Mout minlEm

1. The mass flowrate out, mouV for an E of 95% is wout = 100(1 - 95/100) = 5 lb/h

2. The mass flowrate out, mout, for an E of 99% is mout= 100(1 -99/100) = 1 lb/h

3. The mass flowrate out, wout, for an E of 99.9% is wout = 100(1 -99.9/100) = 0.1 lb/h

4. The mass flowrate out, wout, for an E of 99.99% is mout = 100(1 -99.99/100) = 0.01 lb/h

5. The mass flowrate out, moat, for an E of 99.9999% is

REG.7 CORRECTION TO 7% OXYGEN

The carbon monoxide (CO) concentration in the stack from a solid hazardous waste treatment facility is measured at 20ppmv at a temperature of 175°F. The oxygen concentration in the stack is measured to be 12% by volume on a wet basis. The water content of the stack gas is lOmol %.

1. What is the corrected CO content in the stack gas?

2. Similarly, the particulate concentration in the stack is measured at 20mg/dscf at a stack oxygen concentration of 12% on a dry basis at 70° F. What is the corrected particulate concentration?

REG.7 CORRECTION TO 7% OXYGEN

3. Federal regulations (in SI units) require that corrected particulate concentrations do not exceed 180mg/dscm corrected to 7% 02. Does this stack meet the regulations?

Many current waste regulations require that emissions be corrected to 7% oxygen in the stack on a dry basis. The correction formula provided in the regulations (see 40 CFR Part 264.343) is as follows:

where Pc = corrected concentration PM = measured concentration Y = 02 concentration in stack, dry basis

Solution

Determine the percent by volume of CO and N2 in the stack:

02 concentration = 12% H20 content = 10%

Calculate the 02 concentration in the stack in mole fraction units. Basis 100 mol:

Y = 02 concentration in the stack = 12/90 = 0.1333 = 13.33%

Calculate the corrected CO content in the stack gas in ppm using the equation given above. For CO:

Pc= ^[14/(21 -Y)] = 20[14/(21 - 13.33)] = 36.5 ppm CO (corrected to 7% 02)

Calculate the corrected particulate concentration in the stack in mg/dscf. For particulates:

Determine if the stack meets regulations.

Convert mg/dscf to mg/dscm:

(31.1 mg/dscf)(35.3 ft3/m3) = 1098 mg/dscm Therefore, this stack does not meet regulations.

A hazardous waste incinerator is burning an aqueous slurry of soot (carbon) with the production of a small amount of fly ash. The waste is 70% water by mass and is burned with 0% excess air (EA). The flue gas contains 0.25 grains (gr) of particulates in each 7 ft3 (actual) at 620°F. For regulation purposes, calculate the particulate concentration in the flue gas in gr/acf, in gr/scf, and in gr/dscf.

Solution

Calculate the particulate concentration in the flue gas per actual cubic foot:

Convert actual cubic feet to standard cubic feet:

60° F = 520°R 620° F = 1080°R 7.0 acf (520/1080) = 3.37 scf

Calculate particulate concentration in standard cubic feet:

To determine the volume fraction of water in the flue gas, the reaction equation for the combustion of the waste is

The number of moles of each component becomes

C: 30 lb/(12 lb/lbmol) = 2.5 lbmol H20: 70 lb/(l 8 lb/lbmol) = 3.89 lbmol

Calculate the moles of C02 and H20 in the flue gas:

REG.9 PARTICULATE AND HC1 EMISSIONS 449

The stoichiometric (0% excess air) air requirement for this solid waste can now be determined:

N2: (79/21X2.5 lbmol) = 9.4 lbmol Air: 9.4 + 2.5 = 11.9

The oxygen and nitrogen in the flue gas are therefore

Dividing the moles of H20 by the total moles yields the mole fraction of H20 in the flue gas:

H20 fraction = lbmol H20/(C02 + H20 + N2) lbmol = 3.89/(2.5 + 3.89 + 9.4) = 0.25 (by mole or volume)

The molar quantity of dry gas is obtained by subtracting the moles of H20 from the total moles to yield

Moles dry gas = (2.5 + 3.89 + 9.4) - 3.8 = 12.0 lbmol

The dry volume in dry standard cubic feet (dscf) is obtained by subtracting the volume of H20 from the total volume to yield

The particulate concentration on a dscf basis becomes

REG.9 PARTICULATE AND HC1 EMISSIONS

A hazardous waste incinerator is burning a waste mixture containing solids with 50% excess air at 2100°F with a residence time of 2.5 s. The stack gas flowrate was determined to be 14,280 dscfm. The composition of contaminants in the stack gas is given below:

