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TUGAS INSTALASI LISTRIK II

PERENCANAAN ULANG PADA PANELPOMPA HGF AB

Oleh :

NAMA : SELLI NUR PRAMUDITA

NIM / ABSEN : 1131120101 / 24

KELAS : D3 TL 2A

PROGRAM STUDI TEKNIK LISTRIK

JURUSAN TEKNIK ELEKTRO

POLITEKNIK NEGERI MALANG

2013

SOAL

(No absen genap)



Dari gambar berikut tentukan :

1 Rekapitulasi daya2 Breaking capasity di M1, M2, M3 bila panel tersebut terhubung

dengan trafo 1000 k! , 20k"#00 $ %arak trafo ke panel adalah

300 m$ Rencanakan ulang gambar layout$3 Bila akan dipasang kapasitor di beban secara kelompok, buat

perencanaan instalasinya$ & Motor dibebani '() dari dayaoutputnya*$

# +rounding sistem$

+ambar 1$ ingle line panel pompa -+. !B



+ambar 2$ Rekapitulasi Daya /anel pompa -+. !B



Jawaba

I! Re"a#$%&la'$ Da(a

M)%)* +, -1.P = 22 kW

In = 42 A

Cos ∅ = 0,79

S =22000

0,79=27848,10VA

a! Se%%$ #eaa

= 250 ! 42A

= "05 A # "00 A

• Con$o% &'()*)%+n &'n+(+n -+n $'./+&+$ /+*+( k+$+*o



b! Aal$'a a S%aa*% KA

1HA = "25 ! 42=52.5 A ≈ 5 A

3PUIL +'* 7685+ k+'* /+.+ '.)n$) s+$:

• Con$o% &'()*)%+n s$+n/+.$ 1HA /+*+( PUIL 2000

! L&a' Pea#a Kabel * = "00 (

A= √ 3 l I n

x (V . V )

¿ √ 3 .100. 42

56 (3 x 400 )

¿10,825 mm2

≈16mm2

1+'* -+n /)n+k+n N;; 4 ! 3 " ! "< ((2

: 3PUIL +'* 7685+

k+'*/+.+ '.)n$) s+$:



• Con$o% &'()*)%+n L+s P'n%+n$+. 1+'* &+/+ PUIL 2000

M)%)* +, -2.

P = 22 kW

In = 42 A

Cos ∅ = 0,79

S =22000

0,79 =27848,10VA



a! Se%%$ #eaa

= 250 ! 42A

= "05 A # "00 A


1HA = "25 ! 42=52.5 A ≈ 5 A

3PUIL +'* 7685+ k+'* /+.+ '.)n$) s+$:

! L&a' Pea#a Kabel

* = "50 (

A= √ 3 l I n

x (V . V )

¿ √ 3 .150. 42

56 (3 x 400 )

¿16.24mm2

≈25mm2

1+'* -+n /)n+k+n N;; 4 ! 3 " ! 25 ((2 : 3PUIL +'* 7685+ k+'*

/+.+ '.)n$) s+$:

M)%)* +, -3.

P = 45 kWIn = <65 A

Cos ∅ = 0,2

S =45000

0,82=54878.05 VA

a! Se%%$ #eaa

= 250 ! <65A

= 2"<624 A # "50 A


1HA = "25 ! 86.5=108.125 A ≈ "05 A 3PUIL +'* 7685+ k+'* /+.+

'.)n$) s+$:

! L&a' Pea#a Kabel

* = 200 (

A= √ 3 l I n

x (V . V )



¿ √ 3 .200.86.5

56 (3 x 400)

¿44.59mm2

≈50mm2

1+'* -+n /)n+k+n N;; 4 ! 3 " ! 50 ((2

: 3PUIL +'* 7685+

k+'*/+.+ '.)n$) s+$:

M)%)* DOL -4.

