ejemplos de distribucion

7/26/2019 Ejemplos de Distribucion

1/26

INTEGRANTES :

CHAVEZ TANTARICO,JEISSON M.

CURSO :

SIST. DISTRIBUCINRURAL Y URBANO

TEMA :

PROBLEMASCARACTERISTICAS DE LA

CARGA

ESCUELA PROFESINAL :

INGENIERAMECNICAYELCTRICA.

DOCENTE :

LUIS GONZALES BAZAN.

FECHADEENTREGA :Chic!"#, $% &' (!"# &' )$*+.


2/26

E'(-#).*. A/0(!(#/ 10' #/ &!2#/ &' c!34! 10' 5#/ &!5 '5 ! 2!6! ).*,-'32'5'c' ! 05# &' #/ !i('52!'/ -3i(!3i#/ &' ! c#(-!78! NL 9 NP:NO LIGHT NO PO;ER< " c0"#/ &!2#/ /#5 28-ic#/ &' 05 &8! &' i5=i'35#.R'!i>!3 ! c03=! &' ! c!34! &i!3i! 3'/-'c2i=! -!3! ' hi-#2?2ic# !i('52!-3i(!3i# &!.

L! 43@c! &' ! c!34! -!3! &ich# !i('52! -3i(!3i# 5# 10'&! &' !/i40i'52' (!5'3!


3/26

E'(-#).). A/0(i3 10' ! c!34! (@i(! !50! &' 05 !i('52! -3i(!3i#

'/ &' 2000kW , '5 ' 10' ! -?3&i&! &' -#2'5ci!, '/ &'ci3, c#63' 2#2!, #

I2

R -?3&i&!, '/ &' 80kW -#3 23'/ !/'/. S0-#5i'5 05 !c2#3 &'

-?3&i&! !50! &' 0,15 , &'2'3(i5!3

!< L! -?3&i&! &' -#2'5ci! -3#('&i# !50!.

6< L! -?3&i&! 2#2! !50! &' '5'348! &'6i ! !/ -?3&i&!/ '5 ' c#63' &' #/ci3c0i2#/ !i('52!'/.

S#0ci5

!< D' ! 'c0!ci5 ).)F


4/26

prdida de potencia media=FLSprdida de potencia en horas pico

prdida de potencia media=88kW0.15b

prdida de potencia media=12kW

6< L! -?3&i&! 2#2! !50! &' '5'348! '/

TAELCu=prdida de potencia media8760 h

ao

TAELCu=128760=105,120 kWh

E'(-# )..

H!" + c#5/0(i'/ &' 2i-# 3'/i&'5ci!'/ c#5'c2!&!/ ! 05 23!5/#3(! &'&i/23i60ci5 :DT


5/26

(

i=1

6

9W)x 0,651,1

6x 9Wx0,651,1

31,9W

). A/0(i3 10' ' !i('52! &' ''(-# ). 2i'5' 05 -ic# &' /i/2'(!&' $$$ VA -#3 !/' " -?3&i&! &' c#63' &' $.F '5 ' -ic# &' /i/2'(!.D'2'3(i5' # /i40i'52'!< L! -?3&i&! &' c#63' &' !i('52! '5 ; -#3 !/'6< P?3&i&! 2#2! &' c#63' &' !i('52! '5 ; -#3 !/'/

S#0ci5!< L! -?3&i&! &' c#63' &' !i('52! '5 ; -#3 !/' '/

I2R $.F Pic# &' /i/2'(!

I2

R $.$$F $$$ VA

I2R *F ; -#3 !/'

6< L! -?3&i&! 2#2! &' c#63' &' !i('52! '5 ; -#3 !/'/ '/

I2R *F F ; -#3 !/'/

).F A/0(i3 10' 'i/2'5 ) !i('52!'/ -3i(!3i#/ /0(i5i/23! -#3 05# &'#/ 23!5/#3(!'/ /i20!/ '5 ' NL 9 NP/ '5 !/ #3i!/ &' !