 Compound Inlet (lb/h) Outlet (lb/h) Toluene 860 0.20 Chlorobenzene 450 0.02 Dichlorobenzene 300 0.03 HCl 4.2 Particulates 9.65 Pb 0.05 lb/100 lb (TOC)

TOC = total organic compounds

TOC = total organic compounds

1. Calculate the DREs for toluene, chlorobenzene, dichlorobenzene, and HCl.

2. Calculate the discharge concentrations of Pb and particulates.

3. Is the unit in compliance with present federal regulations? Assume chlorobenzene and dichlorobenzene are the POHCs (principal organics) for this incinerator.

Solution

1. DREs for each compound are calculated as follows: Toluene:

Chlorobenzene:

i0 /4501bin/h-0021bou^= 99996

Dichlorobenzene:

/3001b in/h -0.03 lb out/h\ 100 -„„„„ . „-) = 99.99%

Molar HCl production is calculated based on all CI coming into the incinerator, and assuming that all CI is converted to HCl upon combustion.

REG.9 PARTICULATE AND HC1 EMISSIONS 451 For chlorobenzene, the amount of HC1 produced is calculated assuming:

HCl produced from chlorobenzene

450 lb/h

~ 6(12 lbC/lbmol) + 5(1 lbH/lbmol) + 1(35.45 lb Cl/lbmol) = 4.00 lbmol HCl/h

For dichlorobenzene, the amount of HCl produced is calculated assuming: C6H4C12 + 6.502 6C02 + 2HC1 + H20

HCl produced from dichlorobenzene

~~ 6(12 lb C/lbmol) + 4(1 lb H/lbmol) + 2(35.45 lb Cl/lbmol)

Therefore,

Total amount of HCl produced

= (4.00 + 2.04) lbmol (36.45 lb/lbmol) = 220.2 lb HCl/h

The DRE for HCl is

2. The amount of Pb in the waste mixture is based on the TOC content of the waste as given by:

TOC for toluene (C7H8):

220.2 lb HCl in/h - 4.22 lb HCl out/h 220.2 lb HCl in/h

Mass of C in toluene 7(12)

Toluene TOC emission rate = (0.913)(8601btoluene/h)

TOC for chlorobenzene (C6H5C1, CB):

Chlorobenzene TOC emission rate = (0.64) 4501bCB/h

TOC for dichlorobenzene (C6H4C12, DCB):

Dichlorobenzene TOC emission rate = (0.49) 300 lb DCB/h

The amount of Pb in the flue gas is:

Total emission of TOC = (785 + 288 + 147 lb TOC/h) = 1220 lb TOC/h

Total mass of Pb = (12201bTOC/h)(0.005 lbPB/lbTOC) = 6.1 lb Pb/h

Concentration of Pb in the incinerator flue gas is

The particulate concentration in the flue gas is calculated as follows:

Total emission of particulates = 9.65 lb/h (7000gr/lb)

The concentration of particulates in the incinerator flue gas is

67,550 gr/h

3. The unit is in compliance with present federal regulations for POHCs and particulates but is NOT in compliance for HCl emissions.

REG. 10 COMPLIANCE STACK TEST

A compliance stack test on a facility yields the results below. Determine whether the incinerator meets the state particulate standard of 0.05 gr/dscf. Estimate the amount of particulate matter escaping the stack, and indicate the molecular weight of the stack gas. Use standard conditions of 70°F and 1 atm pressure.

Pitot tube measurements made at eight points across the diameter of the stack provided values of 0.3, 0.35, 0.4, 0.5, 0.5, 0.4, 0.3, and 0.3 in. of H20.

Use the following equations for S-type pitot tube velocity, v (m/s), measurements:

where g = gravitational acceleration 9.81 m/s2 H = fluid velocity head, in. H20 p, = density of manometer fluid, 1000 kg/m3 p,d — density of flue gas, 1.084 kg/m3 h — mean pitot tube reading, in. H20

Volume sampled Diameter of stack Pressure of stack gas Stack gas temperature Mass of particulate collected % moisture in stack gas % 02 in stack gas (dry) % C02 in stack gas (dry) % N2 in stack gas (dry) Pitot tube factor (k)

35 dscf 2ft

7% (by volume) 7% (by volume) 14% (by volume) 79% (by volume) 0.85

Solution

The particulate concentration in the stack is

35 dscf sampled \ g J

Since this does exceed the particulate standard of 0.05 gr/dscf, the facility is not in compliance.

