Design of Facilities for Physical and Chemical Treatment of Wastewater

Material and Energy Balance, Page No.11

Material and Energy Balance

Lecture No. 3

1. Purpose of Material Energy Balance

first step in understanding a process

leads to a better understanding of a process

forces the engineer to be 100% complete

locates imbalances

serves as a starting point for analysis/design

serves as a model for change

2. Flowing Streams of Material

Industrial plants, manufacturing plants and waste treatment plants have at least two things in common:

- Energy usage

- Material flow

Conservation of Mass. If one or more streams of material are flowing into a region of space, a process unit with boundaries, then material must be either flowing out of that region at the same rate or accumulating in the region.

Measurement of Rate of Flow. Units must be correct all the time regardless of conditions such as mass/time or moles/time. Volumetric units would only be correct if the densities of materials are identical and constant.

Example

Given: V=3 m/s, ( = 4. Then the pipe splits into 2-2 pipes.

Find: V and Q in the 2 pipes. (w = 1000 kg/m3. ID-4=10.23cm, ID-2=5.25cm

1. 4 pipe

M4= (VA = (Q ( Q=VA if ( is constant.

M4= 1000 kg/m3 x 3 m/s x [(/4 x (10.23)2 cm2 x (1m/100cm)2]

M4= 24.66 kg/s

2. 2 pipe

A. Q

Q= M/( = 24.66/2 (2-2pipes) kg/s / 1000 kg/m3Q= .01233 m3/s

B. V

V= Q/A = .01233 m3/s / [(/4 x (5.25)2 cm2 x (1m/100cm)2]V= 5.70 m/s

3. Materials Balance Equation

An accounting of all material into and out of an identifiable process area.

eq \F(dM,dt) = Min -Mout

The General Balance for component I is:

(Accumulation rate)i = (input rate)i - (output rate)i + (generation rate)i

We are usually interested in steady state conditions which means that the accumulation rate is zero, eq \F(dM,dt) = 0. Also if there are no reactions, the generation rate is zero:

0 = (input rate)i - (output rate)i + 0

input rate = output rate

4. Choosing Basis and Boundaries

Imagine, picture the region in space to be analyzed.

The boundaries across which mass is flowing must be analyzed

Use flow diagrams

The basis may be a unit of time or a unit of material, i.e. per day or per widget.

The most general approach is to use equations and solve in terms of flow rates. It is convenient and easy to pick a specified time interval (per day) or amount of material (100 lbs) and solve in terms of the mass of material.

Write non-steady state equations.

Label, identify known streams and data

Solve

Example:

Given: Rainwater is collected in barrels for drinking. The intensity of the rainfall is 4.0 cm/hr. ((barrel) = 50cm. The duration of the storm, t, is 2 hrs.

Find: How much water is collected incident to the storm.

(Accumulation rate)i = (input rate)i - (output rate)i + (generation rate)i

Water is not generated within the boundaries of the barrel; the output rate is zero. If there were an open spigot or a leak in the barrel, the output rate would not be zero. The equation reduces to:

(Accumulation rate)i = (input rate)i

(Accumulation rate)i = 4.0cm/hr x (/4 (50)cm2 x 1gm/cm3 x 1kg/1000gm

(Accumulation rate)i = 7.85 kg/hr, t=2 hrs

Accumulation = rate x time = 7.85 kg/hr x 2 hrs

Accumulation = 15.7 kg

5. Steady State (Without chemical reactions)

There are no changes in the parameters with respect to time, eq \F(d(parameter),dt) = 0.

Flow rate, temperature, pressure, liquid levels, etc. are constant.

True steady state is rarely achieved for extended periods of time; it is often approximated to a reasonable degree.,

Designs are usually based on steady state conditions and achieving steady state is a goal of most process operators and operations.

At steady state, the accumulate rate term is zero. In a mathematical solution, the accumulation rate term usually necessitates the use of differential equations; without the accumulation rate term, the solution is usually an algebraic one.