/06'/2!ci5 &' &i/23i60ci5, (#/23! '5 ! 4. ).+. U5# &' #/!i('52!'/ /0(i5i/23! 05! c!34! i5&0/23i! 10' /' -3#&0c'-3i5ci-!('52' '523' Q$$ A.M " ** P.M, c#5 05 -ic# &' )$$$ ; ! !/ F P.(.E #23# !i('52! c!34!/ 3'/i&'5ci!'/ 10' /' -3#&0c' -3i5ci-!('52' '523'!/ + AM " *) PM c#5 05 -ic# &' )$$$ ; ! !/ PM, (#/23! '5 ! 403!).%. D'2'3(i5' # /i40i'52'!< !c2#3 &' &i='3/i&!& '5 ! c!34! c#5'c2!&! '5 ' 23!5/#3(! T.6< Di='3/i&!& &' c!34! &' ! c!34! c#5'c2!&! '5 ' 23!5/#3(! Tc< !c2#3 &' c#i5ci&'5ci! '5 ! c!34! c#5'c2!&! '5 ' 23!5/#3(! T

S#0ci5


6/26

FD i=1

2

D i

Dg

FD 2000+2000

3000 *,

6!3 &!2#/ &!/ '5 ''(-# ).* -!3! ! c03=! &' c!34! &' NL 9 NP/.

N#2' 10' ' -ic# #c033' ! !/ F PM. D'2'3(i5'!< !c2#3 &' c!/' &' c#523i60ci5 -!3! c!&! 05! &' !/ c!/'/ &' c!34!

6< !c2#3 &' &i='3/i&!& -!3! ' !i('52! -3i(!3i#

c< D'(!5&! (@i(! &i='3/ic!&! &' ! c!34! &' 430-#

&< !c2#3 &' c#i5ci&'5ci! &' 430-# &' c!34!

S#0ci5

!< E !c2#3 &' c#523i60ci5 &' c!/' '/

Ci La demandade c!ase en e! momento pico de! sistema

demandamaxima de c!ase no coincidente(indi"i"ua!)

P!3! !0(63! -K6ic#, 3'/i&'5ci! " c!34! c#('3ci!,

Ca!umbrado 0W

100W $


8/26

Cresidencia! 600W

1000W $.+

Ccomercia! 1200W

1200W

*.$

6< P#3 'c0!ci5 ).**, ' !c2#3 &' &i='3/i&!& '/

FD i=1

2

D i

Dg

Y &' 'c0!ci5 ).*Q,Dg C1 D1 C2 D2 C3 D3 ... Cn Dn

S0/2i20"'5 ).*Q &'523# &' ! 'c0!ci5 &' ).**

FD

i=1

n

Di

i=1

n

Cix Di

FD i=1

3

Di

i=1

3

Cix Di

FD 100+1000+1200

0x 100+0.6x1000+1x1200

FD

*.)%Q

c < L! &'(!5&! (@i(! &i='3/ic!&! '/ ! &'(!5&!

(@i(! c#i5ci&'52', 10' '/ Dg . P#3 # 2!52#, &' !

'c0!ci5 ).*, ' !c2#3 &' &i='3/i&!& '/

FD i=1

n

TCDi# DFi

Dg


9/26

D#5&' ! &'(!5&! (@i(!, &' 'c0!ci5 ).*) '/

M@i(! &'(!5&! D'(!5&! c#5'c2!&! 2#2! D

S0/2i20"'5 ! 'c0!ci5 ).*) &'523# &' ! 'c0!ci5 ).*

FD i=1

n

D i

Dg

Dg i=1

n

D i

FD

P#3 # 2!52#, ! &'(!5&! (@i(! &i='3/ic!&! &' ! c!34! &' 430-# '/

Dg i=1

3

D i

FD

Dg 100+1000+1200

1.278

Dg

* Q$$ ;&< E !c2#3 &' c#i5ci&'5ci! -!3! ! c!34! &' 430-#, &''c0!ci5 ).*F, '/

Fc

Dg

i=1

n

D i

O, &' 'c0!ci5 ).*+

Fc 1

FD

Fc 1

1.278

Fc $.%Q)F


10/26

2.7 El factor de carga promedio de una subestacin es 0,65. Determinar el factor

promedio de la prdida de sus alimentadores, si los servicios de la subestacin:

a una !ona urbana.

b una !ona rural.

