2-14. Using the moisture contents(wet basis) of the waste componentsgiven in Table P2-14, calculate theoverall moisture content of the wastehaving the composition also shown.This calculation can be simplified byassuming 100 lb of waste.Table P2-14Moisture content,Percent byComponentpercent (wet basis)weightFood7010Paper633Cardboard58Plastics25Textile104Rubber23Yard waste6018Metals310Miscellaneous692.Calculate the moisture content of thewaste in Problem 2-14 on both a wetand a dry basis. Why the difference?Which numbermakes more sense to you, and why?3.Determine the composition of thewaste in Problem 2-14 if yard waste issepa-rately collected. How would thisaffect the moisture content?

Answer:1. Calculate the Overall Moisture Content (Wet Basis)To calculate the overall moisture…

Answered: 2-14. Using the moisture contents (wet…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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c) Complete the following Table Q2 (c) and calculate the moisture content of solid waste. Table Q2 (c) Compositions Wet percent (%) Moisture Dry Percent (%) 51 80 9 12 31 60 9 6 Food Plastics Rubber Miscellaneous
A truck dumped industrial waste into a pit. They found the truck 3 days ago with a pure pool of waste. Your lab finds the BOD5 of this waste to be 80 mg/L with a k=0.1/day. What was the waste source? Here are the 3 nearby suspects: Winery BOD ultimate = 275 mg/L Vinegar manufacturer BOD ultimate = 80 mg/L Pharmaceutical BOD ultimate = 200 mg/L
The City of Forty-Five (population 465,565) has requested your assistance in evaluating its solid waste collection system. Determine the mean time per collection stop plus the mean time to reach the next stop, the number of pickup locations per load, and the minimum number of trucks the city must own. The City of Forty-Five collection data are as follows: Average truck capacity = 12.5 m3; Average observed compaction ratio = 2.87 Crew size = 2; Number of pickups = 1 /wk (no rear-of-house service) Average number of cans per stop = 2.63/wk at 0.1136 m3/can Average number of residents per stop = 5 Average uncompacted density = 105.76 kg/m3 Average transport time to disposal site including delays and dumping = 1.78 h/trip Average number of trips to disposal site = 2/d; Rest breaks = 2 at 28.0 min Average maintenance downtime = 37.0 min/d; Average workday = 8.00 h Average percent of trucks out of service for major repairs = 15.0% Number of workday per week = 5
Assuming waste not recovered all goes to landfills. Determine the total landfill capacity consumed national wide each year. The specific weights of waste at landfill are as follows. Convert your answer to # of football stadiums filled. Area of football stadium = 1.75 x size of a standard football field Depth of filling = 30 ft Waste category Specific weight (kg/m³) Aluminum 2470 Steel and other metals 7100 Glass 2300 Paper and cardboard 570 Wood 450 All others ~700
Assume that a material recovery facility has a conveyor belt (manual sorting unit) measuring 1,0 m width and an average belt speed of 18 m/min. Waste loading rate of the system is 12000 kg per hour and the average density of the waste on the conveyor belt is 80 kg/m3. Calculate the thickness of the waste on the conveyor belt and comment about this thickness.
If the BOD4 of a waste is 200.0 mg/L and the ultimate BOD is 270.0 mg/L, what is the rate constant at 10⁰C.
How many residences can be served? If we have a compactor truck which has a capacity of 30 yd3, each truck makes 3 trips/day 5 days a week, each residence has 3 people and each person produces 4lbs of waste. The density of the compacted waste is 425 lbs/yd3.
A waste water sample of 2 ml is made upto 300 ml in BOD bottle with distilled water. Initial DO of the sample is 8 mg/l and after 5 days it is 2 mg/1. What is its BOD?
A sample of dry solid waste has the following composition: 531 g carbon (C), 133 g hydrogen (H), and 335 g oxygen (O). How many Litres of air is required to combust the waste? Answer: Check
Example 5-3 Using the typical percentage distribution data in the given table below, estimate the number of Kj/Kg of the remaining solid wastes if 90 percent of the cardboard and 60 percent of the paper were recovered by homeowner Component Percent by weight Organic food wastes 9.0 Paper 34.0 Cardboard 6.0 Plastics 7.0 Textiles 2.0 Rubber Leather Wood 0,5 Yard wastes 18.5 Inorganic Glass 0.5 2.0 8.0 Tin cans 6.0 Aluminum 0.5 Other metal 3.0 Dirt, ash, etc. 3.0 Total Energy kJ/kg Total energy Kj 4720 16747.2 16282 32564 17445 23260 17445 6512.8 18608 139.5 697.5 697.5 6975 42480 570000 97692 227948 34890 11630 8722.5 120486.8 37216 1116 4185 2092.5 20925
B) Estimate the moisture content on a wet base and density of a solid waste sample with the following composition. Based on 1000 kg sample of waste. Component Food waste Percent by mass Moisture centent % Typical density kg/m3 290 45 70 Paper Cardboard Plastics Garden trimming 15 85 50 5 65 60 10 105 Wood 5 240 20 3. Tin cans 15 90
Discuss TWO (2) factors that may influence the rate of solid waste generation

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