔ 9.2

P ≔ 5000

fdp ≔ 0.8

ρCu ≔ 0.017

s ≔ 50

2

D≔1

2

⋅ ――

f ≔ 50

ϕ ≔ acos ((fdp)) = 36.87 ρCu r ≔ ―― = 0.34 ―― s

R ≔ r ⋅ long = 6.8

⎛D⎞ l ≔ ―― + ―― ⋅ ln ⎜―⎟ = 0.001 ―― L ≔ l ⋅ long = 0.023 8⋅ 2⋅ d⎟ ⎜― ⎝⎜ 2 ⎠⎟ P I ≔ ――――― = 180.422 ‾‾ 3 ⋅ VL ⋅ cos ((ϕ)) ΔU ≔ ‾‾ 3 ⋅ ((R ⋅ I ⋅ cos ((ϕ)) + X ⋅ I ⋅ sin ((ϕ)))) = ⎛⎝3.027 ⋅ 10 3 ⎞⎠

X ≔ 2 ⋅ ⋅ f ⋅ L = 7.077

ΔU ⋅ 100 = 15.135 ―― VL

P η ≔ ――――――――⋅ 100 = 86.855 ⎛ ⎞ 3 ⋅ I ⋅ ΔU ⋅ cos ((ϕ))⎠ ⎝P + ‾‾ P Q ≔ ――― ⋅ sin ((ϕ)) = ⎛⎝3.75 ⋅ 10 6 ⎞⎠ cos ((ϕ))

⋅

1 ΔU ≔ ―― ⋅ ((R ⋅ P + X ⋅ Q)) = ⎛⎝3.027 ⋅ 10 3 ⎞⎠ VL P η ≔ ――――――――⋅ 100 = 86.855 ⎛ ⎞ 3 ⋅ I ⋅ ΔU ⋅ cos ((ϕ))⎠ ⎝P + ‾‾

ΔU ⋅ 100 = 15.135 ―― VL

Ejemplo de calculo de caídas de tensión:

 Linea subterranea de distribución trifásica a 50 Hz, con cables unipolares de Al de 70mm2.

 Suministra 4000kw con fdp 0.8 inductivo a 20kV.  Calcular la máxima longitud de línea para que el porcentaje de caída de tensión sea del 1%.

VL ≔ 20

f ≔ 50

ϕ ≔ acos ((fdp)) = 36.87

P ≔ 4000 r ≔ 0.57 ――

fdp ≔ 0.8

s ≔ 70

2

x ≔ 0.13 ――

P I ≔ ――――― = 144.338 ‾‾ 3 ⋅ VL ⋅ cos ((ϕ)) VL ΔUkm ≔ ‾‾ 3 ⋅ ((r ⋅ I ⋅ cos ((ϕ)) + x ⋅ I ⋅ sin ((ϕ)))) = 133.5 ―― l ≔ 0.01 ⋅ ―― = 1.498 ΔUkm

P Q ≔ ――― ⋅ sin ((ϕ)) = ⎛⎝3 ⋅ 10 6 ⎞⎠ ⋅ cos ((ϕ)) 1 ΔUkm ≔ ―― ⋅ ((r ⋅ P + x ⋅ Q)) = 133.5 ―― VL

VL l ≔ 0.01 ⋅ ―― = 1.498 ΔUkm

Se considera una línea con las derivaciones y cargas correspondientes:

VL ≔ 10

fdp ≔ 0.8 r ≔ 1.96 ―― x ≔ 0.391 ――

ia ≔ 10

ib ≔ 5

lAe ≔ 2

lec ≔ 3

ic ≔ 15

id ≔ 5

lcd ≔ 2

ie ≔ 15 ldb ≔ 2

Calculamos la caida de tensión entre A y a por tramos: IAe ≔ ie + ic + id + ib + ia = 50 ΔUAe ≔ ‾‾ 3 ⋅ lAe ⋅ ⎛⎝r ⋅ IAe ⋅ cos ((ϕ)) + x ⋅ IAe ⋅ sin ((ϕ))⎞⎠ = 312.219 Iec ≔ ic + id + ib + ia = 35 ΔUec ≔ ‾‾ 3 ⋅ lec ⋅ ⎛⎝r ⋅ Iec ⋅ cos ((ϕ)) + x ⋅ Iec ⋅ sin ((ϕ))⎞⎠ = 327.83 Ica ≔ ia = 10 ΔUca ≔ ‾‾ 3 ⋅ lca ⋅ ⎛⎝r ⋅ Ica ⋅ cos ((ϕ)) + x ⋅ Ica ⋅ sin ((ϕ))⎞⎠ = 187.332 ΔUAa ≔ ΔUAe + ΔUec + ΔUca = 827.382

