Exercise set

Membrane Filtration and Fouling Control Exercises

Solved membrane filtration exercises for flux, TMP, permeability, recovery, scaling, SDI, cleaning dose, integrity and release gates.

Branch: Environmental Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.2.7 · reviewed

These exercises practise membrane filtration calculations used in water and wastewater operation. They connect flux, sustainable-flux evidence, active membrane area, TMP, permeability, recovery, concentration-factor scaling risk, backwash losses, cleaning response, integrity testing, feed-quality filterability, feed-quality changes and energy to engineering decisions. Assume simplified values unless an exercise states otherwise.

Release Evidence Notes

Use the worked answers as operating evidence only when the calculation basis matches the train configuration, membrane age, active module count, pressure-tap location, temperature-normalization method, feed-water quality, cleaning state and integrity-test procedure. A hydraulic pass is not enough if barrier evidence, cleaning recovery or feed-quality trend data contradicts release.

The most important release gates are sustainable flux, maximum TMP, normalized permeability recovery, recovery limit against concentrate-side chemistry, net production after backwash and cleaning, chemical recipe validity, integrity-test margin and operating response after a feed-quality shock. Each gate should state the measured evidence source, the time window used, the uncertainty allowance and the action required when the result falls between normal operation and a hard failure.

How to Use These Exercises

Use the membrane filtration formula sheet for equations and unit conventions. State whether flow is hourly or daily, whether pressure is gauge pressure, and whether permeability is temperature-normalized. A correct numerical answer is weak if it ignores operating mode or cleaning state.

Engineering Boundary Notes

These exercises use simplified membrane-filtration and fouling-control screens. They do not replace membrane supplier limits, pilot testing, integrity-test procedures, feed-water characterization, scaling review, cleaning validation, concentrate management, chemical safety review or compliance assessment. A calculated filtration margin applies only to the stated train configuration, membrane age, active module count, feed quality, pressure taps, temperature correction, recovery target and cleaning state.

Separate hydraulic capacity from barrier and fouling release. Flux, TMP, permeability, recovery, concentration factor, scaling risk, backwash loss, cleaning response, integrity-test margin, crossflow velocity and SDI filterability can each control a different release decision.

Common Release Mistakes

crediting gross production while backwash, relaxation, cleaning downtime or unavailable modules reduce delivered flow;
comparing permeability trends without temperature normalization and pressure-tap consistency;
increasing recovery before concentrate chemistry, scaling threshold and reject management are checked;
releasing from hydraulic capacity while integrity-test margin or permeate evidence fails;
using a cleaning dose without stock concentration, contact time, residual routing and recipe verification;
scaling pilot flux to full scale without step-test fouling rate, derating and feed-quality evidence.

Scenario Map

Scenario	Main calculation	Engineering decision
Capacity basis	flux, sustainable flux, active area and module availability	Decide whether installed area can sustain the required flow.
Fouling trend	TMP and normalized permeability	Decide whether pressure rise is a flux effect, temperature effect or fouling signal.
Production accounting	recovery, concentration factor, scaling limit and backwash loss	Distinguish advertised gross production from delivered capacity and concentrate-side risk.
Cleaning response	CEB/CIP recovery and cleaning dose	Decide whether reversible fouling, chemical recovery or recipe shortfall controls release.
Barrier release	integrity-test margin and permeate evidence	Decide whether hydraulic capacity is enough to release the train.
Scale-up release	step-flux TMP rise rate and derated design flux	Convert pilot fouling evidence into full-scale capacity and area requirements.
Crossflow hydraulics	channel velocity and Reynolds number	Decide whether recirculation reduction preserves concentrate-side fouling control.
Pretreatment filterability	SDI, plugging factor and guard-banded limit	Decide whether feed quality supports release before higher flux or recovery is credited.

Validation Package Checklist

train configuration, active module count, membrane age, feed quality, pressure taps and cleaning state are documented;
hourly or daily flow basis, gauge pressure, temperature-normalized permeability and recovery target are explicit;
gross flux, sustainable flux, net delivered capacity, backwash loss and unavailable modules are separated;
TMP trend, normalized permeability, SDI, step-flux evidence, crossflow velocity and scaling limit are guarded;
integrity-test margin, permeate evidence, chemical recipe, contact time, residual routing and waste handling are recorded;
feed-quality shock, pilot-to-full-scale derating, concentrate risk and cleaning recovery are reviewed before release;
final release decision states accept, derate flux, improve pretreatment, clean, retest integrity, reduce recovery or hold.

Exercise 1: Required Flux and Membrane Area

A tertiary membrane train must deliver (3600\ \text{m}^3/\text{d}) at peak flow. Installed area is (3000\ \text{m}^2). The allowable design flux for the current feed condition is (45\ \text{L}/\text{m}^2\text{h}). Calculate the required installed flux and the membrane area needed at the allowable flux.

Solution

Convert flow:

3600\ \text{m}^3/\text{d}=150\ \text{m}^3/\text{h}

Installed flux:

\displaystyle J=\frac{150}{3000}=0.050\ \text{m/h}=50\ \text{L}/\text{m}^2\text{h}

Required area:

\displaystyle A_m=\frac{150}{0.045}=3333\ \text{m}^2

Area shortfall:

3333-3000=333\ \text{m}^2

Engineering Comment

The train can meet the flow only by operating above the allowable flux. That may be acceptable briefly during emergency operation, but it is not a sustainable basis without stronger fouling and cleaning evidence.

