contoh perhitungan preliminary design

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CONTOH PERHITUNGAN PRELIMINARY DESIGN (CASE: IPAL RUMAH SAKIT “XYZ”) 1. DESIGN PEROD

IPAL untuk RS “XYZ” direncanakan mulai beroperasi pada tahun 2010 selama 20 tahun.

2. DIAGRAM ALIR PROSES

3. KRITERIA DESAIN

Kecepatan = 0,3-0,9 m/det Saluran Pembawa

Sudut kemiringan = 0,01 m/m

Pembersihan manual (Sumber: Qasim, 1985) Bar Screen

Kecepatan melalui bar (v) = 0,3-0,6 m/det Lebar bar (w) = 4,0-8,0 m Kedalaman bar (D) = 25-50 mm Jarak antar batang = 25-75 mm Slope vertikal = 45O- 60Headloss = 150 mm

O

Headloss max = 800 mm Pembersihan mekanik (Sumber: Qasim, 1985) Kecepatan melalui bar (v) = 0,6-1,0 m/det Lebar bar (w) = 8,0-10,0 m Kedalaman bar (D) = 50-75 mm Jarak antar batang = 10-50 mm Slope vertikal = 75O- 85Headloss = 150 mm

O

Headloss max = 800 mm Faktor bentuk bar (β) Tipe bar Sharp-edged rectangular = 2,42 Rectangular with semicircular upstream face = 1,83 Rectangular with semicircular upstream and downstream face = 1,67 Circular = 1,79 Tear shape = 0,76

td ≤ 10 menit Sumur Pengumpul

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Diameter = 0,3-3 m Screw Pump

Kapasitas = 0,01-3,2 m3

Sudut kemiringan = 30/det

o-38Total head max = 9 m

o

Kecepatan motor = 30-50 rpm

td = 45-90 det (tipikal : 60 det) Grit Chamber (Sumber: Metcalf and Eddy, 1991)

Vh = 0,25-0,4 m/det (tipikal : 0,3 m/det) Vs = 1,0-1,3 m/menit (tipikal : 1,15 m/menit) untuk 65 mesh material = 0,6-0,9 m/menit (tipikal : 0,75 m/menit) untuk 100 mesh material Panjang saluran(L) = 10-20 m

Rectangular Bak Pengendap I (Sumber: Metcalf and Eddy, 1991)

Kedalaman = 3,05-4,6 m (tipikal : 3,66 m) Panjang = 15,24-91,44 m (tipikal : 24,4-39,6 m) Lebar = 3,05-24,4 m (tipikal : 4,88-9,75 m) Flight speed = 0,61-1,22 m/menit (tipikal : 0,91 m/menit) Circular Kedalaman = 3,05-4,6 m (tipikal : 3,66 m) Diameter = 3,05-60,96 m (tipikal : 12,2-45,72 m) Slope dasar = 0,75-2,0 in/ft (tipikal : 1 in/ft) Flight travel speed= 0,02-0,05 m/menit (tipikal : 0,03 m/menit)

Kedalaman = 3,0-5,0 m Tangki Aerasi (Sumber: Qasim, 1985)

Freeboard = 1-1,5 m Lebar:kedalaman = 1 : 1 – 2,2 : 1 Lebar = 3,0-11,0 m

OFR = 8,0-16,0 mSecondary Clarifier (Sumber: Qasim, 1985)

3/m2

Solid loading = 0,5-5 kg/m.hari

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Kedalaman = 3,5-5 m .jam

Kedalaman zona settling = 1,5 m

Solid loading = 40-78 kg/mSludge Thickener

2.hari

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Kons. Solid lumpur = 2,0-8,0 % Sludge vol. Ratio = 0,5-20 dry solid Kedalaman sludge blanket = 0,6-2,4 m Slope bak = 1,4-1,6

HRT pada 20Sludge Digester

oC = 40-78 kg/m2

Solid loading = 1,6-4,8 kg volatile solid/m.hari

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Keb. O.hari

2 utk cell tissue = 2,3 kg O2

Energi utk mixing mech aerator = 19,7-39,5 Kw/10/kg solid destroyed

3 mDiffused air mixing = 0,02-0,04 m

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3/m3

DO residu dlm liquid = 1,0-2,0 mg/l .menit

Reduksi pada VSS = 40-50 %

Tebal pasir = 23,0-30,0 cm Sludge Drying Bed (Sumber: Qasim, 1985)

Tebal kerikil = 20,0-30,0 cm Sludge loading rate = 100-300 kg/m2

Tebal bed = 20,0-30,0 cm .tahun

Lebar bed = 5,0-8,0 m Panjang bed = 6,0-30,0 t pengeringan = 10,0-15,0 hari Uniformity coefficient < 4 Effective size = 0,3-0,75 mm V.air dalam inlet = 0,75 m/det V.air dalam drain = 0,75 m/det

