Mass Balance

Process and Mass balance analysis Gyeongsang National University Department of Biological and chemical Engineering Environmental Engineering Lab Ngoc Thuan Le

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Discussions

• • • • •

Environmental treatment reactors Hydraulic Detention time Study the hydraulic flow characteristics of reactors Mass Balance principles Modeling with tracer elements

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Overview •

The constituents in wastewater are removed by physical, chemical, biological methods. These process occur in a variety of combination in treatment flow diagram.

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Important factors:  The type of reactor (container or tank)  The size of the treatment facilities  Temperature,  and others

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The fundamental for the basis analysis of the wastewater treatment is materials mass balance principle in which an accounting of mass is made before and after reactions.

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Reactors used for the treatment of wastewater

Batch reactor Activated sludge biological treatment, mixing of concentrated solution

Complete-Mix reactor Aerated lagoons, aerated sludge digestion

Plug-Flow reactor Chlorine contact basin, natural treatment systems, Activated sludge biological treatment…

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Complete Mix reactor in series Lagoon treatment systems, used to simulate nonideal flow in plug flow reactor

Packed-bed reactor Trickling filter

Packed-bed upflow reactor Anaerobic treatment system

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Fluidized-bed reactor Upflow sludge blanket reactors, air stripping… Chap.4|Introduction to Process analysis

Detention time  Detention time is the length of time water is retained in a vessel or basin or the period from the time the water enters a settling basin until it flows out the other end.  The theoretical detention time of a container is the same as the amount of time it would take to fill the container if it were empty.  Detention times are normally calculated for the following basins or tanks:  Flash mix chambers (seconds)  Flocculation basins (minutes)  Sedimentation tanks or clarifiers (hours)  Wastewater ponds (days)  Oxidation ditches (hours).

Where:

V τ = Q

τ = hydraulic detention time, T V = volume of the reactor, L3 Q = volumetric flow rate, L3T-1

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Ideal flow in complete-mix and plug-flow reactors

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Ex. Problem The reservoir for the community holds 110,000 gallons. The well will produce 60 gpm. What is the detention time in the reservoir in hours? Solution

τ=

110 ,000 gal 1834 min =1834 min or = 30 .6hrs 60 gal / min 60 min/ hr

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Nonideal flow in complete-mix and plug-flow reactors The flow in complete-mix and plug-flow reactors is seldom ideal a. Temperature differences, a portion of the water can travel to the outlet along the bottom of or across the top of the reactor without mixing completely. b. Wind-driven circulation patterns.

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c. Inadequate mixing, sometimes due to sufficient energy input. d. Poor design, dead zone may develop within the reactor that will not mix with the incoming water. e. Axial dispersion in plug flow reactors

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The Mass-balance principle •

Mass is neither created nor destroyed, but the form of the mass can be altered (e.g., liquid to a gas)

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Precisely, the law of the conservation of mass: “What comes in must

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The mass balance analysis define what occurs within treatment reactors as a function of time.

equal what goes out.”

The two numbers — in (influent) and out (effluent) — must be within 10 to 15% of each other to be considered acceptable. Larger discrepancies may indicate sampling errors or increasing solids levels in the unit or undetected solids discharge in the tank effluent. (Frank R, Spellman)

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Mixer

Accumulation = inflow – outflow + generation (or disappearance)



One or more of the terms can be equal to zero.

Inflow

Outflow

Q, Co

Q, C

V, C

- In a batch reactor in which there is no inflow or outflow, accumulation will be equal to zero. - under steady-state condition, the rate of accumulation is zero 

Generation term: rc = -kC for a decrease in the reactant or, rc = +kC for an increase in the reactant.

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Preparation of the Mass Balance

1. Prepare a simplified schematic or flow diagram of the system or process 2. Draw a system or control volume boundary to define the limit over which the mass balance is to be applied. 3. List all of the pertinent 4. List all of the rate expressions for the biological or chemical reactions that occur within the control volume. 5. Select a convenient basis on which the numerical calculations will be based.