Plausibility Check

The required flow is $150\ \text{m}^3/\text{h}$ , so $3000\ \text{m}^2$ of membrane gives $50\ \text{L}/\text{m}^2\text{h}$ . That exceeds the $45\ \text{L}/\text{m}^2\text{h}$ allowable flux, explaining the $333\ \text{m}^2$ area shortfall.

Exercise 2: Active Area with Offline Modules

The train has (3000\ \text{m}^2) of installed membrane area, but 6 percent of modules are offline after repairs. Required flow is still (150\ \text{m}^3/\text{h}), and allowable flux remains (45\ \text{L}/\text{m}^2\text{h}). Calculate active area, actual flux, required installed area if 6 percent remains offline and gross area shortfall.

Solution

Active area:

A_{\text{active}}=3000(1-0.06)=2820\ \text{m}^2

Actual flux:

\displaystyle J=\frac{150}{2820}=0.0532\ \text{m/h}=53.2\ \text{L}/\text{m}^2\text{h}

Required active area at allowable flux:

\displaystyle A_{\text{active,req}}=\frac{150}{0.045}=3333\ \text{m}^2

Required installed area with 6 percent unavailable:

\displaystyle A_{\text{installed,req}}=\frac{3333}{0.94}=3546\ \text{m}^2

Gross area shortfall:

3546-3000=546\ \text{m}^2

Engineering Comment

Offline modules increase flux on the remaining surface. Capacity claims should use active area, not nameplate area, especially after repairs, isolation, failed integrity tests or module cleaning holds.

Plausibility Check

Taking $6\%$ of area offline leaves $2820\ \text{m}^2$ , so flux must rise above the already high value in Exercise 1. The required installed area also rises to $3546\ \text{m}^2$ , creating a larger shortfall of $546\ \text{m}^2$ .

Exercise 3: TMP and Permeability

A membrane module has feed pressure (P_f=190\ \text{kPa}), concentrate pressure (P_c=170\ \text{kPa}), permeate pressure (P_p=20\ \text{kPa}), and flux (50\ \text{L}/\text{m}^2\text{h}). Calculate TMP and permeability.

Solution

\displaystyle TMP=\frac{190+170}{2}-20=160\ \text{kPa}

\displaystyle K=\frac{50}{160}=0.313\ \text{L}/\text{m}^2\text{h}/\text{kPa}

If (\mu_T/\mu_{20}=1.20):

K_{20}=0.313(1.20)=0.376\ \text{L}/\text{m}^2\text{h}/\text{kPa}

Engineering Comment

Permeability is the decision variable, not TMP alone. TMP can rise because flux is higher, viscosity is higher, fouling is worse or pressure taps are not measuring the intended boundary.

Plausibility Check

The average feed/concentrate pressure is $180\ \text{kPa}$ , and permeate pressure is $20\ \text{kPa}$ , so TMP is $160\ \text{kPa}$ . Dividing flux $50$ by TMP gives permeability near $0.313$ .

Exercise 4: Normalized Permeability Trend

At constant flux (J=42\ \text{L}/\text{m}^2\text{h}), TMP increases from (120\ \text{kPa}) to (170\ \text{kPa}) in 8 days. The viscosity correction (\mu_T/\mu_{20}) is 1.15 at the start and 1.08 at the end. Calculate normalized permeability at both dates, percent loss and average daily normalized loss. A fouling review is required below (K_{20}=0.30\ \text{L}/\text{m}^2\text{h}/\text{kPa}).

Solution

Initial permeability:

\displaystyle K_0=\frac{42}{120}=0.350\ \text{L}/\text{m}^2\text{h}/\text{kPa}

Initial normalized permeability:

K_{20,0}=0.350(1.15)=0.403\ \text{L}/\text{m}^2\text{h}/\text{kPa}

Final permeability:

\displaystyle K_8=\frac{42}{170}=0.247\ \text{L}/\text{m}^2\text{h}/\text{kPa}

Final normalized permeability:

K_{20,8}=0.247(1.08)=0.267\ \text{L}/\text{m}^2\text{h}/\text{kPa}

Normalized permeability loss:

\Delta K_{20}=0.403-0.267=0.136\ \text{L}/\text{m}^2\text{h}/\text{kPa}

Percent loss:

\displaystyle \frac{0.136}{0.403}(100)=33.7\%

Average daily loss:

\displaystyle r_K=\frac{0.136}{8}=0.0170\ \text{L}/\text{m}^2\text{h}/\text{kPa d}

The final value is below the review threshold:

0.267<0.30

Engineering Comment

This trend is stronger evidence than TMP alone because flux is constant and temperature correction is included. The next review should check upstream solids, coagulant dose, backwash effectiveness, recent chemical exposure and module isolation.

Plausibility Check

Normalized permeability falls from $0.403$ to $0.267$ , a loss of $0.136$ . That is about one third of the starting value and below the $0.30$ review threshold, so the fouling review is justified.

