ejercicios de nodos esime zac

8/18/2019 Ejercicios de nodos Esime zac

1/24

Ejercicio 1:

Diagrama original.

Diagrama de admitancias:


2/24

Cálculos:

Admitancias:

ω=500rad

seg

yk =G+ j (ωC − 1

ωL)

y1=G+ j(

−1ωL

)

y1= 1

.25+ j ( −1500 x (1 x 10−3 ) )=4−2 j [simmens ]

y2= j(ωC −

1

ωL)

y2=0+ j (500 x (2 x 10−3)− 1500 x (.5 x 10−3 ) )=−3 j [ simmens ]

y3=G+ j(ωC −

1

ωL)

y3= 1

.5+ j (500 x (1 x 10

−3)− 1500 x ( .25 ) )=2+0.5 j [ simmens ]

Voltajes:

V fc=u1

V y 1=u3

V y 2=u2−u3

V y 3=u2−u1

Admitancias propias de nodo:

Y 11= y

3+2 ( 0 )= y

3=2+0.5 j [simmens]


3/24

Y 22= y

2+ y

3+2( 0 )= y

2+ y

3=−3 j+2+0.5 j=2−2.5 j [simmens ]

Y 33= y

1+ y

2+2( 0 )= y

1+ y

2=4−2 j−3 j=4−5 j [simmens ]

Admitancias mutuas de nodo:

Y 12=− y

3+2( 0 )+1 ( 0)=− y

3=−2−0.5 j [simmens ]

Y 32=− y

2+2 ( 0 )+1(0)=− y

2=3 j [simmens]

Y 13=0

Y 11u

1+Y

12u

2+Y

13u

3=− I fc

Y 21u

1+Y

22u

2+Y

23u

3=0

Y 31

u1+Y

32u

2+Y

33u

3=0

Ecuaciones:

( 2+0.5 j )u1+ (−2−.5 j )u

2+0u

3=−(

−7

√ 2∠0°)

(−2−0.5 j )u1+( 2−2.5 j )u2+(3 j)u3=0

( 0 ) u1+( 3 j ) u

2+(4−5 j)u

3=0

Resolviendo el sistema de ecuaciones se obtiene:

u1=3.6686∠25.209 °

u2=2.3623∠65.225 °

u3=1.1068∠26.565 °

Voltajes:

V fc=3.6686∠25.209°


4/24

V y 1=1.1068∠26.565°

V y 2=2.3623∠65.225°−1.1068∠26.565 °=1.649∠90°

V y 3=2.3623∠65.225 °−3.6686∠25.209°=2.4010∠165.96 °

Ejercicio 2:

Diagrama original



5/24

Cálculos:

f=66.62!"

ω=2πf =2πx (636.62 )=4000 rad

seg

I fc= 5

√ 2∠0 °

#nvertancias:

Γ kl=(−1)l+k cof Lkl

∆ Lkl

∆ Lkl=| L11 L12 L

21 L

22|=|16

−1

−1 5 | x10−

3

∆ Lkl=74 [mHy ]2


6/24

Γ 11=

5 x 10−3

74 [mHy ]2=67.56[ Hy]−1

Γ 22=

15 x10−3

74 x10−6=202.7 [ Hy ]

−1

Γ 12=

(−1 )1+2 x (−1 x 10−3)

74 x 10−6

=13.51[ Hy]−1

$dmitancias:

y1=G

y1=

1

5=0.2[simmens ]

y2=G

y2=

1

5=0.2[simmens ]

y3=G+ j(ωC −

1

ωL)

y3=1

5+ j ((4000∗400 x 10−

3

)− 1

4000 x ( 8 x 10−3 ) ) y3=4.76 x 10

−3−0.0305 j[simmens ]

y4= j(− Γ 11ω )=−67.5674 x103 j=−0.01689 j

y5= j

(

− Γ 2

ω

)=−202.7

4 x10

3 j=−0.05067 j

y4−5= j(− Γ 12ω )=−13.514 x103 j=−0.00338 j

Voltajes:


7/24

V fc=u1

V y 4=u2

V y 5=u3

V y 1=u1−u2

V y 2=u1−u3

$dmitancias %ro%ias de nodo:

Y 11= y1+ y2+ y3+2 ( 0 )=0.2+0.2+4.76 x10−3−0.0305 j

Y 11=0.4047−0.0305 j [simmens]

Y 22= y

1+ y

4+2 ( 0 )=0.2−0.01689 j [ simmens]

