Design of Biological Nutrient Removal Processes

1
Design of Biological Nutrient Removal Processes
2
Nominal Hydraulic Retention Time
(4.2)
Effluent COD Concentration
Ste Sus for filtered sample
(4.4a) Ste Sus fcv Xve fus
Sti fcv Xve for unfiltered sample (4.4b)
3
Process Design Equations
Total volatile suspended solids (mg VSS) M(Xv)
M(Xa) M(Xe) M(Xi)
Refer to Metcalf Eddy pp. 387-388 for
derivation of M(Xa).
4
Carbonaceous Oxygen Demand(mg O/day)
M(Oc) M(Osynthesis) M(Oendogenous
decay)
(4.15)
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Daily Sludge Production M(²Xt)
M(²Xt) Mass of sludge wasted per day
Waste activated sludge M(Xt)/Rs
mg TSS/day mg TSS produced/mg COD load/day
(4.22)
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Nutrient Requirement
N content with respect to VSS (fn) 0.09
0.12 avg. 0.1 P content with respect to VSS
(fp) 0.01 0.03 avg. 0.025 Q  Ns fn
M(Xv)/Rs
(4.13)
9
Nutrient Requirement
N requirement/unit COD, Ns/Sti
(4.23)
fn N content of the volatile solids 0.1 mg
N/mg VSS
P requirement/unit COD, Ps/Sti
(4.24)
?? P content of the active mass (mg P/mg VASS)
0.03 fp P content of the inert and endogenous
mass (mg P/mg VSS) 0.015
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ENBIR - Processes Included

• 1. Carbonaceous removal
• 2. Nitrogenous removal (nitrification)
• 3. Bardenpho process for N removal
• 4. MLE process for N removal
• 5. UCT (VIP) process for N and P removal
• 6. 5 stage Phoredox process for N and P removal
• 7. 3 stage Phoredox (A2/O) process for N and P
  removal
• 8. A/O process for P removal
• 9. Modified oxidation ditch for N and P removal

12
ENBIR - System Requirements

• IBM/PC/XT/AT or 100 compatible computer
• 512 kB RAM
• MS-DOS 3.3 or higher
• 1-Mb hard disk space
• Graphics monitor supported EGA, VGA, CGA,
  Hercules, IBM 8514

13
ENBIR - Files Included

• INSTALL.BAT Batch install file
• E.BAT Run EN.EXE program along with Hi-Screen
  file loading and unloading
• EN.EXE ENBIR execution file
• ENBIR.LBR User interface library file
• CAR.DFT Carbonaceous removal input file
• NIT.DFT Nitrogenous removal input file
• DEN.DFT Denitrification and phosphorus removal
  input file
• PER.DFT Performance evaluation input file

14
ENBIR - Files Included (continued)

• S.DFT Settled wastewater input files
• R.DFT Raw wastewater input files
• .EX1 Example design input files
• ENBIR.OUT Design result output file
• ENBIR.WRK Design data file
• ANOXIC.PRN a-recycle ratio calculation file
• BEG.COM END.COM Load/unload files
• DISPLAY.COM User-interface driver file

15
ENBIR Installation the Program

• Please backup the diskette before installation.
• To install the program
• 1. Type a (or b) ltEntergt
• 2. Type install ltEntergt
• Follow the instructions given in the screen

16
ENBIR Running the Program

• To run the program in the hard disk
• 1. Type cd \enbir ltEntergt
• 2. Type e ltEntergt

17
ENBIR - Summary of Keys

• F1 Displays a help screen
• F2 - F8 Shows a flow chart of available
  biological nutrient removal processes
• Enter Selects an item or a design value
• Arrow keys Move highlighted menu selection or
  data input fields

18
ENBIR More Keys

• ESC Allows the recovery of the original input
  data stored in the data file during data input
• Crtl Break Allows the user to exit from the
  program run

19
Design Example - Carbonaceous Removal
Influent wastewater characteristics (Table 4.3)

• Parameter
  Symbol Raw Settled
• Influent COD conc., mg COD/L Sti 600 360
• Influent TKN conc., mg N/L Nti 48 41
• Influent P conc., mg P/L Pti 10 8.5
• TKN/COD ratio - 0.080 0.114
• P/COD ratio - 0.017 0.024
• Unbiodeg. sol. COD fraction fus 0.05 0.08
• MLVSS/MLSS ratio fi 0.75 0.83
• Temp. - Summer, C Tmax 22 22
• Winter, C Tmin 14 14
• pH of wastewater - 7.5 7.5
• Influent flow, Mil. L/day Q 13.33 13.33

20
Kinetic constants and their temperature
dependency for the steady state carbonaceous
degradation activated sludge model (Table 4.1)

• Temperature Standard
• Constant
  Symbol dependency ? value
• equation (20C)
• Yield coef., mg VSS/mg COD Yh
  Const. 1.000 0.45
• Endog. decay rate, 1/day bh bhTbh20?
  1.029 0.24
• Endog. residue, mgVSS/mgVSS f Const. -
  0.20
• COD/VSS ratio, mgCOD/mgVSS fcv Const.
  - 1.48

T-20
Design sludge age, Rs 3 days
21
Carbonaceous RemovalDesign Procedure

• Calculate
• 1. Supi (Eq. 2.4) or fup, Susi (Eq. 2.3) or fus,
• 2. M(Sti) and/or M(Sbi)
• 3. M(Xv) (Eq. 4.13), M(Oc) (Eq. 4.15) and M(Xt)
  (Eq. 4.14)
• 4. Vp (Eq. 4.16)
• 5. Rhn (Eq. 4.3)
• 6. Ste (Eq. 4.4a or b)

22
Calculations - 1
1. fup 0.13 fus 0.05

• 2. Biodegradable COD in the effluent, Sbe
• where
• Yh heterotrophs yield coe. (mgVSS/mgCOD) bhT
  endogenous respiration rate (1/day) bkT
  biodeg. COD yield rate in effluent (1/day) Rs
  sludge age (day).