Exercise 5: Recovery and Net Production

Feed flow is (190\ \text{m}^3/\text{h}) and permeate flow is (150\ \text{m}^3/\text{h}). Over one day the train produces (3600\ \text{m}^3), uses (180\ \text{m}^3) for backwash and loses (120\ \text{m}^3) equivalent to cleaning downtime. Calculate recovery, concentration factor and net production.

Solution

\displaystyle R=\frac{150}{190}=0.789

or 78.9 percent.

\displaystyle CF=\frac{1}{1-0.789}=4.75

Net production:

Q_{\text{net}}=3600-180-120=3300\ \text{m}^3/\text{d}

Engineering Comment

The plant may advertise (3600\ \text{m}^3/\text{d}) production, but demand planning sees (3300\ \text{m}^3/\text{d}). Recovery also shows that concentrate-side foulant exposure is several times the feed concentration in a simple screen.

Plausibility Check

Recovery is $150/190=78.9\%$ , so the simple concentration factor is greater than one at $4.75$ . Daily gross production is $3600\ \text{m}^3$ , and subtracting backwash plus cleaning loss gives $3300\ \text{m}^3/\text{d}$ net.

Exercise 6: Backwash Interval Tradeoff

A train produces (150\ \text{m}^3/\text{h}) when filtering. Each backwash uses (3.0\ \text{m}^3) of permeate. Compare a 30 minute interval with a 20 minute interval over one day. Cleaning downtime loss is (120\ \text{m}^3/\text{d}) in both cases. Calculate backwash event count, backwash water use, net production and the extra production penalty from the shorter interval.

Solution

Gross daily production:

V_p=150(24)=3600\ \text{m}^3/\text{d}

Events at 30 minutes:

\displaystyle N_{30}=\frac{24(60)}{30}=48

Backwash use at 30 minutes:

V_{bw,30}=48(3.0)=144\ \text{m}^3/\text{d}

Net production at 30 minutes:

Q_{\text{net,30}}=3600-144-120=3336\ \text{m}^3/\text{d}

Events at 20 minutes:

\displaystyle N_{20}=\frac{24(60)}{20}=72

Backwash use at 20 minutes:

V_{bw,20}=72(3.0)=216\ \text{m}^3/\text{d}

Net production at 20 minutes:

Q_{\text{net,20}}=3600-216-120=3264\ \text{m}^3/\text{d}

Extra production penalty:

3336-3264=72\ \text{m}^3/\text{d}

Engineering Comment

Shorter backwash intervals may slow TMP rise, but they also reduce delivered water. The operating decision needs both sides of the tradeoff: permeability trend after the change and net production penalty.

Plausibility Check

Moving from a 30-minute to a 20-minute interval increases events from $48$ to $72$ per day. At $3.0\ \text{m}^3$ per event, that adds $72\ \text{m}^3/\text{d}$ of backwash use and reduces net production by the same amount.

Exercise 7: Capacity at TMP Limit

Current permeability is (0.313\ \text{L}/\text{m}^2\text{h}/\text{kPa}). The maximum allowed TMP is (220\ \text{kPa}), and membrane area is (3000\ \text{m}^2). Estimate maximum capacity and compare it with a required flow of (150\ \text{m}^3/\text{h}).

Solution

J_{\max}=0.313(220)=68.9\ \text{L}/\text{m}^2\text{h}

Q_{\max}=0.0689(3000)=207\ \text{m}^3/\text{h}

Flow margin:

\displaystyle \frac{207-150}{150}=0.38

or about 38 percent.

Engineering Comment

The capacity appears adequate at the current permeability, but this conclusion depends on fouling not accelerating. If TMP is already rising quickly, the apparent margin can disappear before the next cleaning window.

Plausibility Check

At current permeability, $220\ \text{kPa}$ allows $68.9\ \text{L}/\text{m}^2\text{h}$ flux. Over $3000\ \text{m}^2$ , that is $207\ \text{m}^3/\text{h}$ , about $38\%$ above the required $150\ \text{m}^3/\text{h}$ .

Exercise 8: TMP Rise Rate and Cleaning Recovery

At constant flux, TMP rises from (120) to (160\ \text{kPa}) in 10 days. Clean permeability is (0.80), pre-clean permeability is (0.31), and post-clean permeability is (0.62\ \text{L}/\text{m}^2\text{h}/\text{kPa}). Calculate TMP rise rate and cleaning recovery.

Solution

\displaystyle r_{TMP}=\frac{160-120}{10}=4.0\ \text{kPa/d}

Cleaning recovery:

\displaystyle \eta_K=\frac{0.62-0.31}{0.80-0.31}=0.633

or 63.3 percent of recoverable permeability.

Engineering Comment

The cleaning helped, but it did not restore clean condition. The next step is to investigate irreversible fouling, pretreatment breakthrough, chemical strength, contact time, temperature and whether the flux target is too aggressive.

Plausibility Check

TMP rises $40\ \text{kPa}$ over $10$ days, so the rate is $4.0\ \text{kPa/d}$ . Cleaning recovers $0.31$ of the $0.49$ permeability gap to clean condition, which is $63.3\%$ .

Exercise 9: CEB and CIP Recovery Comparison

Clean normalized permeability is (0.80\ \text{L}/\text{m}^2\text{h}/\text{kPa}). Before chemically enhanced backwash, permeability is (0.30). After CEB it is (0.45). After a subsequent CIP it is (0.62). The release threshold is (0.60). Calculate CEB recovery, total recovery after CIP and decide whether CEB alone is enough.

