Exercise set

Enhanced Biological Phosphorus Removal Exercises

Worked EBPR exercises for release, uptake, VFA ratios, fermenter yield, nitrate intrusion, wasting, chemical trim and validation.

Branch: Environmental Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.3.5 · reviewed

These exercises practise first-pass EBPR calculations used in wastewater treatment review. They connect phosphorus load, anaerobic release, uptake, carbon availability, fermenter VFA production, nitrate and DO intrusion, selector contact time, sludge wasting, sidestream return, chemical trim, solids carryover and profile plausibility to operating decisions.

Assume simplified screening values unless an exercise states otherwise. Real EBPR decisions require lab method review, plant-specific trends, profile sampling, chemical dose history, solids separation evidence and responsible engineering judgement.

Release Evidence Notes

Use the worked answers as process evidence only when the sample location, filtration basis, flow window, phosphorus fraction, VFA method, nitrate and DO carryover measurements, selector hydraulic condition, wasting record and chemical-dose history describe the same operating period. A low final phosphorus value is not enough to prove EBPR recovery if chemical trim, solids capture or profile timing explains the result.

The strongest release records separate biological release, aerobic uptake, solids-phase phosphorus export, sidestream return, chemical support and compliance sampling. Each gate should name the profile data, control action, uncertainty or lab-method limitation and the operating response when nitrate, oxygen, VFA shortage, short-circuiting or solids carryover contradicts the expected EBPR pattern.

How to Use These Exercises

For each problem:

state whether phosphorus is total phosphorus or orthophosphate;
keep flow and concentration on the same time window;
separate anaerobic release from final removal;
check carbon, nitrate and dissolved oxygen before blaming PAOs;
end with an operating decision or evidence gap.

EBPR arithmetic is useful only when it explains the process state.

Engineering Boundary Notes

These exercises use simplified EBPR process screens. They do not replace plant-specific process modelling, profile sampling plans, laboratory method validation, fermenter performance testing, chemical-dose review, solids-separation assessment, sidestream management or permit compliance review. A calculated phosphorus margin applies only to the stated sampling locations, time window, filtration basis, VFA method, nitrate and DO carryover condition, selector hydraulics and wasting record.

Separate biological removal from apparent effluent compliance. Anaerobic release, aerobic uptake, PAO carbon access, solids-phase phosphorus export, chemical trim, sidestream return and effluent TSS carryover can each explain a low final phosphorus number. Release evidence should prove which mechanism actually owns the result.

Common Release Mistakes

treating low final phosphorus as EBPR recovery while ferric trim or solids capture is doing the work;
comparing orthophosphate, total phosphorus and solids-bound phosphorus without stating the analytical basis;
blaming PAOs before checking nitrate intrusion, DO carryover, VFA availability and selector short-circuiting;
using fermentate VFA production without yield, reliability, dilution and dosing-capacity evidence;
changing wasting to improve SRT while losing phosphorus export in wasted solids;
accepting profile data that are time-shifted, unfiltered, poorly located or inconsistent with mass balance.

Scenario Map

Exercise	Main Calculation	Operating Question
1	release, uptake and removal load	is the liquid profile consistent with EBPR?
2	VFA ratio and nitrate carbon demand	is carbon being consumed before PAOs can use it?
3	wasted P and sidestream return	does phosphorus actually leave the process?
4	chemical trim molar screen	how much residual P needs chemical support?
5	conservative compliance load	is a nominal result too close to the target?
6	DO carryover load	is oxygen leaking into the wrong zone?
7	rbCOD and VFA basis	is the carbon fraction being interpreted correctly?
8	uptake rate from profile samples	where does uptake slow down?
9	wasting sensitivity	does an SRT change reduce P export?
10	ferric solution feed	does the trim screen become an operational feed rate?
11	implausible profile check	is the data good enough for a biological conclusion?
12	selector HRT and release rate	does contact time support the observed release?
13	effluent TSS phosphorus carryover	is total phosphorus being limited by solids capture?
14	release efficiency per VFA consumed	is PAO selection using carbon effectively?
15	chemical trim versus biological recovery	is ferric dose masking an unrecovered EBPR failure?
16	sidestream struvite recovery and magnesium dose	can recovery reduce recycle phosphorus without hiding the mass-balance gap?
17	fermentate VFA dosing capacity	can available fermentate close the PAO carbon deficit after nitrate demand?
18	primary-sludge fermenter VFA production	can the fermenter produce enough usable VFA-COD after yield and reliability losses?

Validation Package Checklist

sample locations, flow window, filtration basis, phosphorus fraction and VFA method are documented;
anaerobic release, aerobic uptake, PAO carbon access, chemical trim and solids-phase export are separated;
nitrate intrusion, DO carryover, selector HRT, short-circuiting, recycle routing and profile timing are checked;
wasting record, sidestream return, fermenter yield, VFA dosing capacity and chemical-dose history are current;
effluent TSS, solids capture, compliance samples, uncertainty and lab-method limitations are guarded;
failed biological pattern names the action: carbon review, nitrate control, chemical trim hold, wasting change or profile repeat;
final release decision states accept, recover biology, adjust wasting, add VFA, reduce nitrate intrusion, repeat sampling or hold.

Exercise 1: Release, Uptake and Net Signal

An EBPR plant has $Q=18000\ \text{m}^3/\text{d}$ . Influent orthophosphate entering the anaerobic selector is $6.5\ \text{mg/L as P}$ . The anaerobic-zone sample is $20.0\ \text{mg/L as P}$ , and the post-aerobic sample is $0.9\ \text{mg/L as P}$ .

Calculate anaerobic release, uptake, uptake-to-release ratio and overall phosphorus removal load.