"olucin:

a una !ona urbana.

F

(LD)2

FLS=0,3FLD+0,7

0,3 (0,65 )+0,7(0,65)2

0,49

b una !ona rural.


11/26

F

(LD)2

FLS=0,16FLD+0,84

0,16 (0,65 )+0,84 (0,65)2

0,53

).Q A/0(i3 10' ! #3i! &' ! /06'/2!ci5 &' &i/23i60ci5 &' ! NL 9 NPC#(-!5" -3#=''! &' 05 -0'6# !52!/(!, 10' '/ 05 -'10'7! ci0&!&,'-'3i('52! 05! c!34! -ic# !50! &' F$$ ;. L! '5'348! 2#2! !50!/0(i5i/23!&! -!3! #/ ci3c0i2#/ !i('52!'/ -3i(!3i#/ '/ *$ $$$ $$$ ;h.L! &'(!5&! -ic# #c033' '5 0i# # !4#/2# " '/ &'6i ! ! c!34! &' !i3'!c#5&ici#5!.

!< E5c#523!3 ! &'(!5&! !50! &' -#2'5ci! -3#('&i#

6< E5c#523!3 ' !c2#3 &' c!34! !50!S#0ci5

A/0(i3 05! c03=! &' c!34! ('5/0! (#/23! '5 ! 4. ).**

! < L! &'(!5&! !50! &' -#2'5ci! -3#('&i# '/


12/26

A50! $prom energ%atota!anua!

ao

A50! $prom 10

7Wh/ao

8760h /ao

A50! $prom * ** ;

6 < P#3 'c0!ci5 ).+, ' !c2#3 &' c!34! !50! '/

FLD carga promedioanua!

demanda pico anua!

FLD 1141W

3500W

FLD $.)+

O, &' 'c0!ci5 ).Q

!c2#3 &' c!34! !50!

energ%atota! anua!

carga pico anua! x 8760


13/26

!c2#3 &' c!34! !50!

107Wh

ao

3500W #8760

!c2#3 &' c!34! !50! $.)+

). U/' ' &!2# &! '5 ' ''(-# ).Q " /0-#5'3 10' 05! c!34! 50'=! &'*$$ ; c#5 *$$ &' !c2#3 &' c!34! !50! /' =! ! /0(i5i/23!3 &'/06'/2!ci5 #3i!. E IC, # c#/2# &' c!-!ci&!&, &' /i/2'(! &' -#2'5ci!!40!/ !33i6!, 10' '/, h!ci! ' 4'5'3!, &' '/2! /06'/2!ci5 '/ *Q.$$;-#3 ('/. A/0(i3 10' ! '5'348! '523'4!&! ! !i('52! -3i(!3i# c#/2!3@ !-3#='', 10' '/, NL9NP, $.$+;h

!< E5c#523!3 ' 50'=# !c2#3 &' c!34! !50! '5 ! /06'/2!ci56< E5c#523!3 ' c#/2# 2#2! !50! -!3! NL9NP -!3! /'3=i3 '/2! c!34!

S#0ci5

FLD carga promedioanua!

demanda pico anua!

FLD 1141+1003500+100

FLD $.F


14/26

6 < E c#/2# !50! 2#2! " !&ici#5! &' NL9NP -!3! /'3=i3 ! ! c!34! !&ici#5!2i'5' ) c#(-#5'52'/ &' c#/2#/, &' (!5'3! :i< c#/2# !50! &' c!-!ci&!& ":ii< c#/2# &' '5'348! !50!.

