oxygen transfer

5-Oxygen Transfer_F12.doc

1

Activated Sludge - Types of Processes and Modifications

1 Conventional

Influent Effluent

PF Aeration Tank

HRT = 8 - 15 hrs

O2 supply

Return sludge Waste sludge

O2 demand

Tank length

2 Tapered Aeration

Influent Effluent

PF Aeration Tank

Return sludge Waste sludge O2 supply

Tank length O2 demand

3 Step Aeration

Influent

Effluent

PF Aeration Tank

O2 supply


O2 demand

Tank length

4 Completely Mixed

Influent Completely Mixed Aeration Tank Effluent

O2 supply


O2 demand

Tank length

Alternate waste

sludge drawn

off point

Alternate waste

sludge drawn

off point

Alternate waste

sludge drawn

off point

Alternate waste

sludge drawn

off point


2

5 Contact Stabilization

Effluent

Influent

HRT = 3 - 6 hrs


6 Kraus Process

Waste sludge

Effluent

Influent Aeration Tank

Reaeration Tank

HRT = 24 hrs

(Nitrification) Digested supernatant

Digested sludge

Alternate waste

sludge drawn

off point

Alternate waste

sludge drawn

off point

HRT = 20 -

90 min

Aeration alone can account for half of the operation costs at a typical treatment plant (p. 8,

Logan, 2008)

Gas Transfer theory

1. General Gas Transfer Equation

The rate of oxygen transfer

L s

dCK a C C

dt

where Cs = oxygen concentration in the liquid at saturation, mg/L

= f (T, dissolved solids)

C = oxygen concentration in the liquid at time, t

KLa = oxygen transfer rate coefficient, hr-1

= f (T, types of diffuser, depth of aerator, types of mixer, tank geometry)

Cs - C = dissolved oxygen deficit, D, mg/L

C > Cs

C < Cs

t

CsC


3

2. Two-film Theory (Lewis and Whitman, 1923)

Cl = concentration of gas in bulk liquid

Csl = concentration of gas in surface liquid

Cg = concentration of gas in bulk gas

Csg = concentration of gas in surface gas

Gas film control Liquid film control Cg

-for very soluble gas - for not very soluble gas

e.g., NH3 e.g., O2

= Csl = Csg = Cg

Cg

Cl

Csg = Csl = Cl

Cl

=

Air - turbulent (well

mixed body of air)

Liquid film - laminar molecular

layer

Liquid - turbulent (well mixed

body of water)

Gas film - laminar molecular layer -

stagnant mass of air (molecular

diffusion)

Cl

Cg

Csl

Csg

microlayer

resistance (60

um)


4

3. Diffusion

- Gas diffusion (molecular diffusion) through a liquid film

- Ficks first law of diffusion

VC CJ DA

t x

L3 M L

2 M M

------- = ----- L2 ------ = ----

T L3 T L

3 L T

where D = molecular diffusion coefficient, L2 T

-1

A = surface area, L2

x = liquid film thickness, L

Since 2 1

l sC CC C

x x x x

V = film volume between the gas and liquid interface

Assuming V= constant,

2 1 2 1

l s s l s lC C C C C C CJ V DA DA DA

t x x x x x

s l

C DV A C C

t x

L s l

CV K A C C

t

where KL = D/∆x = oxygen transfer rate, LT-1

divided by V yields

L s l

C AK C C

t V

Let a = A/V

L s l

CK a C C

t

L 1

--- ---

T L

The rate of O2 transfer is

controlled by a liquid film

Cg

x1

Cs

x2

Cl

x

x∆


5

where Cs – Cl = concentration gradient, major driving force

KLa = oxygen transfer rate coefficient, T-1

, hr-1

KLa depends on types of gas and liquid (film thickness), increased by mixing intensity, waves

“a” depends on surface area, A, increased by finer bubbles

Bubbles

In general, the rate of oxygen transfer increases with:

a) decreasing bubble size (larger contact area)

b) longer contact time

c) added turbulence

- Gas transfer increases with area A

- A/V increases by producing fine bubbles and/or breaking the surface

Bubble diameter, mm Bubble diameter, mm

10

Bubble

ris

ing v

elo

city

KL

2 10 2

From two observations,

optimum size = 2 mm

< 2 mm clogging problem in diffuser heads (bacterial slime), more maintenance

> 2 mm, tends to lose KL, O2 transfer rate

Smaller bubble size gives slower

velocity, thus more contact time.

Smaller bubble size gives smaller KL, less

- turbulence, less surface breaking


6

Evaluation of KLa value

L s

dCK a C C

dt

0

1

o

C t

LC

s

dC K a dtC C

0( 1) lno

C t

s LCC C K a t

0lno

C t

s LCC C K a t

ln lns s o LC C C C K a t

ln lns s o LC C C C K a t

ln sL

s o

C CK a t

C C

LK a ts

s o

C Ce

C C

Intercept =

Slope = -KLa

t

ln sC C

ln s oC C

Slope = -KLa

t

ln s

s o

C C

C C


7

Oxygen Transfer and Oxygen Requirements

1. Importance

2

m

O

DO

K DO

a. The rate of aerobic microbial metabolism is independent of the DO concentration above a

critical (minimum) value.

b. Below the critical value, the rate is reduced by the limitation of oxygen required for

respiration.

c. Critical DO concentrations reported in the literature for activated-sludge system range

from 0.2 to 2.0 mg/L.