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Application of the Mass Balance analysis

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In a complete-mix reactor, it will be assumed that:

1. The vol. flowrate into and out of the control volume is constant. 2. The liquid within the control volume is not subject to evaporation (constant vol.) 3. The liquid within the control volume is mixed completely. 4. A chemical reaction involving a reactant A is occurring within the reactor. 5. The rate of change in the concentration of the reactant A that is occurring within the control volume is governed by a first-order reaction (rc = -kC) MECALF & EDDY|Wastewater Engineering-treatment and reuse

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Accumulation = inflow – outflow + generation

dC V = QC o − QC + rcV dt

or

dC V = QC o − QC + (−kC )V dt

Steady state simplification

rc = Where:

Q ( Co − C ) V

dC/dt = rate of change of reactant concentration within the control volume, ML-3 T-1 V = volume contained within control vol., L3 Q = volumetric flow rate into and out of control volume, L3T-1 Co = conc. of reactant entering the control vol, ML-3 C = conc. of reactant leaving the control vol., ML-3 rc = first –order reaction (-kC), ML-3 T-1 k = first-order reaction rate coefficient, T-1cc

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Modeling ideal flow in reactors Ideal flow in complete-mix reactor Simplified word statement:

Accumulation = inflow – outflow

dC V = QC o − QC dt Simplifying by noting that Co = 0 yields

dC Q V =− C dt V c

t

dC Q = − dt ∫c C ∫ V 0 o

Where: C= conc. of the tracer in the reactor at time t, ML-3 Co = initial conc. of the tracer in the reactor, ML-3 t = time, T V = reactor volume, L3 Q = volumetric flow rate, L3T-1 τ = theoretical detention time, V/Q, T θ = normalized detention time t/τ, unitless

The resulting expression after integration is

C = Co e −t ( Q / V ) = Co e −t /τ = Co e −θ

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Modeling ideal flow in reactors Ideal flow in plug flow reactor Simplified word statement:

Accumulation = inflow – outflow

∂C ∆V = QC x −QC ∂t

x + ∆x

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Modeling ideal flow in reactors ∂C ∆C   ∆V = QC − Q C + ∆x  ∂t ∆x   A∆x = ∆V, A is the cross sectional area in the x direction

∂C ∆C A∆x = −Q ∆x ∂t ∆x

∂C Q ∆C =− ∂t A ∆x

Taking the limit as ∆x approaches zero yields

∂C Q ∂C ∂C =− = −v ∂t A ∂x ∂x

Where, v =velocity of flow, LT-1 (m/s) Where: ∂C/∂t = constituent concentration, ML-3 (g/m3) ∆V = differential volume element, L3 (m3) t = time T (s) Q = volumetric flowrate, L3T-1 (m3/s) x = some point along the reactor length L, (m) ∆x = differential distance L (m)

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Analysis of nonideal flow in reactors using tracers Need for tracer analysis

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The use of dyes and tracers for measuring the residence time distribution curves is one of the simplest and most successful methods

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Now, to assess the hydraulic performance of full scale reactors a. The assessment of short circuiting in sedimentation tanks and biological reactors b. The assessment of contact time in chlorine contact basins c. The assessment of the hydraulic approach condition in UV reactors d. The assessment of flow patterns in constructed wetlands and other natural treatment systems

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Properties of tracers

a. The tracer should not affect the flow. b. The tracer should be conservative. c. It must be possible to inject the tracer over a short time period. d. Be able to be analyzed conveniently. e. The molecular diffusivity of the tracer should be low. f. Not be absorbed on or react with the exposed reactor surfaces. g. Not be absorbed on or react with the particles in the wastewater. Success in tracer studies: congo red, fluorescein (C20 H12 O5), fluorosilicic acid (H2SiF6), hexafluoride gas (HF6), lithium chloride (LiCl), potassium permanganate (KMnO4)…. MECALF & EDDY|Wastewater Engineering-treatment and reuse

