Exercise set

Sidestream Deammonification Exercises

Worked sidestream PN/A exercises for ammonia load, nitrite target, aeration duty, alkalinity, FA/FNA, anammox activity and release.

These exercises practise sidestream deammonification calculations used during startup, troubleshooting and validation. They focus on load basis, partial nitritation target, oxygen and alkalinity screening, temperature-corrected capacity, free nitrous acid, free ammonia, anammox activity, nitrogen balance, intermittent dewatering, reactor loading and operating decisions.

Assume simplified screening coefficients unless an exercise states otherwise. Real sidestream treatment decisions require site data, biological activity evidence, temperature, pH, alkalinity, reactor configuration, biomass retention, sensor checks, sample timing and responsible engineering review.

Release Evidence Notes

Use the worked answers as release evidence only when ammonia, nitrite, nitrate, pH, alkalinity, temperature, feed-cycle timing, reactor volume, aeration condition, biomass retention and downstream mainstream load all describe the same operating window. A sidestream reactor can look improved locally and still fail release if residual nitrite, pH risk, cold-weather capacity or main-plant reserve is not controlled.

The strongest PN/A release records separate partial nitritation balance, anammox conversion, inhibition risk, oxygen and alkalinity support, intermittent-feed smoothing and downstream impact. Each gate should name the sample timing, as-nitrogen basis, uncertainty allowance, operating action and validation evidence required before ramping feed or returning treated sidestream to the main plant.

How to Use These Exercises

For each problem:

  1. keep all nitrogen species on an as-nitrogen basis;
  2. pair every concentration with the matching flow and time window;
  3. distinguish target nitrite production from allowed effluent nitrite;
  4. state whether the result supports startup, hold, recovery or release;
  5. identify the field evidence needed to validate the calculation.

The most common mistake is solving the arithmetic correctly but making the wrong operating decision. PN/A control is a balance problem, not only an ammonia-removal problem.

Engineering Boundary Notes

These exercises use simplified PN/A and sidestream-deammonification screens. They do not replace startup plans, reactor-specific biological activity testing, toxicity review, control-loop validation, aeration testing, mainstream impact assessment, alkalinity management, solids-retention review or compliance approval. A calculated load-ramp margin applies only to the stated feed window, nitrogen basis, pH, temperature, reactor volume, biomass inventory, aeration state and downstream return condition.

Separate partial nitritation, anammox conversion, inhibition, alkalinity support and mainstream impact before release. Ammonia removal can look acceptable while nitrite accumulation, FNA, free ammonia, cold-weather capacity, biomass washout or return-load impacts make the next ramp unsafe.

Common Release Mistakes

  • treating target nitrite production as allowed residual effluent nitrite;
  • mixing TAN, ammonia as N, nitrite as N and nitrate as N without preserving the basis;
  • ramping load from a single nitrogen balance while temperature, pH, alkalinity and activity evidence are weak;
  • suppressing NOB with free ammonia while crossing into anammox inhibition risk;
  • ignoring intermittent-feed peaks, equalization needs and decant solids loss;
  • returning sidestream flow before checking mainstream nitrogen reserve and compliance margin.

Scenario Map

ExerciseMain CalculationOperating Question
1ammonia load and PN/A targethow much nitrite should partial nitritation produce?
2oxygen and alkalinity screendoes oxygen saving hide pH risk?
3FNA from nitrite and pHis nitrite accumulation becoming inhibitory?
4dissolved inorganic nitrogen balanceis the reactor ready for release?
5conservative load rampis the proposed load step too aggressive?
6free ammonia from pH and TANis NOB suppression becoming biomass inhibition?
7nitrate byproduct checkis nitrate consistent with anammox or NOB drift?
8HRT and volumetric loadingdoes intermittent feed overload the reactor?
9equalization storagecan the batch feed be smoothed into a stable reactor load?
10field oxygen-transfer margindoes the theoretical oxygen saving survive field constraints?
11uncertainty release checkis a passing nitrogen balance still too close to the gate?
12temperature-corrected capacitydoes cooler operation remove the release margin?
13alkalinity supplement dosehow much buffer is needed for a guarded pH margin?
14residual nitrite-to-ammonia ratiois the effluent balanced or nitrite-limited/inhibited?
15mainstream nitrogen impactcan the main process accept the treated sidestream return?
16anammox biomass retention and washoutis decant solids loss eroding the retained biomass inventory?
17intermittent aeration duty cycledoes the blower-on timer match partial-nitritation oxygen demand?
18anammox activity and seed biomassis retained active biomass sufficient for the next load ramp?

Validation Package Checklist

  • feed flow, ammonia, nitrite, nitrate, pH, alkalinity, temperature and sampling window are documented;
  • nitrogen species are kept on a consistent as-nitrogen basis with matching flow and time period;
  • partial nitritation target, anammox conversion, nitrate byproduct and residual nitrite are separated;
  • FNA, free ammonia, pH margin, oxygen-transfer margin, temperature-corrected capacity and alkalinity dose are guarded;
  • reactor volume, feed-cycle timing, equalization, biomass retention, decant solids and seed activity are current;
  • mainstream return load, compliance reserve, sensor checks and validation evidence support any ramp or release;
  • final release decision states accept, hold ramp, add alkalinity, adjust aeration, improve equalization, protect biomass or hold.

Exercise 1: Sidestream Load and PN/A Target

A dewatering centrate sidestream has average flow Q_{side}=95\ \text{m}^3/\text{d} and ammonia nitrogen concentration C_{NH4-N}=780\ \text{mg/L as N}. Use a simplified target ratio R_{NO2/NH4}=1.32.

Find the sidestream ammonia load, target partial-nitritation fraction, target nitrite production and residual ammonia load for anammox.