Example, Steady State, without chemical reactions, no recycle

Given: A baghouse (air pollution control device). Q=100m3/min. The dirty input air is 15.0g/m3 and the cleaned output air is .020g/m3. The state criteria is .90g/m3. Some air is by-passed and the total (by-passed air and treated air) is blended.

Find: A. kg/day of dust. B. Q of air through the baghouse such that the permit requirements are met.

1. General-Sketch the region and draw boundaries, use flow diagram.

2.Choose a basis - per day

3.Write Equations and Solve

Let the air to be by-passed = X

Let the air through the baghouse = Y

A. Equations

AR=IR-OR+GR

Steady state, therefore, AR=0

No dust is being produced, originated, within the boundaries, therefore, GR=0

0=IR-OR+0

There are 2 output streams with dust:

Output rate from baghouse:

z = 100m3/m x 15.0g/m3 - 100m3/m x .90g/m3z = 1410g/min=2030kg/day

For air, from splitter balance:

0=IR-OR

IR = OR

100m3/m x (air = X(air + Y(airAssume (air is constant. If (air is not constant, then the temperatures and pressures must be know and the (air calculated.

For dust, from mixer balance

100m3/m x .9g/m3 = Xm3/m x 15.0g/m3 + Ym3/m x .02g/m3Solving 2 equations for 2 unknowns:

90.0g/m = (100-Y)15.0 + .02Y

Y=94.126m3/m, Q of air through the baghouse

Example, Steady State without chemical reactions, with recycle.

Given: Two liquids produce a third liquid and a gas. The gas is produced at a mass ratio of 1 unit of gas for every thousand units of liquid product.

StreamFlowrateDensity

F11000 L/min1.5 Kg/L

F26000 Kg/min1.2 Kg/L

Product, Pnot given1.3 Kg/L

Recycle, R.5 x PSame as P

Gas, Gnot given1.0 Kg/min

Find: P, G, R (L/min) and the total Q in the reactor.

Note: If the boundaries are drawn around the entire system, it is as if the recycle stream does not exist; the recycle stream does not cross the system boundaries.

Assume: Steady state, AR=0, and nothing is being generated, no chemical reactions, GR=0, therefore,

Input = Output

F1 + F2 = G + P

Note: R does not appear, G=.001P

1000 L/min x 1.5 Kg/L + 6000 Kg/min = .001P + P

P = 5763 L/min x 1.3 Kg/L

P= 7492 Kg/min

G = .001P = .001(7492)

G = 7.49 Kg/min x m3/kg

G= 7490 L/min

R = .5P = .5(7490) = 3746 Kg/min x L/1.3Kg

R = 2882 L/min

Fresh Feed Rate = F1 + F2

Total Feed Rate = F1 + F2 + R = 7492 + 3746

Total Feed Rate = 11,238 Kg/min

6. Steady State (with chemical reaction)

A. General

With a reaction step, the reactor converts at least one reactant to at least one product. Mole balances are in order.

The generation rate refers to the overall rate of generation within the system boundaries which is usually the intrinsic reaction time multiplied by the system volume. The component being used up has a negative rate for example, the elimination of cadmium.

The intrinsic reaction rate is defined as the time rate of change in the number of moles, N, due to the reaction of a given component for unit volume, V. This reaction rate is easily determined in the laboratory.

For A ( B

-rA = rB = eq \F(NB, V) where r = intrinsic reaction rate

If the system is well mixed, constant V, eq \F(NB, V) = CB (mg/l)

rB = eq \F(dCB,dt) Using a simple kinetic model to predict rA and rBrA = -kCAarB = kCBbk = rate constant, temperature dependent

a,b = empirically determined order or exponent

Two common models of ideal chemical reactors: completely mixed and plug flow.

B. Completely Mixed Systems

In a completely mixed system, the contents of the reactor have the same composition everywhere. The reactants flowing into the tank are immediately diluted to the final concentration of the tank.

Example:

Given: A biological process in which there is a 98% conversion of reactants to product.

rA = -kCA , k 10s-1. Qin=75L/s, CA0=.05Mole/L. No volume change.