Y 33= y

2+ y

5+2( 0 )=0.2−0.05067 j [simmens ]

$dmitancias mutuas de nodo:

Y 12=− y

1+2 ( 0 )+1 ( 0)=−0.2 [simmens]

Y 13=− y2+2

(0

)+1

(0

)=−0.2

[simmens]

Y 23=0+2 ( 0 )+1 ( y4−5 )=−0.00338 j [simmens]

&istema de ecuaciones

Y 11u

1+Y

12u

2+Y

13u

3=− I fc

Y 21u

1+Y

22u

2+Y

23u

3=0

Y 31 u1+Y 32 u2+Y 33 u3=0

( 0.4047−0.0305 j )u1+(−0.2 ) u

2+(−0.2)u

3=−(

−5

√ 2∠0° )

(−0.2 )u1+( 0.2−0.01689 j )u

2+(−0.00338 j)u

3=0


8/24

(−0.2)u1+(−0.00338 j )u

2+(0.2−0.05067 j )u

3=0

&olucionando el sistema de ecuaciones:

u1=34.311∠78.579 °

u2=34.047∠84.315 °

u3=33.209∠93.754 °

Voltajes:

V fc=34.311∠78.579°

V y 4=34.047∠84.315°

V y 5=33.209∠93.754 °

V y 1=34.311∠78.579 °−34.047∠84.315°=3.4302∠−4.1475°

V y 2=34.311∠78.579°−33.209∠93.754 °=8.9816∠3.1517 °

Ejercicio :


9/24

Diagrama original:


Cálculos:

Vfv=5

√ 2∟0° [V ]

Invertancias:

Γ kl=(−1)l+k cof Lkl

∆ Lkl

∆ Lkl=| L11 L12 L21 L

22|=| 8 −1−1 2 | x10−3

∆ Lkl=15[mHy ]2

L11=8X10-3 Hy

L22=2X10-3 Hy


10/24

L12=-1X10-3 Hy

Γ11=13333 Hy-1

Γ22=!3333 Hy-1

Γ12="""" Hy

-1

Γ33=120#8 Hy-1

Admitancias

y1=15+ j ( ( 102 )−133.3310 )=15+6.66 j [simmens ]

y2=40+ j(−533.3310 )=40−53.33 j [ simmens ]

y3=20+ j ( ( 52 )−120.4010 )=20+37.95 j [simmens ]

y12=(− j 66.6610 )=−6.66 j [simmens ]

Admitancias propias de nodo

Y 11= y

1+2 ( 0 )=15+6.66 j [ simmens ]

Y 22= y

2+ y

3+2( 0 )=60−15.38 j [simmens]

Y 12=0+2 ( 0 )+1 (+ y12)=−6.66 j [simmes ]

$istema de ecuaciones

Y 11!

1+Y

12!

1=

5

√ 2∟0°

Y 21! 1+Y 22! 2=0

(15+6.66 j )! 1+(−6.66 j )!

2=

5

√ 2∟ 0°


11/24

(−6.66 j )! 1+(60−15.38 j )! 2=0

Los va%ores de & son:

! 1=0.2677−0.1166 j

√ 2=.2064 ∟−23.53 ° [V ]

! 2=

0.00500+0.031 j

√ 2=0.222∟80.83 ° [V ]

Vo%ta'es de cada e%emento:

Vy1= &1¿ .2064 ∟−23.53 ° [V ]

Vy2= &2¿0.022∟80.83° [V ]

Vfc= &1¿0.2064 ∟−23.53 ° [V ]

Vy3= &2¿0.0222∟80.83° [V ]


12/24

Ejercicio ':

Diagrama original:



13/24

Cálculos:

()c=5

√ 2∟ 0 ° [ " ]

Γ kl=(−1)l+k

cof Lkl

∆ Lkl

∆ Lkl=| 8 −4 0−4 8.3 2

0 2 2| x 10−3

∆ Lkl=6.881 0−8[ Hy]3

Γ11=18313 Hy-1

Γ12=11"28 Hy-1

Γ13=-11"28 Hy-1

Γ22=232!"Hy-1

Γ23=-232!" Hy-1

Γ33=*32!" Hy-1

Admitancias:

y1=15+ j (10 2−

183.13

10)=15+1.68 j [simmens]

y2=15+ j (10 2−232.5610 )=15−3.255 j [simmens]

y3=40

+ j(−732.56

10

)=40

−73.26

j [simmens]

y4= j(105− 1108.31 0−3 )=37.95 j [simmens]


14/24

y5= j(

1

108.310−3 )=−12.05 j [simmens ]

y12=−11.63 j [simmens ]

y13=11.63 j [simmens]

y23=23.26[ simmens]


Y 11= y

2+ y

1+2 (+ y12 )=30−24.58 j [simmens]

Y 22= y

3+ y

4+ y

5+ y

2+2 (− y23)=55−97.133 j [simmens]


Y 12=− y

2+2 ( 0 )+1 (− y12+ y23+ y13 )=−15+49.77 j [simmens ]

+cuaciones:

Y 11!