P = 565 kW

In = "269 A

Cos ∅ = 0,

S =5500

0,83=6626.506VA

a! Se%%$ #eaa= 250 !"269 A

= 2625 A # 2 A

! Aal$'a a S%aa*% KA

1HA = "25 ! 12.9=16.125 A ≈ 2< A 3PUIL +'* 7685+ k+'* /+.+

'.)n$) s+$:

e! L&a' Pea#a Kabel

* = 270 (

A= √ 3 l I n

x (V . V )

¿ √ 3 .270.12.9

56 (3 x 400)

¿13.15mm2

≈16mm2

1+'* -+n /)n+k+n N;; 4 ! 3 " ! "< ((2 : 3PUIL +'* 7685+ k+'*

/+.+ '.)n$) s+$:

M)%)* +, -.

P = 765 kW

In = "7 A

Cos ∅ = 0,7

S =7500

0,78=9615.38VA

a! Se%%$ #eaa

= 250 ! "7A

= 4265 A # 40 A




1HA = "25 ! 17=21.25 A ≈ 2< A 3PUIL +'* 7685+ k+'* /+.+

'.)n$) s+$:

! L&a' Pea#a Kabel

* = 00 (

A= √ 3 l I n

x (V . V )

¿

√ 3 .300.17

56 (3 x 400 )

¿13.15mm2

≈16mm2

1+'* -+n /)n+k+n N;; 4 ! 3 " ! "< ((2 : 3PUIL +'* 7685+ k+'*

/+.+ '.)n$) s+$:

Peaa U%aa -M55B.

¿250 × I nterbesar+ I nmotor yanglain

¿(250 × 86.5)+(42+42+12.9+17)

¿330,15 A ≈320 A

Pe*h$%&a B&'ba* III

I ntotal=200,4 A

KHA=125 × I ntotal

= 125 ×200,4=250,4 A ≈287 A

+/) s+. ('nn+k+n &'n+(&+n 25! ((2 ≈ 75 ((2 3PUIL +'* <6<8"

*+&)s+n kon/k$)> '.?(*+% s+$:

• Con$o% &'()*)%+n *+s &'n+(&+n s+. &+/+ PUIL 2000



• Se%%$ K)%a"%)* a TOR

1! U%&" M)%)* 1

Daya Motor = 22 kW

In = 42 A

Se%ela TOR

¿√ 3 =

42√ 3 = 24,24 A # " A

• C+.+ &'()*)%+n OR &+/+ k+$+*o



K)%a"%)*

S'$'*+n ()n)(( kon$+k$o. = ""5 In

""5 6 42 = 4, A # 50 A

• C+.+ &'()*)%+n 1on$+k$o. &+/+ 1+$+*o



2! U%&" M)%)* 2 - P S$*&# AB 2 .

Daya Motor = 22 kW

In = 42 A

Se%ela TOR

¿

√ 3 =

42

√ 3 = 24,24 A # " A

K)%a"%)*

S'$'*+n ()n)(( kon$+k$o. = ""5 In

""5 6 42 = 4, A # 50 A

3! U%&" M)%)* 3 - 5)#*e')* G6 U%a*a .

D+-+ Mo$o. = 45 kW

In = <,5 A

Se%ela TOR

¿√ 3 786

,5

√ 3 7 49,94 A # 45 A

K)%a"%)*

S'$'*+n ()n)(( kon$+k$o. = ""5 In

""5 6 <,5 = 99,5 A # "25 A

4! U%&" M)%)* 4 - TLG G)(a SS U%a*a .

D+-+ Mo$o. = 5,5 kW

In = "2,9 A

Se%ela TOR"00 In 7 "2,9 A # "0 A



K)%a"%)*

S'$'*+n ()n)(( kon$+k$o. = ""5 In

""5 6 "2,9 = "4,5 A # " A

! U%&" M)%)* - M$8e* D1 A%a' .