C#/2# &' c!-!ci&!& !&ici#5! !50!, -#3 # 2!52#

C#/2# !50! &' c!-!ci&!& !&ici#5! *Q ;('/ *) ('/!7# *$$

C#/2# !50! &' c!-!ci&!& !&ici#5! )*,+

C#/2# &' '5'348! !50! *$$; W Q%+$ h!7# $.$+;h

C#/2# &' '5'348! !50! F).F+

E52#5c'/

C#/2# 2#2! !50! !&ici#5! c#/2# !50! &' c!-!ci&!& c#/2# &' '5'348!!50!

C#/2# 2#2! !50! !&ici#5! )*.+F).F+

C#/2# 2#2! !50! !&ici#5! %.*+


15/26

Ejemplo 2.10

A/0(' 10' ! '523!&! &' ! c!34! (@i(! !50! ! 05 !i('52!-3i(!3i# '/ &' ) $$$ X;. U5 -3#43!(! &' c#(-02!! 10' c!c0!

c!8&!/ &' 2'5/i5 " : I2

& R < -?3&i&!/ &'(0'/23!5 10' ! -?3&i&! 2#2!

&' c#63' '5 ' (#('52# &' c!34! &' -ic# '/ I2

& R=100kW . L!

'5'348! !50! 2#2! /0(i5i/23!&! ! '23'(# '(i/#3 &' !i('52! '/F.+* *$+ X;h.

!< M'&i!52' ' 0/# &' ! 'c0!ci5 ).$, &'2'3(i5!3 ' !c2#3 &'-?3&i&! !50!.

6< C!c0!3 ! '5'348! 2#2! &' ! -?3&i&! !50! &' c#63' " /0 =!#3! $.$ X;h

Soluc!"

#$ D' ! 'c0!ci5 ).$, ' !c2#3 &' -?3&i&! !50! '/

FLS=0.3FLD+0.7FLD2

D#5&'

FLD= 5.61x10

6kWh

2000kWx8760h /'r

FLD=0.32

P#3 # 2!52#,

FLS=0.3x0.32+0.7x 0.322

FLS=0.1681

%$ D' ! 'c0!ci5 ).)F

FLs= prdidade potenciamedia

prdida de potencia en horas pico

P'3&i&! &' -#2'5ci! ('&i! $.*+Q* *$$ X;

*+.Q* X;

P#3 # 2!52#,

P?3&i&! 2#2! !50! &' c#63' *+.Q* X; Q%+$ h"3


16/26

*%,$$$ X;h

Y

C#/2# &' ! -?3&i&! 2#2! !50! &' c#63' *%,$$$ X;h $.$+ X;h

QQ)$

E&EMPLO 2.11

S0-#54!(#/ 10' 05# &' #/ DT &' ! /06'/2!ci5 Ri='3/i&' /0(i5i/23! 23'/!i('52!'/ -3i(!3i#/. L#/ $ (i5 &'(!5&!/ !50!'/ (!i(# -#3!i('52! /' '50('3!5 '5 ! /i40i'52' 2!6!, 052# c#5 ' !c2#3 &'-#2'5ci! :P< '5 ! c!34! (@i(! !50! &' 2i'(-#.

D'(!5&!Ai('52! ; P

* *Q$$ $.F) )$$$ $.QF ))$$ $.$

S0-#54!(#/ 10' 05 !c2#3 &' &i='3/i&!& &' *,*F '523' #/ 23'/!i('52!'/ 2!52# -!3! -#2'5ci! 3'! :P< " ! -#2'5ci! 3'!c2i=! :


17/26

FD=1800+2000+2200

D g=1.15

Por lo tanto

Dg=6000

1.15

=5217 kW=$

Para encontrar la energa en amperios kilovoltios, los ngulos PF no

se han determinado. Por lo tanto,$F

1=cos (

1=0.95)(

1=18.2*

$F2=cos(2=0.85) (2=31.79*

$F3=cos(3=0.90) (3=25.84*

Por lo tanto, las diversidades de la potencia reactiva (Q) es

+=i=1

3

$ix tan(

FD

1800x tan 18.2*+2000x tan31.79*+2200x tan25.84 *

1.15

2518.8 k"ar

Por lo tanto,

Dg=($2++2 )

