Tugas Mata Kuliah Konversi Massa Biologi.docx

TUGAS MATA KULIAH KONVERSI MASA POLUTAN SECARA BIOLOGIS

Name

: ANITA PRATIWI

NRP

: 0321 18 5001 0001

Chapter 8 Suspended Growth 8.4 Shown in the following table, with values for convensional wastewater characterization parameters (to be selected by instructor) Wastewater Parameter

Unit

1

2

3

TSS

Mg/L

220

170

90

VSS

Mg/L

200

140

70

BOD

Mg/L

200

160

120

RbCOD

Mg/L

100

40

80

TCOD

Mg/L

500

400

280

sCOD

Mg/L

160

200

180

(Selected unit 1) Assume 

bCOD/BOD ratio equals 1.6



the activated sludge treatment effluent sCOD equals 30.0 mg/L.

Determine : a) The biodegradable COD concentration b) The slowly biodegradable COD concentration c) The non-biodegradable volatile suspended solids (nbVSS) concentration d) The inert total suspended solids (iTSS) concentration)

Solution : a) The biodegradable COD (bCOD) concentration. The bCOD/BOD ratio is given as 1.6. bCOD = 1.6 (BOD) = 1.6 × (200 mg/L) = 320 mg/L b) The slowly biodegradable COD (sbCOD). sbCOD = bCOD – rbCOD = (320 – 100) mg/L = 220 mg c) 1. The nonbiodegradable COD (nbCOD). nbCOD = COD – bCOD = (500 – 320) mg/L = 180 mg/L 2. The nonbiodegradable soluble COD (nbsCOD). The nonbiodegradable soluble COD is equal to the activated sludge system sCOD = 30 mg/L 3. The nonbiodegradable volatile suspended solids (nbVSS) concentration. a. The particulate nonbiodegradable COD (nbpCOD). nbpCOD = nbCOD – nbsCOD = (180 – 30) mg/L = 150 mg/L b. The nbVSS concentration. From Eq. (8-8, 713) VSSCOD =

TCOD− sCOD VSS

=

[ (500 −160) mg/L] (200 mg/L)

nbVSS = 150/1.7 = 88.2 mg/L d) The inert TSS (iTSS) concentration. iTSS = TSS – VSS = 220 – 200 = 20 mg/L

= 1.7 mg COD/mg VSS

8-7. The following information is given for an activated sludge system design: Parameter

Unit

Value

m3/day

10.000

Influent BOD (So)

mg/L

150

Effluen BOD (S)

mg/L

2

Ɵh

Hour

4

Ɵc

Day

6

Y

g VSS / g bCOD

Flowrate (Q)

Wasterwater 1

g VSS / g bCOD

0,40

Wasterwater 2

g VSS / g bCOD

0,50

Wasterwater 3

g VSS / g bCOD

0,30

Y

g VSS / g bCOD

0,5

Fd (Debris yield)

g VSS / g bCOD

0,15

g VSS / g bCOD. Day

0,08

mg/l

40

Kd (Endogeneous Decay ) NbVSS Temperature

o

C

10

(selected wastewater 2) Assume: -

Assume no nitrification occurs due to the SRT selected and low temperature.

-

bCOD = 1,6 BOD

Determine: a) The aeration tank oxygen requirements (kg/day) b) The aeration tank oxygen uptake rate (mg/L.hour) c) The aeration tank biomass concentration (mg/L)

Solution :

Calculate the biomass production (Px bio) bCOD = 1.6 (BOD) bCODinf = 1.6 × (150 mg/L) = 240 mg/L bCOD efl = 1.6 × (2 mg/L) = 3,2 mg/L

Pxvss

=

Q x Y x (So-S) 1+(kd x θc) m3 day

10000

=

+

Q x Y x (So-S) x Fd x kd x θc 1+(kd x θc)

x 0,5 x (0,24 - 0,003) kg/m3 1+(

0,08 day

x 6 day)