D+-+ Mo$o. = 7,5 kWIn = "7 A

Se%ela TOR

¿√ 3 7

17

√ 3 = 9, A # 765 A

K)%a"%)*

S'$'*+n ()n)(( kon$+k$o. = ""5 In

""5 6 "7 = "9,55 A # 25 A



• LA;OU



II! B*ea"$ 5a#a'$%(

.+>o "000 1@+ , 20 k 400

Ja*$a '$'$ a%a' %*a9)Uo = 400

L = 20 k

S = "000 kAU SC =6

• I n=

S ×10−3

U 0

.√ 3

I n=1000×10

−3

400 .√ 3

=1443,376 A=1,44 kA

• I sc= I n×100

U sc

I sc=1443,376 ×100

6=24056,27 A=24,056 kA

• Psc=√ 3 .V L . I sc

¿√ 320 .103.24,056.10

3

¿833,33 MVA

•

1=

Uo2

P sc

¿ 400

2

833,33



¿192m!

• "1=Uo2

P sc

x cos# .10−3

¿ 400

2

833,33 x0,15 .10

−3

¿00,028m!

• $

1=

Uo2

Psc

xsin# .10−3

¿ 400

2

833,33 x0,98 .10−3

¿0,188m!

T*a9)

S = "000 kA

Us = <

Uo = 400

.+>o ()n-+k 3C%+&$'. G <65:

R 2 = 2, ( !

2 = "0, ( !

Kabel 1

1HA" = "25 ! In = "25 ! "44, = "04,225 A

P'. &%+s+ =1804,225

3=601,408 A ≈612 A

1+'* -+n /)n+k+n N;; 4!3"!240 ((2: 3PUIL +'* 7685+ k+'* /+.+

'.)n$) s+$:

ρ = 22,5 3C:

* = 25 (

R =22,5 x 25

4.240=0.59m!

= 0,"2 ! L = 0,"2 ! 25 = ( !

Pe&%&' a(a 1 -M1.

R 4 = 0 m! 4 = 0 m!

R $" = R " R 2 R R 4 = 0,02 2, 0659 0 = 269 ( !

$" = " 2 4= 0," "0, 0 = "64 ( !



Is " =

400

√ 3√ (2.92+13.4882)

= 400

√ 313.79=16.74 kA

REL B&'ba* 1

ρ = 22,5 3C:* = 2 (

1HA s+. = <0",4" A # 70 A

Bs+. -+n /)n+k+n = 35!50: ((2 3PUIL +'* <6<8" *+&)s+n kon/k$)>

'.?(*+% s+$:

R 5 =

22,5 x 2

250=0.18m!

5 = 0,"5 ! L = 0,"5 ! 2 = 06 (

Pe&%&' a(a 2 -M2.

R < = 0 m! < = 0 m!

R $2 = R $" R 5 R < = 269 0.18 0 = 60 ( !

$2 = $" 5 < = "64 06 0 = "679 ( !

Is 2 =

400

√ 3√ (3.082+13.792 )=

400

√ 314.13=16.35kA

Kabel 2

A=

√ 3 l I n

x (V . V )

¿ √ 3 .300.200,4

56 (3 x 400 )

¿154.9mm2

≈185mm2

1+'* -+n /)n+k+n N;; 4!3"!"5 ((2: 3PUIL +'* 7685+ k+'* /+.+

'.)n$) s+$:

ρ = 22,5 3C:

* = 00 (

R 7 =22,5 x 300

4.185=9.12m!

7 = 0,"2 ! 00 = 0,"2 ! 00 = < ( !

REL B&'ba* 2

ρ = 22,5 3C:

* = 2 (

1HA s+. = "25 ! 20064 = 25065 A = 27 A



Bs+. -+n /)n+k+n = 325!: ((2 3PUIL +'* <6<8" *+&)s+n kon/k$)>

'.?(*+% s+$:

R =

22,5 x 2

75=0.6m!

= 0,"5 ! L = 0,"5 ! 2 = 06 (

Pe&%&' Da(a 3 -M3.

R 9 = 0 m! 9 = 0 m!

R $ = R $2 R 7 R R 9 = 60 9.12 0.6 0 = "26 ( !

$ = $2 7 9 = "679 < 06 0 = 50609 ( !

