Parameter penting yang digunakanuntukmenentukan COD yang dapatteruarai :

bs BOD : biodegradable souble BOD

s BOD : souble BOD

nbs BOD : non-souble BOD

nbp COD : non-biodegradable particulate COD

i TSS : inert TSS

rb COD : rapidly biodegradable COD

BOD = s BOD + nbs BOD

COD = BOD + nb COD

Data Perencanaan :

1) Use 2 Tanks 5) AmoniakOxydized = 80% TKN

2) Kedalamankotakpenuh = 6m (y) 6) b Cod = 1,6 (BOD)
3) Decaut depth = 20%. Y
4) SVI = 150 mL/g
Jawaban:
1. Karakteristik design yang diperlukan :
a) b COD ?
b COD = 1,6 (118,7) = 189,29 g/𝑚3
b) nb VSS ?
bs COD = 1,6 (s BOD)
= 1,6 (80) = 128 g/𝑚3
nbs COD = s COD – bs COD
= 177 – 128 = 49 g/𝑚3
nbp COD = COD – b COD – nbs COD
= (508 – 189,92 – 49) g/𝑚3 = 269 g/𝑚3

𝑇 𝐶𝑂𝐷−𝑠 𝐶𝑂𝐷 508−177

VSS cod = = = 2,21 g COD/g VSS
𝑉𝑆𝑆 150
 𝑛𝑏𝑝 𝐶𝑂𝐷 269 g/𝑚3
nb VSS = = 2,21 𝑔 𝐶𝑂𝐷/𝑔 𝑉𝑆𝑆 = 121,72 g/𝑚3
𝑉𝑆𝑆𝑐𝑜𝑑

c) i TSS
i TSS = 𝑇𝑆𝑆0 - 𝑉𝑆𝑆0
= (195 – 150) g/𝑚3
= 45 g/𝑚3
2. Menentukansiklusoperasi SBR
Diketahui :
TC : Total Cycle Time
tf : Time for fill
tA : Aeration / react time Tc = tf + tA + ts + td + tI = 2 tanks
ts : Settle Time
td : Decant Time
tI : Idle Time

Assume :tA = 2 jam tI = 0

ts = 0,5 jam tf = (2 + 0,5 + 0,5) = 3 jam for each tank
td = 0,5 jam Tc = tf + tA + ts + td
= 3 + 2 + 0,5 + 0,5 = 6 jam
24 ℎ/𝑑
a. Jumlahsiklus yang terjadi / tangkihari = 6 ℎ/𝑐𝑦𝑐𝑙𝑒 = 4
4 𝑐𝑦𝑐𝑙𝑒𝑠/𝑑
b. Total sikluskeseluruhan = 2 tanks ( ) = 8 cycles/d
𝑡𝑎𝑛𝑘
𝑚3
0,153 × 86400𝑠/𝑑
c. Fill volume/cycle = 𝑠
= 1652,4 𝑚3 /𝑓𝑖𝑙𝑙
8 𝑐𝑦𝑐𝑙𝑒𝑠/𝑑

3. Menentukan volume tangki 2 overall hydrolic retention time T

 Full Liquid depth = 6 m
 Decant depth = 0,2 × 6 = 1,2 𝑚
𝑉𝑓 /𝑡𝑎𝑛𝑘 1652,4 𝑚3 /𝑓𝑖𝑙𝑙
𝑉𝑇 = = = 5508 𝑚3 / 𝑡𝑎𝑛𝑘
0,3 0,3
𝑚3 ℎ
2 𝑡𝑎𝑛𝑘𝑠 (5508 )(24 )
𝑡𝑎𝑛𝑘𝑠 𝑑
Overall T = 𝑚3
= 20h
0,153 ×86400 𝑠/𝑑
𝑠
4. Menentukan SRT (Sludge Retention Time) Lumpur active konvensional MLSS <
3500
(Px, TSS) SRT :𝑉𝑇 (𝑥𝑀𝐿𝑆𝑆 ) = 5508 𝑚3 × 3500 g/𝑚3 = 19278000 gram
Q 𝑌 (𝑠 −s)SRT Q 𝑌𝑛 (𝑁𝑂𝑥)SRT
(Px, TSS) SRT =[1+𝑏𝐻𝐻 (𝑆𝑅𝑇)](0,85)
𝑂
+ Q (nbVSS) SRT + [1+𝑏𝐻 (𝑆𝑅𝑇)](0,85)
+
(fd)(bh)Q(Yh)(𝑠𝑂 −s)𝑆𝑅𝑇 2
[1+𝑏𝐻 (𝑆𝑅𝑇)](0,85)
+ Q (𝑇𝑆𝑆𝑜 − 𝑉𝑆𝑆𝑜) 𝑆𝑅𝑇