- For conventional and high-rate aeration basin = 0.5 mg/L

- A typical DO for activated sludge operation would be 2.0 mg/L (W. C. King, PE. Exam,

p. 230)

2. Oxygen Transfer Models

Cell membrane

Liquid film

CO2

DO

(Rate of O2 transfer) Rate of O2 utilization

Microbial cell

Bubble

dC/dt = KLa (Cs - C) dC/dt = r

Figure x.x. Schematic diagram of oxygen transfer in activated sludge.

- Oxygen is dissolved in solution and then extracted from solution by the biological cells.

At steady state, [the rate of oxygen transfer] = [the rate of oxygen utilization]

µm

µ

µm/2

Critical DO cinc (0.2 - 2 mg/L)

KO2 DO (mg/L)

20.2 0.5


8

In clean water

The rate of oxygen transfer

L s

dCK a C C

dt

where Cs - C = dissolved oxygen deficit, mg/L

Kla = oxygen transfer rate coefficient, hr-1

= f (T, types of diffuser, depth of aerator, types of mixer, tank geometry)

Cs = oxygen concentration in the liquid at saturation, mg/L

C = oxygen concentration in the liquid at time, t

KLa depends on temperature, types of diffuser/ mixer, depth of aerator,

tank geometry

Effect of temperature on KLa - van’t Hoff-Arrhenius relationship

20

, ,20

T

La T La CK K

6-61 (ME, p. 286)

where K La, T = oxygen mass-transfer coefficient at temperature T, s-1

K La, 20C = oxygen mass-transfer coefficient at 20 C, s-1

Range of θ value = 1.015 - 1.040

Typical θ value = 1.024

In general, the rate of oxygen transfer increases with:

a ) decreasing bubble size

b ) longer contact time

c ) added turbulence

In wastewater

The rate of oxygen transfer from air bubble to wastewater in an aeration tank:

L s

dCK a C C

dt

where dC/dt = rate of oxygen transfer, mg/L/hr

= alpha factor or coefficient (oxygen transfer coefficient) of the wastewater

= beta factor or coefficient (oxygen saturation coefficient) of the wastewater

KLa = oxygen transfer rate coefficient, hr-1

Cs = oxygen concentration at saturation, mg/L

C = oxygen concentration in the liquid at time t, mg/L

Cs - C = dissolved oxygen deficit in wastewater, mg/L


9

The alpha () factor

- The alpha () factor is the oxygen transfer coefficient factor for waste (see Table 5-32, 4th

ME

447)

- is defined as the ratio of the oxygen transfer coefficient in water to that in clean water:

(KLa) in wastewater

α = ---------------------------

(KLa) in clean water

2) The factor is influenced by many conditions related to:

a ) the characteristics of the wastewater (temp, soluble BOD, SS conc)

b ) the aeration equipment (types of aerators, mixing intensity, tank configuration)

3. The magnitude can even change between the influent and effluent ends of plug-flow aeration

tank.

Viessman & Metcalf &

Hammer Eddy

VH ME (p. 429) King (PE Exam, p. 230)

_____________________________________________________________________________

For municipal wastewater 0.7 - 0.9 (0.4-1.1) .3 - 1.2

0.82

Fine-bubble diffusers as low as 0.4 0.4 - 0.8

Mechanical aerator as high as 1.1 0.6 - 1.2

____________________________________________________________________________

The beta () factor

The beta () factor is the salinity-surface tension correction factor (4th

ME 447)

- is defined as the ratio of the DO saturation concentration in the wastewater to that in clean

water:

Cs in wastewater

β = ------------------------

Cs in clean water

The value is influenced by the wastewater constituents (at constant temperature) including:

a ) dissolved solids, salts

b ) dissolved organics

c ) dissolved gases

For municipal wastewater = 0.7 - 0.98, commonly 0.95 (ME, 429)

= 0.95 (King, PE. Exam, p. 230)

= 0.9, seldom less than 0.8 (VH)


10

Dissolved Oxygen Utilization Rate, r (mg O2/L-hr)

The rate of DO utilization by microorganisms in an activated-sludge system can be determined

by placing a sample of mixed liquor in a closed container and measuring the dissolved-oxygen

depletion with respect to time.

(a) (b)

MLVSS

- O2 is used for cell synthesis and

respiration.

I

1) The oxygen utilization rate, r, is the slope of the resultant curve.

2) The oxygen utilization rate, r, depends on the microorganisms ability to metabolize waste

organics based on such factor as:

a ) F/M ratio

b ) mixing conditions

c ) temperature

3) A general range for the oxygen utilization rate “ r “ in the mixed liquor of conventional

and completely mix (high-rate) activated-sludge systems is:

r = 20 - 100 mg/L·hr. (Typical range: 20 - 80 mg/L·hr)

O2 probe

MLVSS

Slope = r = oxygen utilization rate

mgO2/L-hr

(b) End of plant

(a)

Front of plant

Time (hr)

DO (mg/L)

remaining


11

Air and Oxygen

- At sea level and at 20°C, dry air has a density of ~1.2 kg/m3 varying with pressure and

temperature.