Find: CM Volume.

AR=IR-OR+GR

Steady state, therefore, AR=0

0 = Q0CA0 - QeCAe - KCAeVNote: The concentration in the generation term, CAe, is the exit concentration. Q0=Qe.

KCAeV = QCA0 - QCAe = Q(CA0 - CAe )

V =

eq \F(CA0 - CAe, CAe) ( eq \F(CA0, CA0) V =

( eq \F(1-CAe /CA0,CAe/ CA0))

V = ( eq \F(75 L/s,.1s-1)) x eq \F(98,.02) V = 36,750 L

C. Plug Flow Reactor

In ideal plug flow, the flow is one dimensional, the velocity is constant throughout the cross section and dispersion is negligible.

Example:

Given: A plug flow system.

Find: The concentration at the end of the process and the required volume given the information from the previous example.

1. Concentration

AR=IR-OR+GR

AR = QxCAx - Qx+(xCx+(x + rA(VSteady state, therefore, AR=0, assume dQ/dx=0, therefore, Q=Q+(x

Divide by Ax-section(x = (V0/Ax-section(x = QxCAx - Qx+(xCx+(x + rA(V / Ax-section(x

0 = eq \F(Q(CAx - Cx+(x),A(x) + eq \F(rA(V,A(x) , V = Q/A

0 = V eq \F(CAx - Cx+(x,(x) + rAAs (x ( 0, rA = KCAa0 = - V eq \F(dCA,dx) - KCAaV eq \F(dCA,dx) = - KCAa

eq \F(dCA,CA) = - eq \F(K,V)

ln eq \F(CAL,CA0) = -K eq \F(L,V) CAL = CA0e-k(( = reactor residence time, L/ V or V/Q

2. Volume for a Plug Flow System

ln eq \F(CAL,CA0) = -K eq \F(L,V) ln eq \F(.02CA0,CA0) = -.1s-1 (( = 39.1 s

V = Q/( = 75 L/s x 39.1s

V = 2932 L v. 36,750 for the completely mixed system. Plug flow systems require much smaller tanks than comparable completely mixed systems.

D. More Complicated Problem

Given: A chemical or biological contaminant, S, substrate, which is converted to CO2, H2O and more bacteria or biomass, X. The yield of the biomass is .5g of biomass produced per gram of substrate converted. The substrate intrinsic reaction rate,

-rS = .03S1.5 , mg/l.day

Assume:

Completely mixed system

No reactions in the final clarifier or piping

steady state

( = constant, H2O

Find:

1. The conversion rate of the substrate (kg/day)

2. The growth rate of the biomass (kg/day)

3. Concentration of Biomass in Effluent Leaving the Final Clarifier, mg/L

4. The volume of the reactors.

1. Conversion Rate

Balance around the whole system for the 3 components (substrate, biomass, flow) of interest.

AR = IR - OR + GR

Steady state system, therefore, AR=0

Substrate: 0 = QSf - ESe - WSw + rSVBiomass: 0 = 0 - EXe - WXw + rXVFlow: 0 = Q - E -W

0 = Q - E -W, W=.02Q, Q=4.8x106L/day

0 = 4.8x106L/day - E - .02(4.8x106L/day )

E = 4.704x106L/day

W=.02Q = .02(4.8x106L/day )

W= 9.6 x 104 L/day

0 = QSf - ESe - WSw + rSV substrate equation, solve for last term

rSV = -QSf + ESe + WSw

rSV = -4.8x106L/day x 300 mg/L

+ 4.704x106L/day x 20 mg/L x 1Kg/106mg

+ 9.6 x 104 L/day x 20 mg/L x 1Kg/106mg

rSV = -1440 + 94.1 +1.9

rSV = -1344 Kg/day

2. Growth Rate of Biomass

rxV = .50(-rSV) = .50(-(-1344))

rxV = 672 Kg/day

3. Concentration of Biomass in Effluent Leaving the Final Clarifier

At steady state, nothing accumulates, the biomass must be discharged at the same rate as it grows. Using the biomass equation from the previous section.