Data Input :
- nb VSS = 121,72
- if So ≈ So-S So = b COD = 189,92 g/𝑚3
Q = (0,153 𝑚3 /𝑠 . 86400 s/d) / 2 tanks = 6609,6 𝑚3 /𝑡𝑎𝑛𝑘. 𝑑
i TSS = TSSo – VSSo = 45 g/𝑚3
NOx = 0,8 x 359 TKN/𝑚3 = 28 g/𝑚3
- Kinetic Coefficient Y = 0,45 gvss/gbcod
b 12⁰C = 0,12 g/g⁰d(1,04)12−20 = 0,088 g/g.d
Yn= 0,15gVSS/s.NOx

Aerobic : bn. 12⁰C = 0,17 g/g.d. (1,029)12−20 = 0,135 g/g.d

Anoxic :bn. 12⁰C = 0,07 g/g.d(1,029)12−20 = 0,056 𝑔/𝑔. 𝑑
Average: 0,135 (0,33) + (0,056.0,61) = 0,082 g/g.d
Fd : 0,15 g/g. d

16609,6𝑚3 g
( )(0,45 𝑔𝑉𝑆𝑆/𝑔 .𝑏𝐶𝑂𝐷)(189,92 3 )(𝑆𝑅𝑇) 𝑚3 g
𝑑 𝑚
19.278.000 = 𝑔 + 6609,6𝑡𝑎𝑛𝑘 . 𝑑. 121,72 𝑚3 +
[1+(0,088 .𝑑 )(𝑆𝑅𝑇)(0,85)
𝑔

6609,6 𝑚3 g 0,15𝑔 𝑔 0,45gVSS 189,92g

( )(0,15 𝑔𝑉𝑆𝑆/𝑔 .𝑁𝑂𝑥)(28 3 )(𝑆𝑅𝑇) ( )(0,088 .𝑑)( .bCOD)(5508 𝑚3 /𝑑)( )
𝑑 𝑚 𝑑 𝑔 g 𝑚3
𝑔 + 𝑔
[1+(9,082 .𝑑 )(𝑆𝑅𝑇)(0,85) [1+(0,088 .𝑑 )(𝑆𝑅𝑇)(0,85)
𝑔 𝑔

6609,6𝑚3 0,45𝑔𝑉𝑆𝑆
𝑄𝑥𝑌𝑥(𝑆𝑜−𝑆) 𝑥 𝑋189,92 g/𝑚3 𝑘𝑔 564,88
𝑑 𝑔𝑏𝐶𝑂𝐷
 A: = x 103 𝑑 = 1+0,075𝑆𝑅𝑇
1+𝑘𝑑.𝑆𝑅𝑇 1+(0,088.𝑆𝑅𝑇.0,85)
 𝑚3
𝑓𝑑.𝑘𝑑.𝑄.𝑌(𝑆𝑜.𝑆)𝑆𝑅𝑇 0,15.0,088.6609.6 .0,45.189,92𝑆𝑅𝑇 𝑘𝑔 8,77 𝑆𝑅𝑇
 B: = 𝑑
x 103 =
1+𝑘𝑑.𝑆𝑅𝑇 1+(0,88.𝑆𝑅𝑇.0,85) 1+0,075𝑆𝑅𝑇

𝑚3 g 𝑘𝑔
 C : Q. nbVSS : 6609,6 x 121,72 𝑚3 𝑥 = 804,52 kg/hari
𝑑 103 𝑔

𝑚3
 D : Q (TSSo-VSSo) : Q.i TSS = 6609,6 𝑥 121,72 𝑔/𝑚3 . 𝑘𝑔/103 𝑔 = 297,43 kg/hari
𝑑