4. PRELIMINARY SIZING Perhitungan preliminary sizing seperti pada contoh sebelumnya.

5. MASS BALANCE Kualitas influen air limbah BOD5

COD = 147 mg/l = 71 mg/l

TSS = 116 mg/l NH3

Detergen = 0,2566 mg/l -bebas = 0,184 mg/l

Phenol = 0 mg/l Sisa klor (Cl2

Phosphat (ortho) = 0,4121 mg/l ) = 0 mg/l

Baku mutu limbah cair untuk kegiatan rumah sakit

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BOD5

COD = 80 mg/l = 30 mg/l

TSS = 30 mg/l NH3

Detergen = 0,5 mg/l -bebas = 0,1 mg/l

Phenol = 0,01 mg/l Sisa klor (Cl2

Phosphat (ortho) = 2 mg/l ) = 0,5 mg/l

Kuantitas influen air limbah Q.ave = 0,4709 l/det = 40,6879 m3

Q.min = 0,2355 l/det = 20,3440 m/hari

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Q.maks = 0,4709 l/det = 40,6879 m/hari

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Qp = 0,7064 l/det = 61,0319 m/hari

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/hari

a. Mass balance sebelum grit chamber Perhitungan mass balance untuk grit chamber

MBOD

= 2,8888 kg/hari (utk Q.ave) = Q * BOD/1000

= 1,4444 kg/hari (utk Q.min) = 2,8888 kg/hari (utk Q.maks) = 4,3333 kg/hari (utk Qp) MCOD

= 5,9811 kg/hari (utk Q.ave) = Q * COD/1000

= 2,9906 kg/hari (utk Q.min) = 5,9811 kg/hari (utk Q.maks) = 8,9717 kg/hari (utk Qp) MTSS

= 4,7198 kg/hari (utk Q.ave) = Q * TSS/1000

= 2,3599 kg/hari (utk Q.min) = 4,7198 kg/hari (utk Q.maks) = 7,0797 kg/hari (utk Q.p)

b. Mass balance setelah grit chamber Removal BOD5 dan TSS dalam grit chamber kecil. Oleh karena itu, diasumsikan bahwa konsentrasi BOD5

M

dan TSS yang keluar dari grit chamber sama dengan kualitas influen air limbah (Qasim, 1985)

BOD

= 1,4444 kg/hari (utk Q.min) = 2,8888 kg/hari (utk Q.ave)

= 2,8888 kg/hari (utk Q.maks) = 4,3333 kg/hari (utk Qp) MCOD = 5,9811 kg/hari (utk Q.ave)

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= 2,9906 kg/hari (utk Q.min) = 5,9811 kg/hari (utk Q.maks) = 8,9717 kg/hari (utk Qp) MTSS


= 4,7198 kg/hari (utk Q.maks) = 7,0797 kg/hari (utk Qp)

a. Mass balance sebelum bak pengendap I Perhitungan mass balance untuk bak pengendap I

Mass balance sebelum bak pengendap I sama dengan mass balance setelah grit chamber, yaitu: MBOD


= 2,8888 kg/hari (utk Q.maks) = 4,3333 kg/hari (utk Qp) MCOD


= 5,9811 kg/hari (utk Q.maks) = 8,9717 kg/hari (utk Qp) MTSS



b. Mass balance setelah bak pengendap I Removal BOD = 30 % (25-40%, Eddy dan Metcalf, 1991) Removal TSS = 60 % (50-70%, Eddy dan Metcalf, 1991) MBOD = 30 % * M

= 0,8667 kg/hari (utk Q.ave) BOD

= 0,4333 kg/hari (utk Q.min) = 0,8667 kg/hari (utk Q.maks) = 1,3000 kg/hari (utk Qp) MTSS = 60 % * M

= 2,8319 kg/hari (utk Q.ave) TSS

= 1,4159 kg/hari (utk Q.min) = 2,8319 kg/hari (utk Q.maks) = 4,2478 kg/hari (utk Qp) Sludge solid concentration dari bak pengendap I sebesar 5% = 50000 mg/l = 50 kg/m3

Q = M (Tabel 12-8, Eddy dan Metcalf, 1991)

TSS

= 0,0566 m/TSS

3/hari (utk Q.ave)

6

= 0,0283 m3

= 0,0566 m/hari (utk Q.min)

3

= 0,0850 m/hari (utk Q.maks)

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BOD = M/hari (utk Qp)

BOD

= 15,3017 mg/l (utk Q.ave) /Q

= 15,3017 mg/l (utk Q.min) = 15,3017 mg/l (utk Q.maks) = 15,3017 mg/l (utk Qp) TSS = MTSS

= 50 mg/l (utk Q.ave) /Q

= 50 mg/l (utk Q.min) = 50 mg/l (utk Q.maks) = 50 mg/l (utk Qp)

a. Mass balance sebelum tangki aerasi Perhitungan mass balance untuk tangki aerasi

Mass balance sebelum tangki aerasi sama dengan mass balance setelah BP I, yaitu: MBOD




= 2,8319 kg/hari (utk Q.maks) = 4,2478 kg/hari (utk Qp) Q = 0,0566 m3

= 0,0283 m/hari (utk Q.ave)

