exercises in soil physics - ebooks

Naftali Lazarovitch & Arthur W. Warrick (Editors)

Exercises in Soil Physics

GeoEcology textbook

Cover graph: Heterogeneous soil profile and Soil Water Characteristic Information (Fig. 2-16)

© Copyright 2013 CATENA VERLAG GMBH, 35447 Reiskirchen, Germany All rights are reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without prior permission of the publisher. This publication has been registered with the Copyright Clearance Center, Inc. Submission of an article for publication implies the transfer of the copyright from the author(s) to the publisher. Bibliographic information published by Die Deutsche Bibliothek Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the internet at http://dnb.ddb.de.

ISBN 978-3-510-65389-8, US-ISBN 1-59326-264-7ISBN ebook (pdf) 978-3-510-65501-4

© 2013 CATENA VERLAG GMBH, 35447 Reiskirchen, Germany

Contents Preface Acknowledgements CHAPTER 1: Aziz Amoozegar The Solid Phase 1 CHAPTER 2: Markus Tuller Soil Water Relations 48 CHAPTER 3: Paul A. Ferre Saturated Water Flow 81 CHAPTER 4: Gary W. Parkin and David J. Fallow Unsaturated Flow 108 CHAPTER 5: Alex Furman, Uri Shavit and Ravid Rozenzwieg Field Water Processes 147 CHAPTER 6: Francis Casey and Leilah Krounbi Chemical Fate and Transport 190 CHAPTER 7: Colin S. Campbell, Douglas R. Cobos and Gaylon S. Campbell Heat and Energy Transport 230 CHAPTER 8: Scott B. Jones Soil Gasses and Transport 262 CHAPTER 9: Ole Wendroth Soil Variability 292 List of Figures and Tables 333 Appendix 341 Appendix on CD

GeoEcology textbook 2013

Preface This book is designed to complement available soil physics and vadose zone hydrology texts by providing additional practice exercises. Material is included for beginning to graduate level students and may be studied either independently or in conjunction with formal classes. More than 200 problems are presented with detailed answers.

The main features for the book arose from discussions while Lazarovitch was a postdoctoral student at Arizona. At that time, we recognized the desirability of additional exercises to accompany available books which understandably must concentrate on presenting subjects and explaining concepts. Inclusion of detailed solutions presents suffi-cient detail for discussion as well as to clarify concepts, both for students and teachers.

The topics of soil physics are explored in nine categories: solid phase, soil water relations, saturated water flow, unsaturated flow, field water processes, chemical fate and transport, heat and energy transport, soil gases and transport and soil variability. Experts in each of these topics were chosen to complete one of the nine chapters including the questions and the solutions. We worked with the chapter authors in order to provide some commonality of layout and style as well as provide feedback in terms of appropriateness and validity.

After the reader selects a problem of interest, we encourage first solving independently and then to compare results with the given detailed answer. Many questions require short answers, but some require a spreadsheet or program. Liberal references are made to EXCEL of which familiarity is assumed although the reader can use other programs if preferred. Several problems are also logically solved using HYDRUS-1D, STANMOD, ROSETTA and RETC which are available in the public domain. In Chapter 9, use can be made of WINASTSA, also available in the public domain, although some of the presented results were performed with ISATIS, a proprietary program. The appendix of the book summarizes various file names with the associated problems. These files are available in the CD provided with the book and also will be available for downloading in the WebPage of the first editor: http://cmsprod.bgu.ac.il/Eng/Units/bidr/Faculty_Members/Lazarovitch.htm

As most students soon find out, notations and conventions in soil physics take on a variety of forms. Here, each chapter has its own symbol list although, insofar as practical, we encouraged consistent notations across the entire book. Generally soil water potential expressed as energy per unit weight is favored and results expressed in appropriate length units. Soil water content, θ without extra subscripts or adjectives refer to water content on a volumetric basis. Ultimately, notation and rounding of numbers have been left primarily to the chapter authors. We have striven to be rigorous with respect to signs regarding flow, whether


dealing with water, energy, solutes or gases. In order to differentiate between positive and negative flow, careful attention to the coordinate system is required.

The cartoons in the beginning of each chapter and the Albert Einstein quotes at the end of each chapter are added mostly for entertainment. It probably is a surprise to many readers that one of Einstein's early papers dealt with soil erosion (Einstein, A., 1926. The cause of the formation of meanders in the courses of rivers and of the so-called Baer's Law. Read before the Prussian Academy, January 7, 1926).