(SI units)

a) The density of air is 1.205 kg/m3 at 20ºC and 1 atm.

b) O2 content in air is 23% (w/w); i.e., 0.23 kg O2/kg air

2 2

3 3

0.23 1.205 0.27715kg O kg air kg O

kg air m air m air

1 m3 of air contains 0.27715 kg of O2 under the standard conditions (T=20°C, p=1 atm).

(US customary units)

a) Air density = 0.075 lb/ft3

b) O2 content in air is 23% (w/w); i.e., 0.23 lb O2/lb air

2 2

3 3

0.075 0.23 0.0173

1

lb air lb O lb O

ft air lb air ft air

1 ft3 of air contains 0.0173 - 0.0174 lb of O2

Power Requirement

1) Purposes of Aeration

a ) Provide oxygen

- to satisfy microbial oxygen demand, r

b ) Provide mixing

- Mixing requirements range from 0.75 to 1.50 HP per 1000 ft3 of tank volume (King,

PE Exam, p. 230).

2) The aerator power required depends on:

a ) Type of activated-sludge process

b ) BOD loading

c ) Oxygen transfer efficiency of the aerator equipment.

3) Aerator performance

- Aeration systems are compared on the basis of mass of gaseous oxygen transferred to

dissolved oxygen per unit of energy expended:


12

Oxygen Transfer Rate

2 2 2 2lb of O lbs O kg of O kg O or

horsepower-hr HP-hr killowatt-hr kW-hr

Oxygen Transfer Efficiency, OTE (%)

2

2

mass of O dissolved (transferred) in waterOTE(%) =

mass of gaseous O applied

The values specified for efficiency are based on operation in clean water with zero DO

concentration and standard conditions (20°C, 1 atm).

Example Oxygen Transfer Efficiency (%) – Use SI units

62.43 m3 of air is required (need to be applied) per kg of BOD applied to an aeration

tank. The aerator is capable of transferring 1.7 kg of O2 (dissolved) per kg BOD applied. What

is the oxygen transfer efficiency (OTE)?

Assumptions:

The density of air at 20ºC and 1 atm is 1.205 kg/m3. Since air contains 23% O2 (w/w),

(0.23 kg O2/kg air)(1.205 kg air/m3) = 0.27715 kg O2/ m

3 of air

1 m3 of air contains 0.27715 kg of O2 under the standard conditions (T = 20°C, p = 1 atm).

Mass of O2 dissolved (transferred) in water

Oxygen transfer efficiency (OTE) = -----------------------------------------------------

Mass of gaseous O2 applied

O2 transferred (dissolved) = 1.7 kg of O2 / kg BOD applied.

62.428 m3 of air 0.27715 kg O2 17.30 kg of O2

O2 applied = ------------------------- ---------------------- = -------------------------

kg of BOD applied m3 of air kg of BOD applied

1.7 kg of O2 / kg BOD applied

OTE (%) = ------------------------------------------------ (100) = 9.8 %

17.30 lb of O2/ kg of BOD applied

Unit conversion: 1 kg = 2.2046 lb, 1 ft3 = 0.028317 m

3


13

Example Oxygen Transfer Efficiency (%) – US customary units

1000 ft3 of air is required (need to be applied) per lb of BOD applied. The aerator is

capable of transferring 1.7 lb of O2 (dissolved) per lb BOD applied. What is the oxygen transfer

efficiency (OTE)? Note: 1 ft3 of air under the standard conditions (T = 20°C, p = 1 atm)

contains 0.0174 lb of O2.

(Solution)

Assumptions: Air contains 23% O2 (w/w) and air density is 0.0174 lb/ ft3

Mass of O2 dissolved (transferred) in water

Oxygen transfer efficiency (OTE) = -----------------------------------------------------

Mass of gaseous O2 applied

O2 transferred (dissolved) = 1.7 lb of O2 / lb BOD applied.

1000 ft3 of air 0.0174 lb of O2 17.4 lb of O2

O2 applied = ------------------------- ----------------------- = -------------------------

lb of BOD applied ft3 of air lb of BOD applied

1.7 lb of O2 / lb BOD applied

OTE (%) = ------------------------------------------------ (100) = 9.8 %

17.4 lb of O2 / lb of BOD applied

Oxygen Transfer Rate

Cell membrane

Liquid film

CO2

DO

(Rate of O2 transfer) Rate of O2 utilization

Microbial cell

Bubble

dC/dt = KLa (Cs - C) dC/dt = r

r = oxygen utilization rate

a. Under steady-state conditions of oxygen transfer in an activated-sludge system, the rate of

oxygen transfer to dissolved oxygen is equal to the rate of oxygen utilization:

Change in DO in wastewater = O2 transfer rate - O2 utilization rate

L s

dCK a C C r

dt (1)

where r = oxygen utilization rate by microorganisms in activated sludge, mg/L


14

At steady state, dc/dt =0

L

s

rK a

C C

(2)

For clean water (at standard test conditions, 20°C)

L s

dCK a C C r

dt (1)

At steady state, dC/dt = 0 and let r = ro

o L sr K a C C (2)

The DO deficit is maximum when C = 0, thus

or oo L s L

s

rr K a C K a

C … this KLa is the smallest KLa value.