ϕ ≔ acos ((fdp)) = 36.87

lca ≔ 6

Calculamos la caida de tensión entre A y a por nodos: ΔUe ≔ ‾‾ 3 ⋅ lAe ⋅ ⎛⎝r ⋅ ie ⋅ cos ((ϕ)) + x ⋅ ie ⋅ sin ((ϕ))⎞⎠ = 93.666 lAe = 2 lAc ≔ lAe + lec = 5

ΔUc ≔ ‾‾ 3 ⋅ lAc ⋅ ⎛⎝r ⋅ ic ⋅ cos ((ϕ)) + x ⋅ ic ⋅ sin ((ϕ))⎞⎠ = 234.165

lAc ≔ lAe + lec = 5

ΔUd ≔ ‾‾ 3 ⋅ lAc ⋅ ⎛⎝r ⋅ id ⋅ cos ((ϕ)) + x ⋅ id ⋅ sin ((ϕ))⎞⎠ = 78.055

lAc ≔ lAe + lec = 5

ΔUb ≔ ‾‾ 3 ⋅ lAc ⋅ ⎛⎝r ⋅ ib ⋅ cos ((ϕ)) + x ⋅ ib ⋅ sin ((ϕ))⎞⎠ = 78.055

lAa ≔ lAe + lec + lca = 11

ΔUa ≔ ‾‾ 3 ⋅ lAa ⋅ ⎛⎝r ⋅ ia ⋅ cos ((ϕ)) + x ⋅ ia ⋅ sin ((ϕ))⎞⎠ = 343.441

ΔUAa ≔ ΔUe + ΔUc + ΔUd + ΔUb + ΔUa = 827.382 Calculamos la caida de tensión entre A y b por tramos: IAe ≔ ie + ic + id + ib + ia = 50

ΔUAe ≔ ‾‾ 3 ⋅ lAe ⋅ ⎛⎝r ⋅ IAe ⋅ cos ((ϕ)) + x ⋅ IAe ⋅ sin ((ϕ))⎞⎠ = 312.219

Iec ≔ ic + id + ib + ia = 35

ΔUec ≔ ‾‾ 3 ⋅ lec ⋅ ⎛⎝r ⋅ Iec ⋅ cos ((ϕ)) + x ⋅ Iec ⋅ sin ((ϕ))⎞⎠ = 327.83

Icd ≔ id + ib = 10

ΔUcd ≔ ‾‾ 3 ⋅ lcd ⋅ ⎛⎝r ⋅ Icd ⋅ cos ((ϕ)) + x ⋅ Icd ⋅ sin ((ϕ))⎞⎠ = 62.444

Idb ≔ ib = 5

ΔUdb ≔ ‾‾ 3 ⋅ ldb ⋅ ⎛⎝r ⋅ Idb ⋅ cos ((ϕ)) + x ⋅ Idb ⋅ sin ((ϕ))⎞⎠ = 31.222

ΔUAa ≔ ΔUAe + ΔUec + ΔUcd + ΔUdb = 733.716 Calculamos la caida de tensión entre A y b por nodos: lAe = 2

