26 Municipal Waste MUN

MUN.l MUNICIPAL LANDFILLS

Provide answers to the following questions:

1. What are some reasons given for the opposition to the construction of new municipal landfills?

2. Identify several restrictions with respect to location of a new municipal landfill.

3. Identify several restrictions with respect to operations of a new municipal landfill.

4. What is the most significant resistance to the development of a recycling program?

Solution

1. Public opposition to new landfills as displayed by the "not in my backyard" attitude can be attributed to past problems with older landfills. Poor designs and locations have been responsible for the contamination of nearby groundwater sources. Ground contaminations by older landfills require extensive cleanup efforts at large expense.

2. Several restrictions with respect to location of a new municipal landfill are:

a. Landfills must be constructed away from water sources and wetlands.

b. They must be resistant to damage by flooding and should be located away from areas prone to earthquakes, landslides, mudslides, and sinkholes.

c. Municipal landfills are not to be located in areas where the added bird population can interfere with local air traffic.

3. Several restrictions with respect to operations of a new municipal landfill are:

a. Municipal landfills must be operated to minimize the amount of hazardous waste that can be deposited.

b. Landfills must be covered daily with a fresh layer of soil to control local pest and animal populations.

c. Landfills must monitor methane emissions and must have a plan to deal with excess emissions.

d. No discharge, accidental or intentional, is allowed from a landfill to a local water body.

e. Landfills must restrict any unauthorized dumping and prohibit burning of waste.

4. The most significant resistance to development of a recycling program is operating costs associated with collection, processing, and disposal of residual matter. Other operating costs include the cost for new equipment, labor, and utilities. Due to these costs, materials such as plastics, cardboard, and paper use are often more cost effective as virgin material rather than as recycled products.

MUN.2 MUNICIPAL WASTE MANAGEMENT

Answer the following two-part question:

1. Federal, state, and local governments have adopted an integrated approach to municipal waste management. Describe the three waste management techniques that support this approach.

2. What are the roles of the federal, state, and local governments in establishing regulations for municipal solid waste disposal facilities?

Solution

1. The three waste management techniques include:

a. Reduce the overall toxicity of solid wastes by reducing its volume.

b. Increase recycling efforts to prevent losing renewable resources.

c. Improve landfill design and management.

2. The role of the federal government with respect to the regulation of municipal landfills is to set minimum national standards for states and local governments and to evaluate and monitor the effectiveness of state and local agencies in managing solid waste facilities. It is the responsibility of state and local governments to implement and enforce the national standards and the needs of the local community.

MUN.3 AMOUNT OF WASTE GENERATED BY A MUNICIPALITY

A municipality in the Midwest has a population of 50,000 and generates 100,000 yd3 of municipal waste annually. The waste is made up of 30% compacted waste and 70% uncompacted waste. Assume that the waste has a density of 1,000 lb/yd3

compacted, and 400 lb/yd3 uncompacted. How many pounds of waste are generated by this city each year? By each person each year?

Solution

Based on the waste densities given in the problem statement, the following generation rates are determined:

Waste generated/yr = (0.3)(100,000yd5)

yJ 1,0001b yd3 4001b

Per capita generation rate =

58,000,000 lb/yr

50,000 people = 1160 lb/person • yr = 3.2 lb/person • day

MUN.4 CHARACTERISTICS OF MUNICIPAL SOLID WASTE

Besides economic considerations and site specifications, what are the major characteristics of municipal solid waste that are employed for the selection of the treatment and disposal processes listed below?

