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March 2008, Vol. 20, No.3
Certification Quiz
Test Your Knowledge of Advanced Wasterwater Treatment and Reuse
True or False Questions:
- In most cases, advanced wastewater treatment may not be required to meet criteria required for reuse water.
- Using a membrane bioreactor (MBR) for treatment provides similar effluent quality as conventional activated sludge followed by effluent filtration.
- When low-pressure membranes, microfiltration, or ultrafiltration are used, lower dosages of disinfectant are required for disinfection.
- The most commonly used advanced wastewater treatment technologies are low-pressure membranes, microfiltration or ultrafiltration, media filters, and cloth filters.
- Reuse water is allowed to supplement directly the potable water supply in some states.
- Reverse osmosis is most commonly used to remove high concentrations of total dissolved solids from tertiary treated wastewater effluent.
- Microfiltration and ultrafiltration are most commonly used to remove high concentrations of total dissolved solids from tertiary treated wastewater effluent.
Multiple Choice Questions:
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What type of membranes are used in membrane bioreactors?
A. Microfiltration only.
B. Ultrafiltration only.
C. Nanofiltration only.
D. Microfiltration or ultrafiltration.
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Which process provides the best effluent quality for water reuse?
A. Conventional activated sludge process with media filters.
B. Trickling filters.
C. Membrane bioreactor.
D. Aerated lagoons.
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When media filters are used for advanced wastewater treatment, the most critical limiting parameter is flux, which is calculated using which of the following parameters?
A. Flow, media height, footprint.
B. Flow, media height, footprint, and freeboard.
C. Flow and footprint.
D. Flow, footprint, and filter-cycle between backwashes.
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Four 5-ft × 5-ft square media filters are used as a tertiary treatment. Using the following data, calculate the filtration flux in gal/min•ft².
|
Feed flow
|
Number of filters
|
Media height
|
Filter freeboard
|
|
400 gal/min
|
4
|
6.0 ft
|
4.0 ft
|
A. 25.0 gal/min•ft².
B. 16.66 gal/min•ft².
C. 10.0 gal/min•ft².
D. 4.0 gal/min•ft².
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To produce irrigation water with a permitted maximum of 500 mg/L of total dissolved solids, a portion of a 600-gal/min effluent flow, which has 1000 mg/L of total dissolved solids, was treated via reverse osmosis (RO). Using the data in the chart below, calculate the needed flow of the RO system.
|
Tertiary effluent flow
|
Number of RO trains
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RO recovery
|
RO flux
|
RO system salt rejection
|
|
600 gal/min
|
4
|
90%
|
12
gal/ft²•d
|
> 95%
|
A. 600 gal/min.
B. 300 gal/min.
C. 150 gal/min.
D. 135 gal/min.
Questions were developed by Val Frenkel, director of membrane technologies for Kennedy/Jenks Consultants (San Francisco) and reviewed by the Association of Boards of Certification (Ames, Iowa) Validation and Examination Committee.
Answer Key:
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1. False, because practically all secondary treatments besides MBR cannot achieve treatment goals for reuse water. While MBR is considered secondary treatment, it provides the effluent quality of tertiary treatment.
2. True.
3. True.
4. True.
5. False. No state or country allows using tertiary effluent for direct potable supply. A few states and countries allow using tertiary effluent as an indirect supplement for the potable water supply.
6. True.
7. False. Microfiltration and ultrafiltration are low-pressure membranes. They remove suspended and colloidal matter from the water but cannot remove dissolved solids. Some dissolved components can be removed when using chemicals along with microfiltration and ultrafiltration membranes, but overall total dissolved solids cannot be removed.
8. d.
9. c.
10. c.
11. d. The active filtration area of each filter = 5 ft × 5 ft = 25 ft². The total active filtration area for four filters = 25 ft² × 4 = 100 ft². The filtration flux is the feed flow ÷ the total active filtration area. So, 400 gal/min ÷ 100 ft² = 4.00 gal/min•ft².
12. b. To simplify calculations, we are assuming that RO recovery is close to 100% and salt rejection is close to 100% as well. That is, the RO effluent total dissolved solids concentration is close to nil. If RO effluent flow is X, the bypass flow with the feed total dissolved solid concentration of 1000 mg/L is (600 – X). We can write the mass balance formula for the mixing point downstream of RO treatment as follows: (600 gal/min – X) 1000 mg/L + X × 0 = 500 mg/L × 600 gal/min, So, 600,000 gal•mg/min•L – 1000X mg/L = 300,000 gal•mg/min•L. And, X = 300 gal/min.
References:
Water Environment Federation (2006). Membrane Systems for Wastewater Treatment. New York: WEF Press McGraw-Hill.
Water Environment Federation (2005). Biological Nutrient Removal (BNR) Operation in Wastewater Treatment Plants, Manual of Practice No. 29. New York: WEF Press, McGraw–Hill.
Water Environment Federation and American Society of Civil Engineers–Environmental and Water Resources Institute (2005). Biological Nutrient Removal Operation Study Guide. Alexandria, Va.: Water Environment Federation.