When the test is conducted under standard conditions at T = 20°C,

,20

,20

oLa

s

rK

C (3)

KLa is a function of temperature. For a given temperature T,

20

, ,20

T

La T LaK K (4)

Substituting (3) into (4) yields

20

,

,20

ToLa T

s

rK

C

(5)

For wastewater, at given temperature T,

,La T s T

dCK C C r

dt (7)

Substituting (5) into (7) yields

20

,20

Tos T

s

dC rC C r

dt C (8)


15

Under the steady state conditions of oxygen transfer in an activated-sludge system, the rate of

oxygen transfer is equal to the rate of oxygen utilization.

At steady state, dC/dt = 0, the oxygen utilization rate is given as

20

,20

ToT s

s

rr C C

C

(9a)

or

20

,20

T sT o

s

C Cr r

C

(9b)

DO at saturation, Cs

Empirical formula for DO at saturation, Cs

Cs = 14.652 - 0.41022(T) + 0.007910 (T)2 - 0.000077774 (T)

3

Example: Determine the O2 saturation concentration at T = 30C,

Cs = 14.652 - 0.41022(30) + 0.007910 (30)2 - 0.000077774 (30)

3 = 7.3645

Cs = 7.4 mg/L


16

Oxygen Requirement, OR, W (lb O2 / day or kg O2 / day)

Multiplying the oxygen utilization rate by the aeration tank volume, V, (or volume of MLVSS)

yields the oxygen requirement, OR, W:

Oxygen requirement = (Oxygen utilization rate) (Volume of aeration tank or volume of MLVSS)

OR = W = r V

Units:

32 22 3 3

g O 24 hrs 1 kg OO Requirement (OR, W) = m of MLVSS

m of MLVSS hr day 10 g day

kg

2 22

mg O 24 hrs 1 lb lb OO Requirement (OR, W) = L of MLVSS

L of MLVSS hr day 453,600 mg day

2 24 8.34 /

/

lb O mg hr lb MGMG

day L hr day mg L

Multiplying Eq (9b) by V yields

20

,20

T sT T o

s

C CW r V r V

C

Let Wo = ro V

20

,20

T sT o

s

C CW W

C

where

WT = rT V = the amount of oxygen required under the process condition

= mass transfer rate of oxygen under the process condition, lb O2/d

Wo = ro V = the amount of oxygen transferred under standard test condition

= mass transferre rate of oxygen at standard test conditions (i.e., tap water at 20°C),

lbO2/d, kgO2/d

θ = 1.024

Cs,20 = 9.08, 9.17, or 9.2

20(1.024)9.2

T sT o

C CW W

or

20(1.024)9.17

T sT o

C CW W


17

Total Oxygen Requirement, WT

The total oxygen requirement is composed of:

a) Oxygen required for the CBOD removal

b) Oxygen required for the NBOD removal

a) Oxygen required for CBOD removal, WB (lb O2/d) or (kg O2/d)

WB = r V = r Q θ

b) Oxygen required for NBOD, WN (lb O2/d) or (kg O2/d)

WN = 4.6 (∆TKN) or WN = 4.57 (∆TKN) (mg/L)

= 4.6 ∆TKN) Q θ or = 4.57 ∆TKN) Q θ (kg/d) or (lb/d)

where ∆TKN = TKN that is converted to nitrate, mg/L

4.6 or 4.57 = conversion factor for amount of oxygen required for complete oxidation of

TKN

Note: TKN = Total Kjeldahl nitrogen = (Org-N) + (NH3 -N) + (NH4+-N)

c) Total oxygen required, WT

The total amount of oxygen required on average conditions can be estimated using the following

formula:

WT = WB + WN

Total O2 required = O2 required for CBOD removal + O2 required for NBOD removal

1.42 4.57T

Q So SW Px Q No N

f

where 1.42 = conversion factor for cell tissue to BODL

f = 0.68 = factor to convert BOD5 value to BODL (BOD5/BODL = 0.68)

Px = net mass of VSS (cells) produced

4.57 = conversion factor for amount of oxygen required for complete oxidation of TKN.


18

Air Requirement, w

- based on mass and volume:

Air mass requirement = mass air flow rate = weight of flow of air, w (kg air/s, lb air/s)

Air volume requirement = air flow rate, QA (m3 air/s, ft

3 air /s)

1) Air mass requirement, w (kg air/s, lb air/s)

(SI unit)

2

220.23

( )O

kg O

OR kg Airswkg OX OTE s

kg Air

where w = air mass requirement, kg air/s

OR = O2 requirement, kg O2/s

XO2 = O2 content in (make-up) air = 23 kg O2 /kg Air or 0.27715 kg O2/ m3 of air

OTE or FOTE = field O2 transfer efficiency, fraction

(US customary unit)