1. Composting

2. Incineration

3. Sanitary landfill

Solution

The characteristics relevant to the treatment and disposal options listed in the problem statement are summarized below:

1. Composting: Biodegradable/nonbiodegradable fraction, moisture content, carbon-to-nitrogen ratio

2. Incineration: Heating value, moisture content, inorganic fraction, metal content, alkali earth metal content

3. Sanitary landfill: Biodegradable to nonbiodegradable fraction, liquid content, hazardous material content

MUN.5 CALCULATION OF AVERAGE WASTE DENSITY

An analysis of a solid waste generator has revealed that the waste is composed (by volume) of 20% supermarket waste, 15% plastic-coated paper waste, 10% polystyrene, 20% wood, 10% vegetable food waste, 10% rubber, and 10% hospital waste. What is the average density of this solid waste in lb/yd3? Use the following as discarded waste densities (in lb/yd3) for each of the components of the generator's waste: supermarket waste, 100; plastic-coated paper waste, 135; polystyrene, 175; wood, 300; vegetable food waste, 375; rubber-synthetics, 1,200; and hospital waste,

Based on the volume percent composition and waste densities given in the problem statement, the following average discarded waste density is estimated to be

Average density = (0.2)(100) + (0.15)(135) + (0.15)(175)

+ (0.2)(300) + (0.1)(375) + (0.1)(1200) + (0.1)(100) = 294 lb/yd3

MUN.6 MOISTURE CONTENT

The design of garbage collection vehicles differs from county to county and country to country based on location, local culture, etc. Many factors influence the design of these vehicles, with moisture content being one of the most important considerations in the hauling of municipal waste. The general formula for calculating the moisture content of solid waste is as follows:

where A = weight of sample as delivered, kg B = weight of sample after drying, kg

Use the data provided below on waste component moisture content and the weight composition of Ruocco County, Italy, waste to determine the average moisture content of the municipal solid waste. Base the calculation on a 100-kg sample.

100.

Solution

MUN.6 MOISTURE CONTENT 491

Typical Moisture Content of Municipal Solid Waste

Moisture Content

Component

(wt %)

Food waste

70

Paper

6

Garden waste

60

Plastic

2

Textiles

10

Wood

20

Glass

2

Metals

3

Composition of Ruocco County, Italy, Solid Waste on a Weight Percent Basis

Composition

Component

(wt %)

Food waste

20

Paper

22

Garden waste

18

Plastic

3

Textiles

7

Wood

25

Glass

2

Metals

The dry weight of each component can be determined using the following equation:

Dry weight = (Discarded weight)(100 - %Moisture)/100

The discarded weight for each component can be determined using the following equation:

Discarded weight = (Total waste weight)(wt % of component)

Finally, the average moisture content of the waste is determined as:

Avg. % moisture = [(Discarded weight — Dry weight)/100 kg](100%)

Based on the %moisture content data and waste composition data provided in the problem statement, and assuming a total waste weight of 100 kg, the calculations presented below result for each waste component given in the problem statement.

Composition, Weight Percent, Component Discarded Weight, and Component Dry Weight for the Municipal Solid Waste from Ruocco County, Italy

Component Discarded

Component Dry

Component

Wt %

Weight (kg)

Weight (kg)

Food waste

20

20

(1.0-0.7)(20) = 6.00

Paper

22

22

(1.0 - 0.6)(22) = 20.68

Garden waste

18

18

(1.0 - 0.6X18) = 7.20

Plastic

3

3

(1.0-0.02X3) = 2.91

Textiles

7

7

(1.0-0.1X7) = 6.30

Wood

25

25

(1.0 - 0.2)(25) = 20.00

Glass

2

2

(1.0 - 0.02)(2) =1.96

Metals

3

3

(1.0 — 0.03)(3) = 2.91

Total

100

67.96 = 68

MUN.7 DENSITY OF A COMPACTED WASTE

The Jefferson County Solid Waste Management Corporation analysis of its solid waste includes the following major components: weight, volume, and compaction factors. Determine the density of well-compacted waste as delivered to a landfill. {Note: waste components and compaction factors vary from place to place and may not be identical to those shown below for Jefferson County.)

Composition, Weight, Volume, and Compaction Factor of Solid Waste from Jefferson County, Mississippi

Component

Weight (kg)

Volume as Discarded (m3)

Compaction Factor

Food waste

250

1.00

0.33

Paper

300

3.75

0.15

Garden waste

250

1.18

0.20

Plastic

50

0.80

0.10

Textiles

60

9.61

0.15

Wood

50

0.34

0.30

Glass

20

0.10

0.40

Metals

20

0.10

0.30

Total

1000

Note: Compaction factor represents the ratio of the resultant volume to the original volume.