Naftali Lazarovitch Arthur W. Warrick April 2013 Naftali Lazarovitch French Associates Institute for Agriculture and Biotechnology of Drylands Jacob Blaustein Institutes for Desert Research Ben-Gurion University of the Negev Midreshet Ben-Gurion, 84990, Israel [email protected]

Arthur W. Warrick, Professor Emeritus Soil, Water and Environmental Sciences, University of Arizona Tucson, AZ 85721 [email protected]


Acknowledgements We appreciate the help and effort of a number of people leading to formation and completion of this book. Aziz Amoozegar was involved in discussions with the second editor at a very early stage. At the outset we solicited questions from acquaintances in the soil physics community, drawing heavily on those from Division S-1 from the Soil Science Society of America. Their generous and unselfish response together with exams etc. we had already collected, amounted to several hundred questions. The questions were, in turn, provided to the chapter authors who could use or ignore them as they felt appropriate. Regrettably, we didn't keep a list of the original contributors who we now offer our sincere thanks even though anonymously by necessity.

Leilah Krounbi, a former master student of the first editor, provided critical insight from the point of view of a student. She also contributed clarity and consistency while solving most of the questions of the book.

During the years of the book preparation, several drafts were presented to the soil physics class at the Albert Katz International School for Desert Studies at the Ben-Gurion University of the Negev. The students throughout these years facilitated feedback which was very helpful. In advance, we thank readers for pointing out mistakes directly to us in order to maintain a readily available and up-to-date errata. We would also enjoy hearing ideas for additional questions.

Zoey Gannot drew the cartoons shown at the beginning of each chapter. She exhibited great patience with our limited guidance and was very supportive in creating some levity to balance the otherwise serious tenor of the overall presentation.

This book required many hours of preparations and discussions between the editors as well as with the chapter leaders. Our heartfelt thanks are given to our families for their patience and understanding while we were carrying on.

CHAPTER 1 The Solid Phase

Aziz Amoozegar Department of Soil Science, North Carolina State University

PO Box 7619 Raleigh, North Carolina 27695-7619, USA

[email protected]

2 Amoozegar


Introduction

Soils are composed of mineral particles and organic materials forming a porous system in which the pores are filled with gas (air) and liquid (water). The mineral portion of the solid phase is generated from its respective parent materials and is composed of particles of various sizes, ranging from colloidal clay to rock fragments. The organic fraction is composed of active organisms, plants, and animal residues.

The mineral soil particles can be classified based on their physical (size and shape) or chemical characteristics (composition and crystalline structure). Using only size, soil particles are placed into four general categories: rock, sand, silt, and clay. In the United States Department of Agriculture (USDA) classification system, rock fragments are larger than 2 mm, sand particles are between 2 and 0.05 mm (50 µm), silt particles are between 50 and 2 µm, and clay particles are ≤ 2 µm in diameter. [NOTE: In separating particles into these general categories we assume all particles ≤ 2 mm are spherical.] In this system, the relative proportions of sand-, silt-, and clay-sized particles (expressed as percent on a mass basis) are used to place soils into one of 12 textural classes. This is commonly accomplished using a textural triangle for ease of visualization. The textural class of a soil can be estimated by field texturing or measured in the laboratory by a number of techniques, including the hydrometer and pipette methods (Gee and Or, 2002). In general, the pipette method gives more accurate results for silt and clay contents, and involves simpler calculations for determining the percentages of the three particle size classes than the hydrometer method. The hydrometer method, on the other hand, is easier to perform and allows measurement of a larger number of samples in one setting.

Many soil properties, including hydraulic conductivity and water retention, depend on the size and arrangement of soil pores, which are strongly dependent on the soil particle size distribution. As a result, soil texture is a useful measurement by which to estimate many soil properties that are fundamental to agricultural and environmental issues. Soil texture can also provide useful information pertaining to a number of problems. For example, the rate of settling of soil particles in a suspen-sion depends on soil particle sizes, with larger particles generally settling much faster than smaller particles. This difference plays a significant role in the formation of soils in wet environments (e.g., riverbeds, lakes, flood plains). Similarly, the ability of soil particles to become or remain sus-pended impacts the severity of erosion, with implications for agricultural fields, construction sites, forest roads, and many other settings. Further-more, colloidal particles suspended in surface waters are a major source of pollution impacting not only aquatic organisms (e.g., fish), but also our use of these water resources. Finally, the surface area of a soil, per unit mass, increases with decreasing particle size. Since many chemical

Exercises in Soil Physics: The solid phase 3


reactions in the soil occur on the surfaces of soil particles, the specific surface area of a soil can provide valuable knowledge about the behavior of organic and inorganic chemicals in the soil.

