Chapter 3Material & Energy Balance Eskedar T. (AAiT-CED)

Materials and Energy Balances Tools for quantitative understanding of the

behavior of environmental systems. For accounting of the flow of energy and

materials into and out of the environmental systems.

Materials and Energy Balances

Material Balance Energy Balance

production, transport, and fate modeling

Pollutant Energy

Unifying Theories The law of conservation of matter states that

(without nuclear reaction) matter can neither be created nor destroyed.

We ought to be able to account for the “matter” at any point in time.

The mathematical representation of this accounting system is called a materials balance or mass balance.

Cons

erva

tion

of M

atte

r

Unifying Theories The law of conservation of energy states that

energy cannot be created or destroyed. Meaning that we should be able to account

for the “energy” at any point in time. The mathematical representation of this

accounting system we use to trace energy is called an energy balance.

Cons

erva

tion

of E

nerg

y

Material Balances The simplest form of a materials balance or

mass balance

Accumulation = input – output

Accumulationinput

outputEnvironmental System

(Natural or Device)

The control volume

Accumulation

Control Volume

Food topeople

Consumergoods

Solid Waster

Examples of Control Volume

Time as a factor

Mass rate of accumulation = Mass rate of input – Mass rate of output

Example 3.1Selam is filling her bathtub but she forgot to put the plug in. if the volume of water for a bath is 0.350 m3 and the tap is flowing at 1.32 L/min and the drain is running at 0.32 L/min, how long will it take to fill the tub to bath level? Assuming Selam shuts off the water when the tub is full and does not flood the house, how much water will be wasted? Assume the density of water is 1,000 kg/m3

Solution

VaccumulationQin = 1.32 L/min Qout = 0.32 L/min

We must convert volumes to masses.Mass = (volume)(density)Volume = (flow rate)(time) = (Q)(t)

Solution From mass balance we have Accumulation = mass in –mass out (Vacc)() = (Qin)()(t) - (Qin)()(t) Vacc = (Qin)(t) – (Qin)(t) Vacc = 1.32t – 0.32t 350L = (1.00 L/min)(t) t= 350 min or 5.833 hr The amount of wasted water is

Waste water = (0.32)(350) = 112 L

Efficiencyrate)tion)(flow(concentra

timeMass

Mass flow rate =

outoutinin QcQcdtdM

Mass balance

inin

outoutinin

inin QCQCQC

QCdtdM

Efficiency of a system

inmassoutmassinmass

OR

Example The air pollution control equipment on a municipal waste incinerator

includes a fabric filter particle collector (known as a baghouse). The baghouse contains 424 cloth bags arranged in parallel, that is 1/424 of the flow goes through each bag. The gas flow rate into and out of the baghouse is 47 m3/s, and the concentration of particles entering the baghouse is 15 g/m3. In normal operation the baghouse particulate discharge meets the regulatory limit of 24 mg/m3.

Calculate the fraction of particulate matter removed and the efficiency of particulate removal when all 424 bags are in place and the emissions comply with the regulatory requirements. Estimate the mass emission rate when one of the bags is missing and recalculate the efficiency of the baghouse. Assume the efficiency for each individual bag is the same as the overall efficiency for the baghouse.

Solution

Accumulation =particle removal

Hopper

Cout = 24 mg/m3

Qout = 47 m3/s

Cin = 15 g/m3

Qin = 47 m3/s

Baghouse

Cont’doutoutinin QcQc

dtdM

= (15,000mg/m3)(47m3/s)-(24mg/m3)(47m3/s)=703,872 mg/sThe fraction of particulates removed is

=

The efficiency of the baghouse is

η==99.84%

Cont’d

Bypa

ss

= ?

Baghouse

Cin = 15 g/m3

QBypass =(1/424)( 47 m3/s)Cemission= ?Qemission= 47 m3/s

Cout= ?Qout= (423/424)47 m3/s

Cin = 15 g/m3

Qin =( )( 47 m3/s)

Control Volume

Cont’dA control volume around the baghouse alone reduces the number of unknowns to two:Because we know the efficiency and the influent mass flow rate, we can solve the mass balance equation for the mass flow rate out of the filter.