Solution

Ammonia load:

L_{NH4}=Q_{side}C_{NH4-N}(0.001)
L_{NH4}=95(780)(0.001)=74.1\ \text{kg N/d}

Target partial-nitritation fraction:

\displaystyle f_{PN}=\frac{R_{NO2/NH4}}{1+R_{NO2/NH4}}
\displaystyle f_{PN}=\frac{1.32}{1+1.32}=0.569

Target nitrite production:

L_{NO2,target}=0.569(74.1)=42.2\ \text{kg NO}_2\text{-N/d}

Residual ammonia:

L_{NH4,rem}=74.1-42.2=31.9\ \text{kg N/d}

Ratio check:

\displaystyle \frac{42.2}{31.9}=1.32

Engineering Comment

The target nitrite production is not an effluent limit. It is the approximate nitrite supply that anammox must consume. If 42 kg/d of nitrite is produced but remains in the effluent, the process is not balanced.

Plausibility Check

The target fraction (1.32/(1+1.32)=0.569) keeps ammonia and nitrite on a consistent as-nitrogen basis. The ratio check (42.2/31.9=1.32) confirms that the partial-nitritation target is internally balanced, but it does not prove that anammox biomass will actually consume the nitrite.

Exercise 2: Oxygen and Alkalinity Screen

Using the result from Exercise 1, compare full nitrification oxygen demand with partial-nitritation oxygen demand. Then estimate alkalinity demand if 7.14\ \text{kg as CaCO}_3 is consumed per \text{kg NO}_2\text{-N} produced. The sidestream alkalinity is 3200\ \text{mg/L as CaCO}_3.

Solution

Full nitrification oxygen screen:

O_{conv}=4.57L_{NH4}
O_{conv}=4.57(74.1)=339\ \text{kg O}_2/\text{d}

Partial nitritation oxygen screen:

O_{PN}=3.43L_{NO2,target}
O_{PN}=3.43(42.2)=145\ \text{kg O}_2/\text{d}

Screened oxygen reduction:

S_O=339-145=194\ \text{kg O}_2/\text{d}

Alkalinity demand:

A_{PN}=7.14L_{NO2,target}
A_{PN}=7.14(42.2)=301\ \text{kg/d as CaCO}_3

Available alkalinity load:

L_{Alk}=95(3200)(0.001)=304\ \text{kg/d as CaCO}_3

Alkalinity margin:

M_{Alk}=304-301=3\ \text{kg/d as CaCO}_3

Engineering Comment

The oxygen screen looks attractive, but the alkalinity margin is almost zero. This startup should not be released from nominal oxygen saving alone. It needs pH and alkalinity trends during the actual dewatering schedule.

Plausibility Check

The oxygen saving is large, about (194\ \text{kg O}_2/\text{d}), but the alkalinity margin is only (304-301=3\ \text{kg/d as CaCO}_3). That is roughly a one percent margin against alkalinity demand, so any concentration error, batch variability or pH depression could erase the apparent operating headroom.

Exercise 3: Free Nitrous Acid During Nitrite Accumulation

A PN/A reactor effluent has C_{NO2-N}=160\ \text{mg/L as N} and pH=6.70. Use pK_a=3.25 for a screening calculation. The same nitrite concentration is later observed at pH=7.10 after alkalinity correction.

Calculate free nitrous acid on an as-nitrogen basis for both pH values.

Solution

Free nitrous acid fraction:

\displaystyle f_{HNO2}=\frac{1}{1+10^{pH-pK_a}}

At pH=6.70:

\displaystyle f_{HNO2}=\frac{1}{1+10^{6.70-3.25}}=0.000355

Free nitrous acid:

C_{FNA,N}=160(0.000355)=0.0568\ \text{mg/L as N}

At pH=7.10:

\displaystyle f_{HNO2}=\frac{1}{1+10^{7.10-3.25}}=0.000141
C_{FNA,N}=160(0.000141)=0.0226\ \text{mg/L as N}

Relative reduction:

\displaystyle \frac{0.0568-0.0226}{0.0568}=0.602

Engineering Comment

Raising pH reduces the FNA screen by about 60\% at the same nitrite concentration. That does not solve the nitrogen balance by itself. The reactor still needs lower nitrite residual or stronger nitrite consumption.

Plausibility Check

The calculation changes only pH while holding nitrite at (160\ \text{mg/L as N}), so the drop from 0.0568 to (0.0226\ \text{mg/L as N}) is a speciation effect, not a removal improvement. The result supports alkalinity/pH correction as inhibition mitigation, but not as proof that nitrite accumulation has been solved.

Exercise 4: Nitrogen Balance and Release Decision

A sidestream reactor receives Q=100\ \text{m}^3/\text{d} at C_{NH4,in}=900\ \text{mg/L as N}. Reactor effluent is:

  • C_{NH4,out}=420\ \text{mg/L as N};
  • C_{NO2,out}=120\ \text{mg/L as N};
  • C_{NO3,out}=45\ \text{mg/L as N}.

Estimate dissolved inorganic nitrogen removal and decide whether the reactor is ready for release.

Solution

Influent ammonia load:

L_{DIN,in}=100(900)(0.001)=90.0\ \text{kg N/d}

Effluent ammonia:

L_{NH4,out}=100(420)(0.001)=42.0\ \text{kg N/d}

Effluent nitrite:

L_{NO2,out}=100(120)(0.001)=12.0\ \text{kg N/d}

Effluent nitrate:

L_{NO3,out}=100(45)(0.001)=4.5\ \text{kg N/d}

Effluent dissolved inorganic nitrogen:

L_{DIN,out}=42.0+12.0+4.5=58.5\ \text{kg N/d}

Apparent removal:

L_{DIN,rem}=90.0-58.5=31.5\ \text{kg N/d}

Percent removal:

\displaystyle \eta_{DIN}=\frac{31.5}{90.0}=0.35=35\%

Engineering Comment

This is not a release result. Nitrate is not excessive, but ammonia and nitrite are both high. The likely decision is hold-and-correct: stop further ramp-up, check pH/alkalinity, biomass retention, sample timing and DO exposure, then validate recovery across several feed cycles.