Is =

400

√ 3√ (12.82+50.092 )=

400

√ 351.69=4.47 kA

Kabel 3

#&1 1'( mm2* ! 4#0 mm2 3PUIL +'* 7685+ k+'* /+.+ '.)n$)

s+$:5cu 22,(l 2 m

R10 %. l

A 22,5 .2

740 0,06 m7

810 0,12 l 0,12 2 0,2# m7

REL B&'ba* 3

ρ = 22,5 3C:* = 2 (

1HA s+. = "25 ! 20064 = 25065 A = 27 A

Bs+. -+n /)n+k+n = 325!: ((2 3PUIL +'* <6<8" *+&)s+n kon/k$)>

'.?(*+% s+$:

R "" =

22,5 x 2

75=0.6m!

"" = 0,"5 ! L = 0,"5 ! 2 = 06 (

Pe*h$%&a Kabel a B*ea"$ 5a#a'$%( #aa Ma'$,a'$ M)%)*

1! Paa )%)* +, -1.1+'* -+n /)n+k+n N;; 4 ! 3 " ! "< ((2 : 3PUIL +'* 7685+ k+'* /+.+

'.)n$) s+$:

ρ = 22,5 3C:

* = " (

R "2+ =22,5 x 1

4.16=0.35m!

"2+ = 0,"2 ! L = 0,"2 ! " = 06"2 ( !

Pe&%&' Da(a M)%)* 1 -M4a.



R "+ = 0 m! "+ = 0 m!

R $5 = R $ R "0 R "" R "2+ R "+ = "26 0,06 9 0.35 0= "62" ( !

$5 = $ "0 """2+ "+= 50609 0,2# 06"2 0= 50645 ( !

Is M4+ =400

√ 3√ (13.212+50.452 )= 400

√ 352.15=4.42 kA

Kabel -R 14a;14a.

N;; 4!3" ! "< ((2: A = <4 ((2

= 22,5

* = "00 (

R"4+ = %. l

A =22,5 .100

64 = 5,2 (

"4+ = 0,"2 ! * = 0,"2 ! "00 = "2 (

2! Paa )%)* +, -2.

• 1+'* -+n /)n+k+n N;;4 ! 3 " ! 25 ((2 : 3PUIL +'* 7685+ k+'* /+.+

'.)n$) s+$:

ρ = 22,5 3C:

* = " (

R "2 =22,5 x 1

4.25=0.225m!

"2 = 0,"2 ! L = 0,"2 ! " = 06"2 ( !

Pe&%&' Da(a M)%)* 2 -M4b.

R " = 0 m! " = 0 m!

R $5 = R $ R "0 R "" R "2 R " = "26 0,06 9 0.225 0= "605 ( !

$5 = $ "0 """2 "= 50609 0,2# 06"2 0= 50645 ( !

Is M4 =

400

√ 3√ (13.0852+50.452 )

= 400

√ 352.12=4.43 kA

Kabel -R 14b;14b.

N;; 4!3" ! 25 ((2: A = "00 ((2

= 22,5

* = "50 (

R"4 = %. l

A =22,5 .150

100 = ,75 (

"4 = 0,"2 ! * = 0,"2 ! "50 = " (

3! Paa )%)* +, -3.



• 1+'* -+n /)n+k+n N;;4 ! 3 " ! 50 ((2 : 3PUIL +'* 7685+ k+'* /+.+

'.)n$) s+$:

ρ = 22,5 3C:

* = " (

R "2 =22,5 x 1

4.50 =0.11m!

"2 = 0,"2 ! L = 0,"2 ! " = 06"2 ( !

Pe&%&' Da(a M)%)* 3 -M4.

R " = 0 m! " = 0 m!

R $5 = R $ R "0 R "" R "2 R " = "26 0.11 9 0.225 0= "6"4 ( !

$5 = $ "0 """2 "= 50609 0,2# 06"2 0= 50645 ( !

Is M4 =400

√ 3√ (13.142+50.452 )= 400

√ 352.13=4.42 kA

Kabel -R 14;14.

N;; 4!3" ! 50 ((2: A =200 ((2

= 22,5

* = 200 (

R"4 = %. l

A =22,5 .200

200 = 22,5 (

"4 = 0,"2 ! * = 0,"2 ! 200 = 24 (

4! Paa )%)* DOL -4.