The actual particulate emission rate is the product of the stack flowrate and the stack flue gas particulate concentration. The stack flowrate is calculated from the velocity measurements provided in the problem statement using the second velocity equation given:

V V s2 / \1.084kg/m / = 0.85(21.4) Vh = 0.85(21.4)(0.6142) = 11.2 m/s = 36.75 fps

Stack flowrate = v (cross-sectional area)

Dry volumetric flowrate = (1 — 0.07) x 6924 acfm = 6439dacfm

Correct to standard conditions of 70°F and 1 atm pressure:

Standard volumetric flowrate

= 5631 dscfm

Particulate emission rate = 0.0706 gr/dscf(5631 dscfm) = 398 gr/min

= 398 gr/min (j^jj^) = 0.0569 lb/min = (0.0569 lb/min)(1440min/day) = 81.9 lb/day

The molecular weight of flue gas is based on the mole fraction of the flue gas components. The flue gas is 7% water and 93% other components by volume. On a dry basis, the flue gas molecular weight is

The average molecular weight of the stack gas on an actual (wet) basis is then

Average MW = 0.07 water I ---1 + 0.93 other components ( —-— I

23 Characteristics (CHR)

CHR.1 ELEMENTS OF SOLID WASTE DISPOSAL

1. Identify and describe the six elements of solid waste disposal.

2. What are the four elements included in the hierarchy for the planning of waste management systems? How can each element affect the effectiveness of solid waste management systems?

Solution

1. The six elements of solid waste disposal are:

a. Waste generation: Activities that result in the creation of solid waste.

b. Waste handling and separation: Handling of solid waste prior to contain-erization.

c. Collection: The gathering and transport of solid waste from point sources.

d. Separation, processing and transformation: Preparation and treatment of solid waste.

e. Transfer and transport: The transport of solid waste to final disposal sites.

f. Disposal: The final fate of solid wastes.

2. The four elements included in the hierarchy for the planning of waste management systems are:

a. Source reduction: Involves reducing the amount and toxicity of wastes generated at a point source. This element is the most effective means of reducing the quantity of solid waste.

b. Recycling: Reduces demand on natural resources and solid waste handling systems.

c. Waste transformation: Physical, chemical, and/or biological treatment that is designed to improve the efficiency of a solid waste management operation.

d. Ultimate disposal: Identified as the final means of dealing with solid wastes. Several methods of disposal can be implemented to deal with various types of solid wastes.

CHR.4 MOST COMMON AIR POLLUTANTS

The above hierarchy is based on pollution prevention principles (see part VI, Pollution Prevention problems, additional details).

CHR.2 SOLID WASTE MANAGEMENT PLAN

1. What factors should be considered in the design of a solid waste management plan?

2. Describe the challenges that presently face the implementation of proper solids waste management programs.

Solution

1. First, the factors that need to be considered in the design of a solid waste management plan are processing strategy and technology, plan flexibility, and plan monitoring. Several programs and treatment technologies are available for solid waste disposal. Second, a management plan must be designed with the ability to cope with future changes in the types and amounts of solid waste. Finally, solid waste management plans require constant monitoring and evaluation for optimal performance.

2. One challenge facing proper solid waste management implementation is to change established public habits that promote consumption. Another challenge is the promotion of waste reduction and recycling practices at point sources. A third challenge involves making landfills safer to the surrounding public and environment. A fourth challenge is the development of new technologies that reduce demand for natural resources and reduce solid waste generation.

CHR.3 EFFECTS OF LAND POLLUTION ON ANIMALS

How can land pollution affect the health of plants and animals? Solution

The effects of land pollution on animals are kidney and liver damage, brain and nerve damage, acid burns, cancer, and genetic disorders. Pollutants can infect local water supplies and act as fuels for potential fires. Land pollutants affect plants by hindering growth.

CHR.4 MOST COMMON AIR POLLUTANTS

What are the three most common air pollutants? How does each pollutant affect human health?

Solution

1. Sulfur dioxide: Irritates the human respiratory system and promotes respiratory diseases.

2. Carbon monoxide: Reduces the capacity of blood to carry oxygen to cells.

3. Nitrogen oxides (i.e., NO, N02): Attacks lung tissue.

CHR.5 EFFECT OF TEMPERATURE ON WASTE STORAGE

The physical state of hazardous material in a storage or transport container is an important factor in considering the fate or transport into the environment during an accidental spill or discharge situation. Define the conditions of container temperature (Tc) and ambient temperature (Ta) that will maintain the status of a chemical with the following physical characteristics. Note that MP and BP are the melting point and boiling point of the chemical, respectively.