Solution

Anaerobic release:

\Delta P_{rel}=C_{ana}-C_{in}

\Delta P_{rel}=20.0-6.5=13.5\ \text{mg/L as P}

Uptake:

\Delta P_{up}=C_{ana}-C_{out}

\Delta P_{up}=20.0-0.9=19.1\ \text{mg/L as P}

Uptake-to-release ratio:

\displaystyle R_{up/rel}=\frac{19.1}{13.5}=1.41

Overall removal load:

L_P=Q(C_{in}-C_{out})(0.001)

L_P=18000(6.5-0.9)(0.001)=100.8\ \text{kg P/d}

Engineering Comment

The profile is chemically plausible for EBPR: release is followed by stronger uptake. The review is not finished until wasting, solids capture and sidestream return are checked.

Plausibility Check

Release is (13.5\ \text{mg/L as P}) and uptake is (19.1\ \text{mg/L as P}), so uptake exceeds release with a ratio of 1.41. That pattern is plausible for EBPR, but the liquid-phase removal load of (100.8\ \text{kg P/d}) still has to close against wasted solids and return-stream phosphorus.

Exercise 2: VFA Screen With Nitrate Intrusion

The same selector has $COD_{VFA}=140\ \text{mg/L}$ . A nitrate-rich recycle leak adds $Q_{intr}=2400\ \text{m}^3/\text{d}$ at $C_{NO3-N}=8.0\ \text{mg/L as N}$ . Use the Exercise 1 release value.

Calculate $R_{VFA/P}$ , nitrate load and rough COD demand for denitrification using $2.86\ \text{kg COD/kg NO}_3\text{-N}$ .

Solution

VFA to release ratio:

\displaystyle R_{VFA/P}=\frac{COD_{VFA}}{\Delta P_{rel}}

\displaystyle R_{VFA/P}=\frac{140}{13.5}=10.4\ \text{kg COD/kg P}

Nitrate intrusion load:

L_{NO3}=Q_{intr}C_{NO3-N}(0.001)

L_{NO3}=2400(8.0)(0.001)=19.2\ \text{kg N/d}

Carbon demand:

L_{COD,NO3}=2.86L_{NO3}

L_{COD,NO3}=2.86(19.2)=54.9\ \text{kg COD/d}

Engineering Comment

The VFA ratio alone may look acceptable, but nitrate intrusion consumes carbon before PAOs can use it. The operating response should check recycle routing, RAS nitrate, selector nitrate profile and whether the VFA signal is measured at the correct point.

Plausibility Check

The VFA-to-release ratio is (10.4\ \text{kg COD/kg P}), but the nitrate leak imposes (54.9\ \text{kg COD/d}) of denitrification demand. The arithmetic shows why an apparently adequate VFA concentration can still be operationally weak if oxidized nitrogen reaches the anaerobic selector first.

Exercise 3: Wasted Phosphorus and Sidestream Return

Waste sludge flow is $Q_w=260\ \text{m}^3/\text{d}$ , waste sludge solids are $X_w=7200\ \text{mg/L}$ , and phosphorus fraction in wasted solids is $f_P=0.045$ . A dewatering sidestream returns $Q_{side}=180\ \text{m}^3/\text{d}$ with $C_{P,side}=95\ \text{mg/L as P}$ .

Calculate wasted phosphorus load, sidestream phosphorus return and the net phosphorus removal after sidestream return.

Solution

Wasted phosphorus:

L_{P,WAS}=Q_wX_wf_P(0.001)

L_{P,WAS}=260(7200)(0.045)(0.001)=84.2\ \text{kg P/d}

Sidestream phosphorus return:

L_{P,side}=Q_{side}C_{P,side}(0.001)

L_{P,side}=180(95)(0.001)=17.1\ \text{kg P/d}

Net removal after this sidestream return:

L_{P,net}=84.2-17.1=67.1\ \text{kg P/d}

Engineering Comment

The liquid profile in Exercise 1 showed about $100.8\ \text{kg P/d}$ removed across the process boundary. Wasting and sidestream return account for only $67.1\ \text{kg P/d}$ net on this simplified basis. That mismatch should trigger a mass-balance review, not an immediate biological conclusion.

Plausibility Check

The net solids-side export after sidestream return is (84.2-17.1=67.1\ \text{kg P/d}). Compared with the (100.8\ \text{kg P/d}) liquid removal signal, the gap is (33.7\ \text{kg P/d}), so the exercise correctly points to boundary, sampling, solids capture or return-load review.

Exercise 4: Chemical Trim Requirement

After EBPR recovery, final effluent total phosphorus is $0.45\ \text{mg/L as P}$ at $Q=18000\ \text{m}^3/\text{d}$ . The operating target is $0.20\ \text{mg/L as P}$ . Estimate the phosphorus trim load and a ferric molar requirement using $R_{Fe/P}=1.8$ .

Solution

Phosphorus trim load:

L_{P,trim}=Q(C_{P,current}-C_{P,target})(0.001)

L_{P,trim}=18000(0.45-0.20)(0.001)=4.5\ \text{kg P/d}

Moles of phosphorus:

\displaystyle n_P=\frac{L_{P,trim}}{30.97}

\displaystyle n_P=\frac{4.5}{30.97}=0.145\ \text{kmol P/d}

Ferric molar screen:

n_{Fe}=1.8(0.145)=0.261\ \text{kmol Fe/d}

Engineering Comment

This is only a chemical screen. A final dose also depends on reagent strength, mixing, pH, alkalinity, floc capture, sludge production and whether weak EBPR biology is being masked by chemical addition.

Plausibility Check

Only the excess above target is trimmed: (0.45-0.20=0.25\ \text{mg/L as P}), or (4.5\ \text{kg P/d}). The molar conversion gives (0.145\ \text{kmol P/d}) and (0.261\ \text{kmol Fe/d}), which is a stoichiometric screen, not yet a pump setting or jar-test dose.

Exercise 5: Conservative Compliance Gate

A plant reports $C_{P}=0.20\ \text{mg/L as P}$ at $Q=18000\ \text{m}^3/\text{d}$ . Use a relative flow allowance $u_Q=0.05$ and concentration allowance $U_C=0.03\ \text{mg/L as P}$ . Calculate the conservative phosphorus load.