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Treatment processes involving mass transfer Table 4-9 Principal applications of mass transfer operations and processes in wastewater treatment Type of reactor

Phase equilibria

Application

Absorption

Gas ---> liquid

Addition of gases to water (e.g., O2, O3, CO2, Cl2, SO2), NH3 scrubbing in acid

Adsorption

Gas ---> solid

Removal of organics with activated carbon

Liquid ---> solid

Removal of organics with activated carbon, dechlorination

Solid ---> liquid

Sediment scrubbing

Solid ---> gas

Reactivation of spent activated carbon

Liquid ---> gas

Drying of sludge

Desorption Drying (evaporation)

Gas stripping (also known Liquid ---> gas as desorption)

Removal of gases (e.g.,O2,CO2, H2S, NH3, volatile organic compounds, NH3 from digester supernatant)

Ion exchange

Selective removal of chemical constituents, demineralization

Liquid ---> solid

Adapted from Crittenten (1999), McCabe and Smith (1976) and Montgomery (1985)

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MASS BALANCE USING BOD REMOVAL

From: Mathematics Manual for water and wastewater treatment plant operators

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• Step 1: BODin = influent BOD x flow x 8.34

1g/m3 = 8.34 lb/Mgal

• Step 2: BODout = effluent BOD x flow x 8.34 • Step 3: BOD pounds removed = BODin – BOD out • Step 4: Solids generated (lb) = BOD removed (lb x factor) • Step 5: Solids removed = sludge pumped (gpd) x % solids x 8.34 • Step 6: Effluent solids (mg/L) x flow (MGD) x 8.34 MECALF & EDDY|Wastewater Engineering-treatment and reuse

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Example Problem A conventional activated biosolids system with primary treatment is operating at the levels listed below. Does the mass balance for the activated biosolids system indicate a problem?

Solution • BODin = 166 mg/L x 11.40 MGD x 8.34 = 15,783 lb/day • BODout = 25 mg/L x 11.40 MGD x 8.34 = 2377 lb/day • BOD removed = 15,783 lb/d – 2377 lb/d = 13,406 lb/day • Solids produced = 13,406 lb/day x 0.7 lb solids/lb BOD = 9384 lb solids/day • Solids removed = 6795 mg/L x 0.15 MGD x 8.34 = 8501 lb/day • Difference = 9384 lb/day – 8501 lb/day = 883 lb/day, or 9.4% of solids produced. These results are within the acceptable range. MECALF & EDDY|Wastewater Engineering-treatment and reuse

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MASS BALANCE FOR SETTLING TANKS • Step 1: Solids in = pounds of influent suspended solids • Step 2: Pounds of effluent suspended solids • Step 3: Biosolids solids out = pounds of biosolids solids pumped per day • Step 4: Solids in — (solids out effluent + biosolids solids pumped)

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Ex, Problem A settling tank receives a daily flow of 4.20 MGD. The influent contains 252 mg/L suspended solids, and the unit effluent contains 140 mg/L suspended solids. The biosolids pump operates 10 min/h and removes biosolids at the rate of 40 gpm. The biosolids content is 4.2% solids. Determine if the mass balance for solids removal is within the acceptable 10 to 15% range.

Solution • Step 1: Solids in = 252 mg/L x 4.20 MGD x 8.34 = 8827 lb/d • Step 2: Solids out = 140 mg/L x 4.20 MGD x 8.34 = 4904 lb/d • Step 3: Biosolids solids = 10 min/hr x 24 hr/day x 40 gpm x 8.34 x 0.042 = 3363 lb/d • Step 4: Balance = 8827 lb/day – (4904 lb/day + 3363 lb/day) = 560 lb, or 6.3%

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Thank you for your attention!

%h =

Settled sludge volume (ml/l) *1000 suspended solids (mg/l)

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