Solution

CEB recovery against the clean reference:

\displaystyle \eta_{\text{CEB}}=\frac{0.45-0.30}{0.80-0.30}=0.300

or 30.0 percent.

Total recovery after CIP:

\displaystyle \eta_{\text{CIP,total}}=\frac{0.62-0.30}{0.80-0.30}=0.640

or 64.0 percent.

Incremental recovery from CIP after CEB:

\displaystyle \eta_{\text{CIP,incremental}}=\frac{0.62-0.45}{0.80-0.45}=0.486

or 48.6 percent of the remaining recoverable gap after CEB.

CEB alone fails the release threshold:

0.45<0.60

CIP meets it:

0.62>0.60

Engineering Comment

The result suggests that hydraulic or enhanced backwash removes only part of the foulant. CIP improves the train enough for the stated permeability threshold, but release still needs integrity and water-quality evidence.

Plausibility Check

CEB raises permeability from $0.30$ to $0.45$ , below the $0.60$ threshold. CIP raises it to $0.62$ , so only the CIP result passes the stated release threshold.

Exercise 10: Pressure-Decay Integrity Margin

A simplified pressure-decay integrity test allows a maximum decay rate of (5.0\ \text{kPa/min}). Test A falls from (100) to (92\ \text{kPa}) in 2 minutes. Test B after a module repair falls from (100) to (87\ \text{kPa}) in 2 minutes. Calculate both decay rates and pass/fail margins.

Solution

Test A decay rate:

\displaystyle r_A=\frac{100-92}{2}=4.0\ \text{kPa/min}

Test A margin:

\displaystyle M_A=\frac{5.0-4.0}{5.0}=0.20

or 20 percent. Test A passes the simplified criterion.

Test B decay rate:

\displaystyle r_B=\frac{100-87}{2}=6.5\ \text{kPa/min}

Test B margin:

\displaystyle M_B=\frac{5.0-6.5}{5.0}=-0.30

or negative 30 percent. Test B fails the simplified criterion.

Engineering Comment

Integrity testing is a barrier check, not a fouling-capacity check. A train can have acceptable permeability and still be held from release if the barrier evidence fails.

Plausibility Check

Test A decays $8\ \text{kPa}$ in two minutes, or $4.0\ \text{kPa/min}$ , below the limit. Test B decays $13\ \text{kPa}$ in two minutes, or $6.5\ \text{kPa/min}$ , so it fails by $30\%$ of the limit.

Exercise 11: Feed-Quality Shock and Fouling Window

Feed flow is (190\ \text{m}^3/\text{h}). Normal feed TSS is (12\ \text{mg/L}), but a pretreatment upset raises it to (38\ \text{mg/L}) for 6 hours. During normal operation TMP rises at (3.0\ \text{kPa/d}); after the upset it rises at (7.0\ \text{kPa/d}). Current TMP is (175\ \text{kPa}), and the limit is (220\ \text{kPa}). Calculate excess solids load during the event and time to the TMP limit at both rates.

Solution

Excess TSS concentration:

\Delta C=38-12=26\ \text{mg/L}

Excess solids load:

\Delta L=190(26)10^{-3}=4.94\ \text{kg/h}

Excess solids mass over 6 hours:

\Delta M=4.94(6)=29.6\ \text{kg}

Time to limit at normal TMP rise:

\displaystyle t_{\text{normal}}=\frac{220-175}{3.0}=15.0\ \text{d}

Time to limit after upset:

\displaystyle t_{\text{upset}}=\frac{220-175}{7.0}=6.43\ \text{d}

Engineering Comment

The solids event is short, but the fouling window changes materially. The review should check pretreatment alarms, feed turbidity, particle counts, coagulant or polymer state, backwash recovery and whether the train should be derated until permeability stabilizes.

Plausibility Check

The feed shock adds $26\ \text{mg/L}$ TSS at $190\ \text{m}^3/\text{h}$ , or $4.94\ \text{kg/h}$ . Over six hours that is $29.6\ \text{kg}$ , and the higher fouling rate shortens the time to TMP limit from $15.0$ to $6.43$ days.

Exercise 12: Pumping Power Penalty

A train operates at (Q=150\ \text{m}^3/\text{h}), hydraulic pressure rise (\Delta p=160000\ \text{Pa}), and pump efficiency (0.70). Calculate pumping power. Then estimate power if fouling requires (220000\ \text{Pa}) at the same flow.

Solution

Convert flow:

\displaystyle Q=\frac{150}{3600}=0.0417\ \text{m}^3/\text{s}

Initial power:

\displaystyle P=\frac{0.0417(160000)}{0.70}=9.5\ \text{kW}

Fouled power:

\displaystyle P=\frac{0.0417(220000)}{0.70}=13.1\ \text{kW}

Power increase:

13.1-9.5=3.6\ \text{kW}

Daily energy increase:

\Delta E_d=3.6(24)=86.4\ \text{kWh/d}

Annual energy increase at 60 percent duty:

\Delta E_y=3.6(24)(365)(0.60)=18922\ \text{kWh/y}

Engineering Comment

The electrical penalty is not the only cost. Higher pressure can shorten membrane life, increase leak risk, trigger alarms and hide the need for upstream fouling control.