Note: Compaction factor represents the ratio of the resultant volume to the original volume.

MUR8 REQUIRED LANDFILL AREA

Solution

The compacted volume of each component in the landfill can be determined using the following equation:

Compacted volume = (Discarded Volume)(Compaction factor)

Based on the weight, discarded volume, and compaction factor data provided in the problem statement, the calculations presented in the table below result for each waste component. Finally, the average density of the well-compacted mixed waste delivered to the landfill can be determined as:

Average density = (Discarded weight)/(Total compacted volume)

Results are provided below.

Composition, Weight, Volume, Compaction Factor, and Compacted Volume of Solid Waste from Jefferson County, Mississippi

Component

Weight (kg)

Volume as Discarded (m3)

Compaction Factor

Compacted Volume (m3)

Food waste

250

1.00

0.33

0.33

Paper

300

3.75

0.15

0.56

Garden waste

250

1.18

0.20

0.24

Plastic

50

0.80

0.10

0.08

Textiles

60

9.61

0.15

1.44

Wood

50

0.34

0.30

0.10

Glass

20

0.10

0.40

0.004

Metals

20

0.10

0.30

0.018

Total

1000

2.77

MUN.8 REQUIRED LANDFILL AREA

Estimate the required landfill area for a community with a population of 260,000. Assume that the following conditions apply:

2. Compacted specific weight of solid wastes in landfill = 830 lb/yd3

3. Average depth of compacted solid wastes = 60 ft

Solution

Determine the daily solid wastes generation rate in tons per day:

Generation rate

2000 lb/ton

The required area is determined as follows:

Volume required/day =

830 lb/yd3 = 2381 yd3/day

The actual site requirements will be greater than the value computed because additional land is required for a buffer zone, office and service building, access roads, utility access, and so on. Typically, this allowance varies from 20 to 40%. Thus, if an allowance of 30% is employed, the daily area requirement becomes

A more rigorous approach to the determination of the required landfill area involves consideration of the contours of the completed landfill and the effects of gas production and overburden compaction.

MUN.9 CALCULATIONS ON A MSW INCINERATOR

A small community in New York incinerates 10,000 lb/h of municipal solid waste (MSW) at a central disposal facility using 100% excess air. The waste has a heating value of 6500 Btu/lb, an ash content of 10 wt %, and a moisture content of 25 wt %.

It may be assumed that 7.5 lb of dry gas is generated for every 10,000 Btu of waste burned in the incinerator. In addition, 0.51 lb of moisture is produced for every 10,000 Btu of waste fired. Air cools the incinerator shell at a flow of 10,000 lb/h (this air is discharged to the atmosphere at 400°F). Assume the air entering the incinerator has a humidity of 0.015 lb of moisture per pound of dry air and the ash produced has a heating value of lOOBtu/lb (assume no fly ash). The radiation heat loss from the shell is 1% of the total heating value of the waste. The temperature of the products of combustion (the flue gas) must not be less than 2000°F. The following information is required:

1. Calculate the inlet and outlet flowrates in lb/h on both a dry and a wet basis.

2. Determine the amount of ash remaining after combustion.

3. Calculate the volatile heating value of the waste.

4. Determine if supplemental fuel must be added to the incinerator.

5. Comment on the ultimate disposition of the remaining solid waste (ash).

Solution

Determine the moisture feed rate in lb/h:

Calculate the dry feed rate to the incinerator in lb/h:

Determine the amount of ash remaining after combustion in lb/h:

The ash may be treated and/or sent to a second landfill.

