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Soluciones de ejercicios prácticos y teóricos....

Description

1K = 1000Ω

2·2 2·2 4·4 2·2 4·4 = 2; + 1 = 2; + 3 = 4; = 2; = 1; 1 + 2 = 3 K 4+4 2+2 2+2 4+4 2+2 UAN UBN

I

3 [Ω] = 6 [V] 3 [Ω] + 2 [Ω] 2 [Ω] = 2 [V] UBN = UAN 4 [Ω] + 2 [Ω] 2 [V] UBN = I= = 0,5 [A] 4 [Ω] 4 [Ω] UAN = 10 [V]

7360 W 230 V 230 V

C = 58 C = 1/ρ

S = 6 mm2 S = 10 mm2 ; S = 16 mm2

S = 25 mm2 3 × 230 V

3 × 230 V P = 7360 W

220 V

I=

Uhorno

Pnom 7360 = = 32 A 230 Unom

  3 = 223,1 V = 230 · 1 − 100

∆U = 230 · 0,03 = 6,9 V

∆U = I · Rcable → Rcable =

∆U 6,9 = 0,2156 Ω = 32 I

lcable = 78 · 2 = 156 m l R = C1 · cable S S=

156 1 1 lcable · · = 12,4738 mm2 = 58 0,2156 C Rcable

2 U nom 2302 = 7,1875 Ω → = 7360 Pnom Ucuadro = I= Rcable + Rhorno 230 = = 31,068 A 0,2156 + 7,1875

Rhorno =

Rhorno = Rhorno + Rcable 7,1875 = 230 = 223,3017 V 7,1875 + 0,2156

Uhorno = Ucuadro

Ucuadro − Uhorno · 100 = Ucuadro 230 − 223,3017 = · 100 = 2,9123 % 230

∆U =

16 mm2

Rcable =

1 156 1 lcable · · = = 0,1681 Ω C Scable 58 16

Ucuadro = Rcable + Rhorno 230 = = 31,2687 A 0,1681 + 7,1875

I=

Rhorno = Rhorno + Rcable 7,1875 = 230 = 224,7436 V 7,1875 + 0,1681

Uhorno = Ucuadro

Ucuadro − Uhorno · 100 = Ucuadro 230 − 224,7436 = · 100 = 2,2854 % 230

∆U =

Pnom 7360 = 18,4752 A I=√ =√ 3230 3Unom

Uhorno

  3 = 230 · 1 − = 223,1 V 100 ∆U = 230 · 0,03 = 6,9 V ∆U = I · Rcable →

→ Rcable

,1 230 √ − 223 √ ∆U 3 3 = = = 0,2156 Ω I 18,4752 lcable = 78 m

R= S=

78 1 1 lcable · · = 6,2376 mm2 = 58 0,2156 C Rcable 10 mm2

6 mm2

/ 3 Rhorno = /I 3 = 18,4752 = 7,1875 Ω 1 · lcable = 1 · 78 = 0,2241 Ω Rcable = C Scable 58 6 U

√

230

√

)

→

Rhorno Rhorno + Rcable 7,1875 = 223,0456 V = 230 7,1875 + 0,2241

→ Uhorno = Ucuadro

Ucuadro − Uhorno · 100 = Ucuadro 230 − 223,0456 = · 100 = 3,0237 % 230 3 · 78 · 6 = 1404

∆U =

2 · 156 · 16 = 4992

4992 − 1404 = 71,88 % 4992

1 C

·

lcable S

730 W

350 W 800 W

1,5 CV

3/4 CV 2875 VA

60 W 40 W 19,263 A

129663,50 W

1 CV = 735, 49875 W

100 W 3450 W

230 V

125 W

10350 W

230 V

5750 W 14490 W

230 V

230 V W

230 V

1 CV = 735, 49875 W 12663,50 W 14490 W 63 A

12663,50 W 3 · 735,49875+ 4 + 1,5 · 735,49875 + 800+

12663,50 = 730 + 350 +

2875 cos ϕ + 25 · 60 + 30 · 4 · 40 → 2828,6278125 = 0,98387 → → cos ϕ = 2875 Q = S · sin ϕ = 2875 · sin ϕ = 514,2942 VAr

ω · C · U 2 = 0,7 · Q → 0,7 · Q 0,7 · 514,2942 = →C= = 21,6623 µF ω · U2 2π50 · 2302 12663,50 W

ω T f ϕ

100 m2 230 V 100 W 60 W

150 W

135 W

40 W 50 W 30 W

700 W

350 W 1000 W

1500 W 90 W

600 W 800 W

550 W

700 W 60 W

40 W

80 W 7W 400 W

45 W 25 W

60 W

1000 W

25 A 16 A

16 A 16 A 16 A

C1 C2 C3 C4 C5

W 200,00 3450,00 5400,00 3450,00 3450,00

Fs 0,75 0,20 0,50 0,66 0,40

Fu 0,50 0,25 0,75 0,75 0,50

A 16 A 25 A 16 A 16 A

Fs Fu

n

C1 C2 C2

10 A 16 A

C3 C4

25 A

n

90 W 700 W C4

C5 7W

C1 100 W 3 · 100 = 300 W 50 W 4 · 50 = 200 W 30 W 2 · 30 = 60 W 60 W 5 · 60 = 300 W 7W 1 · 7 = 7W 25 W 3 · 25 = 75 W 60 W 1 · 60 = 60 W C1 1002 W C2