2

220.23

( )O

lbO

OR lb AirswlbOX OTE s

lb Air

where w = air mass requirement, lb air/s

OR = O2 requirement, lb O2/s

XO2 = O2 content in (make-up) air = 23 lb O2 /lb Air or 0.0174 lb O2/ft3 Air


2) Air volume requirement, QA, ft3 air/s

23

223

0.22715( )

A

O

kg O

OR m AirsQkg OX OTE s

m Air

where QA = air volume requirement, w m3 air/s

OR = O2 requirement, kg O2/s

XO2 = O2 content in (make-up) air = 23 kg O2 /kg air or 0.22715 kg O2/m3 air


(U.S. customary units)

23

223

0.0174( )

A

O

lbO

OR ft AirsQlbOX OTE s

ft Air

where QA = air volume requirement, ft3 air/s

OR = O2 requirement, lb O2/s

XO2 = O2 content in (make-up) air = 23 lb O2 /lb Air or 0.0174 or 0.175 lb O2/ft3 Air



19

Note: Under standard conditions,

100 1 1 100

23 0.075 1440A oQ W

eff

3 3

2

2

1

min min

ft air lbO ft air d

d lbO

3 3

2

2

1ft air lbO ft air d

s d lbO s

where

QA = air flow rate (scfm) required to transfer Wo of oxygen at standard condition

eff = oxygen transfer efficiency of the aerator at standard condition, %

Example

Given: V = 0.5 Mgal (1892.7 m3); Q = 2 Mgal/day (7570.8 m

3/d); So = 200 mg/L; S = 0 mg/L;

The oxygen utilization rate, r = 50 mg/L-hr; O2 transfer efficiency = 10%;

1 ft3 of air contains 0.0174 lb O2 (1 m

3 of air contains 0.27715 kg of O2);

1 gal = 3.7854 x 10-3

m3

Determine: 1) oxygen required, (lb O2/day); (kg O2/day)

2) air mass requirement, mass air flow rate, or weight of flow of air, w (lb/d); (kg/d)

3) air volume required, QA (ft3 /day); (m

3 /day)

4) lb O2 used / lb BOD5 removed; kg O2 used / kg BOD5 removed

(Solutions) SI unit

1) Mass of O2 required, kg /day = r V

50 mg 1892.7 m3 24 hr 1 x10

-3 kg/m

3

= --------- ------------ -------- ---------------- = 2,271 kg O2 /day

L-hr day mg/L

2) Air mass required, w (kg/d)

2

22

2,271 /

0.23(0.1)O

OR kgO dw

kgOX OTE

kg air

2,271 kg O2 /day kg air

w = --------------------- --------------- = 98,750 kg air/day

0.1 0.23 kg O2


20

3) Air volume requirement, QA, m3 air/d

2

223

2,271 /

0.27715(0.1)

A

O

OR kgO dQ

kgOX OTE

m air

2,271 kg O2 /day 1 m3 of air 81,941 m

3 air

Air volume (m3) = ---------------------- ------------------ = ------------------

required per day 0.1 0.27715 kg O2 day

QA

4) kg O2 used 2,271 kg O2 /day 1 mg/L

------------------------ = ------------------------------------- --------------

kg BOD5 removed (200 - 0) mg/L (7,570.8 m3/d) 10

-3 kg/m

3

2,271 kg O2 /day 1.5 kg O2

= ----------------------------- = -------------

1,514 kg BOD5 / day kg BOD5

(U.S. customary units)

1) Mass (lb) of O2 required /day = r V

50 mg 0.5 Mgal 24 hr 8.34 lb/Mgal

= ---------- ------------ -------- ----------------- = 5,000 lb O2 /day

L-hr day mg/L

\

2) Air mass required, w (lb/d)

2

22

5,000 /

0.23(0.1)O

OR lbO dw

lbOX OTE

lb air

5,000 lbO2/d lb air

w = ---------------- ---------------- = 217,391 lb air/day

0.1 0.23 lb O2

3) Air volume requirement, QA, ft3 air/d

3

2

223

5,000 / 2,874,000

0.0174(0.1)

A

O

OR lbO d ft airQ

lbOX OTE d

ft air


21

4) lb O2 used 5,000 lb O2 /day

------------------------ = -----------------------------------------------

lb BOD5 removed (200 - 0) mg/L (2 Mgal/day)(8.34)

5,000 lb O2 /day 1.5 lb O2

= --------------------------- = ------------

3336 lb BOD5 / day lb BOD5

Example

Given: Wastewater temperature, T = 15C; Cs = 10.2 mg/L at 15C; C = 2 mg/L at

15C; Cs = 9.2 mg/L at 20C; = 0.9; = 0.95; OTE = 10 %; = 1.024 for temperature

correction; total oxygen required, WT = 4536 kg/d = 10,000 lb O2/d.

Determine: (1) Mass transfer rate of O2 at standard conditions, Wo

(2) Air flow rate QA (w) required at standard test conditions.

