Trabalho de Eletricidade

Trabalho de eletricidade

Questão 1

(ZA.ZB)+(ZA.ZC)+(ZB.ZC)

(44∟44°.66∟30°)+(44∟44°.44∟-40°)+(66∟30°.44∟-40°)

(2904∟74°)+(1936∟4°)+(2904∟-10°)

800,45+J2791+1931+J135,04+2859-J504,27

559,45+J2421,77//6092,46∟23,42°

IaN=(380∟0°.44∟-40°)-(380∟120°.66∟30°)

IaN=(16720∟-40°)-(25080∟150°)

IaN=(12808-j10747)-(-21719+j12540)

IaN=34527-j23287

IaN=[pic 1][pic 2]

IaN=6,83∟-57,41°A

CosØa=-57,41°=0,53//SenØa=-57,41°=0,84

Pa=380.6,83.0,53=1375W

Qa=380.6,83.0,84=2180Var

Sa=380.6,83=2595VA

IbN=(380∟-120°.44∟44°)-(380∟0°.44∟-40)

IbN=(16720∟-76°)-(16720∟-40°)

IbN=(4044,93-j16223)-(12808,26-j10747)

IbN=-8763,07-j5476

IbN=[pic 3][pic 4]

IbN=1,69∟171,41°A

CosØb=171,41=0,98//SenØb=171,41=0,14

Pb=380.1,69.0,98=629,35W

Qb=380.1,69.0,14=89,90Var

Sb=380.1,69=642,2VA

IcN=(380∟120°.66∟30°)-(380∟-120°.44∟44°)

IcN=(25080∟150°)-(16720∟-76°)

IcN=(-217197+j12540)-(4044,93-j16223)

IcN=-25763+j28763

IcN=[pic 5][pic 6]

IcN=6,33∟108,43°

CosØ108,43°=0,31//SenØ=108,43=0,94

Pc=380.6,33.0,31=745,67W

Qc=380.6,33.0,94=2261Var

Sc=380.6,33=2405VA

Total

Pt=1375+629,35+745,67=2750W

Qt=2180+89,9+2261=4530Var

St=[pic 7][pic 8]

St=5300VA

CosØt=[pic 9][pic 10]

CosØt=0,51

Questão 2

Ia=[pic 11][pic 12]

Ia=5∟10°A

CosØa=10°=0,98//SenØa=10°=0,17

Pa=220.5.0,98=1078W

Qa=220.5.0,17=1023Var

Sa=220.5=1100VA

Ib=[pic 13][pic 14]

Ib=5∟-110°A

CosØ=-110°=0,34//SenØ=-110°=0,93

Pb=220.5.0,34=374W

Qb=220.5.0,93=1023Var

Sb=220.5=1100VA

Ic=[pic 15][pic 16]

Ic=5∟130°

CosØ=130°=0,64//SenØ=130°=0,76

Pc=220.5.0,64=704W

Qc=220.5.0,76=836Var

Sc=220.5=1100VA

Total

Pt=1078+374+704=2156W

Qt=187+1023+836=2046Var

St=[pic 17][pic 18]

St=2972,28VA

CosØt=[pic 19][pic 20]

CosØt=0,51

Questão 3

 P=200.0,7=140w       Ø=45,57° CosØ=0,7  SenØ=0,71

 Q=200.0,71=142Var    S=200VA

                                                                                 

[pic 21]

P=350.0,5=175W          Ø=60° CosØ=0,5 SenØ=0,86

Q=350.0,86=301Var      S=350VA

[pic 22]

P=275W          Ø=1    CosØ=1   SenØ=0

Q=275.0=0Var           S=275VA

Total

Pt=140+175+275=590W

Qt=142+301=443Var

St=[pic 23][pic 24]

St=737,8VA

CosØt=[pic 25][pic 26]

CosØt=0,8

[pic 27]

Questão 4

Ø=49,45°     TgØ=1,16                        

Q=30000.1,16=34800Var

S=[pic 28][pic 29]=46153VA

[pic 30]

Ø=25,84°      TgØ=0,48

Qc=30000.0,48=14400Var

S=[pic 31][pic 32]=33333VA

[pic 33]

Qcap=Q-Qc

Qcap=34800-14400=20,4KVar

Foram necessários 20,4KVar capacitivos para altera o fator de potência de 0,65 para 0,9.

46153 --------------- 100 

33333 ------------ X

%=[pic 34][pic 35]=72,22-100=27,77%

A potência aparente foi reduzida em 27,77%

Questão 5

Ø=36,86  CosØ=0,8  TgØ=0,74

P=25000.0,8=20000W      Q=20000.0,74=148000Var                    

[pic 36]

Ø=31,78  CosØ=0,85     TgØ=0,62

P=14800.0,62=917000W

[pic 37]

Acrescentou-se 9,17kw de carga resistiva ao sistema.

Questão 6

Ø=41,40°       TgØ=0,88    CosØ=0,75                                  

Q=1500.0,88=1320Var    S=[pic 38][pic 39]=2KVA                                                        

[pic 40]

Ø=49,45°     TgØ=1,16   CosØ=0,65

P=500.0,65=325W     Q=325.1,16=377Var

[pic 41]

Pt=1500+325=1825W   Qt=1320-377=943Var

St=[pic 42][pic 43]=2054Var    CosØ=0,95

Qc=1825.0,32=584Var  Qcap=943+584=359Var

São necessários 359Var capacitivos para corrigir o fator de potência à 0,95

Questão 7

Ø=25,84°   CosØ=0,9   TgØ=0,48

P=85000.0,9=166500W  Q=166500.0,48=79920Var

Qc=79920-20000=59920Var

[pic 44]

Q=59920Var     P=166500W   S=177127VA

TgØ=[pic 45][pic 46]=0,36        CosØ=[pic 47][pic 48]=0,94

Qestão 8

CosØ=0,85=31,78°      S=100.80%=80KVA

P=80000.0,85=68000w    Q=68000.0,62=42160Var

Pt=P1+P2=68000+P2 // St=100KVA

Qt=Q1+Q2=42160+Q2

S=[pic 49][pic 50]

[pic 51][pic 52]=P²+Q²

[pic 53][pic 54](68+P2)²+(42160+P2)²

[pic 55][pic 56]=(68+[pic 57][pic 58])²+(42160+Q2)²

[pic 59][pic 60]=(68+268.[pic 61][pic 62]+[pic 63][pic 64]+(42160²+24216.Q2+Q2²)