Activited Sludge Personal

8/13/2019 Activited Sludge Personal

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Basin Volume V 2400 m3

Dry sludge [A] 4 Kg/m3

Cell death rat Kd 0.06 l/day

Substract utili A 0.6

Input stream V 10 m3/min

0.166666667 m3/sInput stream1[BOD] 135 mg/l

1.35E+11 Kg/m3

Input stream2[BOD] 200 mg/l

0.2 Kg/m3

out put strrea [BOD] 15 mg/l

0.015 Kg/m3

out put strrea [BOD] 20 mg/l

0.02 Kg/m3

b)

[B]T1 0.015 kg/m3

[B]1 0.135 kg/m3

[B]T2 0.020 kg/m3 Solver

[B]2 0.200 kg/m3 Left -0.0018

[B]eq 0.005 kg/m3 Right -0.0018

V 2400 m3V 0.1667 m3/s

TB 14400 s 4 h

vB -1

[A] 4 kg/m3

k 0.000208333 m3/kgs

vA 0.6

vad -1

kd 0.06 1/day

kd 6.94444E-07 1/s

TA1 1800000 s 500 h

TA2 847059 s 235 h AA

BBK

1


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c) [A] 6.167 kg/m3

TA 500 h

d) V 1200 m3

TB 7200 s 2

[B]T1 0.02357143

TB 7200

TA1 614634 171

[B]T2 0.00765306

TB 7200

TA2 -2756250 -766

dAdeq k


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eq

B BB

BB

Ak ][][

][][

][

1 0

0


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Given data:

Basin Volume

CSTR 1 VB1 4000 m

CSTR 2 VB2 2000 m

initial input stream Vdotini 250 m/h

Concentration of biomass [A] 4 kg/m

Cell death rate per day kd 0.025 1 /day

Substrate utilisation vA 0.6

vad -1

vB -1

Biological Oxigen Demand [B] 600 mg/l

BOD of CSTR 1 [B] t1 35 mg/l

BOD of CSTR 1+2 [B] t2 27 mg/l

b) residence time CSTR 1 B1 16 [h]

residence time CSTR 1+2 B2 24 [h]

[B]eq 1.0301E-02 Kg/m

[B]eq 10.301 mg/l

K 0.357 m/(kg*h)

h s

c) For CSTR 1 4000m A 235.0 8.46E+05

For CSTR 1+2 6000m A 393.8 1.42E+06

d) Big ==>Small

initial input stream Vdot

ini 250 m/h

Volume of 1st

Reactor VR1 4000 m

Volume of 2nd

Reactor VR2 2000 m

Small ==>Big

initial input stream Vot

ini 250 m/h

Volume of 1st

Reactor VR1 2000 m

Volume of 2nd

Reactor VR2 4000 m

There is no different if stream passes the small basin first, the outlet BOD

refer cre levenspiel


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0212

B

BBB

100

Beq

10

1

BCSTR

q


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[A] 4 kg/m

vb -1

vadd 0

vA 0.6

vad -1

kd 0.025 1/d

B) V flow V flow V [B] [B] [B]T [B]T

m/h m/s m mg/l kg/m mg/l kg/m

250 0.069444 4000 600 0.6 35 0.035

250 0.069444 6000 600 0.6 27 0.027

1/K 10072.13115

K 9.92839E-05

0.010300546

C)

V TA TB

m3 s s

4000 846022 57600

6000 1417555 86400

D) If it Passed through in series

First 4000 then 2000 First

As before the outlet of the first basin is the same so For

[B]1 0.035 kg/m TB1

TA1 846022 s [B]1

TB1 57600 s TA1

For 2000 m3 For

TB2 28800 s [B]2

[B]2 0.012286 kg/m TA2

TA2 -5846175 s TB2

No difference space time remains the same in total


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TB x axis values

s -

57600 2.45226E-06

86400 1.65799E-06

2000 then 4000

2000 m3

28800 s

0.057713568 kg/m

394468.5025 s

4000 m3

0.012286432 kg/m-5846174.77 s

57600

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

1.0E-06 1.5E-06 2.0E-06 2.5E-06

[B]T

Chart Title


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3.0E-06

Series1


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Given Data

V,react = 4000 m

T = constant

[A] = 4 kg / m

kd = 0.025 1 / d

0.00104 1 / hv,A = 0.6 substrate utilisation

v,B = -1

Permitted limit = [BOD] 40 mg / l

0.04 kg / m

a)

b)

V, m / h [B]o(mg/l) [B]mg/l [B]

okg/m

1996 162 680 26 0.68

1999 168 608 25 0.608

2000 228 612 31 0.612

2001 267 604 34 0.604

slope = 1 / k = 2.7245

intercept = [BOD]eq = 0.0082

k = 0.3670 m / (kg * h)

[BOD]eq = 0.0082 kg / m

c)

volume basin 4000 m

[B]o

600 mg / L

0.6 kg / m

[B] 40 mg / L

0.04 kg / mB 12.00 h

V 333.46 m/h

d)

