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Nr.1 Leistungen im Wechselstromkreis

i ( t ) = iˆ cos ( wt + ϕi ) u ( t ) = uˆ cos ( wt + ϕ u ) p (t ) = uˆ ⋅ iˆ cos ( wt + ϕu ) ⋅ cos ( wt + ϕi ) 1 p (t ) = uˆ ⋅ iˆ  cos (ϕ u −ϕ i ) + cos ( 2wt + ϕ u + ϕ i )  2 p (t ) =

2U ⋅ 2 I cos (ϕu − ϕi ) + cos (2wt + ϕu + ϕi ) 2

p (t ) = U ⋅ I  cos (ϕu − ϕi ) + cos ( 2wt + ϕu + ϕi )  cos ( 2wt + ϕu + ϕi ) = cos ( ( 2wt + 2ϕu ) − (ϕu − ϕi ) ) cos ( 2wt + ϕu + ϕi ) = cos ( 2wt + 2ϕu ) ⋅ cos ( ϕu − ϕi ) + sin ( 2 wt + 2ϕu ) ⋅ sin ( ϕu − ϕi ) p (t ) = U ⋅ I cos (ϕu − ϕi ) ⋅ 1+ cos ( 2wt + 2ϕu ) + U ⋅ I sin ( ϕu − ϕi ) ⋅ sin ( 2wt + 2ϕu ) Wirkleistungsanteil: p (t ) = U ⋅ I cos (ϕu − ϕi ) ⋅ 1+ cos ( 2wt + 2ϕu ) Blindleistungsanteil: q ( t ) = U ⋅ I sin (ϕ u − ϕ i ) ⋅ sin ( 2wt + 2ϕ u ) S=

P2 + Q2

S = P + jQ

Nr. 2.1.1

HQL = 125W U = 230 f = 50 Hz I = 1,15 A P = 137W Ohne Kondensator: 137W P P = UI cosϕ   →cosϕ = = = 0,518 U ⋅ I 230V ⋅1,15 A ϕ = 58, 66° ILeitung =1,15 A ILampe =1,15 A Mit Kondensator: QC = U 2 w⋅ C = (230 V ) 2π ⋅ 50 Hz ⋅ 8 µ F = 132,95 VAr 2

tan (ϕ 2 ) = tan (ϕ1 ) −

QC = 0, 67 P

ϕ 2 = 33,82 °   cos (ϕ 2 ) = 0,83 → ILampe = 1,15A

I Leitung =

P cos (ϕ2 ) ⋅U

= 0, 72A

Cx = ? c os (ϕ3 ) = 0,95   →ϕ3 =18,19 ° QC = P ( tan ( ϕ1) − tan ( ϕ3) ) = 137W ( tan ( 58, 66° ) − tan (18,19° ) ) QC = 180Var QC = U 2 ⋅ w C⋅ x ILeitung =

C  → x

P cos ( ϕ3 ) ⋅ U

Q 10,8 =2 C = µF U ⋅w

= 0, 626A

Nr. 2.1.2

IM = 10,5 A P = 1932W Kompensation mit C1:

U = 230V

C1 = 30 µ F

C 2 = 150 µ F

QC 1 = U 2 wC1 = ( 230 V ) 2π ⋅ 50 Hz ⋅ 30µ F = 498, 6 VAr 2

ϕ 1 = 36,87° tan (ϕ2 ) = tan (ϕ1 ) −

498, 6VAr QC1 = tan (36,87 °) − = 0,49 1932W P

→ = ϕ 2 = 26,10 °   cos (ϕ 2 ) 0,897 IM 1 =

P = 9,36 A U ⋅ cos ( ϕ2 )

Kompensation mit C1 und C2 : 2

QC 1,C 2 = U 2 w ( C1 + C2 ) = ( 230V ) 2π ⋅ 50 Hz ⋅180µ F = 2991, 4VAr tan (ϕ3 ) = tan (ϕ1 )−

QC1, C 2 P

= tan (36,87° )−

2991, 4VAr = − 0, 798 1932W

→ = 78 Überkompensation !!! ϕ 3 = −38,59 °   cos (ϕ 3 ) 0, IM 2 =

P = 10, 76 A U ⋅ cos ( ϕ3 )

Leistungsfaktor cos ( ϕ ) = 1 QL = U ⋅ I sin ϕ1 = 230V ⋅ 10,5 A⋅ sin (36,87° ) = 1449VAr QC = QL   →QC =1449VAr 1449VAr QC = = 87,19 µ F 2 U ⋅ w 230V 2 ⋅ 2π ⋅ 50Hz P = = 8, 4 A U