Plausibility Check

The influent dissolved inorganic nitrogen load is (90.0\ \text{kg N/d}), while effluent ammonia plus nitrite alone is (42.0+12.0=54.0\ \text{kg N/d}). A 35 percent removal result is far below a mature release expectation, so low nitrate by itself cannot compensate for the unresolved ammonia and nitrite residuals.

Exercise 5: Conservative Load and Ramp Check

A startup plan proposes increasing sidestream ammonia load from 70 to 90\ \text{kg N/d}. The measured new condition is Q=100\ \text{m}^3/\text{d} and C_{NH4-N}=900\ \text{mg/L as N}. Use u_Q=0.05 and U_C=30\ \text{mg/L} for a conservative release check.

Find the nominal ramp, conservative load and release implication.

Solution

Nominal load ramp:

\displaystyle r_L=\frac{L_{new}-L_{old}}{L_{old}}
\displaystyle r_L=\frac{90-70}{70}=0.286=28.6\%

Conservative load:

L_{cons}=Q(1+u_Q)(C+U_C)(0.001)
L_{cons}=100(1.05)(900+30)(0.001)=97.7\ \text{kg N/d}

The conservative basis is:

\displaystyle \frac{97.7-70}{70}=0.396=39.6\%

Engineering Comment

The nominal ramp looks like a 29\% step, but the conservative basis behaves like a 40\% load step. If the reactor is near inhibition, alkalinity limit or biomass-retention limit, the conservative value is the better release screen.

Plausibility Check

The conservative load (97.7\ \text{kg N/d}) is (7.7\ \text{kg N/d}) above the nominal new load and (27.7\ \text{kg N/d}) above the old load. That makes the apparent ramp closer to 40 percent than 29 percent, so the release decision should use the conservative load when the biology is not yet proven stable.

Exercise 6: Free Ammonia Screen During pH Increase

The centrate in Exercise 1 has total ammonia nitrogen TAN_N=780\ \text{mg/L as N}. Use pK_a=9.25. Compare free ammonia at pH=7.85 and pH=8.20.

Calculate the free-ammonia fraction, free-ammonia concentration and relative increase.

Solution

Free-ammonia fraction:

\displaystyle f_{NH3}=\frac{1}{1+10^{pK_a-pH}}

At pH=7.85:

\displaystyle f_{NH3}=\frac{1}{1+10^{9.25-7.85}}=0.0383
C_{FA,N}=780(0.0383)=29.9\ \text{mg/L as N}

At pH=8.20:

\displaystyle f_{NH3}=\frac{1}{1+10^{9.25-8.20}}=0.0818
C_{FA,N}=780(0.0818)=63.8\ \text{mg/L as N}

Relative increase:

\displaystyle \frac{63.8-29.9}{29.9}=1.13

or 113\%.

Engineering Comment

The pH increase more than doubles the free-ammonia screen. That may help suppress nitrite-oxidizing bacteria, but it may also stress ammonia oxidizers or anammox organisms if exposure is excessive. A good operating decision checks temperature, exposure time, ammonia trend, nitrite trend and biomass activity before raising pH further.

Plausibility Check

A pH increase of only 0.35 units raises the free-ammonia estimate from 29.9 to (63.8\ \text{mg/L as N}). The 113 percent increase is plausible because ammonia speciation is logarithmic with pH, so pH control is a biological-control action, not just a chemistry correction.

Exercise 7: Nitrate Byproduct Versus NOB Drift

Use the residual ammonia from Exercise 1, L_{NH4,rem}=31.9\ \text{kg N/d}. For a simplified anammox nitrate byproduct ratio R_{NO3/NH4}=0.11, estimate expected nitrate. A later sample shows Q=95\ \text{m}^3/\text{d} and C_{NO3-N}=115\ \text{mg/L as N}.

Calculate expected nitrate, observed nitrate, excess nitrate and observed nitrate-to-residual-ammonia ratio.

Solution

Expected nitrate byproduct:

L_{NO3,exp}=0.11(31.9)=3.51\ \text{kg NO}_3\text{-N/d}

Observed nitrate:

L_{NO3,obs}=95(115)(0.001)=10.9\ \text{kg NO}_3\text{-N/d}

Excess nitrate:

L_{NO3,excess}=10.9-3.51=7.39\ \text{kg N/d}

Observed nitrate ratio:

\displaystyle R_{NO3/NH4,obs}=\frac{10.9}{31.9}=0.342

Excess nitrate as a fraction of target nitrite production from Exercise 1:

\displaystyle \frac{7.39}{42.2}=0.175

or 17.5\%.

Engineering Comment

The observed nitrate ratio is roughly three times the simplified anammox byproduct screen. That does not prove NOB drift alone, but it is strong evidence to review DO setpoint, aeration intermittency, biofilm stratification, solids retention, sample timing and whether nitrite is being oxidized instead of consumed by anammox.

Plausibility Check

The expected nitrate byproduct is only (3.51\ \text{kg N/d}), while the observed nitrate is (10.9\ \text{kg N/d}). The excess (7.39\ \text{kg N/d}) is 17.5 percent of the target nitrite production, large enough to indicate process drift rather than routine anammox byproduct scatter.

Exercise 8: Reactor Loading Under Intermittent Dewatering

The sidestream reactor volume is V=180\ \text{m}^3. The daily average sidestream flow is 95\ \text{m}^3/\text{d} and the ammonia load is 74.1\ \text{kg N/d}. Dewatering actually runs for only 8\ \text{h/d}.