The amount of feed that is combusted (volatile) is therefore

Determine the total heating value of the waste charged to the incinerator in Btu/h:

Heating value = (10,000 lb/h)(6500 Btu/lb) = 65.0MBtu/h

Calculate the heating value per lb of volatile waste in Btu/h:

6500 lb/h

Calculate the amount of dry gas produced from combustion in lb/h:

Dry gas = (7.5 lb/10,000 Btu)(65.0MBtu/h) = 48,750 lb/h dry gas

Determine the amount of moisture produced from combustion in lb/h:

Combustion moisture = (0.51 lb/10,000 Btu)(65.0MBtu/h) = 3315 lb/h

Therefore the total amount of gas (wet basis) leaving the incinerator due to combustion in lb/h is

Total combustion gas = 48,750 lb/h dry gas + 3315 lb/h wet gas = 52,065 lb/h

Calculate the stoichiometric air requirement for the incinerator in lb/h:

Stoichiometric air = (52,065 - 6500) lb/h = 45,565 lb/h

Determine the amount of excess air in lb/h:

MUN.9 CALCULATIONS ON A MSW INCINERATOR 497 The total air requirement in lb/h is then

Calculate the amount of moisture in the entering air supply in lb/h:

Inlet air moisture = (0.015 lb/lb dry air) (91,130 lb/h) = 1367 lb/h

The total outlet moisture flowrate in lb/h is obtained by summing the three contributing terms:

Total outlet moisture rate = (2500 + 3315 + 1367) lb/h = 7182 lb/h

The total dry gas flowrate exiting the system in lb/h can also be determined:

Outlet dry gas rate = (48,750 + 45,565) lb/h = 94,315 lb/h

Calculate the heat loss in Btu/h due to the ash discharge;

Calculate the heat loss in Btu/h due to the air used to cool the incinerator shell. The enthalpy of air at 400°F is 81.8Btu/lb:

Cooling air heat loss = 10,000lb/h (81.8Btu/lb) = 0.82 MBtu/h

Determine the heat loss in Btu/h due to radiation:

Calculate the amount of heat absorbed in Btu/h by the humidity of the inlet air supply:

Humidity correction = 960Btu/lb (1367 lb/h) = 1.31 MBtu/h

The total heat loss of the incinerator in Btu/h is then

Total heat loss = (0.1 + 0.82 + 0.65 - 1.31) MBtu/h = 0.26 Btu/h

Calculate the outlet heat content of the flue gas in Btu/h:

Outlet heat content = (65.0 - 0.26) MBtu/h = 64.74 MBtu/h

From tables, the enthalpies of dry air and moisture at 2000°F:

Enthalpy of dry air = 513.0Btu/lb Enthalpy of moisture = 2060.0 Btu/lb

Calculate the outlet heat content of the flue gas at 2000°F:

Outlet heat content at 2000°F = (7182 lb/h)(2060.0 Btu/lb)

Since the outlet heat content of 64.74 MBtu/h is greater than the outlet heat content at 2000° F, supplemental fuel is not necessary.

Municipal solid waste (MSW) is usually incinerated at a central disposal facility. Central disposal facilities are often used for energy generation. While these facilities have largely been popular in Europe, more central disposal facilities are appearing in the United States since the cost and availability of traditional energy sources are no longer reliable.

A secondary combustion chamber is necessary on municipal solid waste incinerators since a primary chamber alone will not provide the necessary time, temperature, and turbulence to destroy the organic waste components to the required destruction and removal efficiency (DRE). The primary chamber serves to volatilize the organic waste fraction while the secondary chamber heats the vaporized organics to an adequate temperature for oxidation. The oxidation process occurs in the 1800-

2200°F range, and the flue gas residence time in the secondary chamber is normally at least 2 s.

Most central disposal facilities utilize mass-burning systems; other facilities use controlled air incineration techniques. Mass-burning technology is described in more detail in the following problem (MUN.10).

