trickling filter.doc

7/23/2019 Trickling filter.doc

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Trickling filter

Dari hasil perhitungan pengolahan tingkat I, diperoleh efluen terakhir sebagai berikut:

* Debit rata-rata, Qr = 0,437 m3/dt

* !Dsisadari " I = #4$%7& mg/'

* ((sisadari " I = 34$& mg/'

Kriteria Desain

)enis = *igh +ate

fisiensi penurunan !D = &0 70 .

!perasi = kontinu

+asio resirkulasi, + =

1edalaman = $2 $4 m

*drauli loading, *' = 0 40 m3/m$hari

!D&loading = 0,4 $4 kg/m3$hari

(loughing = kontinu

5edia filter = batu

"o6er = % 0 k/03$m3

fluen = tidak ternitrifikasi

8Metcalf & Eddy, 20039

Desain Terpilih

+ =

*' = & m3/m$hari

Qr = 0,437 m3/dt

)umlah trikling filter = 4 buah 8 tahap, tahap masing-masing buah9

Q bak tiap tahap = 0$2& m3/dt/bak = 2272$4 m3/hari/bak

Q tiap bak, Qo = 0,0#& m3/dt/bak = #43#$ m3/hari/bak

1edalaman bak = m

=

Perhitungan

!D&efluen pengolahan biologis ang diharapkan & mg/'


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fisiensi penurunan !D&, .,73.00%7&,#4

&%7&,#4.00 =

=

=

i

ei

BOD

BODBOD

8-9 = 0$73

= 0$73

= 0$73

- 0$73 = 0

= = 0$42

;sumsi: 1onstanta remo9, 1& = 0, hari

1oefisien koreksi temperatur, ? = ,02

@emperatur rata-rata ;, @ = & >

'uas permukaan spesifik, (a = 00 m/m3per unit


3/5

D = 0$# m = m

)umlah lahan ang dibutuhkan untuk buah filter

0%$2203$#4 m&&t'tal ===

Tahap (edua

1onsentrasi !D

= 8-9 = 8-0$4298 727$3 kg !D/hari9 = ##$4 kg !D /hari

5enghitung Colume

=

!"

#

E

443$0

00

+

42 = 9%&,$8

4$##

42$0

443$0

00

!+

C = 347$% m3

Diameter tahap

; = C/* =0

4%$347= 73$73 m

; = D

D = 4$2 m = & m

)umlah lahan ang dibutuhkan untuk buah filter

4%$34773$73 m&&t'tal ===

!D 'oading masing-masing filter

a$ filter tahap

!D loading = /C = 727$3 /22= #$& kg /m3hari

b$ filter tahap

!D loading = /C = ##$4 /347$% = $% kg /m 3hari

*idrauli loading masing-masing filter

a$ filter tahap

hidrauli loading =903$#498min/4408

9/4$2272988

3

mhari

harim+= 0$4m3/m menit$

b$ filter tahap


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hidrauli loading =973$7398min/4408

9/4$2272988

3

mhari

harim+= 0$ m3/m menit$

@otal kebutuhan lahan untuk tahap, &$&3&4%$3470%$22 m& =+=

- !D effluent dari @rikling filter tahap I = 8-0$429 A #4$%7& = 4#$3 mg/l

- !D effluent dari @rikling filter II = 8-0$429 A 4#$3 = &$% mg/'

- (( effluent dari @rikling filter I = 8-0$429 A 34$& = 7$#4 mg/l

- (( effluent dari trikling filter II = 8-0$429 A 7$#4 = #$33 mg/l$

)istem Distri*usi

filter terdiri atas 4 lengan dengan ukuran masing-masing lengan:

diameter = 0,& m

panEang = &$& m

diameter orifie= 0,0& m

Earak antar or$ = 0,& m

sistem penggerak mekanis

Dimensi pipa utama

diameter = 0,7& m

panEang = m

Debit tiap lengan, dtm% /0&4%$04

2&$0 3==

)umlah orifie tiap lengan, *uahce$ara+'rifi

pan$ang'rifice 40

&,0

&===

Debit tiap orifie, dtmice$umlah'r'f

enganDe*ittiapl%'r /04$

40

0&4%$0 33===

'uas orifie,4

/03,/0/&,04

/4

/ mD&'r ===

1oefisien disharge, > = 0,2

*eadloss pada orifie

( )( ) ( )

mg&,

%-'r 3$0

2,#03,2,0

04$

4

3

=

=

=

Baktor kekasaran pipa utama, f = 0,0

'uas pipa utama,

44,07&,04

/4

/ mD&p ===


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1eepatan aliran dalam pipa utama, dtm&

%.p /4#,0

44,0

2&,0===

*eadloss pada pipa utama

m

g

.

D

/f-p 3

03%$%

2,#

4#,0

7&,0

&,#0,0

=

=

=

ila F = 0,#2, maka headloss pada pipa distributor/lengan

( ) ( ) m-- 'r $0#2,03$0 ===

'uas penampang pipa distributor,

04#,0&,04

/4

/ mD&dist ===

*eadloss pada penampang pipa distributor

( )

( ) ( ) m

g&,

%- 00#,0

2,#04#,02,0

0&4%$0

=

=

=

*eadloss pada fitting diasumsikan 0 . total headloss, aitu

( ) ( m0----'r- pf 00#,0$003%$%3$0,0.0 3 =+++=+++=

*eadloss total pada sistem distribusi

m0hhhhhh fp'rt't &$00#,0$003%$%3$0 3 =++++=++++=

)istem P'mpa

1eepatan rotasi distribusi, n = rpm

"anEang pipa distributor, l = 2 m = %0 ft

*ead untuk menggerakkan distributor

( ) ( )mft

g

nlhc %,00,

%02,#

%0

%0

==

=

=

*ead statis, *s = & m

*ead pompa, m-hh- sct $7&%,0&$ =++=++=

fisiensi pompa, G = 7& .

Daa pompa, ( ) ( ) ( ) +#dtmm+1-%P 7$407&,0 $72&,0/77,# 33

=

=

=

trickling filter.doc

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