Rock fragments must also be considered when describing soil texture. If the rock fragment of the soil, estimated as a percent of the soil bulk volume, exceeds 15%, a modifying adjective is added to the textural class. For example, for a loam soil containing between 15 and 35% rock fragments that are mostly between 2 and 5 mm in size (fine pebbles or gravel), the modified texture description will be fine gravelly loam, and for the same soil texture with 35 to 60% stones that are mostly between 25 and 60 cm in size, the modified texture description will be stony loam.

Soil particles join together by the aid of a number of factors, including organic materials and chemical binding, to form soil aggregates. Soil structure is primarily a description of these aggregates. Stability of a soil’s aggregates in the presence of water is directly related to how that soil reacts to chemical and physical processes. In addition, the degree of compaction affects soil physical and mechanical characteristics. For example, compaction affects not only the total porosity of the soil, but also the distribution of pore sizes, which controls water flow through the soil and the water holding capacity of the soil. Also, seed germination and root growth are affected by compaction. In many cases, the approximate degree of compaction can be inferred from knowledge of soil bulk density and soil texture.

The following exercises will familiarize you with basic calculations that are useful for describing the solid phase of soils. While these calculations serve as important tools for quantifying selected soil properties and their spatial variability, soil scientists also rely on auxiliary observations to characterize soils in the field. For example, knowledge about likely transitions among soil series, landscape position, and parent material, as well as prior soil management activities (e.g., plowing of the soil for crop production) are used for determining land suitability for a given purpose (e.g., residential housing). In short, the following exercises will prepare you to learn about soil solid phase, but they are no substitute for hands-on soil assessment.

References

Gee, G.W. and Or, D. (2002): Particle-size analysis. p. 255-293. In: J. Dane and C. Topp (ed.), Methods of Soil Analysis, Part 4, Physical Methods. Soil Science Society of America, Madison, WI.

Herrick, J.E. and Jones, T.L. (2002): A dynamic cone penetrometer for measuring soil penetration resistance. Soil Sci. Soc. Am. J. 66: 1320-1324.

Lowery, B. and Morrison, J.E. (2002): Soil penetrometers and penetrability. In: J. Dane and C. Topp (ed.), Methods of Soil Analysis, Part 4, Physical Methods. Soil Science Society of America, Madison, WI, pp. 363-388.

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Nimmo, J.R. and Perkins, K.S. (2002): Aggregate stability and size distribution. In: J. Dane and C. Topp (ed.), Methods of Soil Analysis, Part 4, Physical Methods. Soil Science Society of America, Madison, WI. . p. 317-328.

Rooney, D.J. and Lowery, B. (2000): A profile cone penetrometer for mapping soil horizons. Soil Sci. Soc. Am. J. 64: 2136-2139.

Symbols and Notation am Specific surface area on a mass basis L2 M-1 av Specific surface area on a volume basis L2 L-3 As Surface area L2 Cs Solution concentration M L-3 d Particle diameter L g Gravitational constant L T-2 Mf Mass of field-wet soil M Ms Mass of dry soil M h Depth below water or suspension level L l Thickness of particles L Mw Molecular weight M

t Time T T Temperature °C V Volume L3 Vt Volume of soil L3 Vw Volume of pore space occupied by water L3 z Elevation L θm Gravimetric water content M M-1 θ Water content L3 L-3 η Fluid viscosity M L-1 T-1 ηo Dynamic viscosity of pure water M L-1 T-1 ρ Fluid density M L-3 ρ o Density of pure water M L-3 ρ b Bulk density M L-3 ρ s Particle density M L-3 ρ w Density of water M L-3 φ Porosity L3 L-3



Problem 1-1

A 40-g sample of air-dried soil from the Bt horizon is passed through a 2-mm sieve and analyzed for particle size distribution by the hydrometer method. The amount of organic matter in the sample is negligible. Data for the analysis follow: Mass of water in the air-dried soil 1.2 g Mass of sand in the sample 6.5 g Mass of silt in the sample 19.3 g a. Determine the percent sand, silt, and clay for the sample. Answer: Mass of oven-dried soil in sample = 40 g - 1.2 g = 38.8 g % sand = (100)(6.5 / 38.8) = 16.8% % silt = (100)(19.3 / 38.8) = 49.7% % clay = 100 - (16.8 + 49.7) = 33.5% b. Determine the textural class of the soil using the USDA textural triangle (Figure 1-1).

