Solid Waste Engineering: A Global Perspective, Si Edition
3rd Edition
ISBN: 9781305638600
Author: William A. Worrell, P. Aarne Vesilind, Christian Ludwig
Publisher: Cengage Learning
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Chapter 2, Problem 2.14P
To determine
The overall moisture content of the waste.
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Briefly describe how the moisture content of waste is determined.
(a) Calculate the ultimate BOD of a waste that has a measured 5-day BOD of 20 mg/L, assuming a BOD rate coefficient of 0.15/day measured at 20 degrees C.
(b) Estimate the rate coefficient if the temperature of the waste is raised to 30 degrees C.
(c) What would be the BOD5 at 30 deg. C?
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
Chapter 2 Solutions
Solid Waste Engineering: A Global Perspective, Si Edition
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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 Miscellaneousarrow_forwardFor the sample of solid wastes with the typical comp osition given in the following Table component Food Раper Card Plastic Textile Rubber Leath Garden Wood Glass Tin Non Ferro- Drift waste board trimmin er cans us ferrous -gs metals ashes metals bric k.etc Moisture 15 6 2 2 10 60 20 3 2 3 content % Dry 4.5 37.6 3.8 2.9 1.8 0.5 0.4 4.8 1.6 7.8 5.8 1.0 1.9 3.7 weight, Ib Determine : Moisture content of Textile (X) waste if the Overall moisture content ( 21.9 %).arrow_forwardA BOD test is run using 100 ml of treated waste H20 mixed with 200ml of pure water. The initial DO of the mixture is 9.0 mg/ml. After 5 days, Do id 4.0 mg/ml. What is the 5-day BOD of the waste water?arrow_forward
- Question 1 If the waste with BOD4-30 equal 150 mg/l and K1-20 0.23/day is discharge to a stream At an average temperature 30 C.0 = 1.047 A. compute the BOD would de exerted in 5 days at temperature 20 and 30 C (BODS-30, BODS-10). 8. compute the organic loading in (kg/day) for BOD 5-20 lf the flow rate 35 m3/1 Question 2 if the waste with ultimate BOD, equal 190 mg/L and K 1-20 0.23/day is discharge to a stream at an average temperature 25°C. a. compute the BOD would be exerted In 3 and 5 days at temperature 20 and 25° C, (BOD3-20), BOD 5-20, (BOD 3-25 ), BOD -25 ? b.compute the organic loading in (kg/day) for BOD 5-20 If the flow rate 50 m3/h? Question 3 بحبواب uses a single stage rock-media trickling filter for Secondary treatment, A wastewater treatment plant flow rate 2000m3/day concentration of BOD in raw sewage = 400 mg/! 30% BOD removed in primary sed. volume of filter = 169.65 m3. direct recirculation pump operates at 2.78 m3/min to the filter fined the efficiency of T.F and the…arrow_forwardQ4 Estimate the energy content of a solid waste sample with the composition below and what is the content on a dry basis and on ash free dry basis? Component Percent by mass Energy kJ/kg Food waste Paper Cardboard Plastics Garden trimming Wood Tin cans Total 15 45 10 10 10 5 5 100 Based on 100 kg sample of waste Moisture content% 70 6 5 2 60 20 3 4,650 16,750 16,300 32,600 6,500 18,600 700arrow_forwardA waste with a 5-day BOD (BOD)=200mg/L and k = 0.1 day¹ is discharged to a river at a rate of 1 m/s. The river has flow rate of 9 m/s and a background ultimate BOD (Lo) of 2 mg/L upstream of the discharge. (a) Calculate the ultimate BOD (Le) of the waste. (b) Furthermore, assuming instantaneous mixing after discharge, calculate Lo of the river water after it has received the waste. Ans.: (a) Ultimate BOD of wastewater = 508.3 mg/L, (b) Ultimate BOD of river water after discharge of waste-52.63 mg/L.arrow_forward
- A BOD test is run using 100mL of treated wastewater. The initial DO of the mixture after diluting the waste sample with distilled water is 9.0mg/L. After 5-day incubation, the DO measured was 4.0mg/L. After a long period of time, the DO is 2.0mg/L and no longer seems to be decreasing. Assume that nitrification has been inhibited. Determine the following: a)BOD5 & BODu of the waste sample b) BOD5 of the mixture and c) k THANK YOUarrow_forwardQ2/ Calculate the moisture content on wet and dry basis for the typical composition of municipal solid waste for Kirkuk city Weight percentage Water component content % Food waste 60% 70 Plastic 12% 2 Metal 5% 3 Раper 8% 5 Textiles 4% 10 Glass 3% other 8%arrow_forwardFrom the following date, estimate the unit waste disposal rate per week for a residential area consisting of 1200 homes. The average occupancy is 4 persons per home. The observation location is a local transfer station that receives all of the wastes collected for disposal. The observation period was one week. Data: Number of compactor truck loads = 10 Average size of compactor truck = 20 yd³ Number of flatbed loads = 6 Average flatbed volume = 3 yd³ Number of loads from individual residents' private cars and trucks = 20 Estimated average volume per domestic vehicle = 9 ft³arrow_forward
- Calculate the chemical composition of a typical yard waste with and without water (based on N and S). HINT: Assume 100 lb of waste, 40 % Moisture Contentarrow_forwardAn air particulate control (APC) device has the theoretical efficiency/particle size relationship given in the Table below. Use the particulate size distribution data from Q1 above to calculate the overall removal efficiency of this APC. Size range (um) Efficiency (%) 0-2 10 2-4 25 4-7 45 7-10 70 10-15 85 15-25 95 25-40 97 40-60 98 >60 100arrow_forward(i) During BOD test conducted on a 5% dilution of waste, the following observations were taken. (1) DO of aerated water used for dilution = 3.6 mg/l (ii) DO of original sample = 0.8 mg/l. (iii) DO of diluted sample after 5 days incubation = 0.7 mg/l. Compute: (a) 5 days BOD and (b) Ultimate BOD. Assume deoxygenation constant at test temperature as 0.12 /dayarrow_forward
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