Modul-4 Cara Menetukan Diameter Pipa ... · 2017-04-25 · Hitung Menggunakan Tabel : 1. Hitung...
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Transcript of 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