Plausibility Check

At fixed flow, pumping power scales with pressure rise. Increasing pressure from $160$ to $220\ \text{kPa}$ raises power from $9.5$ to $13.1\ \text{kW}$ , a $3.6\ \text{kW}$ penalty.

Exercise 13: Hydraulic Capacity versus Release Gate

After CIP, (K_{20}=0.62\ \text{L}/\text{m}^2\text{h}/\text{kPa}). The current viscosity correction is (\mu_T/\mu_{20}=1.20). Required flow is (150\ \text{m}^3/\text{h}), active area is (3000\ \text{m}^2), and the hydraulic release TMP limit is (170\ \text{kPa}). The integrity-test limit is (5.0\ \text{kPa/min}), but the measured decay is (6.5\ \text{kPa/min}). Permeate turbidity is (0.19\ \text{NTU}), and the release limit is (0.15\ \text{NTU}). Check hydraulic TMP, integrity margin, turbidity exceedance and release decision.

Solution

Convert normalized permeability back to current-temperature permeability:

\displaystyle K_T=\frac{K_{20}}{\mu_T/\mu_{20}}=\frac{0.62}{1.20}=0.517\ \text{L}/\text{m}^2\text{h}/\text{kPa}

Required flux:

\displaystyle J=\frac{150}{3000}=0.050\ \text{m/h}=50\ \text{L}/\text{m}^2\text{h}

TMP needed at required flux:

\displaystyle TMP_{\text{req}}=\frac{50}{0.517}=96.7\ \text{kPa}

Hydraulic TMP margin:

M_{TMP}=170-96.7=73.3\ \text{kPa}

Integrity-test margin:

M_{\text{integrity}}=5.0-6.5=-1.5\ \text{kPa/min}

Turbidity exceedance:

\Delta NTU=0.19-0.15=0.04\ \text{NTU}

Engineering Comment

The train has hydraulic capacity, but it should not be released because barrier evidence and turbidity both fail. Membrane decisions need the full state: flux, TMP, permeability, cleaning state, integrity evidence and water-quality release criteria.

Plausibility Check

Temperature-corrected permeability gives a required TMP of only $96.7\ \text{kPa}$ , well below the $170\ \text{kPa}$ hydraulic limit. However, integrity margin is $-1.5\ \text{kPa/min}$ and turbidity exceeds the limit by $0.04\ \text{NTU}$ , so release still fails.

Exercise 14: Sustainable Flux Step-Test and Full-Scale Area

A pilot membrane skid is tested on representative feed using four flux steps. Each step is held long enough to estimate the TMP rise rate.

Flux step	TMP rise rate
(35\ \text{L}/\text{m}^2\text{h})	(0.4\ \text{kPa/h})
(40\ \text{L}/\text{m}^2\text{h})	(0.6\ \text{kPa/h})
(45\ \text{L}/\text{m}^2\text{h})	(1.1\ \text{kPa/h})
(50\ \text{L}/\text{m}^2\text{h})	(2.8\ \text{kPa/h})

The site defines the sustainable flux as the highest tested flux with TMP rise rate no more than (1.0\ \text{kPa/h}). For full-scale design, apply a (10%) derating to the pilot sustainable flux. The full-scale train has (3000\ \text{m}^2) active membrane area and must deliver (150\ \text{m}^3/\text{h}).

Find the pilot sustainable flux, derated design flux, full-scale capacity, daily production, flow shortfall and active membrane area required to meet the required flow at the derated design flux.

Solution

The (35) and (40\ \text{L}/\text{m}^2\text{h}) steps pass the TMP rise-rate criterion:

0.4<1.0,\quad 0.6<1.0

The (45) and (50\ \text{L}/\text{m}^2\text{h}) steps fail:

1.1>1.0,\quad 2.8>1.0

The pilot sustainable flux is therefore:

J_{sust}=40\ \text{L}/\text{m}^2\text{h}

Apply the (10%) full-scale derating:

J_{design}=40(1-0.10)=36\ \text{L}/\text{m}^2\text{h}

Convert design flux:

36\ \text{L}/\text{m}^2\text{h}=0.036\ \text{m}^3/\text{m}^2\text{h}

Full-scale capacity with (3000\ \text{m}^2) active area:

Q_{cap}=0.036(3000)=108\ \text{m}^3/\text{h}

Daily production at this flux:

Q_d=108(24)=2592\ \text{m}^3/\text{d}

Flow shortfall against the requirement:

\Delta Q=150-108=42\ \text{m}^3/\text{h}

Relative shortfall:

\displaystyle \frac{42}{150}(100)=28.0\%

Active membrane area required at the derated design flux:

\displaystyle A_{req}=\frac{150}{0.036}=4167\ \text{m}^2

Additional active area required:

4167-3000=1167\ \text{m}^2

The pilot evidence supports a derated design flux of (36\ \text{L}/\text{m}^2\text{h}). At that flux, the installed active area cannot deliver (150\ \text{m}^3/\text{h}); the train needs more area, lower required flow, improved pretreatment or stronger step-test evidence before release at the higher capacity.