• 1+'* -+n /)n+k+n N;;4 ! 3 " ! "< ((2 : 3PUIL +'* 7685+ k+'* /+.+

'.)n$) s+$:

ρ = 22,5 3C:

* = " (

R "2/ =22,5 x 1

4.16=0.35m!

"2/ = 0,"2 ! L = 0,"2 ! " = 06"2 ( !

Pe&%&' Da(a M)%)* 4 -M4.

R "/ = 0 m! "/ = 0 m!

R $5 = R $ R "0 R "" R "2/ R "/ = "26 0.11 9 0.35 0= "62< ( !

$5 = $ "0 """2/ "/= 50609 0,2# 06"2 0= 50645 ( !

Is M4/ =

400

√ 3√ (13.142+50.452 )=

400

√ 352.16=4.43kA



Kabel -R 14;14.

N;; 4!3" ! "0 ((2: A =40 ((2

= 22,5

* = 270 (

R"4/ =

%. l

A =

22,5 .270

40 = "5",9 (

"4/ = 0,"2 ! * = 0,"2 ! 270 = 2,4 (

! Paa )%)* +, -1.

1+'* -+n /)n+k+n N;;4 ! 3 " ! "< ((2 : 3PUIL +'* 7685+ k+'* /+.+

'.)n$) s+$:

ρ = 22,5 3C:

* = " (

R "2' =

22,5 x 1

4.16

=0.35m!

"2' = 0,"2 ! L = 0,"2 ! " = 06"2 ( !

Pe&%&' Da(a M)%)* 1 -M4e.

R "' = 0 m! "' = 0 m!

R $5 = R $ R "0 R "" R "2' R "' = "26 0.11 9 0.35 0= "62< ( !

$5 = $ "0 """2' "'= 50609 0,2# 06"2 0= 50645 ( !

Is M4' =

400

√ 3√ (13.142+50.452 )= 400

√ 352.16=4.43 kA

Kabel -R 14e;14e.

N;; 4!3" ! "< ((2: A = <4 ((2

= 22,5

* = 00 (

R"4' = %. l

A =22,5 .300

64 = "05,5 (

"4' = 0,"2 ! * = 0,"2 ! 00 = < (

III! Pe*eaaa Ka#a'$%)*

1! M)%)* 1

Po$ = 22000 W

Cos = 0679



= 76"0

In = 42 A

In &'('+n+n 5 = 42 ! 065 = 567 A

In = 567 8 76"0 = 262 J 2"69?

S=√ 3 . V . I

¿√ 3.400.35.7

¿24733.69VA

2! M)%)* 2

Po$ = 22000 W

Cos = 0679

= 76"0

In = 42 A

In &'('+n+n 5 = 42 ! 065 = 567 A

In = 567 8 76"0 = 262 J 2"69?

S=√ 3 . V . I

¿√ 3.400.35.7

¿24733.69VA

3! M)%)* 3

Po$ = 45000 W

Cos = 062

= 46920

In = <65 A

In &'('+n+n 5 = <65 ! 065 = 76525 A

In = 76525 8 46920 = <0629 J 42609?

S=√ 3 . V . I

¿√ 3.400.73.525

¿50939.61VA

4! M)%)* 4

Po$ = 5500 W

Cos = 06

= 6900

In = "269 A

In &'('+n+n 5 = 42 ! "269 = "069<5 A

In = "069<5 8 6900 = 96"0 J <6"2?

S=√ 3 . V . I

¿√ 3.400.10.965



¿7596.77 VA

! M)%)*

Po$ = 7500 W

Cos = 067 = 6740

In = "7 A

In &'('+n+n 5 = "7 ! 065 = "4645 A

In = "4645 8 6740 = ""627 J 9604?