Cx = IM

Nr. 2.1.3 Parallelkompensation

P = 1500W U = 230V

cos (ϕ ) = 0, 6 cos ( ϕ neu ) = 0,95

P P = UI cos (ϕ )   →I = =10,87 A U cos (ϕ ) QL = UI sin (ϕ ) = 230 V ⋅10,87 Asin (53,13° ) = 2000 VAr tan (ϕneu ) = C= Ineu

QL − QC  →QC =QL −P tan ⋅ (ϕneu ) 1507 = VAr P

QC = 90, 67µ F U ⋅ 2π ⋅ 50 Hz P = = 6,86 A U ⋅ cos( ϕneu ) 2

Nr.2.1.4 Reihenkompensation P = 1500W U = 230V I =

cos (ϕ ) = 0, 6 cos ( ϕneu ) = 1

1500W P = = 10,87 A U ⋅ cos( ϕ ) 230V ⋅ 0, 6

QL = U ⋅ I sin ( ϕ ) = 230V ⋅ 10,87 Asin ( 53,13° ) = 2000VAr →QL =QC cos (ϕ neu ) = 1   I2 C= = 188µ F 2 π ⋅50 Hz ⋅ QC

Nr. 2.1.5 Duoschaltung

a) cos (ϕ ) =

48W P = = 0, 47 UI 230V ⋅ 0, 44A

→tan (ϕ ) =1,86 ϕ = 61,7 °   QC = 2 ⋅ QL , da die Lampen identisch sind QC = 2 P ⋅ tan ( ϕ ) = 2 ⋅ 48W ⋅ tan ( 61, 7° ) = 178VAr

( 0, 44 A) I2 = = 3,5µ F w⋅ QC 2π ⋅ 50 Hz ⋅ 178VAr 2

C= b)

I1 = 0, 44 Ae− j 61,7° I2 = 0, 44 Ae j 61,7 ° IL = I1 + I2 = 0, 417 e j 0° c) UC =

I 0, 44 A = ≈ 400V wC 2π ⋅ 50Hz ⋅ 3,5µ F

Nr. 2.1.6 Duoschaltung

S = U ⋅ I⋅ = 230V ⋅ 0, 43A = 98, 9VA QL =

S 2 − P2 =

( 98, 9VA)

2

QC = 2 ⋅ QL = 2 ⋅ 85,9VAr = 171,8VAr Q 171,8VAr UC = C = = 399VAr I 0, 43 A 399VAr U XC = C = = 928Ω I 0, 43 A 1 C= = 3,5µ F 2 π ⋅50 Hz ⋅928 Ω Nr. 2.1.7

R = 1Ω X L = 1Ω Z =

R⋅ XL = 0, 707 Ω R+ XL

I = Iwirk 2 + Iblind 2 I = 2 =1, 41 A cos (ϕ ) = 1 QL = QC = 1W Ck = 3,18 mF cos (ϕ ) = 0,9 ∆QL = QC = 0,519W Ck = 1, 65 mF

2

− ( 49W ) = 85, 9VAr

Nr. 2.1.8

1 1 + = 1S − j1S = 2 S e − j 45° 1Ω j1Ω 1 Z1 = Ω e j45 ° 2 Z = Re ( Z 1) = 0,5 Ω Y1 =

I=

U = 2 Ae j 0° Z

UC = I ⋅ XC = 2 Ae j 0° ⋅0, 5 Ωe− j 90° = 1V e − j 90° 1 U Z1 = I ⋅ Z1 = 2 Ae j 0 ° ⋅ Ω e j 45 ° = 2Ve j 45 ° 2 P=

45 45 U Z1 ⋅U *Z1 2Ve j ° ⋅ 2Ve − j ° = = 2W 1Ω R

Nr. 3.1.1

1

f0 =

= 318, 3Hz 2π ⋅ LC U q 230V = = 1,5A I = 20Ω R I 1,5A UC = = = 750V 2π ⋅ f 0 ⋅ C 2π ⋅ 318, 3Hz ⋅ 1µF U L = I ⋅ w ⋅ L = 1, 5A ⋅ 2π ⋅ 318,3Hz ⋅ 250mH = 750V Spannungsüberhöhung: Q=

Uq U

=

750V = 25 30V

g) Wenn die Dämpfung geringer wird, dann gehen die Spannungs- und die Stromüberhöhung gegen unendlich....