Engineering Comment

Sustainable flux is an operating envelope, not a membrane catalog value. A step test should use representative feed, stable pretreatment, realistic recovery, normal backwash settings and enough hold time to reveal fouling acceleration. A full-scale derating protects against feed variability, module age, distribution effects, cleaning limits and measurement uncertainty.

Plausibility Check

The TMP rise rate stays below the criterion through (40\ \text{L}/\text{m}^2\text{h}) and then exceeds it at (45\ \text{L}/\text{m}^2\text{h}), so choosing (40) as the pilot sustainable flux is consistent. Derating to (36) lowers capacity to (108\ \text{m}^3/\text{h}), which explains the large shortfall against a (150\ \text{m}^3/\text{h}) requirement.

Exercise 15: CIP Stock Chemical Dose and Batch Shortfall

A membrane train needs a CIP batch volume of:

V_{CIP}=18.0\ \text{m}^3

The target cleaning concentration is:

C_{target}=500\ \text{mg/L as Cl}_2

The available sodium hypochlorite stock is (12%) by mass as active chlorine equivalent, with density:

\rho_{stock}=1.20\ \text{kg/L}

Operations can add only:

V_{stock,available}=55\ \text{L}

before the batch must start. The minimum acceptable batch concentration for release is:

C_{min}=480\ \text{mg/L as Cl}_2

Calculate the active chemical mass required, the stock volume required, the concentration delivered by the available stock volume, the concentration shortfall, and the release decision.

Solution

Convert the CIP volume:

18.0\ \text{m}^3=18000\ \text{L}

Required active chemical mass is:

m_{req}=C_{target}V_{CIP}

\displaystyle m_{req}=500\frac{\text{mg}}{\text{L}}(18000\ \text{L})=9000000\ \text{mg}

Convert to kilograms:

9000000\ \text{mg}=9.0\ \text{kg}

Each litre of stock contains active chemical mass:

m_{stock/L}=\rho_{stock}(0.12)

m_{stock/L}=1.20(0.12)=0.144\ \text{kg/L}

Required stock volume:

\displaystyle V_{stock,req}=\frac{m_{req}}{m_{stock/L}}

\displaystyle V_{stock,req}=\frac{9.0}{0.144}=62.5\ \text{L}

The available stock volume is short:

\Delta V_{stock}=62.5-55=7.5\ \text{L}

Active mass delivered by (55\ \text{L}) is:

m_{delivered}=55(0.144)=7.92\ \text{kg}

Convert delivered mass to milligrams:

7.92\ \text{kg}=7920000\ \text{mg}

Delivered concentration:

\displaystyle C_{delivered}=\frac{7920000}{18000}=440\ \text{mg/L as Cl}_2

Shortfall against the target:

\Delta C=500-440=60\ \text{mg/L}

Relative shortfall:

\displaystyle \frac{60}{500}(100)=12.0\%

The batch also fails the minimum release concentration:

480-440=40\ \text{mg/L}

The correct decision is to hold the CIP release, add the missing stock chemical if compatible with the recipe and membrane limits, remake the batch, or document a controlled recipe change approved by the membrane supplier and site procedure.

Engineering Comment

Cleaning chemistry should not be checked by stock volume alone. The concentration basis, density, active fraction, age or degradation of the chemical, batch volume, contact time, pH, temperature, membrane compatibility and rinse requirement all affect whether the cleaning step is valid. A low-strength CIP can appear to save chemical while leaving irreversible fouling, biofilm or scaling in place.

Plausibility Check

An (18.0\ \text{m}^3) batch at (500\ \text{mg/L}) needs (9.0\ \text{kg}) of active chemical. A (12%) stock at (1.20\ \text{kg/L}) supplies only (0.144\ \text{kg/L}), so a required volume above (60\ \text{L}) is plausible. The available (55\ \text{L}) gives only (440\ \text{mg/L}), below both the target and the (480\ \text{mg/L}) minimum.

Exercise 16: Crossflow Velocity and Reynolds Fouling Screen

A membrane skid uses recirculation flow to maintain crossflow over the membrane channels. The rack has:

N_c=40\ \text{channels}

Each channel has open flow area:

A_c=1.20\times10^{-4}\ \text{m}^2

and hydraulic diameter:

D_h=2.5\ \text{mm}

The normal recirculation flow is:

Q_{rec}=54\ \text{m}^3/\text{h}

An energy-saving trial reduces recirculation to:

Q_{rec,low}=38\ \text{m}^3/\text{h}

The concentrate-side release rule requires:

v_{min}=2.5\ \text{m/s}

and:

Re_{min}=4000

Use:

\rho=998\ \text{kg/m}^3

and:

\mu=1.2\times10^{-3}\ \text{Pa s}

Calculate total open channel area, crossflow velocity and Reynolds number for both recirculation settings. Decide whether the low-recirculation trial can be released.