1+Y

12!

2=− I#c

Y 21!

1+Y

22!

2=0

$ustituyendo %os va%ores de %as admitancias:

( 30−24.58 j )! 1+(−15+49.77 j )!

2=+5

√ 2∟0°

(−15+49.77 j)! 1+(55−97.135 j )! 2=0

Los va%ores de & son:

! 1=

0.1766+0.00270 j

√ 2=0.1248 ∟ 0.87 ° [V ]


15/24

! 2=

0.0805−0.01692 j

√ 20.058∟−11.87°[V ]

A,ora ue conocemos %os vo%ta'es. procedemos a ca%cu%ar e% vo%ta'e de cadae%emento:

Vfc=&1¿0.1248 ∟0.87 ° [V ]

V/1= &1¿0.1248∟0.87 ° [V ]

V/3=V/#=V/!=&2¿0.0581 ∟−11.87 ° [V ]

V/#=&1 &2¿

0.176620

√ 2∟0.87 °−

0.082258

√ 2∟−11.87 °=0.0693 ∟11.52 ° [V ]

Ejercicio (:

Diagrama original:


16/24


Cálculos:

#nvertancias

L1=8 x10−3[ Hy]

L2=5 x10−3[ Hy]

L1−2=2 x 10−3[ Hy ]

L3=3 x10−3[ Hy]

L4=8 x 10−3[ Hy]

L3−4=1 x 10−3 [ Hy ]


17/24

∆ L12=| L11 L12 L

21 L

22|=|8 22 5| x10−3

∆ L12=36 [mHy ]2

Γ 11=(−1)2∗5 x10−3

36 x 10−6 =138.88 [ Hy ]

−1

Γ 22=(−1)4∗8 x 10−3

36 x10−6 =222.22[ Hy ]

−1

Γ 12=(−1)3∗2 x 10−3

36 x 10−6 =−55.55 [ Hy ]

−1

∆ L34=| L33 L34 L

43 L

44|=|3 11 8| x 10−3

∆ Lkl=23[mHy ]2

Γ 33=(−1)6∗8 x10−3

23 x10−6 =347.82[ Hy]

−1

Γ 44=(−1)8∗3 x 10−3

23 x 10−6 =130.43[ Hy ]

−1

Γ 34=(−1)7∗1 x 10−3

23 x 10−6 =−43.47 [ Hy ]

−1

$dmitancias

y1=G+ j(ωC −

Γ

ω)

y1=15+ j ((10∗2)−138.8810 ) =1()6.112j

y2=G+ j(

− Γ ω )


18/24

y2=40+ j(−222.2210 ) = '*+22.22j

y3=G+ j(ωC −

Γ

ω)

y3=20+ j ((10∗5)−347.8210 ) = 2*)1(.21,j

y4=G+ j (ωC −

Γ

ω)

y4=15+ j((10∗2)−130.43 x10

−3

10 ) = 1()6.-(j

y12= j(

− Γ ω )

y12= j(−−55.5510 ) = (.((j

y34= j (

− Γ ω )

y34= j (−−

43.4710 ) = '.'j

$dmitancias %ro%ias de nodo

Y 11= y

1+2 ( 0 )=15+6.112 j [ simmens]

Y 22= y

2+ y

3+2( 0 )=60−7.002 j [simmens]

Y 33= y

4+2 ( 0)=13+6.95 j [simmens ]

Y 12= y

12=5.55 j [simmens ]

Y 13=0 [simmens ]

Y 23= y

34=4.347 j [simmens ]


19/24

Voltajes

V fc=u1

V f$=u3

V y 1=u1

V y 2=u2

V y 3=u2

V y 4=u3

&istema de ecuaciones

Y 11u

1+Y

12u

2+Y

13u

3=− I fc

Y 21u

1+Y

22u

2+Y

23u

3=0

Y 31

u1+Y

32u

2+Y

33u

3=− I f$

( 15+6.112 j )u1+( 5.55 j )u

2+(0)u

3=−(

−5

√ 2∠0 °)