ΔUe ≔ ‾‾ 3 ⋅ lAe ⋅ ⎛⎝r ⋅ ie ⋅ cos ((ϕ)) + x ⋅ ie ⋅ sin ((ϕ))⎞⎠ = 93.666

lAc ≔ lAe + lec = 5

ΔUc ≔ ‾‾ 3 ⋅ lAc ⋅ ⎛⎝r ⋅ ic ⋅ cos ((ϕ)) + x ⋅ ic ⋅ sin ((ϕ))⎞⎠ = 234.165

lAd ≔ lAe + lec + lcd = 7

ΔUd ≔ ‾‾ 3 ⋅ lAd ⋅ ⎛⎝r ⋅ id ⋅ cos ((ϕ)) + x ⋅ id ⋅ sin ((ϕ))⎞⎠ = 109.277

lAb ≔ lAe + lec + lcd + ldb = 9

ΔUb ≔ ‾‾ 3 ⋅ lAb ⋅ ⎛⎝r ⋅ ib ⋅ cos ((ϕ)) + x ⋅ ib ⋅ sin ((ϕ))⎞⎠ = 140.499

lAc ≔ lAe + lec = 5

ΔUa ≔ ‾‾ 3 ⋅ lAc ⋅ ⎛⎝r ⋅ ia ⋅ cos ((ϕ)) + x ⋅ ia ⋅ sin ((ϕ))⎞⎠ = 156.11

ΔUAa ≔ ΔUe + ΔUc + ΔUd + ΔUb + ΔUa = 733.716

Se considera una línea con las derivaciones y cargas correspondientes:

VL ≔ 380 r ≔ 0.63 ――

x ≔ 0.325 ――

P1 ≔ 15

P2 ≔ 25

P3 ≔ 15

P4 ≔ 25

ϕ1 ≔ acos ((0.8)) = 36.87 Q1 ≔ P1 ⋅ tan ⎛⎝ϕ1⎞⎠ = 11.25

ϕ4 ≔ acos ((0.8)) = 36.87 ⋅

Q2 ≔ 0

⋅

Q3 ≔ 0

⋅

Q4 ≔ P4 ⋅ tan ⎛⎝ϕ4⎞⎠ = 18.75

Calcular la caída de tensión de la línea. En primer lugar, no sabemos cuál es el punto de menor tensión por lo que tendremos que calcular la caída de tensión a todos los extremos: Empezamos por el extremo que coincide con el punto de consumo 4. Calcularemos la caída de tensión que provoca cada consumo y luego aplicaremos el principio de superposición. lA1 ≔ 50 lA2 ≔ 100 lA3 ≔ 150 lA4 ≔ 200

1 ΔU1 ≔ ―― lA1 ⋅ ⎛⎝r ⋅ P1 + x ⋅ Q1⎞⎠ = 1.725 VL 1 ΔU2 ≔ ―― lA2 ⋅ ⎛⎝r ⋅ P2 + x ⋅ Q2⎞⎠ = 4.145 VL 1 ΔU3 ≔ ―― lA3 ⋅ ⎛⎝r ⋅ P3 + x ⋅ Q3⎞⎠ = 3.73 VL 1 ΔU4 ≔ ―― lA4 ⋅ ⎛⎝r ⋅ P4 + x ⋅ Q4⎞⎠ = 11.497 VL ΔUA4 ≔ ΔU1 + ΔU2 + ΔU3 + ΔU4 = 21.096

Ahora calculamos el extremo que coincide con el punto de consumo 3. lA1 ≔ 50 lA2 ≔ 100 lA3 ≔ 250 lA4 ≔ 150

1 ΔU1 ≔ ―― lA1 ⋅ ⎛⎝r ⋅ P1 + x ⋅ Q1⎞⎠ = 1.725 VL 1 ΔU2 ≔ ―― lA2 ⋅ ⎛⎝r ⋅ P2 + x ⋅ Q2⎞⎠ = 4.145 VL 1 ΔU3 ≔ ―― lA3 ⋅ ⎛⎝r ⋅ P3 + x ⋅ Q3⎞⎠ = 6.217 VL 1 ΔU4 ≔ ―― lA4 ⋅ ⎛⎝r ⋅ P4 + x ⋅ Q4⎞⎠ = 8.623 VL ΔUA3 ≔ ΔU1 + ΔU2 + ΔU3 + ΔU4 = 20.709

⋅

El extremo del punto de consumo 1 lo podemos descartar, ya que en ningún caso podrá caer más tensión en el tramo de 150m desde A hasta el punto de consumo 1 que en los 200m que hay desde A hasta el punto de consumo 4 debido a la mayor distancia y a la mayor potencia del punto 4. Podemos también calcular la caída de tensión por tramos. Lo hacemos para el tramo A-4y comprobamos que el resultado es el mismo: lA1 ≔ 50