MUN.10 INCOME FROM AN MSW INCINERATION SYSTEM

A mass-burn incineration system has been chosen to incinerate the municipal solid waste in a community. The incinerator will be located at a central disposal system and will provide steam to a local utility. Two rotary kiln incinerators will continuously incinerate 250 tons/day each of MSW at 1800°F with a waste heating value of 5000 Btu/lb. To generate the steam, a waste heat boiler is to be installed at the discharge of each incinerator. Each boiler generates steam at 150 psia. Feedwater is provided to each boiler at 220°F and the blowdown leaving each boiler is 4% of the feedwater entering each boiler. The exhaust gas temperature from each boiler is 400°F and the boiler radiation loss is 0.5% of the total boiler input. The kilns have slopes of 0.04 ft/ft and length-to-diameter ratios of 5 :1. Steam is sold to the local utility at $2.50 per 1000 lb of steam. A tipping fee (the cost of dumping waste at the facility) of $20.00 per ton is charged. Assume a kiln heat release rate of 25,000 Btu/(h • ft3). The products of combustion leaving each incinerator are as

follows:

C02:

2500 lb/h

H20:

5000 lb/h

N2:

12,500 lb/h

02:

500 lb/h

HCl:

250 lb/h

Ash:

83.5 lb/h

(Ash has a heat content of 522 Btu/lb at 1800°F and 102 Btu/lb at 400°F.)

From the steam tables, enthalpies for the steam, feedwater, and blowdown are as follows:

Steam at 150 psia = 1194 Btu/lb

Feedwater at 220°F = 188 Btu/lb

Blowdown at 150psia and 358°F = 330.7 Btu/lb

1. Calculate the inside diameter of each kiln.

2. Determine the steam generation rate in tons/day, the amount of heat in Btu/h in the steam, and the boiler efficiency.

3. Calculate the yearly income of the facility based on the steam charge and the tipping fee.

4. Recalculate the yearly income if the steam is sold at $4.00 per 10001b of steam and the tipping fee is increased to $25.00 per ton. Discuss the disadvantages of increasing these costs.

Solution

Determine the heat release in Btu/h of each kiln:

Q = (5000 Btu/lb)(250ton/day)(20001b/ton)(day/24 h) = 104,166,667 Btu/h

Calculate the volume of each kiln:

V = (104,166,667 Btu/h)/[25,000 Btu/(h- ft3)] = 4167ft3

Calculate the inside diameter, D, of each kiln:

Determine the length, L, of each kiln:

Determine the heat (enthalpy) content of the flue gas at the entrance and the exit of the boiler:

Heat content = (Mass flowrate)(Enthalpy)

Entrance at 1800°F Exit at 400°F

co2

(2500)(470.9) =

1,177,250 Btu/h

(2500)(75.3) =

188,250 Btu/h

h2o

(5000)(1947) =

9,375,000 Btu/h

(5000)(1213) =

6,065,000 Btu/h

n2

(12,500)(465) =

5,812,500 Btu/h

(12,500)(85) =

1,062,500 Btu/h

o2

(500)(430.7) =

215,350 Btu/h

(500)(76.2) =

38,100 Btu/h

HC1

(250)(349) =

87,250 Btu/h

(250)(64.9) =

16,225 Btu/h

Ash

(83.5)(522) =

: 43,587 Btu/h

(83.5)(102) =

8,517 Btu/h

Enthalpies appearing above in parentheses were drawn from the literature.

Calculate the total mass flow in lb/h and the total energy flow in Btu/h at both the entrance and exit of the boiler.

Total mass flow = 20,830 lb/h

Total energy flow at entrance = 17,071,000 Btu/h

Total energy flow at exit = 7,379,000 Btu/h

Calculate the heat loss:

Write an energy balance on the boiler:

Heat in = Heat out

Flue gas in + Feedwater = Flue gas out + Heat loss + Steam + Blowdown (17,071,000 Btu/h) + (Feedwater rate)(188 Btu/lb)=(7,379,000 Btu/h)

+ (85,360 Btu/hr) + (Steam rate)(l 194 Btu/lb) + (Blowdown rate)(330.7 Btu/lb)

Write a mass balance on the boiler including only the steam, feedwater, and blowdown.