(Solution)

20( )9.2

T s lT o

C CW W

20

,20/

To T

s l s

WW

C C C

SI units:

(1)

215 20

4536 /6,789 /

0.9 1.024 0.95 (10.2 / ) 2 / /9.2 /o

kg dW kg O d

mg L mg L mg L

(2)

2 3

2 2

3

14,536 /

1440 min 138.67

min0.227150.1

A

O

dkgO d

OR m airQ w

X OTE kg O

m air

7-Oxygen Transfer_F10

22

US customary units:

(1)

20( )9.2

T s lT o

C CW W

20

,20/

To T

s l s

WW

C C C

215 20

10,000 /14,966 /

0.9 1.024 0.95 (10.2 / ) 2 / / 9.2 /o

lb dW lbO d

mg L mg L mg L

(2)

2 3

2 2

3

114,966 /

1440 min 6,000

min0.01740.1

A

O

dlbO d

OR ft airQ w

X OTE lb O

ft air

or

100 1 1 100

23 0.075 1440A ow Q W

eff

100 1 1 100

14,966 6025 600023 0.075 1440 10

Aw Q scfm scfm

Aeration Devices

Table 6-14 (ME p. 278) lists the commonly used aeration devices.

Classification of Aerators:

1. Submerged

1) Diffused air 2) Sparger turbine

3) Jet

2. Surface

1) Low-speed turbine

2) High-speed floating 3) Rotor-brush

4) Cascade

Typical devices used for the oxygen transfer (Figure 6-33).

a) fine bubble diffused-air

b) medium bubble diffused-air c) sparger turbine

d) static tube mixer

e) jet reactor f) low-speed turbine

g) high-speed floating aerator

h) rotor-brush aerator.


23

Selection of Aeration Devices (Aerator system)

The selection of an aerator system in process design must consider:

a) Oxygen transfer efficiency (%)

b) Oxygen transfer rate (kg/kW·hr; lb/hp·hr)

c) Effective mixing (Mixing Requirement = 10 - 30 SCF/min per 1000 ft3 aeration tank

volume

d) Flexibility

e) Reliability

f) Maintenance of equipment

g) Costs (capital, operational & maintenance)


24

Estimation of Oxygen Supply Requirements (King, PE exam, p. 229)

Design oxygen requirements (lb O2/ lb BOD5) - Ten States Standards (1978):

a) For activated sludge modifications (other than extended aeration), 1.1 lb O2 / lb BOD5

b) For extended aeration mode of operation, 1.8 lb O2 / lb BOD5

Mechanical Aeration (King, PE Exam, p. 229; ME)

Oxygen transfer rate, N (lb O2 / HP-hr), N, under field conditions:

( 20)

0 1.0249.08

TsatDO DON N

(1)

or

( 20)

0 1.0249.17

TsatDO DON N

(2)

We use Eq (1)

or

( 20)

0 1.0249.17

Twaltt LC CN N

5-62 (4th

ME 447)

where N = oxygen transfer rate under field conditions, kg O2/kW-hr, lb O2 / HP-hr

No = oxygen transfer rate under standard test conditions (at 20ºC, zero DO),

kg O2/kW-hr, lb O2 / HP-hr = a test certifying O2 transfer

= alpher factor = oxygen transfer coefficient factor for waste (see Table 5-32, 4th

ME

447)

= beta factor = salinity-surface tension correction factor, usually 1 (4th

ME 447)

CL = DO = operating DO concentration (mg/L)

DOsat = Cwalt = DO saturation concentration in tap water for the specific temperature and

altitude (mg/L) see Fig 5-68 (4th

ME 447) or Table 7-4 (Handout).

9.17 or 9.08 = DO sat for standard test conditions (mg/L)

T = wastewater temperature (°C)

The field and standard transfer rates for various mechanical aeration devices are shown in

Table 7-23.


25

The theoretical nameplate horsepower is

2 2

2 2

( / ) 1( )

( / )

O Demand lbO hrhp nameplate

O Transfer Rate lbO hp hr n

2 2

2 2

( / ) 1( )

( / )

O Demand kgO hrkW nameplate

O Transfer Rate kgO kW hr n

where n = the number of aerators

Example (16.3, R&R, p. 516) A completely mixed activated sludge plant is located at El. 2000 ft

(610 m).

a) the oxygen demand is 2680 lb/day (1220 kg/d) during the summer when the wastewater

temperature is 82°F (27.8°C).

b) the alpha value is 0.75 and beta is 0.95.

c) the operating DO is 2.0 mg/L

d) four aerators are to be used

e) the manufacture has a test certifying the transfer (No) as 2.2 lb O2 /hp-hr (nameplate hp)

(1.34 kg O2 /kW-h).

Determine the theoretical aerator power per aerator (name plate hp).


26

(Solution)

Temperature, T = (82 - 32) (5/9) = 27.8°C

Saturation DO, Cs = 7.95 mg/L at T = 27.8 °C at El = 0 ft

(Cs = 7.95 at T= 27ºC from Table D-1, 4th

ME 1746)

Note: Cs = 14.652 - 0.41022(T) + 0.007910 (T)2 - 0.000077774(T)

3

Cs = 14.652 - 0.41022(27.8) + 0.007910 (27.8)2 - 0.000077774(27.8)

3

= 7.67 mg/L

1. Altitude correction

Pz

Cs, Pz = Cs, 760 ----------------

760 mmHg

where Pz = Barometric pressure at El = z ft

Cs, 760 = Saturation DO at temperature at T°C at El = 0 ft.