If Rs gt 10.0, Sbe 0.0
23
Calculations - 2
Biodegradable COD in the effluent, Sbe
14-20
bh14 0.24 1.029 0.202

• M(Sti) (Sti - Sbe) Q
  (600-20.9)13,330/10
  00
• 7719 kg/day
• M(Sbi) (1-0.05-0.13) 7719
  6330 kg/day

24
Calculations - 3

• 3.

7999 kg VSS M(Xt) M(Xv)/0.75
10665 kg TSS
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Calculations - 4
3.

• 3361 kg O/day

27
Calculations - 5

• 4. Vp M(Xt)/Xt M(Xv)/Xv
  10665 kg VSS/1500 mg VSS/L
  7.11 ML
• 5. Rhn 7.11 ML/13.33ML/day 0.53 days
• 6. Ste fus Sti 0.05 600
  30 mg/L for filtered
  sample

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Performance Evaluation
(4.13)

• Vp M(Xt)/Xt M(Xv)/fi/Xt
• Reasonings (assumptions)
• 1. Assume that the heterotrops yield coefficient
  is relatively constant over wide ranges of sludge
  age.
• 2. Assume that the endogenous decay constant
  varies significantly over wide ranges of sludge
  age.

(4.16)
30
Performance Evaluation - continued

• Vp M(Xt)/Xt M(Xv)/fi/Xt
• By inserting Eq. 4.13 to Eq. 4.16 and
  rearranging, bhT can be estimated as follows
• M(Xv) Xt  fi Vp
• M(Sti) (Sti-Ste) Q

31
Nitrification Design
32
Effluent Ammonia Concentration, Na
Dividing by Vp ²t
Solving the above eqs. for Na yields
(5.13)
33
Minimum Sludge Age for Nitrification
(5.14)
Applicable for all Nai gt 5 mg N/L
34
Specific Growth Rate
EPA Method
South African Method
for pH lt 7.2
for 7.2 pH lt 8.5
The EPA method gives a higher µnmT value,
resulting in a shorter sludge age and a greater
unaerated mass fraction than the South African
method.
35
Design Considerations

• 1. Fate of the influent TKN
• Nai ammonia (free) and ammonium iron (saline)
  concentration
• Noi biodegradable organic nitrogen
  concentration
• Nui soluble unbiodegradable organic nitrogen
  concentration
• For design, the following information is
  needed.
• 1. Effluent TKN concentration
• 2. Effluent nitrate concentration

36
Design Considerations

• 2. Effluent TKN
• Nt Na No Nu for
  filtered Nt (5.28a)
• Ntv Na No Nu fn Xvf for unfiltered
  Nt (5.28b)
• fn fraction of TKN in VSS 0.1 mg TKN-N/mg
  VSS
• Xvf eff. volatile solids conc. (mg VSS/L)

37
Design Considerations
2. Effluent TKN - cont.
a.
(5.20)
fxt unaerated mass fraction
b. Biodegradable organic nitrogen conc.,
(5.26)
38
Design Considerations
2. Effluent TKN - cont.
c. Sol. unbiodegradable organic nitrogen conc.,
(5.27)
39
Design Considerations

• 3. Nitrification capacity (mass of nitrate
  produced per unit volume of flow, mg NO3-N/L of
  flow)
• This equation assumes that no nitrate is
  present in the influent.
• Nitrification capacity per mg COD applied

(5.29)
(5.30)
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Raw and settled water characteristics for
nitrification(Table 5.2)
Nitrification Design Example

• Parameter Symbol Raw Settled
• Influent TKN conc., mg N/L Nti 48 41
• Influent TKN/COD ratio - 0.080 0.114
• Ammonia fraction of TKN fna 0.75 0.83
• Unbiod. sol. organic N fraction fnu 0.03
  0.04
• N fraction of volatile solids in inf. fn
  0.10 0.10
• pH of wastewater - 7.5 7.5
• Total alkalinity, mg/L as CaCO3 alk 200
  200
• Maximum specific growth rate, 1/d µnm20
  0.36 0.36
• Influent flow, ML/day Q 13.33 13.33

Design sludge age 10 days
42
Design Example - Nitrification

• 1. Nti 41.0 mg N/L
• Nai 0.83 41.0 34.03 mg N/L
• Nui 0.04 41.0 1.64 mg N/L
• Npi 0.10 (0.04 360)/1.48 0.97 mg N/L
• Noi 41.0 - 34.03 - 1.64 - 0.97 4.36 mg N/L
• 2. N required for sludge production
• 7.59 mg N/L

43
Continued

• 3. Rsm Min. sludge age for nitrification
• If Rs lt Rsm No nitrification.
• Thus, effl. nitrate conc. (Nne) 0.0 mg N/L.

