modul-4 cara menetukan diameter pipa ... · 2017-04-25 · hitung menggunakan tabel : 1. hitung...

Q = 0 .2 7 8 5 x C x D 2 ,6 3 x S 0 .5 4

Q =0 .2 7 8 5 x C x D 2 ,6 3 x (� H/ L) 0 .5 4

Q = 1 0 6

Q = 0 .2 7 8 5 x 1 2 0 x (7 9 .2 ) 2 ,6 3 x ( (4 0 / 3 .5 0 0 )x1 0 0 )0 .5 4

1 0 6

�

D = [3 .5 9 x 1 0 6 x Q

]0 .3 8

C x S 0 .5 4

Q = 0 .2 7 8 5 x C x D 2 ,6 3 x S 0 .5 4

1 0 6

D = [[[[3 .5 9 x ( 1 0 ) 6 x 1 .0 5

]]]]0 .3 8

1 2 0 x ( 1 .1 4 ) 0 .5 4

D = 4 9 .7 1 m m

V =(Q/ 1 0 0 0 )

(1 / 4 ) x 3 ,1 4 x (D/ 1 0 0 0 )2

V =(1 .4 8 / 1 0 0 0 )

(1 / 4 ) x 3 ,1 4 x (5 5 .4 / 1 0 0 0 )2

Hit u n g Me n g g u n a k a n Ta b e l :

1 . Hitu n g de b it (h a ri m a ks ) = 1 .1 5 x 0 .9 1 = 1 .0 5 lt r / d tk

2 . Hitu n g S lope � H / L = 4 0 / 3 .5 0 0 = 0 .0 1 1 4 x 1 0 0 = 1 .1 4

3 . S ta rt da ri kolom Q

Ta rik ga ris ve rt ika l ke ba wa h , ca ri h in gga ke te m u a n gka Q u n tu k 1 .0 5

Jika t ida k te rda pa t a n gka 1 .0 5 , la ku ka n in te rpola s i da ri n ila i d i a ta s da n d i ba wa h nya

Ke m u d ia n ta rik ga ris h orizon ta l ke ka n a n , ca ri h in gga ke te m u a n gka 1 .1 4

Jika t ida k te rda pa t a n gka 1 .1 4 , la ku ka n in te rpola s i , bu la tka n ke ka n a n

Te ra kh ir, t a r ik ga ris ve rt ika l ke a ta s . . . . Ca ta t be ra pa d ia m e te r p ipa Ø ” !!

ND 40 1 1/2” 50 2 ” 65 2 1/2”

Q V hL/100 V hL/100 V hL/100

1.00 1.93 0.63 0.27

1.05

1.10 2.30 0.77 0.32

Con toh :

Q 1 .0 0 … 0 ,6 3 . . . . (Ø 2 ”)

Q 1 .1 0 … 0 .7 7 . . . (Ø 2 ”)

, g p p p

Ø 2 ” … 1 .4 8 lt r / d tk

In le t

RESERVOIR

DISTRIBUSIou t le t

TRANS MISI

Q ra ta - ra ta 0 .9 1 lt r / d tk

Ø 2 ½ ” … 2 .3 5 lt r / d tk

overflow

S k e m a t is a s i Alir a n Ma s u k – Ke lu a r Re s e r v o ir

Q ra ta ra ta 0 .9 1 lt r / d tk

Te rja d i ove rflow (a ir d ib u a n g ) p a d a Re s e rvoir

Be s a rn ya ove rflow p a d a kon d is i ra t a - ra ta a d a la h :

Q ov er f low = Q in le t – Q d ist r b r at a- r a t a

S u p a y a t id a k t e r j a d i o v e r flo w , a t u r b u k a a n k a t u b ,

k a t u b - in le t - r e s e r v o ir a t a u k a t u b - o u t le t - s u m b e r

BAGAIMANA KONDIS I ALIRAN DALAM PIPAKALAU DEBIT S UMBER AIR TURUN/ DROPMISAL MENJADI 0 .8 5 lt r / d tk ?