1

2=S

Dg=(52172+2518.82 )

1

2=5793.60 k,A

) Para la ecuacin 2.1!, el "# es

LD=i=1

3

DiDg

60005217=783kW

c) $n la lista trans%ormador dado, es apropiado seleccionar eltrans%ormador con la cali&cacin de '!(0*+! k- /a ue con la

sorecarga de corta duracin 2( ue tiene una capacidad de


18/26

*+!1.2( 3 ((.k-, ue es ma/or ue la demanda mima de(.!4',+0 k- como se encuentra en la parte 5a)

d) 6enga en cuenta ue los ventiladores pla7o sore la cali&cacinsigni&ca ue la cali&cacin re%rigerado por aire %or7ado. Para

encontrar el incremento 5g) por a8o,

(1+g )10=2

Por lo tanto1+g=1.07175

9g=7.175/ao

s

(1.07175)nx 5793.60=9375 k,A

9

(1.07175 )n=1.6182

Por lo tanto,

n= ln1.6182

ln1.07175

0.48130

0.06929=6.946o7aos

Por lo tanto, es ue se instala el nominal del trans%ormador

!(004'!( k-, se cargar a sus %ans en cali&cacin en

aproimadamente ! a8os.

).*) A/0(i3 05 23!5/#3(! &' &i/23i60ci5 28-ic# 10' /i3=' + c!34!/3'/i&'5ci!'/, 10' '/ c!/!/, ('&i!52' + !c#('2i&!/ &' /'3=ici# " ) 23!(#/&' 85'! /'c05&!3i!. S0-#5'3 10' '/2#/ /#5 05 2#2! &' *F$ TD/ " $$3'/i&'5ci!/ /0(i5i/23!&!/ -#3 '/2' !i('52! -3i(!3i#. U/' 4. ).* "2!6! ). P#3 ' (#2i=# &' i0/23!ci5. S0-#54!(#/ 10' 05! 3'/i&'5ci! 28-ic!c#52i'5' 05! /'c!! &' 3#-!, 05 3!54#, 05 3'3i4'3! " 05 -#c# &'i0(i5!ci5 " !-!3!2#/ &i='3/#/.

D'2'3(i5' # /i40i'52'

! < L#/ $ (i502#/ (@i(#/ &' &'(!5&! &i='3/ic!&! '5 ' 23!5/#3(!&' &i/23i60ci5

6 < L#/ $ (i502#/ (@i(#/ &' &'(!5&! &i='3/ic!&! '5 2# '!i('52!

c < U2ii>!3 #/ 28-ic#/ !c2#3'/ &' =!3i!ci5 -#3 h#3! &!&!/ '5 ! 2!6! ).) "c!c0' ! -'10'7! -#3ci5 &' ! c03=! &' &'(!5&! &i!3i! '5 '23!5/#3(! &' &i/23i60ci5, 10' '/, ! &'(!5&! &i='3/ic!&! 2#2! -#3

h#3! &' , F, " + PM, '5 ' 23!5/#3(! &' &i/23i60ci5, '5 ;


19/26

S#0ci5

*.+ ;c!/! -!3! /'c!!

$prom- max $.Q ;c!/! -!3! 3!54'

$.$++ ;c!/! -!3! 3'3i4'3!

$.+* ;c!/! -!3! i0(i5!ci5 " !-!3!2#/ &i='3/#/

A/8

prom& max i

$

i=14

*.+ $.Q $.$++ $.+* .$%+ ;c!/!

Y -!3! + c!/!/

:.$%+ ;c!/!


20/26

prom& max i

$

i=1

4

*.) $.F $.$ $.F) ).) ;c!/!