44
Continued

• 4. Noe Effl. biodegrad. organic N conc.
• M(Sbi) 13.33 ML/day360 mg/L(1-0.08-0.04)
• 4223 kg COD/day
• 0.62 mg N/L

45
Continued

• 5. If Rs gt Rsm
• Nte Effl. TKN conc.
• Nne Effl. nitrate conc.
• Nue Nui 1.64 mg N/L
• Nte Nae Noe Nue
• 1.46 0.62 1.64 3.72 mg N/L
• Nne Nti - Ns - Nte
• 41.0 - 7.59 - 3.72 29.69 mg N/L

46
Continued

• 6. M(On) Oxygen demand for nitrification

47

• 7. Alkalinity check
• Alke Alki - 7.14 Nne
• 200 - 7.14 29.69 lt 0.0 0.0
• The effluent is likely to have a pH lt 6.0.

48
Biological Nitrogen RemovalDesign Example
49
Bardenpho Process Design

• 1. Select wastewater characteristics Sti, Nti,
  fbs, fup, fus, µnm20, Tmax, Tmin.
• 2. Select Rs and Sf.
• 3. Calculate fxm for Tmin from Eq. 5.24.
• 4. With fxm and Rs, calculate Sf for Tmax from
  Eq. 5.24.
• 5. Estimate Nte for Tmax and Tmin from 5.36b.
• 6. With fup and fus, calculate Ns for Tmax and
  Tmin for selected Rs from Eq. 4.23.

50
Bardenpho Process Design

• 7. Calculate Nc from Eq. 5.29.
• 8. Calculate Dpp for Tmax and Tmin with fxdm
  given by Eq. 6.23.
• 9. Select s, Oa, and Ox.
• 10. Calculate fx1min from Eq. 6.21 for Tmax and
  Tmin.
• 11. Calculate fx3min from Eq. 6.29 for Tmax and
  Tmin.
• 12. Select the a-recycle ratio and calculate Nne
  for Tmin.
• 13. If Nne lt 0, set Nne0.

51
Bardenpho Process Design

• 14. With Nne calculate fx1, fx3, and fxdt Eqs.
  6.26 to 6.28.
• 15. Check that fx1 fxmin and fx3 fx3min. If
  not, discard selected a-recycle ratio as invalid.
• 16. Repeat steps 12 to 15 for different a-recycle
  ratios.
• 17. Repeat steps 12 to 16 for Tmax.

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Determination of a0

• 1. Find a0 value for minimum fxdt or Nne at Tmin.
• 2. If there is no minimum fxdt or Nne at Tmin,
  find a0 value for minimum fxdt or Nne at Tmax.
• 3. Once minimum fxdt or Nne is determined at a
  temperature, fix fx1 for the other temperature,
  and then determine a0 and fx3.
• 4. If there is no significant improvement in Nne
  with the increased a-recycle ratio, select a
  practical a-recycle ratio.

53
If complete denitrification is impossible

• For all valid a-recycle, nitrate will be present
  in the effluent and fxdt will be equal to fxdm.
• For N removal only, the Bardenpho process will be
  appropriate only if TKN/COD ratio is lt 0.10 mg
  N/mg COD or Nne is lt 5 to 7 mg N/L.
• If TKN/COD ratio is gt 0.10 mg N/mg COD, the MLE
  process will yield a better nitrogen removal
  efficiency.

54
Design Example - Denitrification
Influent wastewater characteristics

• Parameter
  Symbol Raw Settled
• Influent COD conc., mg COD/L Sti 600 360
• Influent TKN conc., mg N/L Nti 48 41
• Influent P conc., mg P/L Pti 10 8.5
• TKN/COD ratio - 0.080 0.114
• P/COD ratio - 0.017 0.024
• Unbiodeg. sol. COD fraction fus 0.05 0.08
• MLVSS/MLSS ratio fi 0.75 0.83
• Temp. - Summer, C Tmax 22 22
• Winter, C Tmin 14 14
• pH of wastewater - 7.5 7.5
• Influent flow, Mil. L/day Q 13.33 13.33

55
Raw and settled wastewater characteristics
important for denitrification

• Value
• Constant Symbol
  Raw Settled
• Max. specific growth rate of nitrifiers
  µnm20 0.36 0.36
• Denitrification rates at 20C K1 0.720 0.720
• (mg NO3-N/mg VASS/day) K2 0.101 0.101
• K3 0.072 0.072
• Readily biodegradable fraction fbs 0.24 0.33