0 = 0 - EXe - WXw + rXVXe = eq \F(rXV - WXw ,E) = eq \F(672 Kg/day x 106mg/kg - 9.6x 104L/day x 6000mg/L,4.704x106L/day) Xe = 20.4 mg/L

4. Volume

rSV = -1344 Kg/day

-rS = .03S1.5 (given)

-.03(20)1.5 V = -1344 Kg/day

V = 5.01 x 108L

7. Non-Steady State (Transient Processes)

Transient means non-steady state processes.

The accumulation term is not zero.

A calculus solution is often required.

Examples:

Given: An oil tanker off loads 300,000 gal of crude to the refinery. It cannot overfill a tank or let a pump run dry. Q(ship) = 20,000 bbls/hr. Tank diameter=100. The maximum tank height is 48 and pumping starts with the liquid level of the tank at 6.

Find: How long do the refinery personnel have to prepare a second tank?

1bbl=42gallons.

AR = IR - OR + GR

Since no oil is being generated, GR=0; There is no output from the tank, OR=0

AR = IR

eq \F(dV,dt) = 20,000 bbl/hr x 42 gal/bbl x 1ft3/7.48gal

eq \F(dV,dt) = 1.123 x 105 ft3/hr

(V = 1.123 x 105 ft3/hr x (t

(V = eq \F((,4) D2 h == eq \F((,4) (100)2 (48-6)

(V = 330,000 ft3Solving for (t

(t = 330,000 ft3/1.123 x 105 ft3/hr

(t = 2.94 hours

Example:

Given: Waste is discharged to a holding pit which then overflows to river. The waste is normally 10mg/l which is not a problem. Suddenly the waste increases in concentration to 100 mg/l. Q=100,000 L/day. V of the holding pit is 106 L.

Find: The concentration discharging to the river in 10 days.

AR = IR - OR + GR

No waste is being generated in the pit, therefore, GR=0

eq \F(d(CeV),dt) = QCi - QCeDivide both sides of the equation by the volume, V.

eq \F(d(Ce),dt) + eq \F(Q,V) Ce = eq \F(d(QCi),V) ( = Hydraulic Residence Time, HRT = eq \F(V,Q) eq \F(d(Ce),dt) + eq \F(Ce,() = eq \F(Ci,() ( and Ci are constants and do not vary with time. The above equation is a linear 1st order differential equation. The general solution is:

Ce = C0 + (Ci - C0)(1-e-t/()

where C0 is the concentration of the exit stream when the inlet stream suddenly jumps to 100 mg/l. At t=0, Ce=C0, at t=(, Ce=Ci.

Ce = 10mg/l + (100-10)(1-e-10/10)

Ce = 67 mg/l after 10 days

Example:

Given: A rooftop holds rainwater but puddles because of poor drainage. Vmax = 3.5m3, Q(storm) = 7m3/hr. Q (leak) = .0167m3/hr. Q (evaporation) = .0004 m3/hr

Find:

1. After it starts raining, how long before the roof overflows.

2. The rain stops after 1 hour; how long will the water be on the roof.

3. How much water will leak from the time the rain starts until the roof is completely dry?

Assumptions:

1. ( = constant

2. GR=0

3. The roof is dry at the start of the rain, V0=0

Basis=time

A. Mass balance for system, t

AR = IR - OR + GR

No rain is being produced on the roof, GR=0.

eq \F(d((V),dt) = (R - (E - (L ( by (

= dV

t = eq \F(Vf - Vo,R-E-L) = eq \F(3.5-0,7-.0004-.016) = 3.5/6.98363-0,7-.0004 - .016)

t= = .501 hrs

B. Rain on Roof

=

t = eq \F(0-3.5m3,0-.0004 - .016) t= 213 hours

C. Leakage

Total time elapsed = 1+213 = 214 hours

Total V = L(t = .016m3/hr x 214hrsV = 3.43m3HOMEWORK No. 3, Material and Energy Balance

Problems: Solve all problems using the AR = IR - OR + GR approach

3A. Given: 5x106 kg of coal are burned per day in a steam power plant. The coal has an ash content of 12%, 12% remains after combustion. 40% of the ash falls out of the bottom of the furnace. The rest of the ash is carried out of the furnace with the hot gases into air pollution device known as an electrostatic precipitator, ESP. The ESP is 99.5% efficient in removing the ash that it sees.

