ENVE-2110 EXAM III Help Session DCC pm

ABBREVIATIONS Look through lecture notes, book and assignments Examples: (sample test Q1) BOD 5, P, seed, t, DO b,t, DO b,i, f, DO s,t, DO s,I, NOD or NBOD, COD or CBOD, WW, WWTP, TSS, VSS, TKN, NPDES, EPA, CSTR, MW, Q, V, batch reactor, OSHA, VOC, C.V., S.S., SS, v s, type I settling, X,  r g, S, K s,  max  AS, WAS, RAS, SRT,  c  OD, media, TF, RBC

UNITS Following units are considered: mg/L ppm ppb Expressed as CaCO 3, P, N or S Molarity (M) Normality (N) Hardness calculation

EXAMPLES See your notes, book and homework assignments 3 and 4: EXAMPLE: (sample test Q9) Q9: Calculate the normality (N) of 123 mg/L CaCO 3.

SOLUTION:

HARDNESS Notes and book: Understand the meaning A sum of all polyvalent cations Be able to calculate hardness:

EXAMPLE: (sample test II Q10) What is hardness? SOLUTION: Characterization of water that does not lather well. Causes scum and scale. Hardness is the sum of all polyvalent cations in a water sample Typically modeled with Ca 2+ and Mg 2+ only. EX. If a solution contains [SO 4 2- ], [K + ], [Ca 2+ ], [Cl - ], and [Fe 3+ ], what ions contribute to hardness? SOLUTION Hardness = [Ca 2+ ] + [Fe 3+ ]

MASS BALANCE To understand the concepts of mass balance. To understand basic terminology used in predicting concentrations of pollutants in WW systems. To model pollution in a reactor.

LAW OF CONSERVATION OF MATTER Disregarding nuclear reactions, matter can neither be created nor destroyed. The mathematical representation of this law is called a materials balance or mass balance.

SYSTEMS Picking a system: what do you want to know?; what data is available? System boundaries are defined so as to make calculations simple. The system within the boundaries is called the control volume.

Control Volume: what is within the dotted lines Cup 1 Cup 2

MATERIALS BALANCE For environmental processes, the basic equation is: Accumulation = Input – Output - Decay and each of these terms refers to a mass quantity (make sure units match) within the system (memorize this). The system could be the planet or it could be a cell or anything in between.

STEPS Draw a picture Include all flows in Include all flows out Assume: well mixed and S.S Flow in = flow out

WHAT IF THE CHEMICAL IS DEGRADING? chemical transformation of compound to another can occur by a living organism (biodegradation) can occur “abiotically” (chemical, photo-degradation, etc.) Often “first order” – cell death is first order or -k d X

REACTION TERMS Sedimentation Volatilization Bio-uptake Degradation

WASTEWATER TREATMENT SOURCE WATER Different sources from domestic to industrial PRETREATMENT SCREENS GRIT CHAMBER EQUALIZATION TANK PRIMARY TREATMENT SEDIMENTATION SECONDATY TREATMENT BIOLOGICAL TREATMENT AS, TF, PONDS SECONDARY SETTLING TERTIARY TREATMENT OR ADVANCED TREATMENT DISINFECTION FILTRATION NITROGEN CONTROL PHOSPHOROUS REMOVAL DISTRIBUTION

WASTEWATER TREATMENT BE ABLE TO DESIGN: (old exam Qs 10,12-14) Grit removal Equalization tank(s) Primary settling tanks TF, AS or ponds Secondary settling tanks Understand the concepts behind tertiary treatment or advanced treatment Effluent

BOD Know why BOD is important How to calculate BOD 5 Where in the WW treatment processes BOD is removed BOD 5 TEST 5-day test BOD bottle volume typically 300 mL Seed = microbes + nutrients Be able to calculate BOD with and without a seed Be able to calculate Dilution factor = P =volume of WW sample/300 mL (volume of WW + dilution water with or without seed)

TYPICAL BOD VALUES TYPE OF WATEREXPECTED BOD5 (MG/L O2) RIVER<5 BIOLOGICALLY SUITABLE BIOMASS FROM WW CLARIFIED BIOMASS FROM WW OR MIDLY POLLUTED INDUSTRIAL WW HEAVILY POLLUTED INDUSTRIAL WW 1,000-3,000

BOD 5 EXAMPLE Determine the BOD 5 value for a 12 mL WW sample when it is placed in a standard BOD bottle filled with dilution water. The bottle had an initial dissolved oxygen concentration of 1.8 mg/L. A blank run parallel showed no change in DO over the incubation period.