Design sludge age, Rs 25 days
56
Design Example - Continued
1. Wastewater characteristics Sti 600
mg COD/L Nti 48 mg N/L fbs 0.24
fup 0.13 fus 0.05 µnm20 0.36 1/day
µnm14 0.18 µnm22 0.45 bn20 0.04
1/day bn14 0.034 bn22 0.042 Tmax
22C Tmin 14C 2. Rs 25 days Sf at
14C 1.25
57
Design Example - Continued
3. Max. design unaerated sludge mass fraction,
fxm for Tmin
4. Sf for Tmax 2.87 from Eq. above
58
Design Example - Continued
5. Effluent TKN concentration, Nte
59
Design Example - Continued
6. Nitrogen req. for sludge production, Ns
7. Nitrification capacity, Nc
60
Design Example - continued
8. Denitrification potential, Dpp (Eqs. 6.20
6.25)
9. RAS ratio s 1 DO in mixed liquor a
Oa 2.0 mg/L DO in underflow recycle s Os
1.0 mg/L
61
Design Example - Continued
10. Min. primary anoxic sludge mass fraction,
fx1min (Eq 6.21)
11. Min. secondary anoxic sludge mass fraction,
fx3min (Eq 6.29)
62
Design Example - Continued
12. Optimum a-recycle ratio a. Find a0 value
for minimum fxdt or Nne at Tmin. b. If there is
no minimum fxdt or Nne at Tmin, find a0
value for minimum fxdt or Nne at Tmax. c. Once
minimum fxdt or Nne is determined at a
temperature, fix fx1 for the other temperature,
and then determine a0 and fx3. d. If there is
no significant improvement in Nne with the
increased a-recycle ratio, select a practical
a-recycle ratio.
63
Design Example - Continued
64
Design Example - Continued
T 22C raw sewage
a0 5.2 fx1 0.16 fx3 0.19 fxdt 0.35
65
Design Example - Continued
T 14C raw sewage
fx1 0.16 a0 3.6 fx3 0.23 fxdt 0.39
66
Design Example - Continued
Total process volume, Vp
67
Design Example - Continued
Oxygen demand (kg O/day)
1. Carbonaceous oxygen demand, M(Oc)
68
Design Example - Continued
Oxygen demand (kg O/day)
2. Nitrification oxygen demand, M(On)
3. Denitrification oxygen demand, M(Od)
4. Total oxygen demand, M(Ot)
M(Ot) 5,108 1,928 - 1,206 5,830 kg O/day
17 oxygen saved by denitrification
69
The MLE Process Design
Design process similar to the Bardenpho
process. The best denitrification potential
is obtained when the anoxic reactor is just
loaded with nitrate to its denitrification
potential. This will give the optimum mixed
liquor a-recycle ratio and the min. Nne.
Nitrate conc. in the anoxic reactor overflow
0 Nitrate conc. in the aerobic reactor overflow
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The MLE Process Design
Optimum a-recycle ratio, a0
Rearrange the above eq. with respect to a. Then,
where
71
The MLE Process Design
Determine a0
Uneconomical. Thus, use a0 5.0.
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The MLE Process Design
If Nne lt 5 to 6 mg N/L are required, the
Bardenpho process is better because this process
can achieve low Nne with low a-recycle ratio. If
complete denitrification is not necessary, the
MLE process will be an acceptable process.
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Biological Phosphorus Removal (BPR) Design
Examples
74
Design Example

• Guidelines for Process Selection
• 1. If TKN/COD 0.08 mg N/mg COD, the Phoredox
  process. Complete nitrate removal is possible.
• 2. If 0.08 lt TKN lt 0.11, the modified UCT
  process. Complete nitrate removal is not
  possible but nitrate can be excluded from the
  anaerobic reactor.

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Design Example Continued

• 3. If 0.11 TKN lt 0.14, the UCT process. The
  modified UCT process no longer can exclude
  nitrate from the anaerobic reactor. A careful
  a-recycle control required.
• 4. If TKN/COD 0.14, it is unlikely that
  biological excess P removal will be achieved with
  normal municipal wastewaters.

76
A/O Process Design

• 1. Select wastewater characteristics Sti, Nti,
  Pti, fbs, fup, fus, µnm20, Tmax, Tmin.
• 2. Select Rs and Sf then, calculate M(Sti),
  M(Xv), M(Xt), M(Oc), Vp, and Rhn.
• 3. Calculate Sf for Tmax from Eq. 5.24.
• 4. Calculate min. sludge age (Rsm) for
  nitrification from Eq. 5.14.
• 5. If Rs lt Rsm, then go to Step 7.
• 6. Follow nitrification calculation steps.
• 7. Select Oi.
• 8. Calculate ²Sbsa from Eq. 7.2b.
• 9. Calculate Sbsa from Eq. 7.1b.

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A/O Process Design - continued

• 10. Calculate Pf from Eq. 7.5
• 11. Calculate ? from Eq. 7.7.
• 12. Calculate Ps from Eq. 7.6.
• 13. Calculate Pte from Eq. 7.9 (Pte Pti - Ps)
  (mg P/L).
• 14. Is Pte sufficiently low? If not, decrease Rs
  or increase fxa however, improvement will only
  be achieved if Nne 0 if Nne gt 0, changes will
  increase Pte.
• 15. Repeat calculations for Tmax.

78
Design Example

• 1. Wastewater characteristics
• Sti 600 mg COD/L Nti 48 mg N/L fbs
  0.24 fup 0.13 fus 0.05 µnm20 0.36
  1/day ?µnm 1.123 bn20 0.04 1/day ?bn
  1.029 Xt 1500 mg TSS/L MLVSS/MLSS
  0.75 Pti 10.0 mg P/L P content (fp)
  0.015 mg P/mg VSS Tmax 22C Tmin 14C
• 2. Rs 3 days Sf at 14C 1.25
• M(Sti), M(Xv ), M(Xt), M(Oc), Vp, and Rhn
  calculated during the carbonaceous removal design
  will be used.

79
Design Example - continued

• 3.

4.
80
Design Example - continued

• 5. Since Rs lt Rsm (3 lt 8.6), go to Step 7.
• 7. Influent oxygen concentration (Oi) 0.0 mg/L
• 8. ²Sbs 1(8.60.0 3.01.0) 3.00.0 3.0
• 9.
• 57.54 mg COD/L
• 10. Pf (57.54 - 25) 0.1 3.25
• 11. ? 0.35-0.29 exp(-0.2423.25)
• 0.218 mgP/mgVSS

81
Design Example - continued

• 12.
• 31.67 mg P/L
• 13. Pte Pt - Ps 10.0 - 31.67 lt 0 mg P/L
• 14. Is Pte sufficiently low? Yes.
• 15. Repeat calculations for Tmax.