Calculate daily-average HRT and volumetric ammonia loading. Then calculate equivalent instantaneous flow, instantaneous HRT and equivalent instantaneous volumetric loading during the dewatering window.

Solution

Daily-average HRT:

\displaystyle HRT_{avg}=\frac{V}{Q}=\frac{180}{95}=1.89\ \text{d}

Daily-average volumetric loading:

\displaystyle VLR_{avg}=\frac{74.1}{180}=0.412\ \text{kg N}/\text{m}^3\text{/d}

The dewatering fraction of the day is:

\displaystyle f_t=\frac{8}{24}=0.333

Equivalent instantaneous flow:

\displaystyle Q_{inst}=\frac{95}{0.333}=285\ \text{m}^3/\text{d}

Instantaneous HRT:

\displaystyle HRT_{inst}=\frac{180}{285}=0.632\ \text{d}

Equivalent instantaneous ammonia loading:

\displaystyle L_{NH4,inst}=\frac{74.1}{0.333}=222\ \text{kg N/d}

Instantaneous volumetric loading:

\displaystyle VLR_{inst}=\frac{222}{180}=1.23\ \text{kg N}/\text{m}^3\text{/d}

Engineering Comment

The daily average suggests moderate loading, but the reactor sees about three times the loading rate during dewatering. PN/A validation should therefore use feed-cycle data, not only daily averages. Equalization, feed pacing or a hold period may be more useful than changing the biological target.

Plausibility Check

The dewatering window is one third of a day, so instantaneous flow and loading are three times the daily-average basis. The HRT drops from 1.89 d to 0.632 d and VLR rises from 0.412 to (1.23\ \text{kg N}/\text{m}^3\text{/d}), confirming that daily averages hide the actual biological shock.

Exercise 9: Equalization Volume for Batch Feed

A dewatering unit sends centrate at 18\ \text{m}^3/\text{h} for 6\ \text{h/d}. The PN/A reactor feed pump should receive a steady 5\ \text{m}^3/\text{h} while centrate is available and after the batch ends. Estimate the equalization volume needed to store the excess centrate during the batch, the drawdown time after the batch and a tank volume with 20\% freeboard.

Solution

Centrate volume produced during the batch:

V_{batch}=18(6)=108\ \text{m}^3

Reactor feed during the batch:

V_{feed,batch}=5(6)=30\ \text{m}^3

Excess storage needed:

V_{stored}=108-30=78\ \text{m}^3

Drawdown time after the batch:

\displaystyle t_{draw}=\frac{78}{5}=15.6\ \text{h}

Tank volume with 20\% freeboard:

V_{tank}=78(1.20)=93.6\ \text{m}^3

Engineering Comment

The tank is not just hydraulic storage. It protects biomass from shock loading, alkalinity swings and inhibition peaks. The design review should also check mixing, odour, temperature loss, struvite risk, emergency overflow path and whether the feed pump can maintain stable low flow.

Plausibility Check

The mass balance closes: the batch produces (108\ \text{m}^3), while the reactor receives (30\ \text{m}^3) during the batch and (78\ \text{m}^3) during the 15.6 h drawdown. The (93.6\ \text{m}^3) tank includes freeboard, but it still needs operational checks for mixing, overflow and low-flow control.

Exercise 10: Field Oxygen-Transfer Margin

Use the partial-nitritation oxygen demand from Exercise 2, O_{PN}=145\ \text{kg O}_2/\text{d}, and the conventional oxygen screen O_{conv}=339\ \text{kg O}_2/\text{d}. The installed system can deliver OTR_{field}=170\ \text{kg O}_2/\text{d} under current sidestream conditions. A diffuser inspection suggests a possible 20\% field-transfer loss.

Calculate the initial oxygen-transfer margin, fouled OTR, fouled margin and theoretical daily energy saving if field oxygen delivery is 1.6\ \text{kg O}_2/\text{kWh}.

Solution

Initial oxygen-transfer margin:

\displaystyle M_{OTR}=\frac{170-145}{145}=0.172

or 17.2\%.

Fouled oxygen-transfer capacity:

OTR_{fouled}=170(0.80)=136\ \text{kg O}_2/\text{d}

Fouled margin:

\displaystyle M_{OTR,fouled}=\frac{136-145}{145}=-0.062

or a 6.2\% deficit.

Conventional energy screen:

\displaystyle E_{conv}=\frac{339}{1.6}=212\ \text{kWh/d}

Partial-nitritation energy screen:

\displaystyle E_{PN}=\frac{145}{1.6}=90.6\ \text{kWh/d}

Theoretical saving:

S_E=212-90.6=121\ \text{kWh/d}

Engineering Comment

The theoretical energy saving is real as a screening result, but the fouled-field margin fails. A release decision should not claim energy benefit while oxygen transfer is marginal. Validate diffuser condition, blower turndown, DO control stability and nitrite response under the actual aeration pattern.

Plausibility Check

The clean field OTR margin is 17.2 percent, but a 20 percent transfer loss reduces capacity to (136\ \text{kg O}_2/\text{d}), which is (9\ \text{kg O}_2/\text{d}) below the partial-nitritation demand. The (121\ \text{kWh/d}) theoretical saving is therefore not releasable unless field oxygen-transfer evidence remains positive under degraded conditions.

Exercise 11: Uncertainty Release Check

A validation run receives Q=100\ \text{m}^3/\text{d} and measured influent ammonia C_{NH4,in}=900\ \text{mg/L as N}. Effluent results are C_{NH4,out}=70\ \text{mg/L as N}, C_{NO2,out}=20\ \text{mg/L as N} and C_{NO3,out}=12\ \text{mg/L as N}. Use a release threshold of 85\% dissolved inorganic nitrogen removal.