Barometric pressure, p = 760 mmHg at El at 0 ft

Barometric pressure, p = 706 mmHg at El at 2000 ft (Table 16.1, R&R, p. 509)

706 at El 2000 ft

Cs = (7.95 mg/L) -------------------- = 7.39 mg/L at El = 2000 ft

760 at El 0 ft


27

( 20)

0 1.0249.17

TsatDO DON N

where N = oxygen transfer rate under field conditions (lb O2 / HP-hr)

No = oxygen transfer rate under standard test conditions = 2.2 lb O2 / HP-hr

= alpha factor = 0.75

= beta factor = 0.95

DO = operating DO concentration = 2.0 mg/L

DO sat = DO saturation concentration in tap water for the temperature 27.8°C and altitude

2000 ft = 7.39 mg/L

9.17 = DO sat for standard test conditions (mg/L)

T = wastewater temperature = 27.8 °C

(US customary units)

(27.8 20) 21.0872.2 (0.95)(7.39) 2.01.024 (0.75)

9.17

lbOlbN

hp hr hp hr

The theoretical nameplate horsepower is

2 2

2 2

2

2

( / ) 1

( / )

2680

24 125.7 ( )

1.087 4

O Demand lbO hrhp

O Transfer Rate lbO hp hr n

lbO day

day hrhp nameplate

lbO

hp hr

(SI units)

DO saturation value = 7.92 mg/L

The elevation = 610 m = (610 m)(3.281 ft/m) = 2000 ft

Barometric pressure = 706 mmHg

Cs = 7.36 mg/L at 610 m

No = 1.34 kg / kW·hr

(27.8 20)

2

(0.95)(7.39) 2.01.34 1.024 (0.75) 0.662 /

9.17N kgO kW hr


28

The theoretical nameplate (kW) is

2

2

1220

24 119.2 ( )

0.662 4

kgO day

day hrkW kW nameplate

kgO

kW hr

Diffused Aerators

The standard oxygen transfer efficiency for various devices is shown in Table 7-24. The

reported values correspond with a diffuser depth of 15 ft

Power requirements for blowers:

1 2

1

129.7

n

w

w R T pP

n e p

for SI unit

where 29.7 = constant for SI units conversion

1 2

1

1550

n

w

w R T pP

n e p

for U.S. customary units

where 550 = ft∙lb/s∙hp

Pw = power requirement of reach blower, kW, hp

w = mass air flow rate or weight of flow of air, kg/s, lb/sec

R = engineering gas constant for air = 8.314 kJ/k mol K

= 53.3 ft. lb/ lb-air.°R

T1 = absolute inlet air temperature, K, °R (Rankine, °R = °F + 459.6)

p1 = absolute pressure at blower inlet, atm, psi

p2 = absolute pressure at blower outlet, atm, psi

n = (k – 1)/k = 0.283 for air, where k = 1.395 for air

e = compressor/blower efficiency (0.70 to 0.90)

* Mixing requirements for diffused aeration systems range from 10 to 30 standard

ft3/min per 1000 ft

3 of tank volume.


29

Example 7-12 (King, PE exam, p 232)

For a community of 100,000, determine the size of aeration equipment for a ceramic grid

diffused aeration system for a municipal activated sludge treatment facility which provides

secondary treatment.

The field O2 transfer efficiency is estimated to be 15%.

The aeration system must be sized to deliver the peak O2 requirement.

Mixing requirements are normally satisfied if the air supply rate exceeds 10 to 30

SCF/min per 1000 ft3 of aeration basin volume.

Wastewater characteristics are:

Average per-capita flow rate = 100 gal/capita∙day

Peak factor = 2.2

Wastewater concentrations: BOD5 in raw wastewater = 220 mg/L

SS = 220 mg/L

Aeration Tank Volume:

Dimension of aeration basin: water depth = 15 ft; width = 30 ft; length = 185 ft

For 4 units, total volume = 4 (15 ft)(30 ft)(185 ft) = 333,000 ft3

General Assumptions:

35% of BOD in raw wastewater is removed in the primary treatment.

Average O2 supply requirement = 1.1 lb O2 /lb BOD5 (Ten State Standards)

Field O2 Transfer Efficiency (FOTE) for the aeration equipment (diffused aerator) = 15%

Assumptions on the blower:

Absolute pressure at blower inlet (p1) is assumed to equal atmospheric pressure (14.7 psi):

Absolute pressure of blower output (p2) is assumed to exceed static pressure by 3 psi to allow

for head loss in the piping system, filter, and diffuser.