Solution

Total channel area:

A_{tot}=N_cA_c

A_{tot}=40(1.20\times10^{-4})=4.80\times10^{-3}\ \text{m}^2

Normal recirculation flow in SI units:

\displaystyle Q_{rec}=\frac{54}{3600}=0.0150\ \text{m}^3/\text{s}

Normal crossflow velocity:

\displaystyle v=\frac{Q_{rec}}{A_{tot}}

\displaystyle v=\frac{0.0150}{4.80\times10^{-3}}=3.13\ \text{m/s}

Convert hydraulic diameter:

D_h=2.5\ \text{mm}=0.0025\ \text{m}

Normal Reynolds number:

\displaystyle Re=\frac{\rho vD_h}{\mu}

\displaystyle Re=\frac{998(3.13)(0.0025)}{1.2\times10^{-3}}=6500

Normal operation passes both screens:

3.13>2.5,\quad 6500>4000

Low recirculation flow in SI units:

\displaystyle Q_{rec,low}=\frac{38}{3600}=0.0106\ \text{m}^3/\text{s}

Low-recirculation velocity:

\displaystyle v_{low}=\frac{0.0106}{4.80\times10^{-3}}=2.20\ \text{m/s}

Low-recirculation Reynolds number:

\displaystyle Re_{low}=\frac{998(2.20)(0.0025)}{1.2\times10^{-3}}=4570

The low-recirculation trial preserves the Reynolds screen:

4570>4000

but fails the minimum velocity screen:

2.20<2.5\ \text{m/s}

The energy-saving trial should not be released on this evidence. It needs a higher recirculation setpoint, a revised supplier-approved velocity rule, or field evidence showing stable TMP, permeability and feed-channel cleaning at the lower velocity.

Engineering Comment

Crossflow fouling control is not captured by permeate flux alone. Reducing recirculation can save pump energy while increasing concentration polarization, cake deposition or channel plugging. A release check should state active channel count, channel geometry, recirculation flow measurement, viscosity, temperature, suspended solids, pressure drop and whether fouling trend data support the lower shear condition.

Plausibility Check

The low-flow setting is (38/54=70.4%) of normal recirculation, so velocity and Reynolds number should fall by the same fraction when geometry and viscosity are unchanged. The velocity drops from (3.13) to (2.20\ \text{m/s}), and Reynolds drops from about (6500) to (4570), matching that proportional behavior.

Exercise 17: Recovery Limit from Concentrate Scaling Threshold

A membrane train is being pushed to higher recovery to meet peak demand. Feed flow is:

Q_f=200\ \text{m}^3/\text{h}

The required permeate production is:

Q_{p,req}=155\ \text{m}^3/\text{h}

The proposed recovery is:

R=84\%

Feed silica concentration is:

C_f=22\ \text{mg/L}

For the release screen, apply a 15 percent feed-concentration guard:

C_{f,g}=1.15C_f

The concentrate-side scaling threshold is:

C_{limit}=120\ \text{mg/L}

For a highly rejected scaling species, use:

\displaystyle CF=\frac{1}{1-R}

where (CF) is concentration factor and (R) is recovery as a fraction. Compute proposed permeate flow, guarded concentrate concentration and scaling decision. Then find the maximum recovery allowed by the guarded scaling limit and check a revised recovery:

R_{new}=78\%

Solution

Guarded feed concentration:

C_{f,g}=1.15(22)=25.3\ \text{mg/L}

Proposed permeate flow:

Q_p=RQ_f

Q_p=0.84(200)=168\ \text{m}^3/\text{h}

Production margin at the proposed recovery:

168-155=13\ \text{m}^3/\text{h}

Proposed concentration factor:

\displaystyle CF=\frac{1}{1-0.84}=6.25

Guarded concentrate concentration:

C_c=C_{f,g}CF

C_c=25.3(6.25)=158\ \text{mg/L}

Scaling threshold exceedance:

158-120=38\ \text{mg/L}

The proposed 84 percent recovery meets production but fails the scaling screen.

For the maximum recovery:

\displaystyle C_{f,g}\frac{1}{1-R_{max}}\leq C_{limit}

Rearrange:

\displaystyle 1-R_{max}\geq\frac{C_{f,g}}{C_{limit}}

\displaystyle R_{max}\leq1-\frac{C_{f,g}}{C_{limit}}

\displaystyle R_{max}\leq1-\frac{25.3}{120}=0.789

Therefore:

R_{max}=78.9\%

Maximum permeate flow at this guarded recovery is:

Q_{p,max}=0.789(200)=158\ \text{m}^3/\text{h}

This is only:

158-155=3\ \text{m}^3/\text{h}

above the required production.

For the revised recovery:

Q_{p,new}=0.78(200)=156\ \text{m}^3/\text{h}

Revised production margin:

156-155=1\ \text{m}^3/\text{h}

Revised concentration factor:

\displaystyle CF_{new}=\frac{1}{1-0.78}=4.55

Revised guarded concentrate concentration:

C_{c,new}=25.3(4.55)=115\ \text{mg/L}

Scaling margin:

120-115=5\ \text{mg/L}

The revised recovery passes both the simplified scaling limit and the minimum production requirement, but the margins are narrow.

Engineering Comment

Recovery is not only a water-yield target. Higher recovery reduces concentrate volume but raises concentration factor, osmotic pressure, scaling risk and cleaning demand. A release decision should state which species controls the limit, whether rejection is high enough for the simplified concentration-factor model, how feed variability is guarded, whether antiscalant or pH control is validated, and what action follows a feed-quality excursion.

