ESCUELA SUPERIOR POLITECNICA DE CHIMBORAZO ELECTROTECNIA II POR:MILTON JAYA COD:1035
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.