Solution

Conservative flow:

Q_{cons}=Q(1+u_Q)

Q_{cons}=18000(1.05)=18900\ \text{m}^3/\text{d}

Conservative concentration:

C_{P,cons}=0.20+0.03=0.23\ \text{mg/L as P}

Conservative load:

L_{P,cons}=Q_{cons}C_{P,cons}(0.001)

L_{P,cons}=18900(0.23)(0.001)=4.35\ \text{kg P/d}

Engineering Comment

The nominal concentration may look exactly on target, but the conservative load is the better release basis when the decision is close. The plant should verify lab method, flow meter accuracy, solids carryover and sidestream timing before declaring stable compliance.

Plausibility Check

The nominal load at (0.20\ \text{mg/L}) is (3.60\ \text{kg P/d}), while the conservative load is (4.35\ \text{kg P/d}). That is about 21 percent higher, so a concentration exactly on target should not be treated as a robust compliance result without measurement and flow uncertainty review.

Exercise 6: DO Carryover Into the Anaerobic Selector

An internal recycle leak sends $Q_{rec}=3200\ \text{m}^3/\text{d}$ with $DO=1.4\ \text{mg/L}$ into the anaerobic selector. Main influent flow is $Q=18000\ \text{m}^3/\text{d}$ , and the measured VFA concentration is $140\ \text{mg/L as COD}$ .

Calculate the oxygen carryover load, the mixed DO concentration screen and the oxygen load as a fraction of the daily VFA-COD load.

Solution

Oxygen carryover load:

L_{O2}=Q_{rec}DO(0.001)

L_{O2}=3200(1.4)(0.001)=4.48\ \text{kg O}_2/\text{d}

Mixed concentration screen:

\displaystyle DO_{mix}=\frac{Q_{rec}DO}{Q+Q_{rec}}

\displaystyle DO_{mix}=\frac{3200(1.4)}{18000+3200}=0.211\ \text{mg/L}

Daily VFA-COD load:

L_{VFA}=Q(COD_{VFA})(0.001)

L_{VFA}=18000(140)(0.001)=2520\ \text{kg COD/d}

Oxygen load fraction:

\displaystyle f_{O2/VFA}=\frac{4.48}{2520}=0.00178

or about $0.18\%$ of the daily VFA-COD load.

Engineering Comment

The daily mass fraction is small, but the mixed DO screen is not automatically harmless. EBPR selection depends on local anaerobic conditions, so the review should check where the oxygenated flow enters, whether the signal is continuous, and whether ORP, nitrate and phosphate release respond at the same time.

Plausibility Check

The oxygen carryover load is only (4.48\ \text{kg O}_2/\text{d}), or 0.18 percent of the daily VFA-COD load, but the mixed concentration screen is (0.211\ \text{mg/L}). That combination is plausible: a small daily mass can still disturb a local selector if it enters continuously at the wrong hydraulic point.

Exercise 7: rbCOD and VFA Basis Check

A plant removes $5.6\ \text{mg/L as P}$ across the EBPR boundary. Influent readily biodegradable COD is $210\ \text{mg/L}$ , but measured VFA-COD is only $72\ \text{mg/L}$ . Flow is $18000\ \text{m}^3/\text{d}$ .

Calculate the removed phosphorus load, rbCOD load, VFA-COD load, $R_{rbCOD/P}$ , $R_{VFA/P}$ and the VFA fraction of rbCOD.

Solution

Removed phosphorus load:

L_P=18000(5.6)(0.001)=100.8\ \text{kg P/d}

rbCOD load:

L_{rbCOD}=18000(210)(0.001)=3780\ \text{kg COD/d}

VFA-COD load:

L_{VFA}=18000(72)(0.001)=1296\ \text{kg COD/d}

rbCOD to phosphorus ratio:

\displaystyle R_{rbCOD/P}=\frac{3780}{100.8}=37.5\ \text{kg COD/kg P}

VFA to phosphorus ratio:

\displaystyle R_{VFA/P}=\frac{1296}{100.8}=12.9\ \text{kg COD/kg P}

VFA fraction of rbCOD:

\displaystyle f_{VFA/rbCOD}=\frac{1296}{3780}=0.343

or $34.3\%$ .

Engineering Comment

The rbCOD screen looks generous, but only about one third of that fraction is measured as VFA-COD. A strong answer should not treat total rbCOD, VFA and total COD as interchangeable. The next evidence is fermentation performance, primary sludge handling, selector profiles and whether PAOs or GAOs are being favoured.

Plausibility Check

The rbCOD-to-P ratio is (37.5), while the VFA-to-P ratio is (12.9), because measured VFA is only 34.3 percent of rbCOD. The result confirms that “available carbon” depends on the fraction used for the EBPR decision, not simply on a large rbCOD number.

Exercise 8: Aerobic Uptake Rate From Profile Samples

An EBPR profile has $C_{ana}=20.0\ \text{mg/L as P}$ at the anaerobic exit. After $55\ \text{min}$ in the first aerobic section, orthophosphate is $5.8\ \text{mg/L as P}$ . The final aerobic sample after $160\ \text{min}$ is $0.9\ \text{mg/L as P}$ .

Calculate early uptake, late uptake, early uptake rate, late uptake rate and the early-to-late rate ratio.

Solution

Early uptake:

\Delta P_{early}=20.0-5.8=14.2\ \text{mg/L as P}

Early uptake rate:

\displaystyle r_{early}=\frac{14.2}{55}=0.258\ \text{mg/L/min}

Late uptake:

\Delta P_{late}=5.8-0.9=4.9\ \text{mg/L as P}

Late contact time:

t_{late}=160-55=105\ \text{min}

Late uptake rate:

\displaystyle r_{late}=\frac{4.9}{105}=0.0467\ \text{mg/L/min}

Rate ratio:

\displaystyle R_{rate}=\frac{0.258}{0.0467}=5.52

Engineering Comment

Most uptake occurs early in this simplified profile. That can be healthy, but the slowing rate also tells the operator where additional aeration time may have limited value. If final phosphorus is still high, check PAO storage, pH, temperature, chemical trim, solids carryover and sample timing before simply increasing aerobic volume.