P#3 # 2!52#, #/ $ (i502#/ (@i(#/ &' &'(!5&! &i='3/ic!&! '5 2# '!i('52! '/

prom& max i

$

i=1

4

$$ ).) ) $+.+ ;c!/!

c < Si5 '(6!34#, /i #/ $ (i502#/ &' &'(!5&! &i='3/ic!&! (@i(! '5 '23!5/#3(! &' &i/23i60ci5 '5c#523! '5 ! -!32' ! '/ (02i-ic! -#3*F$ -!3! &'2'3(i5!3 #/ $ (i502#/ &' &'(!5&! &i='3/ic!&! (@i(! '52# ' !i('52!, ! 3'/-0'/2! /'38!

*F$ *Q.F ) )%F ;

0' '/ (!"#3 10' ) $+.+ ; '5c#523! -3'=i!('52'. E/2! &i/c3'-!5ci!'/ &'6i ! ! !-ic!ci5 &' !/ !-ic!ci#5'/ &i='3/!/

c < P#3 ! 2!6! ).), #/ !c2#3'/ &' =!3i!ci5 -#3 c!&! h#3! /' -0'&'5'5c#523!3 c#(# $.Q, $.), $. " $.) -!3! /'c!!, 3!54#, 3'3i4'3!, 'i0(i5!ci5 " !-ic!ci#5'/ (i/c'@5'!/, 3'/-'c2i=!('52'. P#3 # 2!52#, '2#2! &' &'(!5&!/ &i='3/ic!&!/ -#3 h#3! '5 ' 23!/#3(! &' &i/23i60ci5-0'&'5 /'3 c!c0!&!/ c#(# /' i5&ic! '5 ! /i40i'52' 2!6!

:*.+ ;c!/!


21/26

E&EMPLO 2.1'

A/0(! 10' F ; !c#5&ici#5! &' !i3' i38! '5 Q$ &' 2i'(-# :cic# &'23!6!# Q$< &03!52' ! h#3! -ic# " -#&38! '/2!3 i(i2! -#3 ! 02ii&!& &'05 c#523# 3'(#2# -!3! 05 cic# &' 23!6!# &' +F. &'2'3(i5!3 # /i40i'52'

E 5K('3# &' (i502#/ &' 05ci#5!(i'52# 5'4 ! 5! &' * h#3! &' c#523#&' ! 05i&!&.

L! c!52i&!& &' c#5/0(# &' '5'348! 3'&0ci &03!52' ! h#3! -ic# /i /'!-ic! 2! c#523# /i(02@5'!('52' ! *$$,$$$ !c#5&ici#5!'/ &' !i3' '52# ' /i/2'(!.

L! c!52i&!& 2#2! &' 3'&0cci5 &' c#5/0(# &' '5'348! &03!52' ' -ic#.

E i(-#32' 2#2! &' ! 3'&0cci5 !&ici#5! '5 ' c#5/0(# &' '5'348! '5 !-!32' :c< /i !/ -?3&i&!/ &' 23!5/(i/i5 " &i/23i60ci5 '5 ' -ic# '/ &' Q.

S#0ci5

:!< L! c!52i&!& &' #-'3!ci5 10' 5'4 '/

(0.80.65 ).(60minh)=90min

:6< L! c!52i&!& &' c#5/0(# &' '5'348! 3'&0ci &03!52' ' -ic# '/(0.80.65 ). (5kW)=0.75kW

:c< L! c!52i&!& 2#2! &' 3'&0cci5 &' '5'348! -!3! *$$,$$$ 05i&!&'/ '// S (t)=Spromedio . [D incontro!ados ( t)Dcontro!ados(t)] . 0

/ S (t)=(5kW).3 [0.80.65 ] .100,000.

/ S (t)=751W

:&< L! c!52i&!& !&ici#5! 2#2! &' 3'&0cci5 &' '5'348! &'6i ! !3'&0cci5 &' !/ -?3&i&!/ &' 23!5/(i/i5 " &i/23i60ci5 '/

(751W) .0.08=61W

A/8, ! 3'&0cci5 2#2! 4'5'3! '/

751W+61W=811W

E/2' ''(-# (0'/23! # !23!c2i=# &' c#523#!3 !c#5&ici#5!'/ &' !i3'-!3! c#(-!78!/ &' /'3=ici# -K6ic#.