𝑋𝑡𝑠𝑠.𝑉
= A+B+C+D
𝑆𝑅𝑇

3500𝑔/𝑚3 .5508 𝑚3 .𝑘𝑔/103 𝑔 564,88 + 8,77𝑆𝑅𝑇+ 804,52 + 60,34𝑆𝑅𝑇 + 297,43 + 22,31𝑆𝑅𝑇

=
𝑆𝑅𝑇 1+0,075𝑆𝑅𝑇

19278 1666,83+91,42𝑆𝑅𝑇
=
𝑆𝑅𝑇 1+0,075𝑆𝑅𝑇

19278+1762394,8 SRT = 1666,83SRT + 91,42 𝑆𝑅𝑇 2

0 = 91,42 𝑆𝑅𝑇 2 − 1760728,8 𝑆𝑅𝑇 − 19278

91,42
0 = 𝑆𝑅𝑇 2 – 19859,8 SRT – 210

𝑚3 0,153𝑚3
Q = 0,153 = .86400s/d = 13219,2𝑚3 /𝑑
𝑠 𝑠

-vol.reactor, Vf

6609,6𝑚3 0,45𝑔𝑉𝑆𝑆
𝑄𝑥.𝑌𝑥.(𝑆𝑜−𝑆) . .189,92𝑔/𝑚3 564882,9
𝑑 𝑔𝑏𝐶𝑂𝐷
A: = 𝑔 = 0,85+0,075𝑆𝑅𝑇
1+𝑘𝑑.𝑆𝑅𝑇 [1+(0,088 .𝑑.𝑆𝑅𝑇)].0,85
𝑔

B : Q (nbVSS).SRT = 6609,6 𝑚3 /𝑑. 121,72𝑔/𝑚3 . 𝑘𝑔/103 𝑔 = 804,5 SRT

6609,6𝑚3 𝑔𝑉𝑆𝑆
𝑄.𝑌𝑛.(𝑁𝑂𝑥)𝑆𝑅𝑇 𝑥0,15 𝑥 𝑆𝑅𝑇 28 𝑔𝑁𝑂𝑥/𝑚3 27760,3 𝑆𝑅𝑇
𝑑 𝑔𝑁𝑂𝑥
C: [1+𝑏𝑛(𝑆𝑅𝑇)]0,85
= 0,082𝑔 = 0,85+0,075𝑆𝑅𝑇
[1+( .𝑑)(𝑆𝑅𝑇)].0,85
𝑔

0,15𝑔 0,082𝑔 8609,6𝑚3 𝑉𝑆𝑆

𝑓𝑑.𝑏ℎ.𝑄.𝑌ℎ(𝑆𝑜−𝑆).𝑆𝑅𝑇 2 𝑥 .𝑑 𝑥 𝑥 189,92 𝑔/𝑚3 𝑥 0,459 𝐶𝑂𝐷
𝑔 𝑔 𝑑 𝑔𝑏
D: [1+𝑏𝑛(𝑆𝑅𝑇)]0,85
= 𝑔
[1+(0,088 .𝑑.𝑆𝑅𝑇)].0,85
𝑔

= 69481,1 𝑆𝑅𝑇 2 /0,85+0,075 SRT

 𝑚3
E : Q (TSSo-VSSo)SRT = 6609,6 . 45 𝑔/𝑚3 . 𝑆𝑅𝑇
𝑑

19.278.000 = 564882,9/(0,85+0,075SRT) + 804,5 SRT + 27760,3 SRT/(0,85+0,07SRT)

+ 6948,1 𝑆𝑅𝑇 2 /(0,85+0,075SRT) + 297432 SRT

SRT = 20,7 jam (sesuai)

(Px,VSS).SRT = 𝑉𝑇 (𝑋𝑀𝐿𝑆𝑆 )