Plausibility Check

Early uptake is (14.2\ \text{mg/L}) in 55 min, while late uptake is (4.9\ \text{mg/L}) over 105 min. The early rate is 5.52 times the late rate, so the profile suggests rapid initial uptake followed by a slower tail rather than a uniform benefit from simply adding more aerobic contact time.

Exercise 9: Wasting Change and Phosphorus Export

The plant in Exercise 3 considers reducing wasting from $260$ to $190\ \text{m}^3/\text{d}$ to increase SRT. Waste sludge solids remain $7200\ \text{mg/L}$ and the phosphorus fraction in wasted solids remains $f_P=0.045$ . Sidestream phosphorus return remains $17.1\ \text{kg P/d}$ .

Calculate the current wasted phosphorus, proposed wasted phosphorus, reduction in phosphorus export and proposed net phosphorus removal after sidestream return.

Solution

Current wasted phosphorus:

L_{P,WAS,current}=260(7200)(0.045)(0.001)=84.2\ \text{kg P/d}

Proposed wasted phosphorus:

L_{P,WAS,new}=190(7200)(0.045)(0.001)=61.6\ \text{kg P/d}

Reduction in phosphorus export:

\Delta L_{P,WAS}=84.2-61.6=22.6\ \text{kg P/d}

Proposed net after sidestream return:

L_{P,net,new}=61.6-17.1=44.5\ \text{kg P/d}

Relative reduction compared with the original net export:

\displaystyle \frac{67.1-44.5}{67.1}=0.337

or $33.7\%$ .

Engineering Comment

Increasing SRT may protect nitrification, but this change sharply reduces phosphorus export in wasted biomass. EBPR control is coupled: a wasting change should be reviewed with final total phosphorus, sludge blanket depth, sidestream release, nitrification margin and chemical trim demand.

Plausibility Check

Reducing wasting from 260 to (190\ \text{m}^3/\text{d}) lowers phosphorus export by (22.6\ \text{kg P/d}). After sidestream return, net export falls from 67.1 to (44.5\ \text{kg P/d}), a 33.7 percent reduction, so the SRT change can plausibly weaken EBPR even if nitrification benefits.

Exercise 10: Ferric Trim Feed Estimate

Use the ferric molar requirement from Exercise 4, $n_{Fe}=0.261\ \text{kmol Fe/d}$ . Assume ferric chloride is dosed as $FeCl_3$ , molecular mass $M_{FeCl3}=162.2\ \text{kg/kmol}$ , solution strength $w=0.40$ by mass and density $\rho=1.42\ \text{kg/L}$ .

Estimate pure ferric chloride mass, solution mass, solution volume and average feed rate in $\text{L/h}$ .

Solution

Pure ferric chloride mass:

m_{FeCl3}=n_{Fe}M_{FeCl3}

m_{FeCl3}=0.261(162.2)=42.3\ \text{kg/d}

Solution mass:

\displaystyle m_{solution}=\frac{42.3}{0.40}=105.8\ \text{kg/d}

Solution volume:

\displaystyle V_{solution}=\frac{105.8}{1.42}=74.5\ \text{L/d}

Average feed rate:

\displaystyle q_{feed}=\frac{74.5}{24}=3.10\ \text{L/h}

Engineering Comment

The conversion from stoichiometric screen to pump setting exposes practical checks: minimum turndown, calibration cylinder results, reagent strength certificate, dose pacing, mixing energy, alkalinity impact, sludge production and whether ferric is masking an EBPR selector problem.

Plausibility Check

The ferric chloride requirement is (42.3\ \text{kg/d}) as pure (FeCl_3), but the 40 percent solution requires (105.8\ \text{kg/d}) of liquid reagent. Dividing by density and 24 hours gives (3.10\ \text{L/h}), so a real release check must confirm that the feed pump can meter accurately at this low rate.

Exercise 11: Implausible Profile Diagnosis

A one-day profile reports influent orthophosphate $C_{in}=6.5\ \text{mg/L as P}$ , anaerobic sample $C_{ana}=14.0\ \text{mg/L as P}$ and a sample labelled “post-aerobic” at $C_{post}=16.5\ \text{mg/L as P}$ . The final effluent total phosphorus sample is $0.8\ \text{mg/L as P}$ at $Q=18000\ \text{m}^3/\text{d}$ .

Calculate anaerobic release, labelled post-aerobic uptake, uptake-to-release ratio and final removal load. Decide whether the profile should be accepted.

Solution

Anaerobic release:

\Delta P_{rel}=14.0-6.5=7.5\ \text{mg/L as P}

Labelled post-aerobic uptake:

\Delta P_{up,labelled}=14.0-16.5=-2.5\ \text{mg/L as P}

Uptake-to-release ratio:

\displaystyle R_{up/rel}=\frac{-2.5}{7.5}=-0.33

Final removal load:

L_P=18000(6.5-0.8)(0.001)=102.6\ \text{kg P/d}

Engineering Comment

The final effluent value suggests removal, but the labelled profile says phosphorus increased after the anaerobic zone. That contradiction is a data-quality finding, not proof of a biological mechanism. Check sample labels, hydraulic lag, grab-sample timing, filtered versus total basis, recycle contamination and lab QA/QC before changing wasting or chemical dose.

Plausibility Check

The labelled post-aerobic “uptake” is negative, (-2.5\ \text{mg/L as P}), giving an uptake-to-release ratio of (-0.33), while final effluent still implies (102.6\ \text{kg P/d}) removal. Those signals cannot all describe the same profile boundary, so the correct conclusion is data-quality hold, not biological diagnosis.

