Determine the moisture content and density (kg/m3) of the sample from the following data:Waste Composition % by Weight % MoistureDensity ContentFood wastes1570290Paper45685Cardboard10550Plastics10265Garden Trimmings1060105Wood520240Tins, cans3390Dirt, ashes20480Based on 100 kg’s sample

To find:- Moisture content Density Of given municipal solid waste. Moisture content = (wet weight -…

Answered: Determine the moisture content and…

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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From the table shows a summary of a liquid limit and plastic limit test. Compute the plastic limit. Liquid Limit Test Test Number 1 3 4 Number of 39 30 23 |12 Blows Weight of Wet Soil + 22.24 21.19 21.27 26.12 Container Weight of Dry Soil + Container 20.10 19.28 19.25 22.60 Weight of Container 12.74 13.24 13.06 13.27 60 2. 6.0
PROBLEM #1 A stream receives a waste outfall which results in oxygen at saturation levels when mixed. The initial condition right after entry of the waste creates a BOD5 of 40 mg/l. KBOD is 0.3/day and KO2 is 1.3/ day. The stream temperature is 12 degrees centigrade. In terms of travel time when does the maximum DO sag occur? What is the minimum DO concentration in the river at the maximum sag point? If the average stream velocity is 3 feet per minute, what is the downstream location of the DO sag minimum relative to the location of the waste stream? Use the relevant equations provided in the textbook to solve this problem. Hint: Consider carefully what was stated in class about the BOD input for this type of problem.
The waste water is flowing at a rate of 0.2 m2/s in a settling tank. The specific gravity of particles in the water is 2.7 and the kinematic viscosity of water is 1.02x10-2 cm?/s. Calculate the required area of tank for removing particles of size 0.07 mm.
I want the solution in less than 15 minutes
What is the moisture contect for 2
18. The results for sieve analysis is shown in the following table: Sieve No. 10 20 40 60 100 % Passing 69.3 59.1 48.3 38.5 28.4 19.8 4.5 Liquid Limit Plastic Limit Non-Plastic (NP) Non-Plastic (NP)
PART B 1. The data are given as follows: Weight of evaporating dish: 53.5430 g Weight of evaporating dish and residue after drying at 103°C: 53.5760 g Weight of filter and residue after drying at 103°C: 1.5530 g Weight of filter and residue after ignition at 550°C: 1.5450 g Volume of sample filtered: 50 mL Volume of sample dried: 80 mL Determine the total solids, volatile suspended solids, suspended solids, fixed suspended solids and total dissolved solids. Assume 60% of suspended solids were loss on ignition.
The tabulated data are obtained for a wastewater sample.Which of these numbers is questionable? Why?
What diameter would a circular clarifier have to be to remove 50 μm particles with 75% efficiency from a 4 MGD (million gallon per day) flow? Assume the particle specific gravity is 1.5 and the temperature is 10°C. Answer: 18 m
Soil . Asap
IL Test sample weight= احتمان Sieve size Weight retained (g) Calculations: Sieve size in. (mm) 3/8" 9.5 mm No.4 4.75 mm No.8 2.36 mm No.16 1.18 mm No.30 0.6 mm No.50 0.3 mm No.100 0.15 mm Pan gm 3/8" No.4 No.8 No.16 No.50 No.100 0.15 mm 9.5 mm 4.75 mm 2.36 mm 1.18 mm mm 0.3 mm 0.0 од 1597 21419 від 50д 3589 174д Weight retained (g) Percent retained Beitel 15-11-2021 Fineness modulus отсивно о u 4/8²/00 = 5.3 15 діни з 2141-44871.3 61-448203 Божнизис 1.66 50 78.63 21.3 80.29 19.71 35 84421 23 92.22 7.78 174448518 98.02 1.98 чудд 4 /100 = Cumulative = No.30 Percent cumulative 0 53 94.7 766 23-4 Percent passing loo 222 Sieve deviation = 262 (5) ASTM Limit 100 95 100 80 100 50_85 25 60 10_30 2_10

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Waste Composition	% by Weight	% Moisture	Density Content
Food wastes	15	70	290
Paper	45	6	85
Cardboard	10	5	50
Plastics	10	2	65
Garden Trimmings	10	60	105
Wood	5	20	240
Tins, cans	3	3	90
Dirt, ashes	2	0	480