7887007,4 = 5508 𝑚3 . 𝑋𝑀𝐿𝑆𝑆

𝑔3
𝑋𝑀𝐿𝑆𝑆 = 1431,9𝑚

 Fraksi MLSS
𝑋𝑀𝐿𝑉𝑆𝑆 1431,9𝑔/𝑚3
= = 0,4
𝑋𝑀𝐿𝑆𝑆 7500 𝑔/𝑚3

 Jumlah𝑁𝐻4 𝑁 𝑢𝑛𝑡𝑢𝑘 𝑜𝑥𝑖𝑑𝑎𝑠𝑖 𝑁𝑂𝑥

NOx = TKNo – Ne – 0,12𝑃𝑥𝑏10 /Q
- 𝑃𝑥𝑏10 = 𝐴 + 𝐵 + 𝐶
= 235127,95 + 16653,15 + 1239205,5 = 1491 kg/d
1491𝑘𝑔
(0,12− .103 𝑔 )
𝑑 kg
- NOx = 35 – 1 – 𝑚3
= 7 g/𝑚3
6609,6
𝑑

 Check Degree Nitrification

𝑚3
a.) -Vf (NOx) = 1652,4𝑐𝑦𝑐𝑙𝑒 . 7 g/𝑚3 = 11566,8 g/fill
 -NH4 before fill = Vs (Ne)
Vs(Ne) = Ne (V-Vf)
= 1 g/𝑚3 – (5508-1652,4) 𝑚3 = 3855,6 g
Total N padaawalsiklus = 11566,8 + 3855,6 = 15422,4 g
15422,49
Luteal consentration : No = = 2,8 g/𝑚3
5508 𝑚3

b.) Reaction Time

𝑁𝑜 ϻ max 𝐴𝑂𝐵 𝑆𝑜
𝐾𝑁𝐻4 ln( 𝑁𝐼 ) + (No-NI) = Xn( ) (𝐾𝑜𝑛𝑜𝑏+𝑆𝑜)t
𝑌𝑛

 Nitrifier Concentration
𝑚3 𝑉𝑆𝑆 7g
Q Yn (NOx) SRT 6609,6 .0,15 .𝑁. .20,7 𝑑
𝑑 𝑔𝑏𝑁𝐻4 𝑚3
Xn = [1+𝑏𝑛(𝑆𝑅𝑇)]𝑉
= 𝑔 = 9,67g/𝑚3
[1+(0,082 .𝑑)(20,7𝑑)]5508 𝑚3
𝑔

 WaktuReaksi
No = 6 g/𝑚3

Ne = 1 g/𝑚3

g 𝑔
6 3 0,52 .𝑑 2
𝑚 3 3 𝑔
0,50ln[ g ] + (6-1) g/𝑚 = 9,67 g/𝑚 ( 𝑔𝑣𝑠𝑠 ) (0,5+2) . 𝑡
1 3 0,15 .𝑁
𝑚 𝑔𝑁𝐻4

5,9 = 26,82 t
5,9
t = 26,82 = 0,22 d ≈ 5,3 jam

 Waktu aerasi = 6jam

 Decant Pumping Rate

Decant Volume = Fill Volume = 1612,4
Td = Decant time = 0,5 jam = 30 menit
1652,4
Pumping rate = = 55,08𝑚3 /𝑚𝑒𝑛𝑖𝑡
30

 Total yang dibutuhkan

Ro = Q(So-S) – 1,42 𝑃𝑥𝑏10 + 4,57 𝑄 (𝑁𝑂𝑥)
𝑚3 𝑚3
= (6609,6 𝑑 − 189,92) − (1,42𝑥1491 𝑘𝑔/𝑑) + (4,57𝑥6609,6 𝑥 7 g/𝑚3
𝑑

= 4513,9 kg/d

-Siklus 8
𝑂2persiklus = 4513

 Waktuaerasi 6 jam
564,24
Average 𝑂2 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 = = 94,04 kg 𝑂2 /𝑗𝑎𝑚
6
5508𝑚3 3500𝑔
(𝑣)(𝑀𝐿𝑆𝑆) (2𝑡𝑎𝑛𝑘𝑠)( )( 3 )(1𝑘𝑔/103 𝑔)
𝑡𝑎𝑛𝑘 𝑚
Px,tss = = = 1862,6 kg/d
𝑆𝑅𝑇 20,7 𝑑
1𝑘𝑔
b COD removed = (6609,6 𝑚3 /𝑑) (189,92 g/𝑚3 )(103 𝑔)

= 1255,3 kg/d