Feedwater rate = Steam rate + Blowdown rate

Solve the mass balance equation above for the steam flowrate in terms of feedwater flowrate only.

Feedwater rate = Steam rate + 0.04(Feedwater rate) Steam rate = 0.96(Feedwater rate)

Substituting the above result into the energy balance equation leads to:

(17,071,000 Btu/h) + (Feedwater rate)( 188 Btu/lb) = (7,379,000 Btu/h) + (85,360 Btu/h) + [0.96 (Feedwater rate)(l 194 Btu/lb)] + [0.04(Feedwater rate) (330.7 Btu/lb)]

Calculate the feedwater mass rate in lb/h:

9,607,000 Btu/h = [(0.96)(Feedwater rate)(1194 Btu/lb)]

+ [(0.04)(Feedwater rate)(330.7 Btu/lb)] - (Feedwater rate)(188 Btu/lb)

9,607,000 Btu/h = 971.5 Btu/lb(Feedwater rate)

Feedwater rate = 9889 lb/h

Determine the total steam generation rate in tons/day from both boilers:

Steam generation rate = 0.96(Feedwater rate)

= 0.96(9889 lb/h)(24 h/day)(ton/2000 lb) = 114 ton/day per boiler

For both boilers,

Total steam generation rate = 2(114 ton/day)

Calculate the amount of heat in Btu/h in the steam from each boiler:

Steam heat amount = (Steam generation rate)(Steam enthalpy) = (9493 lb/h)( 1194 Btu/lb) = 11,330,000 Btu/h

Calculate the efficiency of each boiler:

Boiler efficiency = [(11,330,000 Btu/h)/(17,071,000 Btu/h)](100) = 66.4%

Calculate the yearly income of the facility based on the steam charge:

Steam income = ($2.50/1000 lb steam)(228 ton steam/day) (365 day/yr)(2000 lb steam/ton steam) = $416,100/yr

Calculate the yearly income of the facility based on the tipping fee:

Tipping fee income = ($20.00/ton)(500 ton/day)(365 day/yr) = $3,650,000/yr

Calculate the total yearly income of the facility.

Total yearly income = Steam income + Tipping fee income = $416,100/yr + $3,650,000/yr = $4,066,100/yr

Recalculate the total yearly income of the facility based on the rates given in question 4 of the problem statement and data.

Steam income = ($4.00/1000 lb steam)(228 tons steam/day) (365 day/yr)(2000 lb steam/ton steam) = $65,760/yr

Tipping fee income = ($25.00/ton)(500 tons/day)(365 days/yr)

= $4,562,500/yr Total yearly income = Steam income + Tipping fee income = $665,760/yr + $4,562,500/yr = $5,228,260/yr

Low steam charges and tipping fees are necessary to encourage sales. In the United States, steam production from incineration is not considered conventional although there are successful incineration facilities in the United States that produce steam.

Low rates would attract as many customers as possible. Therefore, it would probably be disadvantageous for a facility to charge excessively high steam and tipping fees.

The two leading methods of generating energy from the incineration of municipal solid waste (MSW) are the mass-burn system and the refuse derived fuel (RDF) system. The mass-burn system incinerates unprocessed MSW to recover energy and the RDF system processes unprocessed MSW into a usable fuel prior to incineration. Both methods use either starved-air modular, stoker grates, rotary kiln, or fluidized-bed units for incineration. While the mass-burn system is currently more widely utilized, both systems may be used for large waste capacities.

Many considerations are necessary when deciding if heat recovery equipment should be included in an incineration facility. Unless a practical use for the recovered heat exists, it is usually not advisable to include heat recovery equipment at a facility since the equipment is expensive. If an incineration facility operates at a large capacity, the generation of power from heat recovery equipment is generally economical. Municipal solid waste incineration facilities are widely used to produce steam for electric power generation.

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Responses

  • haben
    How to calculate the volume of a solid waste composition when percentage mass is given?
    2 years ago
  • Lorenza
    How to calculate specific gravity of municipal waste?
    2 years ago

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