For a conservative check, use U_{in}=30\ \text{mg/L} subtracted from influent ammonia and add U_{NH4}=10, U_{NO2}=8 and U_{NO3}=4\ \text{mg/L} to effluent species. Calculate nominal removal and conservative removal.

Solution

Nominal influent load:

L_{DIN,in}=100(900)(0.001)=90.0\ \text{kg N/d}

Nominal effluent load:

L_{DIN,out}=100(70+20+12)(0.001)=10.2\ \text{kg N/d}

Nominal removal:

\displaystyle \eta_{DIN}=\frac{90.0-10.2}{90.0}=0.887

or 88.7\%.

Conservative influent load:

L_{DIN,in,cons}=100(900-30)(0.001)=87.0\ \text{kg N/d}

Conservative effluent load:

L_{DIN,out,cons}=100[(70+10)+(20+8)+(12+4)](0.001)
L_{DIN,out,cons}=12.4\ \text{kg N/d}

Conservative removal:

\displaystyle \eta_{DIN,cons}=\frac{87.0-12.4}{87.0}=0.857

or 85.7\%.

Margin above the release threshold:

M_{release}=85.7\%-85.0\%=0.7\ \text{percentage points}

Engineering Comment

The nominal run passes comfortably, but the conservative release margin is narrow. A prudent decision is conditional release or hold-for-confirmation rather than full ramp-up. Repeat the balance across multiple dewatering cycles and check lab QA/QC, online sensor drift, flow pacing and downstream nitrogen response.

Plausibility Check

Nominal removal is 88.7 percent, but the conservative calculation drops to 85.7 percent against an 85.0 percent threshold. A 0.7 percentage-point guarded margin is too narrow for full ramp-up from one run, so the operating decision should preserve the pass as conditional evidence rather than treating it as robust release.

Exercise 12: Temperature-Corrected Anammox Capacity

A sidestream PN/A reactor has demonstrated an anammox-supported nitrogen-removal capacity of:

R_{30}=95\ \text{kg N/d}

at 30^\circ\text{C}. During a cooler operating period, reactor temperature falls to 25^\circ\text{C}. Use a simplified temperature factor:

R_T=R_{30}\theta^{T-30}

with:

\theta=1.07

The release target is at least 85\% dissolved inorganic nitrogen removal for the 90.0\ \text{kg N/d} influent load used in Exercise 11. Estimate the cooler-temperature capacity and release margin.

Solution

Temperature-corrected capacity:

R_{25}=95(1.07)^{25-30}
R_{25}=95(1.07)^{-5}=67.7\ \text{kg N/d}

Required removal for the release gate:

R_{req}=0.85(90.0)=76.5\ \text{kg N/d}

Capacity margin:

M_R=67.7-76.5=-8.8\ \text{kg N/d}

Relative margin:

\displaystyle \frac{-8.8}{76.5}\times100=-11.5\%

At the cooler temperature, the simplified capacity screen does not support full release at the same load.

Engineering Comment

This temperature factor is only a screening model, but it shows why a reactor that passed during warm startup may fail during cooler operation. A release decision should check actual temperature trend, activity tests, retained biomass, feed pacing, nitrite residual, pH and whether the downstream main process can absorb a temporary sidestream shortfall.

Plausibility Check

A five-degree drop with \theta=1.07 reduces capacity by roughly one third, from 95 to 67.7\ \text{kg N/d}. Since the release gate requires 76.5\ \text{kg N/d} removal, the negative margin is plausible and should trigger hold, load reduction, warming, added biomass retention or a guarded ramp.

Exercise 13: Alkalinity Supplement Dose for Guarded Release

Exercise 2 estimated partial-nitritation alkalinity demand:

A_{PN}=301\ \text{kg/d as CaCO}_3

and available sidestream alkalinity:

A_{avail}=304\ \text{kg/d as CaCO}_3

The startup team wants a 15\% alkalinity margin above the calculated demand before the next ramp. Estimate the additional alkalinity required as \text{kg/d as CaCO}_3, the equivalent dry sodium bicarbonate dose, and the feed rate if the bicarbonate product is 90\% active by mass.

Use:

1\ \text{kg as CaCO}_3=1.68\ \text{kg NaHCO}_3

Solution

Guarded alkalinity requirement:

A_{guard}=1.15A_{PN}
A_{guard}=1.15(301)=346\ \text{kg/d as CaCO}_3

Additional alkalinity required:

A_{add}=346-304=42\ \text{kg/d as CaCO}_3

Dry sodium bicarbonate equivalent:

m_{NaHCO3}=1.68A_{add}
m_{NaHCO3}=1.68(42)=70.6\ \text{kg/d}

Product feed for 90\% active material:

\displaystyle m_{product}=\frac{70.6}{0.90}=78.4\ \text{kg/d}

Engineering Comment

The dosing screen is not a substitute for jar testing, feed-system calibration or pH control tuning. It should be validated with alkalinity trend, reactor pH, nitrite residual, aeration duty, mixing, chemical storage, feed-point location and downstream sodium or solids constraints before the next load ramp.

Plausibility Check

The original margin was only about 3\ \text{kg/d as CaCO}_3, so a guarded 15 percent margin should require tens of kilograms per day of additional alkalinity. The calculated 42\ \text{kg/d as CaCO}_3 deficit and 78.4\ \text{kg/d} product feed are therefore plausible as a startup buffer screen.

Exercise 14: Residual Nitrite-to-Ammonia Stoichiometry Gate

A PN/A reactor receives the same influent ammonia load as Exercise 1:

L_{NH4,in}=74.1\ \text{kg N/d}

Current effluent loads are:

L_{NH4,out}=28.0\ \text{kg N/d}
L_{NO2,out}=18.0\ \text{kg N/d}
L_{NO3,out}=5.0\ \text{kg N/d}

For a stable anammox polishing step, the site expects the residual nitrite-to-ammonia ratio to remain between:

0.8 \le R_{NO2/NH4,res} \le 1.5

and residual nitrite load to be below:

L_{NO2,max}=12.0\ \text{kg N/d}

Calculate dissolved inorganic nitrogen removal, residual nitrite-to-ammonia ratio, nitrite exceedance and release decision.