Problem Physical State of Material MP/BP Container Conditions

Example Cold or refrigerated solid MP > Ta Tc < MP and Ta a Solid b Warm hot liquid c Cold liquid d Liquid e Hot liquid f Hot or warm compressed gas or vapor over hot liquid g Compressed liquefied gas h Hot or warm compressed gas or compressed liquefied gas

Solution

The following is a summary of the solution to this problem.

Problem

Physical State of Material

MP/BP

Container Conditions

Example Cold or refrigerated solid MP > Ta a Solid MP > Ta b Warm hot liquid BP > Ta c Cold liquid MP < Fa d Liquid BP > Ta e Hot liquid BP > Ta f Hot or warm compressed gas or vapor BP > Ta over hot liquid g Compressed liquefied gas BP < Ta h Hot or warm compressed gas or BP < Ta compressed liquefied gas

Tc > MP and Ta < BP Ta > MP < Ta and BP rc near Ta BP >TC> Ta Tc > BP and Ta

CHR.7 HEAT OF COMBUSTION CALCULATION 459 CHR.6 DECOMPOSITION OF ALKYL DICHLOROBENZENES

The general formula for the alkyl dichlorobenzenes is

Write a balanced, general chemical equation for the decomposition in the presence of oxygen of alkyl dichlorobenzenes.

Solution

The general balanced equation for the complete combustion of the alkyl dichlorobenzenes is as follows:

C„H2„_8C12 + (1.5b - 2.5)02 -»■ nC02 + 2HC1 + (n- 5)H20 where n > 6.

CHR.7 HEAT OF COMBUSTION CALCULATION

Compare the heat of combustion of 1 mol of benzene (C6H6) with the combined heats of combustion of 6 mol of solid carbon and 3 mol of H2.

Solution

The following data are provided by J. Santoleri, J. Reynolds, and L. Theodore (Introduction to Hazardous Waste Incineration, 2nd ed., Wiley-Interscience, New York, 2000):

A//c[c6H6(g)i = -789,080 cal/gmol A//°[C] = -94,052 cal/gmol A//c°[H2] = -68,317 cal/gmol

The heat of combustion of 6 mol of carbon and three moles of hydrogen molecules is

AHC = 6(-94,052 cal/gmol) + 3(-68,317 cal/gmol) = -769,263 cal/gmol

The error involved in such a calculation is

CHR.8 HEATING VALUES ESTIMATED BY DULONG'S EQUATION

Calculate the net heating value (NHV) of methane, chloroform, benzene(g), chloro-benzene, and hydrogen sulfide. This assumes that the water product is in the vapor state. Compare these values with those calculated using Dulong's equation. Calculate the relative percent difference between the "true" NHVs as determined by thermodynamic calculations and the "estimated" values calculated using Dulong's equation. Dulong's equation can be written as follows:

NHV % 14,000 mc + 45,000(mH -¡m0)~ 760(mcl) + 4500(ms)

where ml is the mass fraction of component i.

Solution

The first step in the solution of this problem is to write the balanced oxidation reaction equation for each compound, and from these balanced equations, calculate the standard heat of combustion as follows:

 CH4: CH4 + 202 C02 + 2H20(g) CHC13: CHCI3 + |o2 C02 + iH20 + ±HCl+- C6H6: C6H6 + ^O2- 6C02 + 3H20(g) C6H5C1: C6H5C1 + 702 6C02 + 2H20(g) + HC1 H2S: H2S + |o2 S02 + H20(g)

For the methane reaction, the heat of combustion is calculated from heats of formation, as follows:

AH° = (-94,052 cal/gmol C02) + 2(-57,798cal/gmol H20(g))

Heat of combustion values for the balance of these compounds are calculated in a similar manner and are found in R. H. Perry and D. W. Green, Ed., Perry's Chemical Engineers' Handbook, 7th Edition, McGraw Hill, New York, 1996. The results of these calculations are summarized in the table below using the conversion 1.8cal/g = Btu/lb.