Solving for CoutQout CoutQout =(1- η)CinQin= (1-0.9984)(15,000mg/m3)(47m3/s)() =1,125mg/s

inin

outoutinin

QCQCQC

Cont’d

=CinQin from bypass +CinQin from baghouse-CemissionQemission

Because there is no accumulation in the junction =0 and the mass balance equation CoutQout=CinQin from bypass +CinQin from baghouse

= (15,000mg/m3)(47m3/s)(1/424)+1,125=2788 mg/sThe concentration in the effluent is = = 59 mg/m3

The overall efficiency of the baghouse with missing bag is η== 99.61%

Bypass

From baghouse

Effluent

Exercise A storm sewer is carrying snow melt

containing 1.200 g/L of sodium chloride into a small stream. The stream has a naturally occurring sodium chloride concentration of 20 mg/L. If the storm sewer flow rate is 2.00 L/min and the stream flow rate is 2.0 m3/s, what is the concentration of salt in the stream after the discharge point? Assume that the sewer flow and the storm flow are completely mixed, that the salt is a conservative substance (it does not react) and that the system is at steady state.

Reaction Order and Reactorskinetic reactions : reactions that

are time dependent. Reaction kinetics: the study of

the effects of temperature, pressure, and concentration on the rate of a chemical reaction.

Rate of reaction The rate of reaction, ri, the rate of formation

or disappearance of a substance. +r, r=-KCn

Homogenous reactions. single phase reactions

Heterogeneous reactions : multiphase reactions (between phases surface)

Rate of reaction ri = kf1(T,P);f2([A],[B], …)Rate constant Concentration of reactant

Assuming that the pressure and temperature are constantaA + bB cC

Rate of reaction rA = - k[A]α[b]β = k[C]γ

Order of reactionRate of reaction rA = - k[A]α[b]β = k[C]γ

order of reaction = α + β,the order with respect to reactant A is α, to B is β, and to product C is

γ. rA = -k zero-order reactionrA = -k[A] first-order reactionrA = -k[A2] second-order reactionrA = -k[A][B] second-order reaction

Order of reaction

Types of Reactorsbatch reactors and flow reactors.

fill-and-draw

Unsteady statematerial flows into, through,

and out of the reactor

Flow reactorsIDEAL REACTORS REAL REACTOR

Steady-state conservative system Conserved system: where no chemical or

biological reaction takes place and no radioactive decay occurs for the substance in the mass balance.

Steady-state: Input rate = Output rate Accumulation =0

Steady-state conservative system

Mixture

Wastes

StreamQm

CmQw

Cw

Qs

Cs

Decay rate = 0Accumulation rate = 0

Q = flow rate C = concentration

CsQs + QwCw = QmCm

Including reactions For non-conservative substancesAccumulation rate = input rate – output rare

± transformation rate

Simple completely mixed systemsWith first-order reactions Total mass of substance = concentration x volume

when V is a constant, the mass rate of decay of the substance is

first-order reactions can be described by r = -kC=dC/dt,

ExampleA well-mixed sewage lagoon is receiving 430 m3/d of sewage out of a sewer pipe. The lagoon has a surface area of 10 ha and a depth of 1.0m. The pollutant concentration in the raw sewage discharging into the lagoon is 180 mg/L. The organic matter in the sewage degrades biologically in the lagoon according to first-order kinetics. The reaction rate constant is 0.70 d-1. Assuming no other water losses or gains and that the lagoon is completely mixed, find the steady-state concentration of the pollutant in the lagoon effluent.

Solution

SewageLagoon

Cin = 180 mg/LQin = 430 m3/d

Ceff = ?Qeff = 430 m3/d

Decay

Accumulation=input rate – output rate – decay rateAssuming steady-state condition, accumulation = 0

input rate = output rate + decay rateCinQin = CeffQeff + (K)(Clagoon)(V)

Ceff=1.10mg/ L

Control volume

Batch reactors

0)()(

dtoutd

dtind

kCVdtdM

kCdtdC

VdtdC

dtdM

dM/dt = ?

ktoeCC

Plug-flow reactors

dtCdV

dtoutd

dtind

dtdM )()()(

Mass balance for each plug element

No mass exchange occurs across the plug boundaries, d(in) and d(out) = 0

kCdtdC

VdtdC

dtdM

ktinout eCC

Plug-flow reactors

The residence time for each plug:

koutin eCC Residence time

L=length

QV

AuAL

)()()()(

QVk

uLkk

CC

in

out )()(ln

ExampleA wastewater treatment plant must disinfect its effluent before discharging the wastewater to a near-by stream. The wastewater contains 4.5 x 105 fecal coliform colony-forming units (CFU) per liter. The maximum permissible fecal coliform concentration that may be discharged is 2,000 fecal coliform CFU/L. It is proposed that a pipe carrying the wastewater be used for disinfection process. Determine the length of the pipe required if the linear velocity of the wastewater in the pipe is 0.75 m/s. Assume that the pipe behaves as a steady-state plug-flow system and that the reaction rate constant for destruction of the fecal coliforms is 0.23 min-1.

