oxygen transfer
DESCRIPTION
NOTESTRANSCRIPT
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
Return sludge Waste sludge
O2 demand
Tank length
4 Completely Mixed
Influent Completely Mixed Aeration Tank Effluent
O2 supply
Return sludge Waste sludge
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
5-Oxygen Transfer_F12.doc
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5 Contact Stabilization
Effluent
Influent
HRT = 3 - 6 hrs
Return sludge Waste sludge
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
5-Oxygen Transfer_F12.doc
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)
5-Oxygen Transfer_F12.doc
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-Oxygen Transfer_F12.doc
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
5-Oxygen Transfer_F12.doc
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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
5-Oxygen Transfer_F12.doc
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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
5-Oxygen Transfer_F12.doc
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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
5-Oxygen Transfer_F12.doc
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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)
5-Oxygen Transfer_F12.doc
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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
5-Oxygen Transfer_F12.doc
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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:
5-Oxygen Transfer_F12.doc
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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
5-Oxygen Transfer_F12.doc
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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
5-Oxygen Transfer_F12.doc
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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)
5-Oxygen Transfer_F12.doc
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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
5-Oxygen Transfer_F12.doc
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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
5-Oxygen Transfer_F12.doc
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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.
5-Oxygen Transfer_F12.doc
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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
OTE or FOTE = field O2 transfer efficiency, fraction
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
OTE or FOTE = field O2 transfer efficiency, fraction
(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
OTE or FOTE = field O2 transfer efficiency, fraction
5-Oxygen Transfer_F12.doc
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
5-Oxygen Transfer_F12.doc
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
5-Oxygen Transfer_F12.doc
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.
7-Oxygen Transfer_F10
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)
7-Oxygen Transfer_F10
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.
7-Oxygen Transfer_F10
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).
7-Oxygen Transfer_F10
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
7-Oxygen Transfer_F10
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
7-Oxygen Transfer_F10
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.
7-Oxygen Transfer_F10
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
7-Oxygen Transfer_F10
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)
7-Oxygen Transfer_F10
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
7-Oxygen Transfer_F10
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
7-Oxygen Transfer_F10
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