V

VB


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[B]o

600 mg / L

0.6 kg / m

for [B] 40 mg / L

Volume basin 2000 m

B 2.26 h

Required

[B]1, kg/m [B]2, kg/m [B]2, kg / m Diff V (m / h)

0.15 0.04 0.04 1.9791E-09 886.01

V

VB


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[B]kg/m B (h)([BOD] - [BOD]

o)/

vB*[A]*B

0.026 24.69 0.0066

0.025 23.81 0.0061

0.031 17.54 0.0083

0.034 14.98 0.0095y = 2.724

R =

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.0000 0.0020 0.0040

V

V

B

eq

BB

o

B

B BOD

A

BODBOD

k

BOD

1

eq

B

BBB

BB

Ak ][][

][][

][

1 0

0

oB BODBOD 1


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eqBBB

Ak

1][

][][][][

Ak

BBAkB

BB

eq

BB


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x + 0.0082

0.9973

0.0060 0.0080 0.0100


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V,react = 4000 m

T = constant

[A] = 4 kg / m

kd = 0.025 1 / d

0.001041667 1 / h

v,A = 0.6substrate

utilisation

v,B = -1

Permitted limit = 40 mg / l0.04 kg / m

V, m / h [B]o, mg/l [B]t, mg/l [B]o, kg/m [B]t, kg/m

1996 162 680 26 0.68 0.026

1999 168 608 25 0.608 0.025

2000 228 612 31 0.612 0.031

2001 267 604 34 0.604 0.034

Slope 2.7245265980.0100

Improvement of an activated sludge system

Into an industrial activated sludge system of 4000m (CSTR basin) the volume flow of

concentration has decreased (long duration average values):

volume flow inlet concentration outlet concentra

1996 162m/h 680mg/l 2

1999 168m/h 608mg/l 2

2000 228m/h 612mg/l 3

2001 267m/h 604mg/l 3

Your task is to predict the maximum volume flow related to the permitted limit of 40mg/l

a) without any change in the process and

b) when the basins will be divided in two parts of 2000m each and passed in ser

Assume a multi-component substrate and ideal CSTR conditions in all basins. The conce

basins is to be estimated to 4kg/m and the normal cell death rate to 2,5% per day (kd= 0,

= 0,6). The temperature in the system is constant!

b) Calculate the equilibrium concentration of BOD [BOD]eqand the rate constant k

(graphical solution possible)! The linearized form of the equation for residence ti

eq

BB

o

B

B BOD

A

BODBOD

k

BOD

1

Draw an adequate diagram!


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Intercept 0.008202071

K 0.3670 m / (kg h)

[BOD]eq 0.0082 kg/m3

Inlet [ ] 600 mg/l

0.6 kg/m3

Outlet [ ] 40 mg/l

0.04 kg/m3

12.00 h

Max volume 333.46 m3/h

Volume Basis 2000 m3

4000 m3

Inlet [ ] 600 mg/l

0.6 kg/m3

Outlet [ ] 40 mg/l

0.04 kg/m3

2.26 h

Volume flow 886.00 m3/h

y = 0.366x -

R = 0.99

0.0000

0.0010

0.0020

0.0030

0.0040

0.0050

0.0060

0.0070

0.0080

0.0090

.

0 0.01

c) Calculate for the 4000m basin the maximum volume flow for an estimated inlet

limit of 40mg/l using the two coefficients from b) or take the value from the dia

d) Calculate (iteration) for the two 2000m basins in series the maximum volume f

600mg/l and an outlet limit of 40mg/l using the two coefficients from b)!


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Outlet

concentration B1

[kg/m3]

Outlet

concentration B2

[kg/m3]

Final

concentration

required [kg/m3]

Difference

0.145379 0.040 0.04 0.000


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Residence

timeBig Equation

24.69 0.0066

23.81 0.0061

17.54 0.0083

14.98 0.0095

astewater has increased, while the outlet

tion

6mg/l

5mg/l

1mg/l

4mg/l

in the outlet

ies.

tration of biomass (dry sludge [A]) in the

25d-1

). The substrate utilisation is 60% (A

from the given data by linear regression

me is

11 points

eq

BB

o

B

B BOD

A

BODBOD

k

BOD

1


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0.003

73

0.02 0.03 0.04

Big Equation

Linear (Big Equation)

concentration of 600mg/l and an outlet

ram!

4 points

eq

B

BBB

BB

Ak ][][

][][

][

10

0

low for an estimated inlet concentration of

6 points

eq

B

BBB

BB

Ak ][][

][][

][

10

0

1][

][][][][

Ak

BBAkB

BB

eq

BB


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V,react 4000 m

T constant

[A] 4 kg / m

kd 0.025 1 / d

0.00104 1 / h

v,A 0.6 substrate utilisation

v,B -1

Permitted limit 40 mg / l

0.04 kg / m

a)

b)


okg/m [B]kg/m

1996 162 680 26 0.68 0.026

2001 267 604 34 0.604 0.034

Slope = 1/K 2.76804636

[B]eq 0.007671 kg / mk 0.3613 m / (kg * h)

c)

[B]omg/l [B]

okg/m tB (h) V (m/h)