82
A2/O Process Design

• 1. Select wastewater characteristics Sti, Nti,
  Pti, fbs, fup, fus, µnm20, Tmax, Tmin.
• 2. Select Rs and Sf then, calculate M(Sti),
  M(Xv), M(Xt), M(Oc), Vp, and Rhn.
• 3. Calculate fxm for Tmin from Eq. 5.24.
• 4. With fxm and Rs, calculate Sf for Tmax from
  Eq.5.24.
• 5. With fup and fus, calculate Ns for Tmax and
  Tmin for selected Rs from Eq. 4.23.
• 6. Estimate Nte for Tmax and Tmin.
• 7. Calculate Nc from Eq. 5.29.

83
A2/O Process Design - continued

• 8. Select fxa.
• 9. Calculate Dpp from Eq. 6.25 for Tmax and Tmin
  with fxdm given by Eq. 6.23.
• 10. Select s, Oa, and Os.
• 11. Determine the optimum a-recycle ratio.
• 12. Calculate Nne for Tmin. If Nne lt 0, set
  Nne0.
• 13. Calculate oxygen demand.
• 14. Select Oi.
• 15. Calculate ²Sbsa from Eq. 7.2b.
• 16. Calculate Sbsa from Eq. 7.1b.

84
A2/O Prodess Design - continued

• 17. Calculate Pf from Eq. 7.5.
• 18. Calculate ??from Eq. 7.7.
• 19. Calculate Ps from Eq. 7.6.
• 20. Calculate Pte from Eq. 7.9 (Pte Pti - Ps,
  mg P/L).
• 21. Is Pte sufficiently low? If not, decrease Rs
  or increase fxa however, improvement will only
  be achieved if Nne 0 if Nne gt 0, changes will
  increase Pte.
• 22. Repeat calculations for Tmax.

85
Design Example

• 1. Wastewater characteristics
• Sti 600 mg COD/L Nti 48 mg N/L fbs
  0.24 fup 0.13 fus 0.05 µnm20 0.36
  1/day Pti 10.0 mg P/L K1, K2 K3 0.72,
  0.101, 0.072 mg NO3-N/mg/d ?K1, ?K2 ?K3
  1.20, 1.08, 1.029 P content (fp) 0.015 mg
  P/mg VSS Xt 2500 mg TSS/L MLVSS/MLSS
  0.75 Tmax 22C Tmin 14C
• 2. Rs 15 days Sf at 14C 1.25 Sbe
  0.0 when Rsgt10 days
• M(Sti) (Sti-Sbe)Q 60013,330/1000 7998
  kg/day
• M(Sbi) M(Sti)(1-0.05-0.13) 6558 kg/day

86
Design Example - continued
28174 kg VSS M(Xt)
M(Xv)/0.75 37565 kg TSS
4818 kg O/day
87
Design Example - continued

• Vp M(Xt)/Xt M(Xv)/Xv
  37565 kg VSS/2500 mg VSS/L
  15.03 ML
• Rhn 15.03 ML/13.33ML/day 1.13 days

3. Max. design unaerated sludge mass fraction,
fxm for Tmin (Eq. 5.24b)
88
Design Example - continued
4.
Nitrogen required for sludge production
(Ns) 14.1 mg N/L Inf. TKN conc.
available for nitrification Nti' Nti -
Ns 48 - 14.1 33.9 mg N/L
5.
(5.33)
89
Design Example - continued
6. Effluent TKN concentration, Nte

• Influent ammonia concentration Nai 0.75
  48 36 mg N/L
• Inf. unbiodegradable soluble organic nitrogen
  Nui 0.03 48 1.44 mg N/L
• Inf. N conc. assoicated with unbiode. parti. COD

Inf. biode. organic nitrogen conc. Noi
48 - 36 - 1.44 - 5.27 5.29 mg N/L
90
Design Example - continued

• Noe Effl. biodegrad. organic N conc.
• M(Sbi) 13.33 ML/day600 mg/L(1-0.13-0.05)
• 6558 kg COD/day

Nue Nui 1.44 mg N/L
91
Design Example - continued
Effluent ammonia concentration (Nae)
Nte Nae Noe Nue 1.99 0.46 1.44
3.89 mg N/L
92
Design Example - continued

• 7. Nitrification capacity, (Nc)
  Nc Nti - Nte - Ns 48.0 - 3.89 - 14.1 30.01
  mg N/L
• 8. Anaerobic sludge mass fraction (fxa) 0.1
• Guidelines for fxa selection
• 1. Sti lt 400 mg COD/L, fxa 0.20 - 0.25
  2. 400 Sti 700, fxa 0.15 - 0.20
  3. Sti gt 700 mg COD/L, fxa 0.10 -
  0.15
• 9. Denitrification potential, Dpp (Eqs. 6.20
  6.25)

(fxdm fxm - fxa Eq. 7.8)
93
Design Example - continued

• 10. RAS ratio s 1
• DO in mixed liquor a Oa 2.0 mg/L
• DO in underflow recycle s Os 1.0 mg/L
• 11. Determine optimum a-recycle ratio

where
optimum a-recycle ratio, a0 2.03
94
Design Example - continued
12. Determine effluent nitrate concentration
(Nne)
13. Oxygen demand for nitrification, M(On)
M(On) 4.57 Nc Q 4.57 30.01
13.33 1828 kg O/day
Oxygen saved by denitrification, M(Od)
M(Od) 2.86 (Nc - Nne) Q
2.86 (30.01 - 7.45) 13.33 860 kg
O/day
Total oxygen demand, M(Ot)
M(Ot) 4818 1828 - 860 5786 kg O/day
95
Design Example - continued

• 14. Influent oxygen concentration (Oi) 0.0
  mg/L
• 15. ²Sbs 1(8.67.45 3.01.0) 3.00.0
  67.07
• 16.
• 17. When Sbsa lt 25 mg COD/L, then Pf 0.05
• 18. ? 0.35-0.29 exp(-0.2420.05)
  0.06 mgP/mgVSS

96
Design Example - continued

• 19.
• 4.78 mg P/L
• 20. Pte Pt - Ps 10.0 - 4.58 5.22 mg P/L
• 21. Is Pte sufficiently low? No.
• 22. Repeat calculations by increasing fxa or
  decreasing sludge age.

