ESCUELA SUPERIOR POLITECNICA DE CHIMBORAZO

ELECTROTECNIA II

POR:MILTON JAYA

COD:1035

CIRCUITO CON 4 MOTORES

DATOS:

5%≤∆v≤10%

M1: Pmec=1oHP ; η=o.4 ; fp=0.6; L=50m

M2: Pmec=12HP ; η=0.5 ; fp=0.7; L=70m

M3: Pmec=15HP ; η=0.6 ; fp=0.7; L=80m

M4: Pmec=9HP ; η=0.7 ; fp=0.8 ; L=85m

MOTOR 1

Pmec=7460w

Pele=7460/0.4=18650w

I=P/√3.v.cosƍ=18A

Conductor de calibre #14 AWG; Sección 2.08 mm²

Rl=ρ.l/s =0.42Ω

∆v=√3.Rl.Il.cosƍ=7.8v

BREKER=18x1.25=22.5A

BREKER DE 25A

POTENCIAS:

P=√3.I.V.cosƍ=18706.1w

Q=√3.I.V.senƍ=24629.7VAR

S=√(P²+Q²)=30894.1VA

MOTOR 2

Pmec=8952w

Pele=8952/0.5=17904w

I=P/√3.v.cosƍ=15A

Conductor de calibre #14 AWG; Sección 2.08 mm²

Rl=ρ.l/s =0.5Ω

∆v=√3.Rl.Il.cosƍ=9.1v

BREKER=15x1.25=18.7A

BREKER DE 20A

POTENCIAS:

P=√3.I.V.cosƍ=18186.5w

Q=√3.I.V.senƍ=18446.3VAR

S=√(P²+Q²)=25903.9VA

MOTOR 3

Pmec=11190w

Pele=11190/0.6=18650w

I=P/√3.v.cosƍ=15A

Conductor de calibre #14 AWG; Sección 2.08 mm²

Rl=ρ.l/s =0.6Ω

∆v=√3.Rl.Il.cosƍ=10v

BREKER=15x1.25=18.7A

BREKER DE 20A

POTENCIAS:

P=√3.I.V.cosƍ=18186.5w

Q=√3.I.V.senƍ=18688.9VAR

S=√(P²+Q²)=26077.2VA

MOTOR 4

Pmec=6714w

Pele=6714/0.7=9591.4w

I=P/√3.v.cosƍ=7A

Conductor de calibre #16 AWG; Sección 1.31mm²

Rl=ρ.l/s =1.1Ω

∆v=√3.Rl.Il.cosƍ=10v

BREKER=7x1.25=8.75A

BREKER DE 10A

POTENCIAS:

P=√3.I.V.cosƍ=9695.4w

Q=√3.I.V.senƍ=7296.67VAR

S=√(P²+Q²)=12134.2VA

INTENSIDAD TOTAL

IT=I1+I2+I3+I4

IT=18+15+15+7=55

BREKER PRINCIPAL

55X1.25=68.7A BREKER DE 70A

POTENCIAS TOTALES:

PT=P1+P2+P3+P4=18706.1+18186.5+18186.5+9695.4

PT=64774.5WW

QT=Q1+Q2+Q3+Q4=24629.7+18446.3+18688.9+7296.6

QT=69061.5VAR

ST=SUMATORIA DE S EN FORMA VECTORIAL

S1=30894.1е^J53®= (18592.5+J24673.1)

S2=25903.9е^J46®= (17994.3+J18633.7)

S3=26077.2е^J46®= (18114.7+J18758.3)

S4=12134.2е^J37®= (9690.8+J7302.5)

ST=94647.9е^J47®= (64392.3+J69367.6)

PT=64774.5W

QT=69061.5VAR

ST=94647.9 е^J47®VAR

GRAFICA

Tg23®=Q1/P

Q1=PTg23®=27495.1VAR

QT=Q1+Q2

Q2=QT-Q1=69061.5-27495.1

Q2=41566.4VAR

Q2/3=13855.4VAR

XC=V²/Q2 → XC=72.2Ω

C=1/2∏.f.xc

C=36.7Цf

ESQUEMA MULTIFILAR

MOTOR TRIFASICO

DATOS:

Pmec=3000w

Cosƍ=0.79

f=50Hz

η=0.76

SOLUCION:

Potencia eléctrica:

Pele=Pmec/η

Pele=3947.4w

Intensidad:

I= P/√3.v.cosƍ

I=13.8A

Calibre de conductor:

#14 AWG; sección 2.08mm²

Rl=ρxl/s=o.29Ω

Caída de tensión:

∆v=I x Rl

∆v=4v

220-100%

4-x=10.35v no cumple

Bajo la sección

Rl=ρ x l/s=0.75Ω

∆v=I x Rl

∆v=10.35v

220-100%

10.35-x=4.7% si cumple

Mejoramiento de fp:

S=√3xvxI

S=5258.5VA

Q=S x senƍ

Q=34428.1VAR

Qf =Ptgƍ =1675.6VAR

QC=Q-QF=(34428.1-1675.6)

QC=1767.2VAR

C=QC/2∏.f.v²

C=116.22цF

CIRCUITO DESEQUILIBRADO CON RL

DATOS:

VA=93ej0v

VB=93ej120v

VC=93e-j120v

Rl=0.93Ω

Z1=1.84ej46.1Ω

Z2=1.54e-j46.1Ω

Z3=2.05ej61.5Ω

Impedancias ZA=Z1+Rl=(1.27+j1.32)+0.93=2.55ej31.25Ω

ZB=Z2+Rl=(1.06-j1.4)+0.93=2.27e-j29.25Ω

ZC=Z3+Rl=(0.98+j1.80)+0.93=2.61ej43.57Ω

Admitancias YA=1/ZA=0.39e-j31.05Ω

YB=1/ZB=0.49ej29.25Ω

YC=1/ZC=0.38e-j43.57Ω

Caída de tensión

UNN=(EA.YA+EB.YB+EC.YC)/(YA+YB+YC+YN)

UNN=

[(35.9e-j31.05)+(40.6ej149.25)+(35.03e-j163.57)]/[(0.33-

j0.20)+(0.38+j0.21)+(0.26-j0.27)+1]

UNN=38.5e-j168.47/1.98e-j7.52

UNN=19.44e-j160.98(V)

Intensidades

IA=(EA-UNN)YA

=93-(-18.37-j6.34)x0.39e-j31.05

=110.75ej3.2xo.39e-j31.05=43.19e-j27.85(A)

IB=(EB-UNN)YB

=[(-46.1+j79.8)-(-18.37-j6.34)]x0.44ej29.25

=90.43ej107.84x0.44ej29.25=39.79ej137.09(A)

IC=(EC-UNN)YC

=[(-46.1-j79.8)-(-18.37-j6,34)]0.38e-j48.57

=78.52e-j110.68x0.38e-j43.57=29.84e-j154.25(A)

Caídas de tensión

VA=IA.ZA=115.23ej3.4(V)

VB=IB.ZB=90.32ej107.84(V)

VC=IC.ZC=77.88e-j110.68(V)

Potencias

PA=IA VF cosƍ=4254.7(w)

PB=IB VF cosƍ=3135.6(w)

PC=IC VF cosƍ=1683.7(w)

PT=9073.3(W)

QA=IA VF senƍ=2582.5(VAR)

QB=IB VF senƍ=1756.1(VAR)

QC=IC VF senƍ=1601.7(VAR)

QT=5940.3(VAR)

ST=√(PT²+QT²)

ST=10844.9(VA)

ENCONTRAR RL DEL SISTEMA Y RESOLVER

DATOS:

Z1=10ej40

Z2=8ej45

Z3=12ej50

Vl=208v

VF=120v

INTENSIDAD

I1=Vl/Z1

I1=208ej0/10ej40

I1=20.8e-j40(A)

I2=vl/Z2

I2=208e-j120/8ej45

I2=26e-j165(A)

I3=vl/Z3

I3=208ej120/12ej50

I3=17.3(A)

Calculo calibre conductor

#12AWG; sección 3.31mm²

Rl=0.0175x50/3.31

Rl=0.26Ω

208→100%

x→3.5%=7.28v

Sumamos la Rl con impedancias Za=Z1+Rl=(7.66+j6.42)+0.26 → Za=10.2ej39 Ω

Zb=Z2+Rl=(5.65+j5.65)+0.26 → Zb=8.2ej43 Ω

Zc=Z3+Rl=(7.71+j9.19)+0.26 → Zc=12.2ej49 Ω Admitancias Ya=1/Za=0.09e-j39 Ω

Yb=1/Zb=0.12e-j43 Ω

Yc=1/Zc=0.08e-j49 Ω

Yn=1/0.26=3.84Ω Diferencia de potencial UNN=(EA.YA+EB.YB+EC.YC)/(YA+YB+YC+YN) UNN= (10.8e-j39+14.4e-j163+9.6ej71)/[(0.06-j6.79)+(0.08-j0.08)+(0.05-j0.06)+3.84] UNN=25.4e-j3.9/4.1e-j2.6

UNN=6.2e-j1.3(v)

Intensidades

IA=[(EA-UNN)YA]

=[120-(6.19-j0.14)]0.09e-j39

=113.8ej0.07x0.09e-j39=10.2e-j38.9(A)

IB=[(EB-UNN)YB]

=[(-60-j103.9)-(6.19-j0.14)]0.12e-j43

=123.07e-j122x0.12e-j43=14.7e-j165(A)

IC=[(EC-UNN)YC]

=[(-60+j103.9)-(6.19-j0.14)]0.08e-j49

=123.3ej122x0.08e-j49=9.8ej73(A)

Caídas de tensión

UA=(EA-UNN)=113.8ej0.07(v)

UB=(EB-UNN)=123.6e-j122(V)

UC=(EC-UNN)=123.3ej122(V)

Potencias

PA=IA VF cosƍ=889.1w

PB=IB VF cosƍ=1279.2w

PC=IC VF cosƍ=776.7w

PT=2945w

QA=IA VF senƍ=746.1VAR

QB=IB VF senƍ=1279.2VAR

QC=IC VF senƍ=956.6VAR

QT=2950.9VAR

S=√(P²+Q²)=4169.1VA.