10 enchufes · 16 A = 160 A → PC2 = 160 A ·

C3

1 enchufe · 25 A = 25 A → PC3 = 25 A ·

230 V = 36800 W 230 V = 5750 W C4 PC4 = 48 A · 230 V = 11040 W

3 enchufes · 16 A = 48 A →

C5 64 A · 230 V = 14720 W

4 enchufes · 16 A = 64 A → PC5 = C2 150 W 135 W 60 W 40 W 385 W

PC2 = 385 + 160 + 445 = 990 W C3

80 W 80 W 160 W 400 W 45 W 445 W

1000 W 1500 W 2500 W C4 800 W 600 W 700 W 2100 W C5 350 W 600 W 700 W 550 W 90 W 1000 W 3290 W C1

W

P = Ptot · Fs · Fu = 1002 · 0,75 · 0,50 = 375,75 W 200 W C1 C2 16 A

10 · 16 · 230 = 36800 W PdemandaC1 = Pmax · Fs · Fu = 36800 · 0,20 · 0,25 = 1840 W

IprevistaC1 =

1840 PdemandaC1 = 8A = 230 Unom

W

10 A

8A

16 A

W

1840 W

C2 . C3 5750 W

25 A

PprevistaC3 = 5750 · 0,50 · 0,75 = 2156,25 W 2156,25 230

2500 230

= 9,375 A

10 A

P = 1000 + 1500 = 2500 W → 16 A

= 10,8696 A C4

3 · 16 · 230 =

11440 W PprevistaC3 = Pmax · Fs · Fu = 11400 · 0,66 · 0,75 = 5643 W IprevistaC3 =

3450 W) 800 + 600 + 700 = 2100 W

PprevistaC3 5643 = = 24,53 A 230 Unom

Un ≤ 50 V 50 < Un ≤ 500 V 500 < Un ≤ 1000 V

Un ≤ 75 V 75 < Un ≤ 750 V 750 < Un ≤ 1500 V

P S f.d.p =

P S

f.d.p. cos ϕ

C5 C2

≥ 30 mA

W

Q

VAr

(Um´ax · Im´ax)/2

P =0 I = U/XL S

P Q S2 = 2

2

P +Q

S cos ϕ = 1

VA P f.d.p.

S f.d.p =

cos ϕ

100 kA

P S

B2

230 V

4 kVA

3 kVA f.d.p = 0,8

f.d.p.1 = 

P2 → S2

P2 = f.d.p. p · 4 = 3,2 kW p 2 · S22 = 0,8 2−P 42 − 3,22 = 2,40 kVAr Q2 = S2 2 = f.d.p.3 =



P1 → S1

P1 = f.d.p. p · 3 = 2,25 kW p 1 · S1 = 0,75 2 2 Q1 = S1 − P 1 = 32 − 2,252 = 1,98 kVAr f.d.p.2 =



f.d.p = 0,75

P3 → S3

P3 = f.d.p. p 3 · S3 = 1p· 3,46 = 3,46 kW Q3 = S32 − P32 = 3,462 − 3,462 = 0 VAr

3,46 kW

P = P1 + P2 + P3 = 2,25 + 3,2 + 3,46 = 8,91 kW

Q = Q1 + Q2 + Q3 = 1,98 + 2,40 + 0 = 4,38 kVAr

S=

p

P 2 + Q2 =

p

8,912 + 4,382 = 9,93 kVA

I1 =

S1 3000 = 13,04 A = 230 U1

I2 =

S2 4000 = = 17,39 A 230 U2

S3 P3 3460 = 8,69 A I3 = √ = √ =√ 3230 3U3 3U3

f.d.p. =

P 8,91 = 0,8973 = 9,93 S

R−S−T T

230 V R R

S

S

T

e = 1 % → 230 ×

1 100

= 2,3 V

2l → R = ρ 2l → ∆U = ρ S2l I → S = ρ ∆U I S l → ∆U = ρ l I → S = ρ l I → R = ρS S ∆U 2l 13,04 = 0,9695 mm2 → (1,5 mm2 ) S == ρ ∆U I = 0,0171 2·5 2,3 2l 17,39 = 2,3272 mm2 → (2,5 mm2 ) S == ρ ∆U I = 0,0171 2·9 2,3 l 8 8,69 = 0,5169 mm2 → (1,5 mm2 ) S == ρ ∆U I = 0,0171 2,3

730 W

350 W 800 W

1,5 CV

3/4 CV

2875 VA

60 W 40 W 19, 263 A

12663,50 W 1 CV =

735,49875 W

12663,50 W 14490 W 63 A

12663,50 W 3 · 735,49875+ 4 + 1,5 · 735,49875 + 800+ 2875 cos ϕ + 25 · 60 + 30 · 4 · 40 → 2828,6278125 → cos ϕ = = 0,98387 → 2875 Q = S · sin ϕ = 2875 · sin ϕ = 514,2942 VAr

12663,50 = 730 + 350 +

ω · C · U2 = Q → Q 514,2942 →C= = 30,9461 µF = 2 2π50 · 2302 ω·U 12663,50 W

14490 W 63 A

185 m2 40 A 10 A 10 A 25 A

25 A

30 mA

25 A 32 A

185 m2 > 160 m2

C1 C2 C3 C4 C5

C1 C2 C3 C4 C5

10 A ≥ 10 A

16 A ≥ 10 A

25 A ≥ 25 A

20 A  25 A

C4

16 A  32 A

C5

100 m2

200 m2

n > 21

15,3 + (n − 21) · 0,5

100 m2

5750 W 160 m2 9200 W



12 · 5750 + 2 · 9200 11,3 14



= 70, 544 kW...