97
Phoredox Process Design

• 1. Select wastewater characteristics Sti, Nti,
  Pti, fbs, fup, fus, µnm20, Tmax, Tmin.
• 2. Select Rs and Sf then, calculate M(Sti),
  M(Xv), M(Xt), M(Oc), Vp, and Rhn.
• 3. Calculate fxm for Tmin from Eq. 5.24.
• 4. With fxm and Rs, calculate Sf for Tmax from
  Eq. 5.24.
• 5. With fup and fus, calculate Ns for Tmax and
  Tmin for selected Rs from Eq. 4.23.
• 6. Estimate Nte for Tmax and Tmin.
• 7. Calculate Nc from Eq. 5.29.

98
Phoredox Process Design

• 8. Select fxa.
• 9. Calculate Dpp from Eq. 6.25 for Tmax and Tmin
  with fxdm given by Eq. 6.23.
• 10. Select s, Oa, and Os.
• 11. Calculate fx1min from Eq. 6.21 for Tmax and
  Tmin.
• 12. Calculate fx3min from Eq. 6.29 for Tmax and
  Tmin.
• 13. Select the a-recycle ratio and calculate Nne
  for Tmin.
• 14. If Nne lt 0, set Nne0.
• 15. With Nne calculate fx1, fx3, and fxdt from
  Eqs. 6.26 to 6.28.

99
Phoredox Process Design

• 16. Check that fx1 fxmin and fx3 fx3min. If
  not, discard selected a-recycle ratio as invalid.
• 17. Repeat steps 13 to 16 for different a-recycle
  ratios.
• 18. Repeat steps 13 to 17 for Tmax.
• 19. Fix fx1 and fx3 by procedure given in Chapter
  6, Section 6.2. This fixes Nne and hence Nns.
• 20. Select Oi.
• 21. Calculate ²Sbsa from Eq. 7.2b.
• 22. Calculate Sbsa from Eq. 7.1b.
• 23. Calculate Pf from Eq. 7.5.

100
Phoredox Process Design

• 24. Calculate ??from Eq. 7.7.
• 25. Calculate Ps from Eq. 7.6.
• 26. Calculate Pte from Eq. 7.9 (Pte Pti - Ps,
  mg P/L).
• 27. Is Pte sufficiently low? If not, decrease Rs
  or increase fxa at expense of fxdm however,
  improvement will only be achieved if fxdt lt fxdm,
  i.e., Nne lt 0 if fxdt fxdm, i.e., Nne gt 0,
  changes will increase Pte.
• 28. Repeat calculations for Tmax.
• 29. Can required Pte be achieved with the
  Phoredox process? If not, select a UCT process.

101
Design Example

• 1. Wastewater characteristics
• Sti 600 mg COD/L Nti 48 mg N/L fbs
  0.24 fup 0.13 fus 0.05 µnm20 0.36
  1/day Pti 10.0 mg P/L K1, K2 K3 0.72,
  0.101, 0.072 mg NO3-N/mg/d ?K1, ?K2 ?K3
  1.20, 1.08, 1.029 Xt 4000 mg
  TSS/L MLVSS/MLSS 0.75 P content (fp) 0.015
  mg P/mg VSS Mass fraction of the reaeration
  reactor (fmr) 0.05 Tmax 22C Tmin 14C
• 2. Rs 25 days Sf at 14C 1.25

102
Design Example - Continued
3. Max. design unaerated sludge mass fraction,
fxm for Tmin (Eq. 5.24b)
4. Sf for Tmax 2.87 from Eq. above
103
Design Example - Continued
5. Effluent TKN concentration, Nte
(from Eqs. 5.35a, 5.31, and 5.32, respectively)
6. Nitrogen req. for sludge production, Ns (Eq.
4.23)
104
Design Example - Continued
7. Nitrification capacity, Nc (Eq. 5.29)

• 8. Anaerobic sludge mass fraction (fxa) 0.1
• Guidelines for fxa selection
• 1. Sti lt 400 mg COD/L, fxa 0.20 - 0.25
• 2. 400 Sti 700, fxa 0.15 - 0.20
• 3. Sti gt 700 mg COD/L, fxa 0.10 - 0.15

105
Design Example - Continued
9. Denitrification potential, Dpp (Eqs. 6.20
6.25)
(fxdm fxm - fxa Eq. 7.8)
10. RAS ratio s 1 DO in mixed liquor a
Oa 2.0 mg/L DO in underflow recycle s
Os 1.0 mg/L
106
Design Example - Continued
11. Min. primary anoxic sludge mass fraction,
fx1min (Eq 6.21)
12. Min. secondary anoxic sludge mass fraction,
fx3min (Eq 6.29)
107
Determination of a0 (same as the
Bardenpho process)

• 1. Find a0 value for minimum fxdt or Nne at Tmin.
• 2. If there is no minimum fxdt or Nne at Tmin,
  find a0 value for minimum fxdt or Nne at Tmax.
• 3. Once minimum fxdt or Nne is determined at a
  temperature, fix fx1 for the other temperature,
  and then determine a0 and fx3.
• 4. If there is no significant improvement in Nne
  with the increased a-recycle ratio, select a
  practical a-recycle ratio.