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3


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BOD = 15,3017 mg/l (utk Q.ave) /hari (utk Qp)

= 15,3017 mg/l (utk Q.min) = 15,3017 mg/l (utk Q.maks) = 15,3017 mg/l (utk Qp) TSS = 50 mg/l (utk Q.ave) = 50 mg/l (utk Q.min) = 50 mg/l (utk Q.maks) = 50 mg/l (utk Qp)

b. Mass balance setelah tangki aerasi Q = Qinfluen-Qsebelum tangki aerasi

= 40,6313 m3/hari (utk Q.ave)

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= 20,3156 m3


3


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M/hari (utk Qp)

BOD = MBOD sblm BP.I – M = 2,0222 kg/hari (utk Q.ave)

BOD stlh BP.I

= 1,0111 kg/hari (utk Q.min) = 2,0222 kg/hari (utk Q.maks) = 3,0333 kg/hari (utk Qp) MTSS = MTSS sblm BP.I – M = 1,8879 kg/hari (utk Q.ave)

TSS stlh BP.I

= 0,9440 kg/hari (utk Q.min) = 1,8879 kg/hari (utk Q.maks) = 2,8319 kg/hari (utk Qp) BOD = MBOD

= 0,0498 mg/l (utk Q.ave) /Q

= 0,0498 mg/l (utk Q.min) = 0,0498 mg/l (utk Q.maks) = 0,0498 mg/l (utk Qp) TSS = MTSS

= 0,0465 mg/l (utk Q.ave) /Q

= 0,0465 mg/l (utk Q.min) = 0,0465 mg/l (utk Q.maks) = 0,0465 mg/l (utk Qp)

a. Mass balance untuk liquid line Input data

Perhitungan mass balance untuk secondary clarifier

Q = 40,6313 m3


3


3


3


= 0,0498 mg/l (utk Q.min) = 0,0498 mg/l (utk Q.maks) = 0,0498 mg/l (utk Qp) TSS = 0,0465 mg/l (utk Q.ave) = 0,0465 mg/l (utk Q.min) = 0,0465 mg/l (utk Q.maks) = 0,0465 mg/l (utk Qp) MBOD


= 2,0222 kg/hari (utk Q.maks)

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= 3,0333 kg/hari (utk Qp) MTSS



Perhitungan BODair

= 0,0050 mg/l (utk Q.ave) = 10% * BOD

= 0,0050 mg/l (utk Q.min) = 0,0050 mg/l (utk Q.maks) = 0,0050 mg/l (utk Qp) Px = Yobs * Q * (So-Se

Y

)

obs )(1 ckdY

θ×+ =

Y = 0,5 g biomass/g substrat (0,4-0,6) kd = 0,06 per hari θc = 10 hari (3-15 hari) Yobs

Q = 40,6313 m = 0,3125 g/g

3


3


3


3

So (BOD/hari (utk Qp)

in

Se (BOD)= 0,0498 mg/l

air

Px = 0,5687 g MLVSS/hr = 0,7109 g MLSS/hr (utk Q.ave) )= 0,0050 mg/l

= 0,2844 g MLVSS/hr= 0,3555 g MLSS/hr (utk Q.min) = 0,5687 g MLVSS/hr = 0,7109 g MLSS/hr (utk Q.maks) = 0,8531 g MLVSS/hr = 1,0664 g MLSS/hr (utk Qp)

b. Mass balance untuk solid line Input data Q = 40,6313 m3


3


3


3


= 0,0498 mg/l (utk Q.min) = 0,0498 mg/l (utk Q.maks) = 0,0498 mg/l (utk Qp) TSS = 0,0465 mg/l (utk Q.ave) = 0,0465 mg/l (utk Q.min) = 0,0465 mg/l (utk Q.maks) = 0,0465 mg/l (utk Qp)

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MBOD





Perhitungan Qw

= 7,1093.10 = Px/TSS (TSS = 8000-12000, diambil TSS = 10000)

-5 m3

= 3,5546.10/hari (utk Q.ave)

-5 m3

= 7,1093.10/hari (utk Q.min)

-5 m3

= 1,0664.10/hari (utk Q.maks)

-4 m3

BOD/hari (utk Qp)

in * Qin

= 0,001011 mg/hari (utk Q.min) = 0,002022 mg/hari (utk Q.ave)

= 0,002022 mg/hari (utk Q.maks) = 0,003033 mg/hari (utk Qp) Qair = Qin-Q = 4,0631.10

w

1 m3

= 2,0316. 10/hari (utk Q.ave)

1 m3

= 4,0631. 10/hari (utk Q.min)

1 m3

= 6,0947. 10/hari (utk Q.maks)

1 m3

BOD/hari (utk Qp)

air * Qair

= 1,0111 mg/hari (utk Q.min) = 2,0222 mg/hari (utk Q.ave)

= 2,0222 mg/hari (utk Q.maks) = 3,0333 mg/hari (utk Qp)

6. TATA LETAK (LAYOUT) 7. PROFIL HIDROLIS

contoh perhitungan preliminary design

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