Pip a Ø 2 ” dg n Q 1 .4 8 lt r / d tk Pip a Ø 2 ” d g n Q 0 .8 5 lt r / d tk

Pe n u ru n a n d e b it s u m b e r a ir m e n ga kiba tka n a lira n a ir

d a la m p ip a m e n ja d i t id a k p e n u h ,

Kon d is i in i m e n g a kib a tka n p ip a m e n ja d i s e p e r t i

s a lu ra n te rb u ka b ia s a d a n t id a k b e r t e k a n a n la g i !

P I P A DI S TRI BUS I

Sistem Distribusi

Merepresentasikan porsi bagian terbesar untuk investasi awal (45% sampai dengan 70% biaya) – reservoir, pompa, klorinasi, bak pelepas tekan

SetiapSetiap sambungansambungan seharusnyaseharusnyakk tt

SetiapSetiap sambungansambungan seharusnyaseharusnyakk tt

IdealnyaIdealnya,, dirancangdirancang dapatdapatmempertahankanmempertahankan tekanantekanan minimumminimumselamaselama2424 jamjam setiapsetiap harihari

IdealnyaIdealnya,, dirancangdirancang dapatdapatmempertahankanmempertahankan tekanantekanan minimumminimumselamaselama2424 jamjam setiapsetiap harihari

menggunakanmenggunakan metermetermenggunakanmenggunakan metermeter

Sistem Distribusi

JaringanJaringan distribusidistribusi dapatdapat berupaberupajaringanjaringan ““percabanganpercabangan”” atauatau “loop”“loop”;;sistemsistem “loop”“loop” disarankandisarankan untukuntukefisiensiefisiensi hidrolikhidrolik..

JaringanJaringan distribusidistribusi dapatdapat berupaberupajaringanjaringan ““percabanganpercabangan”” atauatau “loop”“loop”;;sistemsistem “loop”“loop” disarankandisarankan untukuntukefisiensiefisiensi hidrolikhidrolik..

UkuranUkuran pipapipa didesaindidesain berdasarkanberdasarkankebutuhankebutuhan puncakpuncak;; tekanantekanan minimumminimumuntukuntuk mencegahmencegah aliranaliran balikbalik;; headhead yangyangtrsediatrsedia daridari pompapompa atauatau reservoirreservoir;;minimumminimum tekanantekanan yangyang diterimaditerima oleholehkonsumenkonsumen

UkuranUkuran pipapipa didesaindidesain berdasarkanberdasarkankebutuhankebutuhan puncakpuncak;; tekanantekanan minimumminimumuntukuntuk mencegahmencegah aliranaliran balikbalik;; headhead yangyangtrsediatrsedia daridari pompapompa atauatau reservoirreservoir;;minimumminimum tekanantekanan yangyang diterimaditerima oleholehkonsumenkonsumen

RENCANA SISTEM BPSAB TIRTA MAYA( J umlah J iwa yang akan dilayani 1.100 orang)

Res v

A

200 jiwaElev + 60

300 jiwaElev + 40 m

C

B

L = 500 m

L = 1000 m

L = 300 m

Elev + 90 m

BPSAB Tirta Ma ya :

• Ju m la h jiwa : 1 .1 0 0• Re n ca n a Ca ku pa n : 1 0 0 %

JalurPipa

JumlahPenduduk L D Pipa

KehilanganTekan

Tinggi GarisTekan (HGL)

(Jiwa) (m) (inchi) (m) (m)

A – B 1.100 300 ? ? ?

B – C 300 500 ? ? ?

B – D 600 1.000 ? ? ?

600 jiwaElev + 20 m

D

• Fa ktor ja m pu n ca k (Qjp ) : 1 .7 5• Da ta la in :

Lih a t pa da ga m ba r !

SKEMATIK DISTRIBUSI KEBUTUHAN AIR

Res v

A

200 jiwaElev + 60

300 jiwaElev + 40 m

CB L = 500 m

L = 300 m

Elev + 90 m

600 jiwaElev + 20 m

D

L = 1000 m

La n g k a h P e n y e le s a ia n :

1 . Hitu n g ke bu tu h a n a ir u n tu k Blok B, Blok C da n Blok D pa da Ja m Pu n ca k (Q jp ) !