What is the volume of a standard BOD bottle? 300 mL What is a blank? BOD bottle that contains only dilution water or dilution water with seed. Allows you to see if there is any contribution to oxygen demand from dilution water only. What is dilution water? Dilution water is what is added to BOD bottle to top off to 300 mL. It contains trace elements for bacterial metabolism. May also contain inoculum of microorganisms (not always required) called “seed”. How long is a typical BOD test? 5 days

Calculate the BOD 5 of WW:

BOD 5 with “seed” Understand the equation: (check your notes and book)

DESIGN of WW

PRETREATMENT BAR RACKS No design criteria Know what they are Purpose Trash racks, manually cleaned racks and mechanically cleaned racks Grit Chamber Know why they are used Depth: 2-5 m L:W ratio: 3:1 to 5:1  min

Equalization Not a treatment process. Use to improve the effectiveness of both secondary and tertiary treatment processes. WW does not flow in at a constant flow rate. Equalizing the flow using a separate tank. No specific design criteria other than functional, in-line or off-line.

PRIMARY TREATMENT Primary settling tanks (clarifier) 1. Number tanks > or = 2 2. Circular or rectangular 3. Detention time = hrs 4. Depth usually = 15 ft 5. L:W ratio = 3:1 to 5:1 6. Sludge removal 1. 62% removal of TSS 2. 35% removal of BOD

SECONDARY TREATMENT 1. Biological treatment 1. AS 2. TF 3. Ponds 2. Secondary clarifier

Activated Sludge (AS) Has flow in Has RAS in Has flow out Has RAS out

EXAMPLE: Design a secondary settling tank dimensions for a rectangular settling tank, area and volume assuming the return activated sludge flow of 1.3 MGD and flow in 3.5 MGD. Calculate the total flow using mass balance. Draw a schematic of the settling tank with RAS State your assumptions! Show your work!

SOLUTION Flow balance, S.S. assumption: Flow IN = Flow OUT Question asked total flow?

DESIGN If v 0 =1,200-1,700 gal/ft 2 d Assume 1,300 gal/ft 2 d Assume  hrs HEIGHT:

VOLUME: Assume 2 tanks V(per tank)=26,738 ft 3

L:W RATIO: (3:1 to 5:1) Assume 4:1 L=4W

TANK DIMENSIONS ITEMDIMENSIONS L86 ft W21.5 ft h14.5 ft V26,738 ft3

TRICKLING FILTER TF: Bed of coarse material Rocks, plastic, stones WW applied on the top Circular 80% removal (varying depending on the media) Depth<6ft Area=Q/Q loading rate or Q/0.09 gal/min ft 2 D 2 =4A/ 

TF EXAMPLE Determine the area and loading rate of a low-rate TF that has a diameter of 35.0 m and depth of 1.5 m. The flow rate is 1,900 m 3 /d and the influent BOD 5 is 150 mg/L.

LOADING RATE:

PONDS Aerobic ponds Anaerobic ponds Facultative ponds

SECONDARY CLARIFIER SAME AS PRIMARY CLARIFIER

TERTIARY TREATMENT Disinfection Filtration Nitrogen control Phosphorous removal

Disinfection Chlorine is added to WW Cl 2 + H 2 O = HOCl + H + + Cl - results in a pH drop (decrease) Detention time = 15 min Depth = 3 m (typically) To deactivate reproduction of microbes (not necessarily kill them)

FILTRATION Can be different types Surface filters Deep bed filters Membrane filters Carbon filters More or equal to 2

BATCH REACTOR Assumptions: Well-mixed Uniform composition Constant volume Noting in or out 1-st order 2-nd order See sample exam Q9 Be able to recognize, draw and label

CSTR Assumptions: Completely mixed Uniform concentration throughout the reactor Water enters and leaves the reactor at a given volumetric flow rate of Q (volume/time) Mass or quantity = C i V Average r = r Able to define terms with units Be able to recognize, draw and label

During S.S. conditions: See examples from sample exam – Q 2