Find: 1.) Draw a schematic 2.) The mass emission rate into the atmosphere

Assumptions:

1. Since there is no accumulation of coal, the system is steady state, AR=0.

2. There is no chemical or biological reaction is which a material is being generated or destroyed.

3. Basis = time

3B. Given: A stream flowing at 10 m3/s has a treatment plant flowing into it at 5.0m3/s. The concentration of chlorides in the stream is 20.0 mg/l and the plants concentration is 40.0 mg/l.

Find: The concentration of chlorides downstream of the confluence.

Assume:

1. The chloride is a conservative substance, i.e. no chemical reactions, meaning that GR=0. This is probably a poor assumption in the long term, but is probably accurate at the point where the two streams combine.

2. The two sources become completely mixed. This is probably another poor assumption. Mixing is a high energy endeavor. If the river is the River Wild good mixing will ensue; if the river is the Mississippi, placidly plodding along, the two streams will not readily combine.

3. The process is steady state meaning AR = 0. In plane English, water is not being accumulated, stored, in the process. If it started raining and/or the waste treatment plant changed its flow, this would not be steady state and there would be an accumulation term. The nature of the physical change for AR(0 would be the river rising on its banks. This would be a flood condition. In the other direction the river would decrease in volume, a drought condition.

4. Q refers to flow; C refers to concentration.

3C. Given: A clarifier-thickener system to separate solids and liquids.

StreamFlow Rate

L/sSolids

mg/l

A1003000

B9515

C6000

D50

E

A

Find: Complete the table.

Assume:

1. Ignore the effects of solids contents on the density of the stream, i.e. all specific gravities are water, sg=1. Lousy assumption in the real world, but makes the problem easier at this stage. Constant density, ( = constant.

2. Steady state, AR=0. No sludge is being stored within the system boundaries; what goes in, goes out.

3. No generation of sludge within the system, GR=0. No sludge is being created or destroyed. No chemical changes. No biological changes.

4. Q refers to flow; C refers to concentration.

5. Basis = time

3D. Given: A couple is building a beach on a lake in front of their house. Every weekend they add 5 yd3 of sand. It is windy and he losses 2 ft3/day of sand as he spreads the sand. He needs 75 yd3 to complete the project.

Find: How many extra weekends must be work because of he wind?

Assumptions:

1. The sand is of constant density, ( = constant

2. No sand is being created or destroyed or changed into something else. No chemical or biological reactions. GR=0.

3. Two days in a weekend

4. Basis = time

3E. Given: A lake with a volume of 10x106 m3. Its fed by a stream, Q=5.0m3/s and the concentration of chromium is 10 mg/l. There is also a sewage outfall, Q=.5m3/s and the concentration of chromium is 100 mg/l with a reaction coefficient of .20/day.

Find: The steady state concentration of chromium in the lake.

Assumptions:

1. Complete mixing, a completely mixed system. This means that the mixing is instantaneous and the concentration in the lake is the same as the concentration of the mix leaving the lake.

2. Since a steady state condition is desired, AR=0

3. Since there is a reaction, GR(0.

3F. Given: The same lake as above which was found to have a steady state concentration of 3.5 mg/l. The condition of the lake was deemed unacceptable and to solve the problem it was decided to divert the sewage flow around the lake, thus eliminating the sewage as a chromium source. This problem is very complicated and need not be attempted; it is FYI, for your information.

Find: 1.) Find the new steady state concentration of chromium.2.) Find the concentration of chromium after 1 week. Assume a completely mixed system.

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