Solution

Effluent dissolved inorganic nitrogen:

L_{DIN,out}=L_{NH4,out}+L_{NO2,out}+L_{NO3,out}
L_{DIN,out}=28.0+18.0+5.0=51.0\ \text{kg N/d}

Dissolved inorganic nitrogen removed:

L_{DIN,removed}=74.1-51.0=23.1\ \text{kg N/d}

Percent removal:

\displaystyle \eta_{DIN}=\frac{23.1}{74.1}=0.312=31.2\%

Residual nitrite-to-ammonia ratio:

\displaystyle R_{NO2/NH4,res}=\frac{18.0}{28.0}=0.643

This is below the lower balance gate:

0.643<0.8

Nitrite exceedance:

\Delta L_{NO2}=18.0-12.0=6.0\ \text{kg N/d}

The reactor fails both the residual stoichiometry screen and the residual nitrite load limit. The likely decision is hold-and-correct, not ramp or release.

Engineering Comment

Low residual nitrite-to-ammonia ratio can mean partial nitritation is under-producing nitrite, anammox is not receiving the right substrate balance, sampling windows are mismatched, or ammonia conversion is simply too weak. The high residual nitrite load still matters: even if the ratio is low, the treated sidestream can return a damaging nitrite pulse to the main plant.

Plausibility Check

The reactor removes only 31.2 percent of influent dissolved nitrogen, so a release failure is expected. The residual ratio (18/28=0.643) is too low for the stated balance gate, while the nitrite load is also (6.0\ \text{kg N/d}) above the local limit. These two facts together show an unbalanced and incomplete PN/A state.

Exercise 15: Mainstream Nitrogen Impact From Treated Sidestream Return

A treated sidestream is returned to the head of the main wastewater plant. The main plant influent total nitrogen load before sidestream return is:

L_{TN,main}=620\ \text{kg N/d}

The treated sidestream still contains:

L_{NH4,side}=18\ \text{kg N/d}
L_{NO2,side}=9\ \text{kg N/d}
L_{NO3,side}=4\ \text{kg N/d}

The main plant has a daily total nitrogen treatment capacity of:

L_{TN,cap}=670\ \text{kg N/d}

and the operating rule requires at least:

M_{min}=7\%

capacity margin after sidestream return. Calculate sidestream nitrogen return, combined load, actual capacity margin and release decision.

Solution

Sidestream nitrogen return:

L_{TN,side}=L_{NH4,side}+L_{NO2,side}+L_{NO3,side}
L_{TN,side}=18+9+4=31\ \text{kg N/d}

Combined main-plant load:

L_{TN,total}=L_{TN,main}+L_{TN,side}
L_{TN,total}=620+31=651\ \text{kg N/d}

Absolute capacity margin:

M_N=L_{TN,cap}-L_{TN,total}
M_N=670-651=19\ \text{kg N/d}

Relative capacity margin:

\displaystyle M_{rel}=\frac{19}{670}=0.0284=2.8\%

Minimum required reserve:

L_{reserve,min}=0.07(670)=46.9\ \text{kg N/d}

Reserve shortfall:

\Delta L_{reserve}=46.9-19=27.9\ \text{kg N/d}

The combined load is below absolute treatment capacity, but it does not preserve the required 7 percent operating margin. The sidestream should not be released at this quality without improved PN/A performance, load pacing, storage, or a documented downstream contingency.

Engineering Comment

Sidestream release is a plant-wide decision. A sidestream reactor can look improved locally while still consuming the main plant’s nitrogen-removal reserve. The release package should show sidestream ammonia, nitrite and nitrate loads, the main plant load forecast, wet-weather sensitivity, aeration capacity, alkalinity and downstream compliance margin.

Plausibility Check

The sidestream adds (31\ \text{kg N/d}), increasing the main plant load from (620) to (651\ \text{kg N/d}). The plant is still below the nominal (670\ \text{kg N/d}) capacity, but only by (19\ \text{kg N/d}), or 2.8 percent. That is much smaller than the 7 percent reserve rule, so the rejection is due to margin, not arithmetic overload.

Exercise 16: Anammox Biomass Washout and Retention Gate

A sidestream PN/A reactor relies on retained anammox granules. The reactor volume is:

V=180\ \text{m}^3

The retained anammox-active volatile suspended solids concentration is estimated as:

X_a=2.8\ \text{kg VSS/m}^3

During decant, the reactor discharges:

Q_{dec}=100\ \text{m}^3/\text{d}

with suspended biomass concentration:

X_{dec}=120\ \text{mg VSS/L}

The site release rule allows biomass washout no greater than:

f_{loss,max}=0.8\%\ \text{per day}

and requires biomass retention time at least:

BRT_{min}=120\ \text{d}

A screen retrofit is expected to reduce decant suspended biomass to:

X_{dec,screen}=35\ \text{mg VSS/L}

Calculate retained biomass inventory, current daily biomass loss, current washout fraction and biomass retention time. Then repeat the washout and retention checks after the screen retrofit.