108
Design Example - Continued
109
Design Example - Continued
T 22C raw sewage
a0 4.2 fx1 1.4 fx3 0.21 fxdt 0.35 Nne
0.0
110
Design Example - Continued
T 14C raw sewage
fx1 1.4 a0 3.0 fx3 0.24 fxdt 0.38 Nne
0.0
111
Design Example - Continued

• 20. Influent oxygen concentration (Oi) 0.0
  mg/L
• 21. ²Sbs 1 (8.60.43 3.01.0) 3.00.0
  6.44
• 22.
• 55.8 mg COD/L
• 23. Pf (55.8 - 25) 0.1 3.08
• 24. ? 0.35-0.29 exp(-0.2423.08)
  0.212 mg P/mgVSS

112
Design Example - Continued

• 25.
• 9.10 mg P/L
• 26. Pte Pt - Ps 10.0 - 9.10 0.90 mg P/L
• 27. Is Pte sufficiently low? Yes.
• 28. Repeat calculations for Tmax.
• Process volume For the same wastewater and
  conditions, the total volume of a Phoredox
  process will be the same as that of the Bardenpho
  process.

113
Design Example - Continued

• In the Phoredox process, complete denitrification
  should be achieved to obtain the estimated P
  removals. However, this cannot be guaranteed
  because there is no safety factor on the
  denitrification. To introduce a safety factor,
  two options are open
• 1. Increase the primary anoxic sludge mass
  fraction (fx1) at the expense of a reduction in
  fxa, which results in a reduction in P removal
  (at 20C, a reduction from 0.1 to 0.07 reduces
  the P removal from 8.8 to 7.7 mg P/L)
• 2. Increase the sludge age, resulting in a larger
  unaerated sludge mass fraction (fxm) but a larger
  process volume and the reduced P removal (an
  increase in Rs from 25 to 30 days reduces the P
  removal from 8.8 to 7.6 mg P/L).

114
The UCT Process

• In the UCT process design, the N and P removal
  aspects can be dealt with separately and the P
  removal in the processes can be calculated
  without first having to determine the effluent
  nitrate concentration.
• First, design the UCT process then, check is
  made as to whether this configuration can be
  converted to a modified process. It is
  recommended that where it is possible to convert
  the UCT process to a modified one, consider the
  option of incorporating the Phoredox/modified
  UCT/UCT combination.

115
UCT Process Design

• 1. Select wastewater characteristics Sti, Nti,
  Pti, fbs, fup, fus, µnm20, Tmax, Tmin.
• 2. Select r (usually 1.0) and Nnr and Or (say 1
  mg/L each).
• 3. Calulate Sbsa from Eqs. 7.1a and 7.2a.
• 4. Select fxa (between 0.1 and 0.25).
• 5. Calculate Pf from Eq. 7.5 and ? from Eq. 7.7.
• 6. Select Rs.
• 7. Calculate Ps from Eq. 7.6 for Tmax and Tmin.
• 8. Calculate Pte from Eq. 7.9 for Tmax and Tmin.

116
UCT Process Design

• 9. Is Pte adequate? If not, repeat steps 4 to 8
  for different fxa and/or Rs.
• 10. Select Sf.
• 11. Calculate fxm for Tmin from Eq. 5.24.
• 12. With fxm and Rs, calculate Sf for Tmax from
  Eq. 5.25.
• 13. Estimate Nte for Tmax and Tmin from Eq. 4.23.
• 14. With fup and fus, calculate Ns for Tmax and
  Tmin.
• 15. Calculate Nc from Eq. 5.29.
• 16. Calculate Dpp from Eq. 6.25 for Tmax and Tmin
  with fxdm given by Eq. 7.8.

117
UCT Process Design

• 17. Select s, Oa, Os.
• 18. Calculate a0 from Eq. 6.30 and Nne from Eq.
  6.31.
• 19. Are Pte and Nne adequate? If Pte is too
  high, increase fxa or reduce Rs - these changes
  will increase Nne if Pte is too low, decrease
  fxa or increase Rs - these changes will decrease
  Nne.
• 20. Repeat from steps 4 to 17 until required or
  optimal Pte and Nne is obtained.

118
Design Example

• 1. Wastewater characteristics
• Sti 600 mg COD/L Nti 48 mg N/L fbs
  0.24 fup 0.13 fus 0.05 µnm20 0.36
  1/day Pti 10.0 mg P/L K1, K2 K3 0.72,
  0.101, 0.072 mg NO3-N/mg/d ?K1, ?K2, ?K3
  1.20, 1.08, 1.029 P content (fp) 0.015 mg
  P/mg VSS Mass fraction of the reaeration
  reactor (fmr) 0.05 Tmax 22C Tmin 14C
• 2. r 1, Nnr 1.0 mg/L, Or 1.0 mg/L

119
Design Example - Continued

• 3. ²Sbs 1 (8.61.0 3.01.0) 3.00.0
  11.6
• 53.2 mg COD/L
• 4. fxa 0.1
• 5. Pf (53.2 - 25) 0.1 2.82
• ??????? 0.35-0.29 exp(-0.2422.82) 0.203
  mgP/mgVSS
• 6. Rs 25 days

120
Design Example - Continued

• 7. Ps 600
• 8.77 mg P/L
• 8. Pte Pti - Ps 10.0 - 8.77 1.23 mg P/L
• 9. If Pte adequate? Yes.
• 10. Sf 1.25 at Tmin.
• 11.