Blok B :Ju m la h Pe n du du k = 2 0 0 jiwa (a s u m s i ke bu tu h a n a ir : 6 0 lt r / jiwa / h a ri)Ke bu tu h a n a ir ra t a - ra ta Q = (2 0 0 x6 0 ) / (2 4 x6 0 x6 0) = 0 .1 3 9 lt r / d tkKe bu t a ir pa da ja m pu n ca k (Q jp ) = 1 .7 5 x 0 .1 3 8 = 0 .2 4 lt r / d tk

Blok C :Ke bu t . a ir Qjp u n tu k Blok C dgn 3 0 0 jiwa = 0 .3 6 lt r/ d tk

Blok D :Ke bu t . a ir Qjp u n tu k Blok D dgn 6 0 0 jiwa = 0 .7 3 lt r/ d tk

Tota l Blok B , C da n D : Tota l Qjp u n tu k Blok B , C da n = 1 .3 4 lt r / d tk

2 . Be ra pa Dia m e t e r p ip a u n t u k j a lu r A – B (pa n ja n g L = 3 0 0 m ) ?

• De b it ja m pu n ca k ja lu r p ipa A – B = 1 .3 4 lt r / d tk

• Be ra pa d ia m e te r p ipa j a lu r A- B ?

Tria l a n d e rror, coba -coba m a s u kka n be be ra pa u ku ra n d ia m e te r p ipa

(Gunakan tabel kehilangan tekanan ! )

o Ke h ila n ga n te ka n a n dgn Pipa Dia . 2 5 m m (Ø 1 ”)

D 25 30 40 50= 3 0 0 / 1 0 0 x 2 7 .3 3 = 8 1 .9 9 m

Tota l h L (m a yor + m in or) = 1 .1 * 8 1 .9 9 = 9 0 .1 9 m

o Ke h ila n ga n te ka n a n de n ga n Pipa Dia . 3 0 (Ø 1 ¼ ”)

= 3 0 0 / 1 0 0 x 9 .2 8 = 2 7 .8 4 m

Tota l h L = 1 .1 x 2 7 .8 4 = 3 0 .6 2 m

o Ke h ila n ga n te ka n a n dgn Pipa Dia . 4 0 m m (Ø 1 ½ ”)

= 3 0 0 / 1 0 0 x 3 .1 3 = 9 .3 9

Tota l h L = 1 .1 x 9 .3 9 = 1 0 .3 3

o Ke h ila n ga n te ka n a n dgn Pipa Dia . 5 0 m m (Ø 2 ”)

= 3 0 0 / 1 0 0 x 1 .0 2 = 3 .0 6

Tota l h L = 1 .1 x 3 .0 6 = 3 .3 7

D 25 30 40 50

Q

1.30 3.13

1.40

3 . Be ra pa S is a Te k a n d i B ?

Tria l a n d e rror, coba -coba m a s u kka n be be ra pa u ku ra n p ipa

Da ta : Ele va s i A = + 9 0 m , Ele va s i B = + 6 0 ,

Mis a l t in gg i a ir d i Re s e rvoir = 1 m ,

m a ka : Ga ris Tin gg i Te ka n a n (HGL) d i A = 9 0 + 1 m = 91 m

Tin gg i Te ka n a n te rs e d ia u tk ja lu r A-B = 9 0 – 6 0 = 3 0 . . 3 0 m + 1 m = 3 1 m

Un tu k pe m a s a n ga n p ipa de n ga n Dia . 2 5 m m ,

HGL d i B = 9 1 – 9 0 .1 9 = 0 .8 1 m

Diam. pipa hL pipa A - B HGL di B Sisa Tekan di B

Ø 25 ( 1” ) 90.19 91 – 90.19 = 0.81 0.81 – 60 = - 59.19

Ø 30 (1 ¼”) 30.62 91 - 30.62 = 60.38 60.38 - 60 = 0.38

Ø 40 ( 1 ½”) 10.33 91 – 10.33 = 80.67 80.67 - 60 = 20.67

Ø 50 (2’) 3.37 91 – 3.37 = 87.63 87.63 - 60 = 27.63

Jika d igu n a ka n p ipa Dia . 2 5 m m (Ø 1 ”) , t e ka n a n d i B a ka n n e ga t if ( - 5 9 .1 9 m ) !!