T1 = inlet air temperature = 100°F = 100°F + 460 = 560°R (Rankine)

e = compressor / blower efficiency = 0.80 (ranging from 0.70 to 0.90)

Steps:

1. Calculate average flow rate, Qave (MGD)

2. Calculate BOD loading to the secondary treatment process (lb BOD5/d)

3. Calculate oxygen requirement (lb O2 /d) at average flow and peak flow

4. Calculate air requirement (lb air/min)

5. Calculate volume of air supply rate (SCF/min) based on air requirement

6. Calculate volume of air supply rate based on mixing requirement, (SCF/min)/1000 ft3

7. Determine design air supply rate based on step 5) and 6) using conservative approach (lb

air/s)

8. Calculate blower horsepower


30

(Solution)

100 gal 1 Mgal

1 ) Average flow rate, Qave = (---------------)(100,000 people) (----------) = 10 Mgal/day

Capita ∙ day 106 gal

2) BOD loading to the secondary treatment process (lb BOD5/d)

Assumption: BOD5 removal in the primary clarifier = 35% (see Table 7-9)

8.34 lb/Mgal

BOD loading to the aeration tank = (220 mg/L)(10 Mgal/day)( ------------------) (1- 0.35)

mg/L

= 11926.2 lb BOD5 /day

3) O2 requirement (lb O2 / day)

Average O2 supply requirements = 1.1 lb O2 / lb BOD5 [Ten States Standards (1978)]

At average flow

11926.2 lb BOD5 1.1 lb O2

O2 requirement = ---------------------- ------------ = 13,118.8 lb O2 / day

day lb BOD5

= 13,120 lb O2 / day

At peak flow - the aeration system must be sized to deliver the peak O2 requirement.

Using the peak factor of 2.2:

O2 requirement (peak design) = (2.2) (13,120 lb O2 / day) = 28,900 lb O2/day

4) Mass air flow requirement, w (lb air/min):

Note: The standard transfer efficiency of a ceramic grid, diffused air system is estimated to

be 30% (see Table 7-24).

The O2 transfer efficiency under field conditions (FTE) is estimated to be 15%

The O2 content (mass fraction) of air (XO2) is 0.23 lb O2/lb air.

2

2( )( )O

Ow

X FTE

where w = air (mass) requirement (lb air / sec)

O2 = biological oxygen requirement (lb/sec)= 28,900 lb O2 / day

XO2 = oxygen content in make-up air (mass faction = 0.23)


31

FTE = field transfer efficiency of oxygen (decimal) = 0.15

(28,900 lb O2 /day)(day/1440 min)

w = -------------------------------------------- = 582 lb air /min

(0.23 lb O2/lb air) (0.15)

5) Air volume requirement, QA, ft3 air/d (volumetric air supply rate):

(582 lb air /min)

Air (SCF/min) = -------------------------------- = 7,760 ft3 (SCF) air / min

0.075 lb air / ft3 air (SCF)

SCF = standard cubic foot

or

32

223

28,900 / 1 /1440min 7,689

0.0174 min(0.15)

A

O

lbO d dOR ft airQ

lbOX OTE

ft air

6) Mixing requirement:

- Mixing requirements are normally satisfied if the air supply rate exceeds 10 to 30

SCF/min per 1000 ft3 of aeration basin volume.

Since the volume of the aeration basin = 333,000 ft3,

7,760 ft3 (SCF) / min 23.3 ft

3 (SCF) / min

Air supply = -------------------------- = -------------------------- OK

333,000 ft3 1000 ft

3

7) Design air supply rate (mass flow rate)

For purpose of the design, a conservative approach is assumed; that is,

air supply rate = 30 ft3 (SCF)/min per 1000 ft

3 of aeration basin volume (see (6)).

Mass air flow rate, w

30 ft3 /min 0.075 lb air min

w = -------------- (333,000 ft3)(----------------) (---------) = 12.5 lbs air/sec

1000 ft3 ft

3 60 sec


32

8) Blower horsepower

1 2

1

1550

n

w R T pHP

n e p

where HP = horsepower of compressor/blower

w = mass air flow rate (lb/sec)

R = gas constant = 53.3 ft. lb / lb-air

T1 = inlet air temperature (°R, where °R = °F + 459.6), Rankine

p1 = absolute pressure at blower inlet (psi)

p2 = absolute pressure at blower outlet (psi)

n = 0.283 for air

e = compressor/blower efficiency (0.70 to 0.90)

Assumptions on the blower:

T1 = inlet air temperature = 100°F = 100°F + 459.6 = 560°R

e = compressor/blower efficiency = 0.80 (0.70 to 0.90)

Absolute pressure at blower inlet (p1) is assumed to equal atmospheric pressure (14.7

psi):

p1 = absolute pressure at blower inlet = 14.7 psi (atmospheric pressure)

Absolute pressure of blower output (p2) is assumed to exceed static pressure by 3 psi to

allow for head loss in the piping system, filter, and diffuser.

absolute pressure = atmospheric pressure + gage pressure

p2 = p1 + P2 + P3

Inlet Headloss Head due to

in piping water depth

system

(15 ft) 62.4 lb ft 2

p2 = 14.7 psi + 3 psi + --------- --------- ----------- = 24.2 psi

ft3 144 in

2

where 15 ft = water depth

p1

=

15 ft

p2


33

The total blower horsepower:

0.283

(12.5 / sec)(53.3)(560 ) 24.21 454

(550)(0.283)(0.80) 14.7

lb R psiHP HP

psi

Six units at 100 HP are recommended to provide flexibility to match air supply rates with

experienced demand: i.e., 5 + 1 down for maintenance = 6 units

oxygen transfer

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