Exercise 12: Anaerobic Selector HRT and Release Rate

An anaerobic selector has volume $V_{sel}=2200\ \text{m}^3$ . Influent flow is $Q=18000\ \text{m}^3/\text{d}$ , and return activated sludge entering the selector is $Q_{RAS}=4500\ \text{m}^3/\text{d}$ .

Across the selector, orthophosphate increases from $6.5$ to $20.0\ \text{mg/L as P}$ , and VFA as COD decreases from $140$ to $78\ \text{mg/L}$ .

Estimate selector hydraulic residence time, phosphorus release rate and VFA uptake rate.

Solution

Selector flow:

Q_{sel}=Q+Q_{RAS}

Q_{sel}=18000+4500=22500\ \text{m}^3/\text{d}

Hydraulic residence time:

\displaystyle \theta=\frac{V_{sel}}{Q_{sel}}

\displaystyle \theta=\frac{2200}{22500}=0.0978\ \text{d}

Convert to hours and minutes:

\theta=0.0978(24)=2.35\ \text{h}=141\ \text{min}

Phosphorus release:

\Delta P_{rel}=20.0-6.5=13.5\ \text{mg/L as P}

Release rate:

\displaystyle r_{P,rel}=\frac{13.5}{141}=0.0957\ \text{mg/L/min as P}

VFA decrease:

\Delta COD_{VFA}=140-78=62\ \text{mg/L as COD}

VFA uptake rate:

\displaystyle r_{VFA}=\frac{62}{141}=0.440\ \text{mg/L/min as COD}

Engineering Comment

The selector provides a little over two hours of nominal contact time and shows both phosphate release and VFA consumption. That is a coherent EBPR selection signal, but the conclusion still depends on short-circuiting, RAS nitrate, dissolved oxygen, mixing, sample timing and whether the measured VFA decrease is biological uptake rather than dilution or sampling bias.

Plausibility Check

A $2200\ \text{m}^3$ selector receiving $22500\ \text{m}^3/\text{d}$ should have residence time near one tenth of a day, or a little over two hours. Dividing a $13.5\ \text{mg/L}$ release by 141 minutes gives a release rate near $0.1\ \text{mg/L/min}$ , which matches the calculation.

Exercise 13: Effluent Suspended-Solids Phosphorus Carryover

An EBPR plant reports final effluent soluble orthophosphate of $0.09\ \text{mg/L as P}$ . Effluent total suspended solids are $8.0\ \text{mg/L}$ , and the phosphorus fraction in escaped biological solids is estimated as $f_P=0.025\ \text{kg P/kg TSS}$ .

Flow is $Q=18000\ \text{m}^3/\text{d}$ , and the total phosphorus target is $0.20\ \text{mg/L as P}$ .

Estimate the particulate phosphorus concentration, estimated total phosphorus concentration and total phosphorus load. Decide whether the result is likely compliant.

Solution

Particulate phosphorus from escaped solids:

C_{P,TSS}=TSSf_P

C_{P,TSS}=8.0(0.025)=0.20\ \text{mg/L as P}

Estimated total phosphorus:

C_{TP}=C_{ortho}+C_{P,TSS}

C_{TP}=0.09+0.20=0.29\ \text{mg/L as P}

Total phosphorus load:

L_{TP}=QC_{TP}(0.001)

L_{TP}=18000(0.29)(0.001)=5.22\ \text{kg P/d}

Target load at $0.20\ \text{mg/L as P}$ :

L_{target}=18000(0.20)(0.001)=3.60\ \text{kg P/d}

Load ratio:

\displaystyle R_L=\frac{5.22}{3.60}=1.45

The estimated total phosphorus is above the target, even though soluble orthophosphate alone is low.

Engineering Comment

Good EBPR biology can still fail a total phosphorus permit if solids capture is weak. The operating review should check final clarifier blanket depth, effluent TSS trend, filtration performance if present, sludge settleability, hydraulic peaks, coagulant trim, sample fractionation and whether reported phosphorus is filtered orthophosphate or unfiltered total phosphorus.

Plausibility Check

Eight milligrams per litre of escaped solids with 2.5 percent phosphorus contributes $0.20\ \text{mg/L as P}$ by itself. Adding $0.09\ \text{mg/L}$ soluble orthophosphate gives $0.29\ \text{mg/L}$ total phosphorus, so a target of $0.20\ \text{mg/L}$ is not plausibly met unless the solids estimate is wrong or solids capture improves.

Exercise 14: Phosphorus Release Efficiency per VFA Consumed

An EBPR profile shows VFA-COD entering the anaerobic selector at:

C_{VFA,in}=150\ \text{mg/L as COD}

and leaving at:

C_{VFA,out}=84\ \text{mg/L as COD}

Orthophosphate rises from:

C_{P,in}=6.0\ \text{mg/L as P}

to:

C_{P,ana}=18.0\ \text{mg/L as P}

The site expects at least:

0.16\ \text{mg P released/mg COD consumed}

for a healthy selector profile under this operating mode. Calculate VFA consumed, phosphorus released, release efficiency, VFA consumed per phosphorus released and the profile decision.

Solution

VFA consumed:

\Delta COD_{VFA}=C_{VFA,in}-C_{VFA,out}

\Delta COD_{VFA}=150-84=66\ \text{mg/L as COD}

Phosphorus released:

\Delta P_{rel}=C_{P,ana}-C_{P,in}

\Delta P_{rel}=18.0-6.0=12.0\ \text{mg/L as P}

Release efficiency:

\displaystyle \eta_{rel}=\frac{\Delta P_{rel}}{\Delta COD_{VFA}}

\displaystyle \eta_{rel}=\frac{12.0}{66}=0.182\ \text{mg P/mg COD}

VFA consumed per phosphorus released:

\displaystyle R_{COD/P}=\frac{66}{12.0}=5.5\ \text{mg COD/mg P}

The profile passes the local release-efficiency screen because:

0.182>0.16

Engineering Comment

This check asks whether consumed VFA is producing a credible phosphate-release response, not merely whether VFA exists. A weak efficiency can point to nitrate or DO intrusion, GAO competition, sample timing error, dilution, poor fermentation or PAO stress. A passing value still needs downstream uptake and final total phosphorus evidence.