Solution

Retained active biomass inventory:

M_a=VX_a
M_a=180(2.8)=504\ \text{kg VSS}

Convert current decant solids:

120\ \text{mg/L}=0.120\ \text{kg/m}^3

Current daily biomass loss:

M_{loss}=Q_{dec}X_{dec}
M_{loss}=100(0.120)=12.0\ \text{kg VSS/d}

Current washout fraction:

\displaystyle f_{loss}=\frac{M_{loss}}{M_a}
\displaystyle f_{loss}=\frac{12.0}{504}=0.0238=2.38\%\ \text{per day}

Current biomass retention time:

\displaystyle BRT=\frac{M_a}{M_{loss}}
\displaystyle BRT=\frac{504}{12.0}=42.0\ \text{d}

The current decant condition fails both release gates:

2.38\%>0.8\%

and:

42.0<120\ \text{d}

After the screen retrofit:

35\ \text{mg/L}=0.035\ \text{kg/m}^3

Screened biomass loss:

M_{loss,screen}=100(0.035)=3.5\ \text{kg VSS/d}

Screened washout fraction:

\displaystyle f_{loss,screen}=\frac{3.5}{504}=0.00694=0.694\%\ \text{per day}

Screened biomass retention time:

\displaystyle BRT_{screen}=\frac{504}{3.5}=144\ \text{d}

The screened case passes both gates:

0.694\%<0.8\%

and:

144>120\ \text{d}

The reactor should not be ramped under the current decant washout condition. The screen retrofit can support release only after field evidence confirms lower decant solids, stable granule inventory and no new hydraulic short-circuiting or clogging risk.

Engineering Comment

PN/A capacity depends on retained slow-growing biomass, not only on influent ammonia and effluent nitrogen species. A reactor can have acceptable chemistry for a short validation period while gradually losing anammox granules. Release evidence should include decant suspended solids, granule inventory trend, screen condition, hydraulic shear, sludge wasting, activity tests and recovery response after feed interruptions.

Plausibility Check

The current decant solids concentration is more than three times the screened value, so biomass loss should also be more than three times larger at the same decant flow. The retention time rises from (42) to (144\ \text{d}), which is consistent with reducing daily loss from (12.0) to (3.5\ \text{kg/d}).

Exercise 17: Intermittent Aeration Duty Cycle for Partial Nitritation

A sidestream PN/A reactor is being controlled by an intermittent aeration timer. The daily sidestream ammonia load is:

L_{NH4}=86\ \text{kg N/d}

The target fraction converted by partial nitritation is:

f_{PN}=0.57

Use the partial-nitritation oxygen demand:

Y_{O2,PN}=3.43\ \text{kg O}_2/\text{kg NH}_4\text{-N oxidized}

When the blower is on, field testing gives an actual oxygen transfer rate of:

OTR_{on}=18\ \text{kg O}_2/\text{h}

The current timer runs the blower for:

t_{on,current}=14\ \text{h/d}

The release review applies a transfer uncertainty allowance of:

u_{OTR}=15\%

Calculate the target nitrited ammonia load, oxygen required, nominal blower-on time, guarded blower-on time, current oxygen excess and equivalent extra nitritation capacity. Decide whether the current timer supports release.

Solution

Target nitrited ammonia load:

L_{PN}=f_{PN}L_{NH4}
L_{PN}=0.57(86)=49.0\ \text{kg N/d}

Oxygen required for the target partial nitritation:

L_{O2,req}=Y_{O2,PN}L_{PN}
L_{O2,req}=3.43(49.0)=168.1\ \text{kg O}_2/\text{d}

Nominal blower-on time:

\displaystyle t_{on,nom}=\frac{L_{O2,req}}{OTR_{on}}
\displaystyle t_{on,nom}=\frac{168.1}{18}=9.34\ \text{h/d}

Nominal aerated duty cycle:

\displaystyle D_{nom}=\frac{9.34}{24}=0.389=38.9\%

Guarded oxygen transfer rate:

OTR_{guard}=OTR_{on}(1-u_{OTR})
OTR_{guard}=18(0.85)=15.3\ \text{kg O}_2/\text{h}

Guarded blower-on time:

\displaystyle t_{on,guard}=\frac{168.1}{15.3}=11.0\ \text{h/d}

Guarded duty cycle:

\displaystyle D_{guard}=\frac{11.0}{24}=45.8\%

Current oxygen transfer at the measured field rate:

L_{O2,current}=18(14)=252\ \text{kg O}_2/\text{d}

Oxygen excess above target:

\Delta L_{O2}=252-168.1=83.9\ \text{kg O}_2/\text{d}

Equivalent extra nitritation capacity:

\displaystyle \Delta L_{N,PN}=\frac{83.9}{3.43}=24.5\ \text{kg N/d}

The current 14-hour timer is longer than the guarded requirement of about (11.0\ \text{h/d}). It may over-aerate the reactor, push nitrite toward nitrate through NOB activity, or increase residual nitrite exposure if anammox consumption is limiting. The release decision should be hold-and-retune rather than ramp: set the timer near the guarded requirement, then validate ammonia, nitrite, nitrate, pH, DO and anammox activity over several feed cycles.

Engineering Comment

Field oxygen-transfer capacity is necessary but not sufficient. PN/A control also needs the right amount of oxygen in the right part of the cycle. Too little aeration leaves ammonia; too much aeration can overproduce nitrite, favor NOB drift or suppress anammox through oxygen exposure. A timer change should be tied to load, field OTR, DO trend, nitrogen species, pH and feed-cycle timing.

Plausibility Check

Oxidizing 57 percent of an (86\ \text{kg N/d}) sidestream load gives about (49\ \text{kg N/d}), so oxygen demand near (168\ \text{kg O}_2/\text{d}) is plausible. At (18\ \text{kg O}_2/\text{h}), that is less than 10 hours of blower-on time. A 15 percent transfer allowance moves the target to about 11 hours, so the 14-hour timer contains about three extra hours of aeration and the equivalent of roughly (25\ \text{kg N/d}) extra nitritation capacity.