 Compound NHV (gal/gmol) MW (g/gmol) NHV (cal/g) NHV (Btu/lb) ch4 191,759 16 11,985 21,593 CHCIj 96,472 119.5 807 1,453 c6h6 717,886 78 9,204 16,583 c6h5ci 714,361 112.5 6,350 11,441 h2s 123,943 34 3,645 6,567

CHR.9 VOST AND SEMI-VOST SAMPLING METHODS 461

 Compound Mass %C Mass %h Mass %0 Mass %c1 Mass %s nhv (Btu/lb) ch4 0.75 0.25 0 0 0 21,750 chci3 0.10 0.0084 0 0.891 0 1,101 c6h6 0.92 0.08 0 0 0 16,480 c6h5ci 0.64 0.04 0 0.32 0 10,520 h2s 0 0.06 0 0 0.94 6,930

Based on these calculations, the difference between the thermodynamically based NHV values and those estimated using Dulong's equation can be summarized as:

Compound NHVthermo (Btu/lb) NHVDu,ong (Btu/lb) % Difference

CH4 21,593 21,750 0.73

CHC13 1,453 1,101 6.7

C6H6 16,5883 16,480 0.62

C6H5C1 11,441 10,520 8.05

H2S 6,567 6,930 5.53

CHR.9 VOST AND SEMI-VOST SAMPLING METHODS

Principal organic hazardous constituents (POHCs) are monitored using the volatile organic sampling train (VOST) and semivolatile organic sampling train (Semi-VOST). The VOST method is intended for POHCs with boiling points from 30 to 100°C, while the Semi-VOST is designed for POHCs with boiling points > 100°C.

1. For the following POHCs determine whether one would expect them to be detected by the VOST or Semi-VOST method:

Carbon tetrachloride Hexachlorobenzene Decachlorobiphenyl 1,1,1 -Trichloroethane Hexachlorobutadiene

Heptachlor epoxide Chlordane Pentachlorophenol Kepone

Tetrachloroethylene

2. Will the distribution of compounds between methods be absolute, i.e., all or nothing of a given chemical in one method or another? Why or why not?

3. These protocols allow for an accuracy of ±50% in final POHC concentrations. If the true value for a given POHC was just small enough to produce a destruction and removal efficiency (DRE) of 99.99%, what is the tolerance of the measured DRE that could arise for a single determination based upon the allowable analytical tolerance? Assume a fixed mass flow into the incinerator of l.Okg/h.

Solution

1. Examining the boiling point (BP) data for the compounds listed and comparing these values to the criteria of 30°C <BP < 100°C for VOST and BP> 100°C for Semi-VOST, the following distribution can be given:

VOST

Semi-VOST

Carbon tetrachloride 1,1,1 -Trichloroethane Tetrachloroethylene

Hexachlorobenzene

Heptachlor epoxide

Chlordane

Decachlorobiphenyl

Pentachlorophenol

Kepone

2. The distribution will not be mutually exclusive. Some of the less volatile compounds from the VOST procedure will be collected in the Semi-VOST system and vice versa. Compounds will have substantial vapor pressures below their boiling points so the dividing line between analytical methods is not distinct.

3. If the true value of the DRE is 0.9999, then the true value of the mass flow out of the incinerator is wout = (0.0001)(1.0kg/h) = 0.1 g/h

If the analytical tolerance is the true value ±50%, then the range of the mass flow out of the incinerator will be 0.05 g/h to 0.15 g/h. Because the DRE is defined as

then the allowable measured range of the true DRE will be 99.985-99.995% for a true DRE of 99.99%.

CHR. 10 RANKING OF POHCS

1. The Environmental Protection Agency (EPA) proposes that the difficulty of incinerating individual POHCs can be predicted based upon the heat of combustion value for each POHC. Using this rationale, rank the 10 POHCs from Problem CHR.9 in order of most to least difficult to incinerate.

CHR.10 RANKING OF POHCS

2. Comment on the suitability of using a thermodynamic parameter, such as heat of combustion, for predicting incinerability under reaction conditions with reaction times of only «2 5. Propose an alternative concept that might be more suitable if data were available?

Solution

1. The difficulty of incinerating individual POHCs from Problem CHR.9 in order of most to least difficult to incinerate are:

 kcal/g Supposedly most Carbon tetrachloride 0.24 difficult to incinerate Tetrachloroethylene 1.19 Hexachlorobenzene 1.79 1,1,1 -Trichloroethane 1.99 Pentachlorophenol 2.09 Hexachlorobutadiene 2.12 Kepone 2.15 Supposedly least Decachlorobiphenyl 2.31 difficult to incinerate Heptachlor epoxide 2.70 Chlordane 2.71

2. A thermodynamic property may be appropriate for predicting equilibrium conditions, but the rapid reactions in an incinerator will be governed by kinetic considerations as well. Consequently, a measure of incinerability based on reaction energy and destruction kinetics would be a more relevant indicator of the difficulty of combustion for a given chemical.

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Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable.

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