Plausibility Check

VFA falls by (66\ \text{mg/L}) while phosphorus rises by (12\ \text{mg/L}). A release efficiency of (0.182\ \text{mg P/mg COD}) means each (5.5\ \text{mg COD}) consumed is associated with about (1\ \text{mg P}) released. That is above the stated local screen but should not be used alone as a compliance conclusion.

Exercise 15: Chemical Trim Masking Biological Recovery

After a nitrate-intrusion event, final total phosphorus is controlled with ferric trim. Before biological recovery, the plant has:

C_{TP,bio}=0.92\ \text{mg/L as P}

at:

Q=18000\ \text{m}^3/\text{d}

The compliance target is:

C_{TP,target}=0.30\ \text{mg/L as P}

Ferric trim removes an estimated:

L_{P,chem}=8.5\ \text{kg P/d}

After selector recovery work, the biological effluent concentration before ferric trim improves to:

C_{TP,bio,recovered}=0.58\ \text{mg/L as P}

Calculate the total phosphorus load before recovery, the phosphorus load that must be removed to reach the target before recovery, the final concentration after the current ferric trim, the recovered biological load, the remaining trim load after recovery and the biological recovery fraction. Decide whether the ferric trim is masking an unrecovered EBPR problem.

Solution

Biological effluent load before recovery:

L_{P,bio}=QC_{TP,bio}(0.001)

L_{P,bio}=18000(0.92)(0.001)=16.56\ \text{kg P/d}

Target load:

L_{P,target}=QC_{TP,target}(0.001)

L_{P,target}=18000(0.30)(0.001)=5.40\ \text{kg P/d}

Required trim before recovery:

L_{P,trim,required}=16.56-5.40=11.16\ \text{kg P/d}

Final load after current ferric trim:

L_{P,final}=16.56-8.5=8.06\ \text{kg P/d}

Final concentration after current ferric trim:

\displaystyle C_{P,final}=\frac{L_{P,final}}{Q(0.001)}

\displaystyle C_{P,final}=\frac{8.06}{18}=0.448\ \text{mg/L as P}

This still exceeds the target:

0.448>0.30

Recovered biological load:

L_{P,bio,recovered}=18000(0.58)(0.001)=10.44\ \text{kg P/d}

Remaining trim load after recovery:

L_{P,trim,recovered}=10.44-5.40=5.04\ \text{kg P/d}

Biological load reduction from recovery:

\Delta L_{bio}=16.56-10.44=6.12\ \text{kg P/d}

Recovery fraction relative to the original required trim:

\displaystyle f_{recovery}=\frac{6.12}{11.16}=0.548=54.8\%

The biological recovery is meaningful but incomplete. Ferric trim alone was not enough before recovery, and the recovered process still needs about (5.04\ \text{kg P/d}) of trim to reach the target. The plant should not declare EBPR fully recovered only because chemical dosing reduces the final number.

Engineering Comment

Chemical trim can be useful, but it can hide whether the biological selector has actually recovered. The correct review separates biological effluent load before chemical addition, chemical removal, final compliance load and recovery evidence. Otherwise operators may increase ferric dose while nitrate intrusion, poor VFA generation or weak PAO selection remains unresolved.

Plausibility Check

At (18000\ \text{m}^3/\text{d}), each (0.10\ \text{mg/L as P}) is (1.8\ \text{kg P/d}). Dropping biological TP from (0.92) to (0.58\ \text{mg/L}) therefore saves (6.12\ \text{kg P/d}), which matches the calculation. The current ferric trim leaves (0.448\ \text{mg/L}), so it cannot be mistaken for a compliant or fully recovered EBPR state.

Exercise 16: Sidestream Struvite Recovery and Magnesium Dose

A dewatering sidestream returns:

Q_{side}=180\ \text{m}^3/\text{d}

with soluble phosphorus:

C_{P,side}=95\ \text{mg/L as P}

A struvite recovery unit is proposed to reduce the residual soluble phosphorus to:

C_{P,res}=25\ \text{mg/L as P}

Use a magnesium dosing target:

R_{Mg/P}=1.2\ \text{mol Mg/mol P removed}

The reagent is magnesium chloride hexahydrate:

MgCl_2\cdot6H_2O

with molecular mass:

M_{MgCl2}=203.3\ \text{kg/kmol}

The delivered solution is:

w=0.32

by mass, with density:

\rho=1.30\ \text{kg/L}

Use the wasted phosphorus from Exercise 3:

L_{P,WAS}=84.2\ \text{kg P/d}

Compare original sidestream return, recovered sidestream return, magnesium-solution feed rate and the revised net phosphorus export.