S=√ 3 . V . I

¿√ 3.400.14.45

¿10011.25VA

I %)%al

= 3262 J 2"69?: 3262 J 2"69?: 3<0629 J 42609?: 396"0 J <6"2?: 3""627 J

9604?:

In $o$+* = "760< J "0"60"7? = "70627 8 <69 0

• 5)' < %)%al

= os 8"38<69:

Cos $o$+* = 0605

$o$+* = <69 0

• S%)%al = S" S2 S S4 S5

= 2476<9 2476<9 5096<" 759<677 "00""625

= "045960" A

• P%)%al = S$o$+* ! Cos $o$+*

= "045960" ! 0605

= 709650 W

•

&' =√ S'

2− P'

2

&' =√ (108459.01)2−87309.50

2

&' =139234.706VA"



Pe*eaaa N$la$ Ka#a'$%)*

• Cos 2 = 0695

2 = "6"90

• tan∅2=

&2

P'

tan18.190=

&2

87309.50

&2¿28689.02 VA"

• &C =&' −&

2

&C =139234.706−28689.02

¿110545.69VA"

• S

2= P'

cos∅2

S2=28689.02

0.95

S2=30198.97 VA

• &C =V

2

$ C

$ C =V

2

&C

$ C =

4002

110545.69

$c=1,45!

• C =

1

(.$c

C = 1

2.3,14 .50.1,45



C =2.196 x10−3

) =2196 *)

• I C = &C

√ 3 . U

I C =110545.69

√ 3 .400

¿159.56 A

• C+=1

3 √ 3

I C

U . (

C+=1

3 √ 3

159.56

400 .2.3,14 .50

¿7,33 .10−4

) =733 *)

• Pe&%&' Da(a -6&'e. = ",5 ! I= ",5 ! "5965<

= 2964 A # 275 A

C+.+ &'()*)%+n .'$)n &'($s /+-+ k+&+s)$o.

+'* <6" P'($s D+-+ n$k 1+&+s)$o.

D+-+

R'+k$)

> 3kAR

:

In 3+.s

no()n+*8

A: >+s', 400

S'$$)n I.

()n)(( P'($s

D+-+ 3A:

7,5 "" "5

"0 "4 20

"5 22 29

20 29 9

0 4 59

40 5 79

50 72 9

<0 7 ""

70 "0" "7

0 ""5 "57

"00 "44 "9<

"20 "7 2<

"40 202 275



• K)%a"%)* = ",< ! I

= ",< ! "5965<

= 2"7,00 A # "9< A

• Kabel = ",5 ! I

= ",5 ! "5965<

= 2964 A # 250 A # 240 ((2

Con$o% &'()*)%+n .'$)n k+'* n$k k+&+s)$o.



; sebelum dipasang

kapasitor 140$24 !

; setelah diberi kapasitor

159.56 A



=! G*)&$ S$'%e Pael P)#a G6 AB



LAMPIRAN

PUIL UNTUK MENNETUKAN LUAS PENAMPANG KABEL



PUIL UNTUK MENENTUKAN LUAS PENAMPANG BUSBAR



5APTER G UNTUK MENENTUKAN U' DAN NILAI R DAN ;

PADA TRA6O

KATALOG TOR



KATALOG KONTAKTOR



KATALOG M55B -She$e* M55B.



UNTUK PERITUNGAN INSTALASI PADA KAPASITOR



• K)%a"%)* S#e$al -K)%a"%)* Kh&'&'.

• Pe&%&' Da(a &%&" Ka#a'$%)*

+'* <6" P'($s D+-+ n$k 1+&+s)$o.

D+-+

R'+k$)> 3kAR

:

In 3+.s

no()n+*8A: >+s', 400

S'$$)n I.

()n)(( P'($s

D+-+ 3A:

7,5 "" "5

"0 "4 20

"5 22 29

20 29 9

0 4 59

40 5 79

50 72 9

<0 7 ""

70 "0" "70 ""5 "57

"00 "44 "9<

"20 "7 2<

"40 202 275

C+$+$+n 8 R'*+) '.()s n$k ).)$ .'+k'. /)s'$'* ",< In

K R+$' n$k k+'* ",5 In

• Kabel

+'* <62 P'n%+n$+. n$k )ns$+*+s) k+&+s)$o.

busul besar

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