Plausibility Check

At 84 percent recovery, the reject stream is only 16 percent of feed flow, so a concentration factor of (1/0.16=6.25) is expected. Applying that to guarded feed silica pushes the concentrate above the threshold. Reducing recovery to 78 percent lowers concentration factor to about 4.55, which explains why the guarded concentration falls just below (120\ \text{mg/L}) while production barely remains above the required flow.

Exercise 18: SDI Pretreatment Release Gate

A membrane train is being considered for higher recovery after upstream pretreatment maintenance. A silt density index test uses the standard 15-minute basis:

T=15\ \text{min}

The initial time to collect the fixed sample volume is:

t_i=31\ \text{s}

After 15 minutes of filtration, the collection time is:

t_f=58\ \text{s}

Use:

\displaystyle SDI_T=\left(1-\frac{t_i}{t_f}\right)\frac{100}{T}

The release rule requires:

SDI_{15}\leq3.0

and applies a method/field guard:

U_{SDI}=0.20

Calculate plugging factor, SDI, guarded SDI and release decision. Then check a revised pretreatment setting where the final collection time improves to:

t_{f,rev}=48\ \text{s}

with the same initial time and guard.

Solution

Plugging factor is:

\displaystyle PF=1-\frac{t_i}{t_f}

\displaystyle PF=1-\frac{31}{58}=0.466=46.6\%

The 15-minute SDI is:

\displaystyle SDI_{15}=0.466\frac{100}{15}=3.10

Guarded SDI is:

SDI_{g}=SDI_{15}+U_{SDI}

SDI_g=3.10+0.20=3.30

The guarded value exceeds the release rule:

3.30>3.0

The higher-recovery release should be held. The feed may look hydraulically acceptable in a short run, but this SDI result says pretreatment filterability is not yet strong enough for the proposed operating envelope.

For the revised final collection time:

\displaystyle PF_{rev}=1-\frac{31}{48}=0.354=35.4\%

Revised SDI:

\displaystyle SDI_{15,rev}=0.354\frac{100}{15}=2.36

Guarded revised SDI:

SDI_{g,rev}=2.36+0.20=2.56

Revised SDI margin is:

M_{SDI,rev}=3.0-2.56=0.44

The revised pretreatment setting passes this feed-filterability gate with a modest margin. Release should still require stable feed turbidity, particle count or upstream filter differential pressure evidence over the relevant operating window.

Engineering Comment

SDI is a pretreatment and filterability screen, not a direct capacity guarantee. A passing SDI does not prove sustainable flux, scaling control or membrane integrity, but a failing SDI is a strong warning against increasing recovery or flux. The release record should state sample location, test pressure, filter type, feed temperature, turbidity, coagulant state, upstream filter condition, replicate spread and whether the SDI result represents the operating mode being released.

Plausibility Check

The initial test slows from (31) to (58) seconds, so nearly half of the fixed-volume flow capacity is lost during the test. An SDI just above 3 is plausible for a marginal pretreatment condition, and adding a 0.20 guard makes the failure clear. Improving the final collection time to (48) seconds lowers plugging factor and moves the guarded SDI to (2.56), which explains why the revised setting can pass while the original setting cannot.

Review Checklist

Before accepting an answer, check that it states:

the flow period and membrane area boundary;
whether installed area or active area is being used;
whether sustainable flux comes from representative step-test evidence or only an assumed design value;
whether crossflow velocity and Reynolds number use active channel geometry, recirculation flow and current viscosity;
the TMP pressure taps and basis;
whether permeability is normalized for temperature;
whether recovery changes concentrate-side exposure;
whether recovery is checked against scaling, osmotic, antiscalant or concentrate-disposal limits;
whether SDI or equivalent feed-filterability evidence supports the flux or recovery release;
whether net production includes backwash and cleaning losses;
whether a shorter backwash interval improves fouling enough to justify production loss;
whether TMP rise is compared at similar flux;
whether cleaning recovery is judged by permeability;
whether CEB, CIP and integrity testing answer different questions;
whether chemical dose uses batch volume, stock density, active fraction and concentration basis;
whether release requires both hydraulic capacity and barrier evidence;
whether energy screens use SI units and realistic efficiency.

Common Mistakes

Comparing daily flow to hourly flux or mixing (m^3/d), (m^3/h) and (L/m^2h) without a unit conversion.
Using installed area when offline modules, isolated racks or blanked fibers reduce the active area actually available.
Calling pump discharge pressure TMP without checking the correct membrane-side pressure taps and gauge-pressure basis.
Accepting clear permeate as proof of capacity while ignoring normalized permeability, TMP rise, recovery loss and integrity-test margin.
Reducing recirculation for energy savings without checking crossflow velocity, Reynolds number, pressure drop and fouling trend response.
Raising recovery to meet production while ignoring concentration factor, feed uncertainty and scaling thresholds.
Releasing higher flux or recovery while SDI, particle count, turbidity or pretreatment differential pressure still indicates unstable feed quality.
Interpreting pressure rise as fouling without checking whether flux, viscosity, temperature or feed solids changed at the same time.
Dosing cleaning chemicals by stock volume alone without batch volume, active fraction, stock density, chemical age, pH and membrane compatibility.
Treating CEB or CIP recovery as proof of membrane integrity even though hydraulic cleaning and barrier validation answer different release questions.
Releasing a train after one good cycle when the fouling trend, feed shock history or post-cleaning permeability shows unstable operation.

REF