Solution

Original sidestream phosphorus return:

L_{P,side}=Q_{side}C_{P,side}(0.001)

L_{P,side}=180(95)(0.001)=17.1\ \text{kg P/d}

Recovered residual sidestream return:

L_{P,res}=180(25)(0.001)=4.5\ \text{kg P/d}

Phosphorus removed in the sidestream:

L_{P,rem}=17.1-4.5=12.6\ \text{kg P/d}

Moles of phosphorus removed:

\displaystyle n_P=\frac{L_{P,rem}}{30.97}

\displaystyle n_P=\frac{12.6}{30.97}=0.407\ \text{kmol P/d}

Magnesium required:

n_{Mg}=R_{Mg/P}n_P

n_{Mg}=1.2(0.407)=0.488\ \text{kmol Mg/d}

Magnesium chloride hexahydrate mass:

m_{salt}=n_{Mg}M_{MgCl2}

m_{salt}=0.488(203.3)=99.2\ \text{kg/d}

Solution mass:

\displaystyle m_{soln}=\frac{m_{salt}}{w}

\displaystyle m_{soln}=\frac{99.2}{0.32}=310\ \text{kg/d}

Solution volume:

\displaystyle V_{soln}=\frac{m_{soln}}{\rho}

\displaystyle V_{soln}=\frac{310}{1.30}=238\ \text{L/d}

Average feed rate:

\displaystyle q_{feed}=\frac{238}{24}=9.9\ \text{L/h}

Net phosphorus export before recovery:

L_{P,net,old}=84.2-17.1=67.1\ \text{kg P/d}

Net phosphorus export after recovery:

L_{P,net,new}=84.2-4.5=79.7\ \text{kg P/d}

Net export improvement:

\Delta L_{P,net}=79.7-67.1=12.6\ \text{kg P/d}

The struvite unit reduces sidestream recycle phosphorus and improves net export by the removed sidestream load, but it does not close the full mass-balance gap from Exercise 3.

Engineering Comment

Sidestream recovery can support EBPR by reducing return phosphorus, but it is not proof that the biological selector is healthy. The release decision should check pH, ammonia availability, magnesium residual, scaling risk, crystal separation, recycle timing, reagent feed calibration and whether the main EBPR profile still shows anaerobic release followed by aerobic uptake.

Plausibility Check

The sidestream concentration drop is (95-25=70\ \text{mg/L as P}). At (180\ \text{m}^3/\text{d}), that is (12.6\ \text{kg P/d}), so the same number should appear as both sidestream recovery and net export improvement. A feed rate near (10\ \text{L/h}) is credible for a reagent dose near (238\ \text{L/d}).

Exercise 17: Fermentate VFA Dosing and PAO Carbon Deficit

An EBPR plant is evaluating primary-sludge fermentate addition to recover anaerobic phosphorus release. Average influent flow is:

Q=22000\ \text{m}^3/\text{d}

The target anaerobic phosphorus release is:

\Delta P_{rel,target}=9.0\ \text{mg/L as P}

The operating target for available VFA is:

R_{VFA/P,target}=7.0\ \text{kg COD/kg P}

Measured influent VFA is:

COD_{VFA,in}=45\ \text{mg/L}

A nitrate leak into the selector creates a denitrification demand based on:

L_{NO3-N}=35\ \text{kg N/d}

Use:

2.86\ \text{kg COD/kg NO}_3\text{-N}

for the nitrate carbon demand. Available fermentate has:

COD_{VFA,ferm}=4200\ \text{mg/L}

and the practical feed limit is:

Q_{ferm,max}=70\ \text{m}^3/\text{d}

Calculate target VFA load, existing VFA available after nitrate demand, VFA deficit, required fermentate flow, maximum fermentate contribution, actual VFA-to-P ratio at maximum feed and the release decision.

Solution

Target release load:

L_{P,rel,target}=Q\Delta P_{rel,target}(0.001)

L_{P,rel,target}=22000(9.0)(0.001)=198\ \text{kg P/d}

Target VFA-COD load:

L_{VFA,target}=R_{VFA/P,target}L_{P,rel,target}

L_{VFA,target}=7.0(198)=1386\ \text{kg COD/d}

Measured influent VFA-COD load:

L_{VFA,in}=QCOD_{VFA,in}(0.001)

L_{VFA,in}=22000(45)(0.001)=990\ \text{kg COD/d}

Nitrate carbon demand:

L_{COD,NO3}=2.86L_{NO3-N}

L_{COD,NO3}=2.86(35)=100.1\ \text{kg COD/d}

Existing VFA available after nitrate demand:

L_{VFA,net}=990-100.1=889.9\ \text{kg COD/d}

VFA deficit before fermentate:

\Delta L_{VFA}=1386-889.9=496.1\ \text{kg COD/d}

Fermentate VFA-COD concentration on a mass basis is:

4200\ \text{mg/L}=4.2\ \text{kg/m}^3

Required fermentate flow:

\displaystyle Q_{ferm,req}=\frac{496.1}{4.2}=118.1\ \text{m}^3/\text{d}

Maximum fermentate VFA contribution:

L_{VFA,ferm,max}=70(4.2)=294\ \text{kg COD/d}

Residual VFA deficit at maximum feed:

\Delta L_{VFA,res}=496.1-294=202.1\ \text{kg COD/d}

Actual VFA-to-P ratio at maximum feed:

\displaystyle R_{VFA/P,actual}=\frac{889.9+294}{198}=5.98\ \text{kg COD/kg P}

The available fermentate improves the selector carbon balance but does not meet the (7.0\ \text{kg COD/kg P}) target. The recovery claim should fail unless nitrate intrusion is reduced, fermentate capacity or concentration is increased, supplemental carbon is justified, or the plant demonstrates a lower site-specific VFA requirement with profile evidence.

Engineering Comment

Fermentate dosing is not only a concentration adjustment. The decision depends on fermentate strength, pump capacity, storage, odor control, pH, alkalinity, sidestream nitrogen, solids carryover, dose timing and whether nitrate consumes the carbon before PAOs can use it. A release record should separate measured influent VFA, nitrate COD demand, fermentate contribution and the observed anaerobic release response.

Plausibility Check

The target release load is (198\ \text{kg P/d}), so a (7.0\ \text{kg COD/kg P}) screen requires nearly (1400\ \text{kg COD/d}). The plant already has about (990\ \text{kg COD/d}), but nitrate removes roughly (100\ \text{kg COD/d}), leaving a deficit near (500\ \text{kg COD/d}). At (4.2\ \text{kg COD/m}^3), that needs about (118\ \text{m}^3/\text{d}) of fermentate, so a (70\ \text{m}^3/\text{d}) pump limit cannot close the gap.