580 0.58 11.56 346.06

600 0.6 11.99 333.70

620 0.62 12.41 322.19640 0.64 12.84 311.45

d)

BB

B

B

A

BODBOD

k

BOD

1

eq

B

BBB

BB

Ak ][][

][][

][

1 0

0

V

VB


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V,each 1330 m

V,total 3990 m

[B]t 0.04 kg / m

V, m / h t,B, h [B]2, kg/m [B]1, kg/m [B]0, kg/m

800 1.66 0.1177 0.3819 1.2811

1000 1.33 0.1021 0.2837 0.8142

1200 1.11 0.0918 0.2265 0.5769

V

VB

kBB ][][ 0


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600

700

800

900

1000

1100

1200

1300

0.0000 0.5000 1.0000 1.5000

Volumeflow,m/h

Inlet concentration, kg/m

Volume flow vs. Inlet concentration

First Basin

Second Basin

Third Basin

eqBB BBA ][][][ 0


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T constant

[A] 4 kg / m

kd 0.025 1 / d

0.00104 1 / hv,A 0.6

v,B -1

[BOD] 20 mg / l

0.02 kg / m

a)

b)


okg/m [B]kg/m B (h)

([BOD] -

[BOD]o)/

vB*[A]*B

160 680 20 0.68 0.020 0.00 #DIV/0!

180 600 20 0.6 0.02 0.00 #DIV/0!

220 580 20 0.58 0.02 0.00 #DIV/0!

slope = 1 / k = #DIV/0!

intercept = [BOD]eq = #DIV/0!

k = #DIV/0! m / (kg * h)

[BOD]eq = #DIV/0! kg / m

substrate utilisation

eq

BB

o

BB BOD

A

BODBOD

kBOD

1


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Given Data

V,react 5000 m

T constant

[A] 4 kg / m

kd 0.025 1 / d

0.00104 1 / h

v,A 0.6

v,B -1

Permitted limit = [BOD] 20 mg / l

0.02 kg / m

a)

b)


okg/m

1995 220 580 30 0.58

1998 250 520 32 0.52

2001 270 510 33 0.51

2004 300 470 35 0.47

slope = 1 / k = 7.9484

intercept = [BOD]eq = -0.0174

k = 0.1258 m / (kg * h)

[BOD]eq = -0.0174 kg / m

c)

Total

volume basin 5000 m V

[B]o

400 mg / L volume basin

0.4 kg / m B

[B]2 20.00 mg / L [B]o

0.02 kg / m

B 20.19 h [B]1 = [B]o2

V 247.68 m/h

substrate utilisation

0


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d)

V 272.68 m/h

volume basin 2500 m

B 9.17 h

[B]o

400.00 mg / L

0.40 kg / m

[B]2 0.05695 kg / m

56.95 mg / L

diff [B]2 0.00570 kg / m

5.70 mg / L

with additionally flow

influence of the bypass flow on the outlet concentration

eq

B

BBBAk ][][][ 0

][ k

B


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[B]kg/m B (h)([BOD] - [BOD]

o)/

vB*[A]*B

0.030 22.73 6.05E-03

0.032 20.00 6.10E-03

0.033 18.52 6.44E-03

0.035 16.67 6.53E-03

First

247.68 m/h

2500 m

10.09 h

400 mg / L

0.4 kg / m

0.05125 kg / m

51.25 mg / L

y = 7.9484x - 0.

R = 0.829

0.029

0.030

0.031

0.032

0.033

0.034

0.035

0.036

6.00E-03 6.10E-03 6.20E-03 6.30E

V

V

B

eq

BB

o

B

B

BOD

A

BODBOD

k

BOD

1

101

00

1

AK

BBAKB

BB

eqBB

V

V

B


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1][

][][][

Ak

BBA

BB

eq

BB


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0174

-03 6.40E-03 6.50E-03 6.60E-03


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Given Data

input stram 15.6 m3/min 0.26 m3/sec

T constant

[A] 4 kg / m

kd 0.025 1 / d

0.00104 1 / h 2.89352E-07 1/sv,A 0.6 substrate utilisation

v,B -1

[BOD] 25 mg / l

0.025 kg / m

[BOD] 500 mg/l

0.5 kg/m3

lab

[A] 1000 mg/l

1 kg/m3

y axis x axis

time in minseconds [BOD] in mg/l in kg/m3

30 1800 135 0.135 0.000203 slope 634.17 1/K

60 3600 40 0.04 0.000128 int 0.0064 BOD EQ

120 7200 12 0.012 6.78E-05 K 0.001577 1/s

1440 86400 10 0.01 5.67E-06

Tau A 2142.636312 sTau B 4048.800403 s

Volume of basiin 1052.688 m3

BBOD

dAdeqA

AkBBK

1

eqB

BBB

BB

AKV

V

0

0

1


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0.135

0.01

y = 634.17x + 0.0064

R = 1

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0 0.00005 0.0001 0.00015 0.0002 0.00025

Series1

Linear (Series1)

eq

BB

o

BBOD

A

BODBOD

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