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 4, Problem 4.17P
To determine
Grate spacing and horsepower required by the shredder
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1. Three samples of aggregates are sieved to obtain the grading curve. A set of different sieves is
stacked with the coarsest mesh at the top. The aggregate is then placed in the top sieve and the
stack is placed in a sieve shaker. The raw data is as follows:
Sieve size (mm)
19
9.5
4.75
2.36
1.18
0.6
0.3
0.15
Pan
Mass of sieve (g)
670
620
620
570
570
520
520
520
490
For each sample:
Plot the grading curve.
a.
b.
Identify the type of gradation.
c. Determine the fineness modulus.
Mass of sieve + aggregates retained (g)
Sample B
mass of the empty measure
mass of measure + water
Sample A
670
770
820
740
720
770
720
570
525
mass of measure + rodded aggregates
mass of measure + loose aggregates
670
1820
620
570
570
670
720
620
544
2. The rodded and loose bulk density of an aggregate sample is to be determined. The following
are acquired during the experiment:
- 12.15 kg
- 27.15 kg
- 34.60 kg
- 25.16 kg
Sample C
670
620
620
670
770
720
620
620
506
An aggregate sample was collected in a container and weighed as 528.7 g. The sample was oven dried and theoven dry weight was measured as 506.4 g. If the empty weight of the container was 79.6 g, determine the MC of theaggregates
7. Make a spread sheet blend template program to perform a blend gradation analysis
for up to four stockpiles with the ability to produce automatically semilog or power
0.45 gradation charts. Use the table to demonstrate the template, and calculate the
fineness modulus for the aggregate D.
Aggregate
Blend Percent
(%)
Sieve size (mm)
37.5
25
19
12.5
9.5
4.75
2.36
1.18
0.60
0.30
0.15
0.075
A
20
100
100
95
89
70
10
2
N|N
2
2
2
2
2
B
25
100
100
100
100
82
55
15
55
32
2
1
с
25
Percent Passing
100
100
100
100
100
100
69
55
35
22
15
6
D
30
100
100
100
100
100
100
86
54
34
25
14
5
Chapter 4 Solutions
Solid Waste Engineering: A Global Perspective, Si Edition
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- this title "name" for test in highway lab. Standard Test Method for Ductility of Bituminous Materials and this is reading measurment during laboratory highway. calculations test reading 1 = 100+ reading 2 = 100+ reading 3 = 100+ please i want calcualtion and discussionarrow_forwardThree aggregates are to be mixed together in the following ratio:Aggregate A: 35%Aggregate B: 40%Aggregate C: 25%For each aggregate, the percent passing a set of five sieves is shown in Table PDetermine the percent passing each sieve for the blended aggregate.arrow_forward1. Why do you think there is a need to agitate the aggregates inside the cylinder mold by rodding? 2. Determine the unit weight of the aggregates from the video. a. Dry Loose Unit Weight of the Fine Aggregate b. Dry Rodded Unit Weight of the Fine Aggregate c. Dry Loose Unit Weight of the Coarse Aggregate d. Dry Rodded Unit Weight of the Coarse Aggregate 3. A coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained. a. Volume of Bucket = 0.0094m³ %3D b. Weight of empty bucket = 8,391 g c. Weight of bucket filled with dry rodded coarse aggregate = 25,355 g %3D Calculate the dry-rodded unit weight. 4. Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained. a. Volume of Bucket = 0.0142m³ %3D b. Weight of empty bucket = 9,207 g c. Weight of bucket filled with dry rodded coarse aggregate: i. Trial 1= 34,745 g ii. Trial 2 = 34,065 g…arrow_forward
- 1. In the bulk unit weight experiment why should the aggregate be placed in layers and then rodded instead of filling up the container and then rodding it.arrow_forwardLaboratory specific gravity and absorption tests are run on two coarse aggre-gate sizes, which have to be blended. The results are as follows: Aggregate A: Bulk specific gravity = 2.491; absorption = 0.8,Aggregate B: Bulk specific gravity = 2.773; absorption = 4.6,a. What is the specific gravity of a mixture of 60% aggregate A and 40%aggregate B by weight?b. What is the absorption of the mixture?arrow_forwardthe Nominal Maximum aggregate Size (N.M.S) for the aggregate gradation shown below is equal to Sieve size Passing % 3/4 inch 100 1/2 inch 3/8 inch 99 89 No.4 72 No.8 65 No.30 22 No.200 Your answer 4-arrow_forward
- Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained: Volume of bucket ½ ft3 Weight of empty bucket Weight of bucket filled with dry rodded coarse aggregate 20.3 lb Trial 1 = 69.6 lb Trial 2 = 68.2 lb Trial 3 = 71.6 lb a. Calculate the average dry-rodded unit weight b. If the bulk dry specific gravity of the aggregate is 2.620, calculate the per- cent voids between aggregate particles for each trial.arrow_forwardThe aggregates are placed in a cylindrical container weighing 0.3 kg with a base circumference of 0.18π ft. and a height of 1.4 ft. If the Total weight of the aggregates and the container is 27.5 kg, find the volume of the container.arrow_forwardCoarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained:Volume of bucket = 1/3 ft3Weight of empty bucket = 18.5 lbWeight of bucket filled with dry rodded coarse aggregate = 55.9 lba. Calculate the dry-rodded unit weightb. If the bulk dry specific gravity of the aggregate is 2.630, calculate the percent voids in the aggregate.arrow_forward
- Laboratory specific gravity and absorption tests are run on two coarse aggre- gate sizes, which have to be blended. The results are as follows: Aggregate A: Bulk specific gravity = 2.491; absorption = 0.8% Aggregate B: Bulk specific gravity = 2.773; absorption = 4.6% a. What is the specific gravity of a mixture of 60% aggregate A and 40% aggregate B by weight? b. What is the absorption of the mixture? TABLE P5.32 Aggregate A D Blend Blend 20% 15% 25% 40% Sieve Size, mm Percent Passing 37.5 100 100 100 100 25 100 100 100 100 19 95 100 100 100 12.5 89 100 100 100 9.5 50 85 100 100 4.75 10 55 100 100 2.36 2 15 88 98 1.18 2. 5 55 45 0.6 2 3 35 34 0.3 2 2 22 25 0.15 2 15 14 0.075 2 1 6 5arrow_forwardExplain what is setting time, slump value and workability of conrcete having Plasticiser Superplasticizer Hyper plasticizerarrow_forward5.17 Calculate the sieve analysis of the following aggregate and plot on a semilog gradation paper. What is the maximum size? What is the nominal maximum size? Sieve Size 25 mm (1 in.) 9.5 mm (3/8 in.) 4.75 mm (No. 4) 2.00 mm (No. 10) 0.425 mm (No. 40) 0.075 mm (No.200) Pan Amount Cumulative Amount Cumulative Percent Retained, g Retained, g 0 47.1 239.4 176.5 92.7 73.5 9.6 Retained Percent Passingarrow_forward
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