Exercise 18: Anammox Activity and Seed-Biomass Ramp Gate

A sidestream PN/A reactor is preparing for the next load ramp. The target dissolved nitrogen removal for the ramp is:

L_{N,target}=85\ \text{kg N/d}

The site requires a 5 percent capacity reserve above the target:

F_s=1.05

The currently retained anammox-active volatile suspended solids inventory is:

M_{AX}=360\ \text{kg VSS}

A batch activity test at the current process temperature gives:

SAA=0.22\ \text{kg N removed/kg VSS d}

Use an operating availability factor for hydraulics, mixing, substrate contact and test-to-field uncertainty:

\eta_a=0.80

Calculate guarded removal requirement, current effective capacity, capacity margin, required retained active biomass and the additional retained biomass needed. Then check a seed addition of:

M_{seed}=230\ \text{kg active VSS}

with expected retention through startup of:

\eta_{ret}=0.75

Solution

Guarded removal requirement is:

L_{N,guard}=F_sL_{N,target}
L_{N,guard}=1.05(85)=89.3\ \text{kg N/d}

Current effective anammox removal capacity is:

L_{N,cap}=M_{AX}SAA\eta_a
L_{N,cap}=360(0.22)(0.80)=63.4\ \text{kg N/d}

Capacity margin against the guarded requirement is:

M_N=63.4-89.3=-25.9\ \text{kg N/d}

The reactor should not be ramped on the current retained activity basis.

Required retained active biomass is:

\displaystyle M_{AX,req}=\frac{L_{N,guard}}{SAA\eta_a}
\displaystyle M_{AX,req}=\frac{89.3}{0.22(0.80)}=507\ \text{kg VSS}

Additional retained active biomass needed is:

\Delta M_{AX}=507-360=147\ \text{kg VSS}

Retained seed biomass after startup loss is:

M_{seed,ret}=M_{seed}\eta_{ret}
M_{seed,ret}=230(0.75)=173\ \text{kg VSS}

Revised retained active biomass is:

M_{AX,rev}=360+173=533\ \text{kg VSS}

Revised effective removal capacity is:

L_{N,cap,rev}=533(0.22)(0.80)=93.8\ \text{kg N/d}

Revised guarded margin is:

M_{N,rev}=93.8-89.3=4.5\ \text{kg N/d}

The seed addition can support a conditional ramp release on this activity basis, but the margin is modest and should be verified with nitrogen species, retained granule inventory, decant solids loss and repeated activity testing.

Engineering Comment

Startup release is not proven by influent load alone. Slow-growing anammox activity must be present, retained and hydraulically available to the sidestream. A seed calculation should state whether activity was measured at the process temperature, whether the retained inventory is active VSS rather than total solids, how much seed is expected to wash out, and whether substrate contact and inhibition conditions match the activity test.

Plausibility Check

An activity of (0.22\ \text{kg N/kg VSS d}) means each (100\ \text{kg}) of active retained biomass provides only about (22\ \text{kg N/d}) before field availability losses. With a 0.80 availability factor, the current (360\ \text{kg}) inventory can remove only about (63\ \text{kg N/d}), well below the guarded ramp target. Retaining about (173\ \text{kg}) of seed raises inventory to roughly (533\ \text{kg}), which is enough to pass but not enough to justify an unguarded ramp.

Review Checklist

Before accepting an answer, check that it includes:

  • the flow and concentration time window;
  • nitrogen units stated as \text{as N};
  • target nitrite production separated from measured effluent nitrite;
  • residual nitrite-to-ammonia balance checked before release;
  • oxygen and alkalinity interpreted together;
  • intermittent aeration duty cycle checked against load, field OTR and nitrogen-species trends;
  • temperature checked before assuming warm-weather biological capacity still applies;
  • pH included when nitrite or ammonia inhibition is discussed;
  • nitrate interpreted with ammonia and nitrite, not alone;
  • reactor loading checked on the actual feed-cycle basis;
  • equalization and flow pacing considered for intermittent sidestreams;
  • biomass retention, decant solids and granule washout checked before feed ramping;
  • anammox activity, retained active biomass and seed retention checked before load ramping;
  • alkalinity supplement calculations tied to pH trend and chemical feed validation;
  • field oxygen-transfer margin separated from theoretical oxygen saving;
  • downstream mainstream nitrogen impact checked when treated sidestream returns to the main plant;
  • an operating decision such as ramp, hold, recover or release;
  • field evidence needed to validate the arithmetic.

An answer can be numerically correct and still be weak if it stops before the decision. For PN/A work, the calculation should explain whether the reactor balance is improving, drifting, inhibited or too uncertain to release.

Common Mistakes

  • Using target nitrite production as if it were allowed effluent nitrite.
  • Ignoring alkalinity and pH after calculating theoretical oxygen savings.
  • Treating low nitrate as proof of healthy anammox without checking ammonia, nitrite and dissolved inorganic nitrogen together.
  • Calculating FNA or free ammonia without the pH and temperature basis that controls speciation.
  • Carrying warm-weather capacity into cooler operation without a rate correction and margin check.
  • Adding alkalinity without validating chemical feed control, pH trend and downstream sodium or solids constraints.
  • Accepting ammonia removal without total nitrogen evidence, residual nitrite control and mainstream return impact.
  • Ramping feed while decant solids loss is washing out retained anammox biomass.
  • Ramping load from nominal reactor volume without checking measured anammox activity and retained active biomass.
  • Using daily averages for intermittent centrate feed when peak feed cycles overload the reactor.
  • Leaving an aeration timer unchanged after load changes while interpreting excess oxygen as release margin.
  • Claiming theoretical oxygen saving without checking field oxygen-transfer rate, blower capacity and DO control.
  • Ramping feed from nominal loads while ignoring uncertainty, sample timing, inhibition history and recovery evidence.

A strong exercise answer ends with the operating decision and the validation evidence still required.

REF

See also