121
Design Example - Continued

• 12.

13. Effluent TKN concentration, Nte
(from Eqs. 5.35a, 5.31, and 5.32, respectively)
14. Nitrogen req. for sludge production, Ns (Eq.
4.23)
122
Design Example - Continued
15. Nitrification capacity, Nc (Eq. 5.29)
16. Denitrification potential, Dpp (Eqs. 6.20
6.25)
(fxdm fxm - fxa Eq. 7.8)
17. s 1.0 Oa 2.0 mg O/L Os 1.0 mgO/L
123
Design Example - Continued
18.
Uneconomical. Thus, use a0 5.0.
19. Are Pte and Nne adequate? Yes.
124
Comparison of Phoredoxand UCT Processes

• For 25-day sludge age and fan of 0.1 for raw
  wastewater
• The Phoredox and UCT processes achieved the same
  excess P removal, but the Phoredox process was
  better than the UCT process for the N removal.
• The Phoredox process is less flexible in that an
  unexpected increase in TKN/COD ratio above 0.08
  will result in a deterioration in excess P
  removal.

125
Comparison of Phoredoxand UCT Processes -
Continued

• For 25-day sludge age and fan of 0.1 for settled
  wastewater (TKN/COD 0.114)
• The Phoredox and UCT processes produced similar
  effluent nitrate concentrations, but the Phoredox
  process was poorer than the UCT process for the P
  removal.
• The difference arises from the fact that in the
  Phoredox process, the exclusion of nitrate from
  the anaerobic reactor becomes impossible while in
  the UCT process, this is possible up to TKN/COD
  ratios of about 0.13 (depending on the fbs
  fraction).

126
Redesign of the UCT Process

• The actual P removal attained in the process for
  the settled wastewater is insufficient to reduce
  the effluent P concentration below 1 mg P/L. To
  improve the P removal, Rs needs to be reduced and
  fxa increased.

127
Redesign of the UCT Process, continued

• The difference arises from the fact that in the
  Phoredox process, the exclusion of nitrate from
  the anaerobic reactor becomes impossible while in
  the UCT process, this is possible up to TKN/COD
  ratios of about 0.13 (depending on the fbs
  fraction).

128
The Modified UCT Process

• The equivalent nitrate load is calculated from
  the effluent nitrate concentration for the UCT
  process before subdivision and is given by s(Nne
  Os/2.86).
• The factor by which Dpd1 is greater than s(Nne
  Os/2.86) is called the safety factor (Sfd). It
  is recommended that this factor be greater than
  1.2.
• In general, the modification is acceptable for
  TKN/COD lt 0.12 with a readily biodegradable COD
  fraction (fbs) around 0.25.

129
The Modified UCT Process - Continued

• The conversion to the modified UCT process is
  done by subdividing the anoxic sludge mass
  fraction in two subfractions the first is
  usually allocated a sludge mass fraction of fxd1
  of 0.10 and the second, fxd2, having the
  remaining sludge mass fraction (fxd2 fxdm -
  fxd1).
• For successful operation of the modified UCT
  process, the denitrification potential of fxd2
  (Dpd1) must be greater than its equivalent
  nitrate load imposed by the s-recycle at Tmin.

130
The UCT Process - Process Volume

• The MLSS is not uniform throughout the process
  thus, the reactor volume fractions are not equal
  to the sludge mass fractions. There are two
  methods
• 1. Calculate the reactor volume fractions from
  the mass fractions for a fixed total process
  volume.
• 2. Calculate the reactor volumes from the sludge
  mass fractions for a fixed aerobic reactor sludge
  concentration. This method will yield a larger
  process volume than Method 1.

131
The UCT Process - Process Volume

• The first method
• 1. The anaerobic volume fraction, fva
• fva fxa (1r)/(rfxa)
  (7.10)
• 0.1 (11)/(10.1)
  0.182
• The anoxic and aerobic volume fractions, fvd/fxd
• fvd/fxd fvb/fxb 1-fxa/(rfxa)
  r/(rfxa) (7.11)
• 1-0.1/(10.1) 0.91
• where first subscripts v and x refer to volume
  and mass fractions, respectively and second
  subscripts a, d, and b refer to anaerobic,
  anoxic, and aerobic reactors, respectively.

132
The UCT Process - Process Volume

• 2. For a selected mean process MLSS conc., Xt
  (7.13)
• Xt Xtbr/(rfxa) 40001/(10.1)
  3636 mg TSS/L
• The total process volume, Vp
• Vp M(Xt)/Xt 56090/3636 15425 m
• The anaerobic sludge conc., Xta
• Xta Xt fxa/fva 3636 0.1/0.18 2000 mg
  TSS/L
• The anoxic sludge conc., Xtb
• Xtd Xtb Xt/1-fxa/(rfxa)
• 3636/1-0.1/(10.1)
  4000 mg/L

3
133
The UCT Process - Process Volume

• 3. The anaerobic reactor volume, Va
• Va fvaVp 0.18215425 2807 m3
• The primary anoxic reactor volume, Vd1
• Vd1 fxd1r/(rfxa) Vp 0.11/(10.1)15425
  1402 m3
• The secondary anoxic reactor volume, Vd2
• Vd2 fxd2r/(rfxa) Vp 0.281/(10.1)15425
  3926 m3
• The aerobic reactor volume, Vb
• Vb fxbr/(rfxa) Vp 0.521/ (10.1)15425
  7292 m3