Exercise 18: Primary-Sludge Fermenter VFA Production Gate

An EBPR plant has already accounted for influent VFA and nitrate demand. The remaining additional VFA-COD needed at the anaerobic selector is:

\Delta L_{VFA,req}=430\ \text{kg COD/d}

A primary-sludge fermenter receives:

Q_{PS}=95\ \text{m}^3/\text{d}

with volatile-solids concentration:

X_{VS}=26\ \text{kg VS/m}^3

Bench testing supports a VFA yield:

Y_{VFA}=0.17\ \text{kg VFA-COD/kg VS fed}

Soluble VFA recovery to usable fermentate is:

\eta_s=0.82

and the operating availability factor for storage, transfer, downtime and off-spec batches is:

\eta_a=0.90

Calculate volatile-solids feed load, raw VFA-COD production, delivered usable VFA-COD, residual deficit, and the minimum VFA yield needed at the existing sludge feed. Then check a revised operating case with:

Q_{PS,rev}=105\ \text{m}^3/\text{d}

and:

Y_{VFA,rev}=0.24\ \text{kg VFA-COD/kg VS fed}

using the same recovery and availability factors.

Solution

Volatile-solids feed load is:

L_{VS}=Q_{PS}X_{VS}

L_{VS}=95(26)=2470\ \text{kg VS/d}

Raw VFA-COD production is:

L_{VFA,raw}=Y_{VFA}L_{VS}

L_{VFA,raw}=0.17(2470)=420\ \text{kg COD/d}

Delivered usable VFA-COD is:

L_{VFA,use}=L_{VFA,raw}\eta_s\eta_a

L_{VFA,use}=420(0.82)(0.90)=310\ \text{kg COD/d}

Residual VFA deficit is:

\Delta L_{VFA,res}=430-310=120\ \text{kg COD/d}

The current fermenter production basis does not close the EBPR carbon gap.

Minimum yield at the existing sludge feed is:

\displaystyle Y_{VFA,min}=\frac{\Delta L_{VFA,req}}{L_{VS}\eta_s\eta_a}

\displaystyle Y_{VFA,min}=\frac{430}{2470(0.82)(0.90)}=0.236\ \text{kg VFA-COD/kg VS fed}

The current yield is below that screen:

0.17<0.236

For the revised case:

L_{VS,rev}=105(26)=2730\ \text{kg VS/d}

Raw revised VFA-COD production is:

L_{VFA,raw,rev}=0.24(2730)=655\ \text{kg COD/d}

Delivered revised VFA-COD is:

L_{VFA,use,rev}=655(0.82)(0.90)=484\ \text{kg COD/d}

Revised margin is:

M_{VFA,rev}=484-430=54\ \text{kg COD/d}

or:

\displaystyle \frac{54}{430}=12.6\%

The revised fermenter case passes this VFA-production gate with modest margin. The release should still require profile confirmation that anaerobic phosphorus release improves after the fermentate reaches the selector.

Engineering Comment

Fermentate availability is not proven by pump capacity or a single concentration sample. The upstream fermenter must receive enough volatile solids, convert them at a credible yield, preserve soluble VFA through separation and storage, and deliver usable carbon during the same operating window as EBPR demand. A production pass can still fail if nitrate intrusion consumes the carbon, fermentate strength swings, solids carryover fouls dosing equipment or odor and pH constraints limit runtime.

Plausibility Check

The base feed supplies about (2500\ \text{kg VS/d}). At a (0.17) yield, raw VFA production should be roughly (400\ \text{kg COD/d}), and recovery plus availability losses reduce that to about (300\ \text{kg COD/d}). That is clearly short of a (430\ \text{kg COD/d}) requirement. Raising both sludge feed and yield increases delivered VFA to about (484\ \text{kg COD/d}), so the revised pass is plausible but not generous.

Review Checklist

Before accepting an EBPR exercise answer, check that it states:

phosphorus fraction and sample basis;
flow window and sample location;
release, uptake and final removal separately;
selector contact time, recycle flow contribution and release-rate plausibility;
VFA consumed, phosphorus released and release efficiency as separate checks;
carbon availability and nitrate/DO interference;
fermentate strength, feed limit and nitrate COD demand before crediting VFA dosing;
fermenter sludge feed, VFA yield, recovery and availability before crediting fermentate production;
rbCOD, VFA and total COD basis;
wasting and sidestream return boundary;
sidestream recovery stoichiometry, reagent strength and residual recycle load;
whether chemical trim is supporting or masking biology;
whether biological recovery is separated from chemical phosphorus removal;
whether effluent total phosphorus includes suspended-solids phosphorus carryover;
profile timing and data plausibility;
validation evidence still needed.

Common Mistakes

Calling anaerobic phosphorus release a removal result instead of checking downstream uptake and final total phosphorus.
Ignoring nitrate or DO intrusion in the anaerobic selector before blaming PAO activity or VFA supply.
Using total COD as if it were immediately available VFA, or treating all rbCOD as the same biological substrate.
Crediting fermentate dosing from concentration alone without checking pump capacity, nitrate demand and actual anaerobic release response.
Crediting fermentate production without checking volatile-solids feed, yield, soluble recovery, storage losses and off-spec downtime.
Comparing filtered orthophosphate to total phosphorus compliance without accounting for suspended-solids phosphorus.
Calculating chemical trim before checking whether solids capture, wasting, sidestream return or biological recovery is the controlling problem.
Treating struvite recovery as pure phosphorus removal without checking magnesium dose, pH, ammonia, crystal separation and residual recycle load.
Reducing wasting to protect biomass without checking whether phosphorus export from the system is falling.
Accepting a low soluble phosphorus value as proof of EBPR recovery when ferric trim may be masking the biological profile.
Trusting an impossible profile order, a single final effluent number or a stale sample window without matching profile evidence and operating history.

A strong solution links every number to a process decision.

REF