10000

+

m3 x day

kg

1+(

0,08 day

x 6 day)

= 800,68 kg/day + 57,65 kg/day = 858,33 kg/day

a)

Determine the oxygen requirements (Ro) Ro

= Q x (So-S) – 1,42 x Px bio = 10.000 m3/day x (0,24 - 0,003) kg/m3 – 1,42 x 858,33 kg/day = 1151,17 kg/day

b)

Determine the oxygen uptake rate (Ru) The formula is: Ru

=

𝑹𝒐 𝑽

First, calculate the Volume of aeration tank V

= Q x Ɵh 1 day

= 10.000 m3/day x 4 hours x 24 hours = 1.666,7 m3

Ru

=

1151,17 kg/day 1.666,7 m3

= 0,691 kg/m3.day = 28,79 mg/L.hours

0,08

0,5 x (0,24-0,003) 3 x 0,15 x x 6 day day m

c)

Determine the biomass concentrations (X) Px x Ɵc X

V

=

1666,7 m3

=

X

= 3,089 kg/m3

858,33 kg/day x 6 day X

= 3089 mg/L

8.12 An activated sludge system is operating with the conditions describes below for system 1,2, or 3 to be selected by instructor. The conditions include the average influent flow, aeration tank volume and MLSS concentration, return activated sludge flow recycle ratio and TSS concentration, and the secondary clarifier effluent TSS concentration. Wasting is from the return activated sludge line leaving the bottom of the secondary clarifier. Activated Sludge System 2 3 1

Parameter

Unit

Flowrate

m3/d

4000

10000

5000

m3

2000

4000

5000

Aeration tank MLSS concentration

mg/L

3000

3500

3000

Clarifier effluent TSS concentration

mg/L

10.0

10.0

10.0

unitless

0.5

1.0

0.75

mg/L

9000

7000

7000

Aeration tank volume

Return sludge recycle ratio Return sludge TSS concentration (Selected unit 1) Determine :

a) What should be the average daily waste sludge rate in m3/d to maintain an SRT of 10 d? b) The Plant operater decides to waste 1/10th of the aeration volume each day to maintain the SRT. What is the waste volume in m3/d and what is the actual SRT ?

Solution : a) Determine the average daily waste sludge rate in m3/d from the return activated sludge line for a 10 d SRT. Solve for Qw in Eq. (8-27) in Table 8- 10. SRT = Qw =

V×(X) (Q- Qw ) × (Xe ) + Q × XR

,

V × (X) / SRT- Qw× Xe XR - Xe

From data in table, [

Qw=

(2000 m3 )×(3000 g /m3 ) ] - (4000 10 d

m3 / d) × (10 g /m3 )

((9000 -10) g /m3 )

Qw = 62.3 m3/d b) Determine the actual SRT when 1/10th of the aeration volume is wasted. In this case the wasting rate, Qw, equals V/10 at a solids concentration = X. Substitute Qw with V/10 into Eq. (8-27).

SRT = SRT =

V(X) 𝑉 𝑉 [𝑄− ]𝑋𝑒 +[ ]𝑋𝑅 10 10

(2000 m3 )(3000 g /m3 ) [(4000m3 / d)−

SRT = 3.3 d

2000 m3 ](10 g /m3 10

)+[

2000 m3 ](9000 g /m3 10

)

Chapter 9 Attached Growth 9.1 A 20-m diameter plastic packing trickling filter, containing cross-flow plastic packing at a 6,1 – m depth with a specific surface area of 100 m2/m3, receives domestic wastewater after primary treatment. The average flowrate is 390, 440, or 490 m3/h (value to be selected by instructor) and the BOD concentration is 150 mg/L. Determine and compare the effluent BOD concentration and percent BOD removal at 20oC and 15oC. Assume an n value of 0,5 and recirculation ratio equal to zero.