4 . Be ra pa Dia m e t e r p ip a j a lu r B - C ? da n S is a t e k a n d i C ?

• De b it ja m pu n ca k u n tu k ja lu r B-C = 0 .3 6 lt r/ d tk

• Ele va s i d i B = + 6 0 m , Ele va s i d i C = + 4 0 m , Pa n ja n g ja lu r B-C = 5 0 0 m

• Jika d ip ilih ja lu r p ipa A – B m e n ggu n a ka n p ipa Ø 1 1 / 2 ” , m a ka :

S is a Te ka n d i B= 2 0 .6 7 , Tin gg i Ga ris Te ka n a n (HGL) d i B= 6 0 + 2 0 .6 7 = 8 0 .6 7 m

• Tria l a n d e rror, coba -coba m a s u kka n u n tu k be be ra pa d ia m e te r

Dia pipa hL pipa B C (m)Tinggi Garis Tekan, HGL

Sisa Tekan (m) di CDia pipa hL pipa B- C (m) (m) di C Sisa Tekan (m) di C

Ø 25 ( 1” ) 2.45 x 500/100 x 1,1 = 13.48 80.67 – 13.48 = 67.19 67.19 – 40 = 27.19

Ø 30 (1 ¼”) 0.83 x 500/100 x 1.1 = 4.57 80.67 - 4.57 = 76.1 76.1 – 40 = 36.1

Ø 40 ( 1 ½”) 0.28 x 500/100 x 1.1 = 1.54 80.67 – 1.54 = 79.13 79.13 – 40 = 39.13

Ø 50 (2’) 0.09 x 500/100 x 1.1 = 0.50 80.67 – 0.50 = 80.17 80.17 – 40 = 40.17

Angka dari tabel, interpolasi

5 . Be ra pa Dia m e t e r p ip a j a lu r B - D ? da n S is a t e k a n d i D ?

• De b it ja m pu n ca k u n tu k ja lu r B-C = 0 .7 3 lt r/ d tk

• Ele va s i d i D = + 4 0 m , Ele va s i d i B = + 6 0 m , pa n ja n g = 1 .0 0 0 m

• Jika ja lu r p ipa A – B m e n ggu n a ka n p ipa Dia . 1 2 ” , m a ka :

S is a Te ka n d i B = 2 0 .6 7 m , Tin gg i Te ka n a n (HGL) d i B = 6 0 + 2 0 .6 7 = 8 0 .6 7 m

• Tria l a n d e rror, coba -coba m a s u kka n u n tu k be be ra pa d ia m e te r

Ti i G i T k HGLDia pipa hL pipa B- D

Tinggi Garis Tekan, HGL (m) di D Sisa Tekan (m) di D

Ø 25 ( 1” ) 8.68 x 1000/100 x 1.1 = 95.48 80.67 – 95.48 = - 14.81 - 14.81 – 20 = - 34.81

Ø 30 (1 ¼”) 2.95 x 1000/100 x 1.1 = 32.45 80.67 – 32.45 = 48.22 48.22 – 20 = 28.22

Ø 40 (1 ½”) 0.99 x 1000/100 x 1.1 = 10.89 80.67 – 10.89 = 68.78 68.78 – 20 = 48.78

Ø 50 (2’) 0.32 x 1000/100 x 1.1 = 3.52 80.67 - 3.52 = 77.15 77.15– 20 = 57.15

Jalur

Pipa

JmhPddk L D Pipa

Khlgn TekanhL

Tinggi GarisTekanan

,HGL

SisaTekan

(Jiwa) (m) (inchi) (m) (m) (m)

A – B 1.100 300 1 ½ “ 10.33 80.67 20.67

B – C 300 500 1 “ 13.48 67.19 27.19

B – D 600 1.000 1 ¼ “ 32.45 48.22 28.22

Garis Tinggi Hidrolis (HGL)Jalur Pipa A-B-Dp

A

Tin

ggi

(M)