Solution

Using the steady-state solution on the mass-balance equation, we obtain

Solving for the length of pipe, we have ln(4.44x10-3)=-0.23min-1

L=1,060m

Cout=2,000 CFU/LU=0.75 m/s

Cin=4.5x105 CFU/LU=0.75 m/s

min)/60)(/75.0(min23

/105.4/2000ln

ln

15 ssm

LLCFUxLCFU

uLK

CC

in

out

ExampleA contaminated soil is to be excavated and treated in a completely mixed aerated lagoon. To determine the time it will take to treat the contaminated soil, a laboratory completely mixed batch reactor is used to gather the following data. Assuming a first-order reaction, estimate the rate constant, k, and determine the time to achieve 99 % reduction in the original concentration.

Time (d) Waste Concentration (mg/L)1 28016 132

Solution Using the 1st and 16th day, the time interval t=

16-1 = 15 dkt

o

t eCC

)15(

/280/132 dkeLmgLmg

Solving for k, we have k = 0.0501 d-1

To achieve 99 % reduction the concentration at Time t must be 1 – 0.99 of the original concentration

)(05.001.0 t

o

t eCC t = 92 days

Step function response

0Time

Step increase0

TimeStep decrease

0Time

pulse/spike increase

Conc

entra

tion,

Cin

Batch reactorCo

ncen

tratio

n, C

in

0Time

1/kTime

Decay

1/kTime

Formationkt

o

t eCC

CMFR ConservativeCo

ncen

tratio

n, C

in

0Time

Influent concentration

Conc

entra

tion,

Cin

C1

C0

TimeEffluent concentration

0 t=

C1

C0

-0.37C0 + 0.63C1

C0

Conc

entra

tion,

Cin

Conc

entra

tion,

Cin

C0

0.37C0

CMFR Conservative For balanced flow (Qin = Qout) and no

reaction, the mass balance becomes

outoutinin QcQcdtdM

Where M = CV. The solution is

tCtCCt exp1exp 10

Where = V/Q

CMFR Conservative Flushing of nonreactive contaminant from a

CMFR by a contaminant-free fluid Which means Cin = 0 and the mass balance

becomesoutoutQc

dtdM

Where M = CV. The initial concentration is C0=M/V

tCCt exp0For time t 0 we obtain

ExerciseBefore entering an underground utility vault to do repairs, a work crew analyzed the gas in the vault and found that it contained 29mg/m3 of H2S. Because the allowable exposure level is 14 mg/m3 the work crew began ventilating the vault with a blower. If the volume of the vault is 160 m3 and the flow rate of contaminant-free air is 10 m3/min, how long will it take to lower the H2S level to that will allow the work crew to enter? Assume the manhole behaves as CMFR and that H2S is nonreactive in the time period considered.

CMFR NonConservative For balanced flow (Qin = Qout) and first-order

reaction the mass balance becomes

VkCQCQCdtdM

outoutoutinin

Where M = CV. By dividing with Q and V we have

outoutin kCCCdtdC

1

CMFR NonConservative

0Time

0Time

Conc

entra

tion,

Cin Co Co

For stead-state conditions dC/dt=0

kCC o

out

1 kCC o

out

1ORDecay Formation

Conc

entra

tion,

Ceff

CMFR NonConservative

VkCQCdtdM

outoutout 0

A step decrease in influent concentration (Cin=O)for non-steady-state conditions with first-order decay

Where M = CV. By dividing with Q and V we have

outCkdtdC

1

t

kCCC o

oout 1exp

CMFR NonConservative

0Time

Influent concentration

0Time

Effluent concentration

C0

Conc

entra

tion,

Cin

Conc

entra

tion,

Cin

C0

ExampleA chemical degrades in a flow-balanced, steady-state CMFR according to first-order reaction kinetics. The upstream concentration of the chemical is 10 mg/L and the downstream concentration is 2 mg/L. Water is being treated at a rate of 29 m3/min. The volume of the tank is 590 m3. What is the rate of decay? What is the rate constant?

Solution For a first-order reaction, the rate of decay, r =-

kC, thus we have to solve for kC from

VkCQCQCdtdM

outoutoutinin

For steady-state, dM/dt = 0 and for balanced flow, Qin = Qout

3

3

580min)/29)(/2/10(

mmLmgLmg

VQCQCkCr outoutinin

r=kC=0.4

Solution… For a first-order reaction in a CMFR

kCC o

out

1

The mean hydraulic detention time ismin20

min/29580

3

3

mm

QV

Solving for the rate constant we get1min20.0

min201)/2//10(1)/(

LmgLmgCCk outo

Mini Project

Report format20 to 30 pages

Cover page: Title and Group member names

Table of contentReferences

Last Date of Submission: June , 2011