22/26

E&EMPLO 2.1(

A/0('/ 10' ! c!34! '/ ('&i&! / ='c'/ c#5 05 ('&i &' =!2i#/ h#3!/' c0! 2i'5' 05 ('23# c#5/2!52' &' %.) " #/ /i40i'52'/ &!2#/ /#5 #62'5i/

D'2'3(i5! ! &'(!5&! i5/2!52@5'! " ! &'(!5&! ('&i!

S#0ci5

D'(!5&! i5/2!!&! &' ! -3i('3! 'c203!

D*3,6x rx g

T

D*3,6x 32x 7.2

59

D**.$FQ;

D'(!5&! i5/2!!&! &' ! -3i('3! 'c203!

D)3,6x rx g

T

D)3,6x 27x7.2

40

D**%.+;

L! &'(!5&! -3#('&i

DP3D1+D2

2

DP314.058+17.496

2

DP3*F.%%%;


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E&EMPLO 2.1)

A/0(! 10' #/ &!2#/ &!/ '5 ' ''(-# ).*% /#5 #/ 3'/02!/ &' !c!34! ('&i&! c#5 [!22h#3! ('23#/ " &' #/ i5/230('52#/ &'23!5/#3(!'/. S0-#5'3 10' ! 50'=! c#5/2!52' (?23ic! '/ *.Q " 10' #/3!&i#/ &' ! CT/ " PT/ 0/!/ /#5 )$$ " *, 3'/-'c2i=!('52'. D'2'3(i5' !/&'(!5&!/ i5/2!52@5'!/ -!3! !(6!/ 'c203!/ " ! &'(!5&! -3#('&i#.

S#0ci5

D1=3.6thCTR$TR

T

D1=3.6321.82001

59

D1=702.9kW

Y

D2=3.6271.82001

40

D2=874.8kW

A/8, ! &'(!5&! -3#('&i# '/

Dpromedio=D1+D2

2

Dpromedio=702.9874.8

2

Dpromedio=788.9kW

E&EMPLO 2.1*:


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"upongamos #ue la carga se mide con $%& ' (ar%& metros, transformadores de medida

' #ue se obtienen los siguientes datos.

Lecturas en W-h Lecturas en Var-h

)* "et 2* "et )* "et 2* "et

+evoluciones del disco 20 0 )0 20-ntervalo de tiempo para las

revoluciones del disco50 60 50 60

"uponga #ue los nuevos constantes medidor son ),2 ' #ue las proporciones de la ts '

/ts utili!ados son 0 ' 20, respectivamente. Determinar lo siguiente:

a 1as demandas instantneas en 3$.

b 1a demanda promedio en 3$.

c 1as demandas instantneas en 3(4+.

d 1a demanda promedio en 3(4+.e 1a demanda promedio en 3(4.

Solucin:

a 1as demandas instantneas en 3$ son:

D1=3,6.20.1,2.80.20

50

D1=2764,8 kW

D2=3,6.30.1,2.80.20

50

D2=3456 kW

b 1a demanda promedio en 3$ es:

Da"=D1+D2

2

Da"=2764,8+3456

2


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Da"=3110,4 kW

c 1a demandas instantneas en 3(4+ son:

D1=3,6. r . h .CTR.$TR

T

D1=3,6.10.1,2.80.20

50

D1=1382,4 kW

D2=3,6.20.1,2.80.20

60

D2=2304 kW

d 1a demanda promedio en 3(4+ es:

Da"=D1+D2

2

Da"=1382,4+2304

2

Da"=1843,2 kW

e 1a demanda promedio en 3(4 es:


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D a" kW

(Da" kW)2+

D a"=

1843,2

(3110,4)2+

Da"=

Da"=3615,5 kW

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