( 5.55 )u1+ ( 60−7.002 j )u2+(4.34 j )u3=0

( 0)u1+( 4.34 j )u

2+(15+6.95 j )u

3= I f$

u3=

2

√ 2∠0 °

(4.34 j )u3=(4.34 j )∗( 2√ 2∠0 ° )=

8.68 j

√ 2

( 15+6.95 j )u3=( 15+6.95 j )∗( 2√ 2∠0°)=

30+13.9 j

√ 2

&istema de ecuaciones:


20/24

( 15+6.112 j )u1+( 5.55 j )u

2+(0)u

3=−(

−5

√ 2∠0 °)

( 5.55 )u1+ ( 60−7.002 j )u

2+0=

−8.68 j

√ 2

( 0)u1+( 4.34 j )u

2− I f$=

−30+13.9 j

√ 2

(15+6.112 j 5.55 j 0−5.55 j 60−7.002 j 00 4.34 j −1)=−973.59−261.69 j

( 5

√ 2−5.55 j 0

−8.68 j

√ 260−7.002 j 0

−30+13.9 j√ 2

4.34 j −1)△

=0.229460−0.08710

u1=.24544∠−20.787 °

(15+6.112 j

5

√ 20

−5.55 j −8.68 j

√ 20

0 −30+13.9 j

√ 2−1)

△=.080920∠−77.668°=u

2


21/24

(

15+6.112 j −5.55 j 5

√ 2

−5.55 j 60−7.002 j −8.68 j

√ 2

0 4.34 j −30+13.9 j

√ 2

)△=23.723∠24.676 °= I f$

Voltajes:

V fc=24544∠−20.787 °

V f$= 2

√ 2∠0°

V y 1=24544∠−20.787 °

V y 2=.080920∠−77.668 °

V y 3=.080920∠−77.668 °

V y 4= 2

√ 2∠0°

Circuito 6:

Cálculos:


22/24

Vfv=5

√ 2∟0° [V ]

%=2

πf =2

π 318.35

=2000

H&

(mpedancias de% circuito:

' 1=2+ j ( 2000210−3 )=2+4 j [()ms ]

' 2=2+ j ( 2000110−3 )=2+2 j [()ms ]

' 3= j( −1

2000 .2510−3 )=−2 j [()ms ]

' 4=4 [()ms]

' 5=1[()ms ]

Admitancias:

y1=

1

' 1

=0.1−0.2 j [simmens]

y2=

1

' 2

=0.25−0.25 j [simmens]

y3=

1

' 3

=0.5 j [simmens ]

y4=

1

' 4

=0.25 [simmens ]

y5=

1

' 5

=1[simmens ]

+% vo%ta'e da cada uno es:

Vfv= !

1=

5

√ 2∟0° [V ]


23/24

Vy#=&2

Vy!=&3

ero saemos ue e% vo%ta'e en e%ementos en para%e%o es e% mismo. por %o tanto:

Vfv= &1 =&2 =&3=5

√ 2∟ 0 °[V ]


Y 11= y

2+ y

1+2 ( 0 )=0.35−0.45 j [simmens]

Y 22= y

3+ y

4+ y

1+2 ( 0 )=0.35−0.3 j [simmens]

Y 33= y

5+ y

3+ y

2+2( 0 )=01.25+0.25 j [simmens]


Y 12=− y

1+2 ( 0 )+1 ( 0)=−0.1+0.2 j [simmens ]

Y 13=− y

2+2( 0 )+1 ( 0)=−0.25+0.25 j [simmens ]

Y 23=− y3+2( 0 )+1 ( 0 )=−0.5 j [simmens ]

+cuaciones:

Y 11!

1+Y

12!

2+Y

13!

3= I#$

Y 21

! 1+Y

22!

2+Y

23!

3=0

Y 31!

1+Y

32!

2+Y

33!

3=0

omo so%o tenemos una inc45nita. so%o uti%i6amos %a ecuaci4n 1:

Y 11!

1+Y

12!

2+Y

13!

3= I#$

(0.35−0.45 j)( 5

√ 2)+(−0.1+0.2 j)(

5

√ 2)+(−0.25+0.25 j )(

5

√ 2)= I#$


24/24

0= I#$

ejercicios de nodos esime zac

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