Solution : The average flowrate is 390 m3/h (selected) 1. Determine effluent concentration at 20oC for an average flowrate of 390 m3/h a. k20 for design conditions using Eq. (9-20). D 0,5 S1 0,5

k2= k1 [ 1 ] D2

[S ] 2

6,1 0,5 150 0,5

= 0,210 [6,1]

[150]

= 0,210 (L / s)0,5 /m2 b. The hydraulic application rate. 𝑄

q= 𝐴 where, Q

= (390 m3/h)× (103 L/1 m3) ×(1 h/3600 s) = 108.3 L/s

A

= 𝜋D2/4 = [(𝜋/4)(202)] = 314.2 m2

q

= (108.3 L/s)/(314.2 m2) = 0.345 L/m2 • s

c. Effluent concentration using Eq. (9-15). Se Si

-kD

=exp [ qn ] n

Se =Si e-k D/q

0,5

= (150 mg / L) [e-(0,210)(6,1)/0,345 ] = 16.9 mg / L

d. Percent BOD removal [(150 -16.9) mg / L]

%removal=

(150 mg / L)

(100%) =88.7

2. Determine effluent concentration at 15oC. a. Correct k2 for design conditions using Eq. (9-16).

kT =k20 (1,035)(T-20) 𝑘15 = 0,210(1,035)(15−20) =0,177 b. The hydraulic application rate is the same as in Step 1b. q = 0.345 L/m2 • s c. Effluent concentration using Eq. (9-15). n

Se =Si e-k D/q

0,5

= (150 mg / L)[e−(0,177)(6,1)/0,345 ] = 23,9 mg / L

d. Percent BOD removal

%removal=

[(150 -23,9) mg/L] (150 mg/L)

(100%) =84,1

3. In comparing BOD removal, the removal rate is reduced at the lower temperature (88.7 vs. 84.1 %).

9.8 Two 20-m diameter plastic tower trickling filters containing 6,1 m of conventional plastic media (100 m2 area/m3 volume) receive a primary clarifier effluent at an average flowrate of 11,200 m3/d. The TKN concentration is 24 mg/L and the BOD concentration is 150, 130, or 120 mg/L (value to be selected by instructor). The temperature is 18oC. Evaluate the trickling filter BOD loading and determine the nitrogen removal efficiency due to nitrification. Solution : The BOD concentration is 150 mg/L (selected) 1. The BOD loading for a wastewater with a concentration of 150 mg/L. BOD loading = (11,200 m3 /d) × (150 g/m3 ) × (1 kg/103 g) = 1680 kg/d π D2 )= 4

Surface area of filter = A =2 (

2 [

π(20)2 ]= 4

628 m2

(1680 kg/d)

BOD surface loading =

628 m2

= 2.68 kg/m2 • d 2. The nitrogen removal efficiency by estimating the effluent NH4-N concentration using Eq.(9-23). NH4-Ne=20.81(BOD)L 1,03 (NH4 -NL )1,53 (lv)-0,38 (T)-0,12 a. Determine the media surface area. Media volume = A(D) = (628 m2 ) × (6.1 m) = 3830.8 m3 Surface area

= (3830.8 m3 )×(100 m2 /m3 ) = 383.080 m2

b. Determine specific loading rates for Eq. (9-23) terms. BODL =

(11,200 m3 /d)(150 g BOD/m3 ) = 4.386 g BOD/ m2 .d 383,080 m2

NH4 -NL = lv =

(11,200 m3 /d)(24 g TKN/m3 ) = 0.70 g N/ m2 .d 383,080 m2

(11,200 m3 /d)×(10 L / 1m3 ) = 29.2 L// m2 .d 383,080 m2

c. Determine the effluent NH4-N concentration and removal efficiency. NH4 -Ne = 20.81×(4.386)1,03 ×(0.70)1,52 ×(29.2)-0,38 ×(18)-0,12 = 11.6 mg / L Nitrogen removal efficiency

11.6) mg / L] = (100)×[(24 = 51.7% (24 mg/ L)