1 ΔUA1 ≔ ―― lA1 ⋅ ⎛⎝r ⋅ ⎛⎝P1 + P2 + P3 + P4⎞⎠ + x ⋅ ⎛⎝Q1 + Q2 + Q3 + Q4⎞⎠⎞⎠ = 7.914 VL

l12 ≔ 50

1 ΔU12 ≔ ―― l12 ⋅ ⎛⎝r ⋅ ⎛⎝P2 + P3 + P4⎞⎠ + x ⋅ ⎛⎝Q2 + Q3 + Q4⎞⎠⎞⎠ = 6.19 VL

l23 ≔ 50

1 ΔU23 ≔ ―― l23 ⋅ ⎛⎝r ⋅ ⎛⎝P3 + P4⎞⎠ + x ⋅ ⎛⎝Q3 + Q4⎞⎠⎞⎠ = 4.118 VL

l34 ≔ 50

1 ΔU34 ≔ ―― l34 ⋅ ⎛⎝r ⋅ ⎛⎝P4⎞⎠ + x ⋅ ⎛⎝Q4⎞⎠⎞⎠ = 2.874 VL ΔUA4 ≔ ΔUA1 + ΔU12 + ΔU23 + ΔU34 = 21.096

Cálculo de caída de tensión en redes en anillo: VL ≔ 20

ϕ ≔ −36.87

l1 ≔ 5

l2 ≔ 10

i1 ≔ 5

i2 ≔ 10

r ≔ 1.96 ―― x ≔ 0.391 ―― l3 ≔ 5 i3 ≔ 5

l4 ≔ 5 i4 ≔ 15

l5 ≔ 10

l6 ≔ 10

i5 ≔ 10

iTOTAL ≔ i1 + i2 + i3 + i4 + i5 = 45

L1 ≔ l1 = 5

L2 ≔ l1 + l2 = 15

L5 ≔ l1 + l2 + l3 + l4 + l5 = 35

L3 ≔ l1 + l2 + l3 = 20

L4 ≔ l1 + l2 + l3 + l4 = 25

LTOTAL ≔ l1 + l2 + l3 + l4 + l5 + l6 = 45

⎛⎝i1 ⋅ L1 + i2 ⋅ L2 + i3 ⋅ L3 + i4 ⋅ L4 + i5 ⋅ L5⎞⎠ Ix ≔ iTOTAL − ―――――――――――― = 22.778 LTOTAL ⎛⎝i1 ⋅ L1 + i2 ⋅ L2 + i3 ⋅ L3 + i4 ⋅ L4 + i5 ⋅ L5⎞⎠ Iy ≔ ―――――――――――― = 22.222 LTOTAL Iab ≔ Ix − i1 = 17.778 Ied ≔ Iy − i5 = 12.222

Ibc ≔ Ix − i1 − i2 = 7.778

Icd ≔ Ix − i1 − i2 − i3 = 2.778

El punto alimentado por ambos extremos y por tanto el de minima tensión es d.

3 ⋅ l1 ⋅ ⎛⎝r ⋅ Ix ⋅ cos ((ϕ)) + x ⋅ Ix ⋅ sin ((ϕ))⎞⎠ = 263.028 ΔUoa ≔ ‾‾ ΔUab ≔ ‾‾ 3 ⋅ l2 ⋅ ⎛⎝r ⋅ Iab ⋅ cos ((ϕ)) + x ⋅ Iab ⋅ sin ((ϕ))⎞⎠ = 410.58 ΔUbc ≔ ‾‾ 3 ⋅ l3 ⋅ ⎛⎝r ⋅ Ibc ⋅ cos ((ϕ)) + x ⋅ Ibc ⋅ sin ((ϕ))⎞⎠ = 89.814 ΔUcd ≔ ‾‾ 3 ⋅ l4 ⋅ ⎛⎝r ⋅ Icd ⋅ cos ((ϕ)) + x ⋅ Icd ⋅ sin ((ϕ))⎞⎠ = 32.077

ΔUd ≔ ΔUoa + ΔUab + ΔUbc + ΔUcd = 795.499