Jarak (M)B D

RENCANA SISTEM BPSAB TIRTA MAYA( J umlah J iwa yang akan dilayani 1.300 orang)

Res v

A

+ 110

200 jiwaElev + 60

300 jiwaElev + 40 m

C

B

L = 500 m

L = 1000 m

L = 300 m

Elev + 90 m

BPSAB Tirta Ma ya :

• Ju m la h jiwa : 1 .3 0 0• Re n ca n a Ca ku pa n : 1 0 0 %

SOAL :

JalurPipa

JumlahPenduduk L D Pipa

KhlgnTekan

TinggiGarisTekan(HGL)

SisaTekan

(Jiwa) (m) (inchi) (m) (m) (m)

A – B 1.300 300 ? ? ? ?

B – C 300 500 ? ? ? ?

B – D 800 1.000 ? ? ? ?

D - E 200 400 ? ? ?

600 jiwaElev + 20 m

D

• Fa ktor ja m pu n ca k (Qjp ) : 1 .7 5• Da ta la in :

Lih a t pa da ga m ba r !

200 jiwaElev + 30 m

E

L = 400 m

Ha r g a P ip a & Va lv e ( S e p t ‘0 9 )

HARGA

Pe r Pa n ja n g 4 m ’ 6 m ‘

No In ch i m mPAV

(Ma s p ion )PVC

(Ru cika )Pipa Be s i Ba ll Va lve Ga te va lve

AW , S NI A loka l-1 loka l-2

1 ½ 1 6 1 7 ,0 0 0 2 2 ,0 0 0 9 1 ,5 0 0 1 0 6 ,0 0 0 4 0 ,0 0 0 7 0 ,0 0 0

2 ¾ 2 0 2 0 ,5 0 0 2 6 ,0 0 0 1 1 8 ,5 0 0 1 3 5 ,0 0 0 5 2 ,0 0 0

3 1 2 5 2 5 ,5 0 0 3 8 ,0 0 0 1 8 4 ,0 0 0 1 8 5 ,0 0 0 8 2 ,0 0 0 1 2 0 ,0 0 0

4 1 ¼ 3 0 3 6 ,0 0 0 5 1 ,0 0 0 2 3 6 ,0 0 0 1 6 5 ,0 0 0

5 1 ½ 4 0 4 7 ,0 0 0 6 6 ,0 0 0 2 7 5 ,0 0 0 1 8 5 ,0 0 0 2 1 0 ,0 0 0

6 2 5 0 7 0 ,0 0 0 9 4 ,0 0 0 3 7 4 ,0 0 0 5 4 3 ,0 0 0 2 9 8 ,0 0 0 3 1 5 ,0 0 0

7 2 ½ 6 5 9 0 ,0 0 0 1 2 0 ,0 0 0 5 3 0 ,0 0 0

8 3 7 5 1 2 8 ,0 0 0 1 8 5 ,0 0 0 6 8 0 ,0 0 0 2 ,2 4 1 ,0 0 0 9 7 0 ,0 0 0 9 1 5 ,0 0 0

9 4 1 0 0 1 8 7 ,0 0 0 2 8 3 ,0 0 0 9 8 3 ,0 0 0 3 ,4 7 2 ,0 0 0 1 ,7 5 4 ,0 0 0 2 ,7 0 7 ,0 0 0

Dia m e te r Lu a r(m m )

Uku ra n Ke te ba la n Din d in g Pipa (m m )

S -1 0 S – 1 2 ,5

Ku a t Te ka n

T = 1 0 kg / cm 2 T = 1 2 ,5 kg / cm 2 T = 1 0 kg / cm 2

3 2 1 ,6 - -

4 0 1 ,9 - -

5 0 2 ,4 - -

6 3 - 3 ,0 2 ,4

S t a n d a r S N I

6 3 3 ,0 2 ,4

7 5 - 3 ,6 2 ,9

9 0 - 4 ,3 3 ,5

1 1 0 - 5 ,3 4 ,2