Fig. 1-1: USDA textural triangle for determining soil textural classes

Answer: Use the textural triangle (Figure 1-1). Any two particle fractions can be used to identify a soil; let’s consider how to use % sand and % clay as an

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example. Find the % sand along the bottom axis and extend a line parallel to the silt axis. (Note the direction of increase, right to left, along the sand axis). Find the % clay along the left axis and extend a line parallel to the sand axis. Find the point of intersection of these lines and read the soil texture of the region that includes this point. The textural class for this soil is silty clay loam.

Problem 1-2

A large sample collected from the A horizon is air dried for particle size distribution analysis. A 200-g representative subsample of the air-dried soil is crushed by a mortar and pestle before being passed through a 2-mm sieve. Exactly 20 g of the sieved soil is treated with hydrogen peroxide (H2O2) prior to chemical and mechanical dispersion. Following chemical and mechanical dispersion, the soil suspension is passed through a 0.053 mm sieve and analyzed for silt and clay content by the pipette method. The data for the analysis are presented below. Mass of sieved (≤ 2-mm diameter) air-dried soil 161.3 g Gravimetric water content 0.025 g g-1 Mass of air-dried soil for particle size analysis 20 g Organic matter content in material passed through 2-mm sieve 4.2% Mass of sand in the 20-g air-dried sample 9.17 g Mass of clay in the 20-g air-dried sample 5.80 g a. Determine the percent sand, silt, and clay for the sample. Answer: Mass of oven-dried soil in 20 g air-dried sample = 20/(1 + 0.025) = 19.51 g Mass of organic material (OM) passing 2-mm sieve = (19.51)(0.042) = 0.82 g Mass of mineral particles in the 20-g air-dried sample = 19.51 - 0.82 = 18.69 g Percent sand in sample = (100)(9.17 / 18.69) = 49.1% Percent clay in sample = (100)(5.8 / 18.69) = 31.0% Percent silt in sample = 100 - (49.1 + 31) = 19.9% b. Name the textural class of the soil. Answer: The textural class of the sample is sandy clay loam (consult the textural triangle provided for the previous problem).



c. We also know that the original 200 g soil sample had a bulk volume of 130 cm3 and we assume that the rock fragments (composed of fine pebbles) have a density of 2.65 g cm-3. Should we add a rock fragment texture modifier? Answer: Mass of rock in the sample = 200 g - 161.3 g = 38.7 g Rock density = 2.65 g cm-3 Volume of rocks = 38.7 / 2.65 = 14.6 cm3 Bulk volume of sample = 130 cm3 % rock fragments in the sample = (100) (14.6 / 130) = 11.2% Since the volume fraction of rock fragment is less than 15%, we should not add a rock fragment textural modifier to the textural class for this soil. It should be noted that the best way to estimate the volume percent of rock fragments is by field visual determination. This allows for sampling over a larger volume, which is more likely to give a representative estimate of the proportion of rock fragments in the soil.

Problem 1-3

The average values for particle size distribution of the mineral fraction of soil samples collected from the 20 to 25 cm depth in two adjacent agricultural fields are given below.

Table 1-1: Average values for particle size distribution

a. Draw the cumulative particle size distribution curves (also known as the “percent finer” curves) for the two fields. For this, construct a loga-rithmic x-axis for the particle diameter that ranges from 0.0001 to 2 mm.

Particle Equivalent Diameter Field 1 Field 2 mm % %

< 0.0001 8 3 0.0001 - 0.001 15 9 0.001 - 0.01 21 16 0.01 - 0.04 17 13 0.04 - 0.1 14 14

0.1 - 0.25 13 17 0.25 - 0.5 6 15 0.5 - 1.0 3 9 1.0 - 2.0 3 4

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Plot the percent by mass of the solids that are smaller than the upper value of each diameter range. Answer:

Fig. 1-2: Cumulative particle size distributions for the two soil samples

b. Using the USDA classification system, determine the textural class of the soil in each field. Answer: Plot vertical lines extending from the diameters that define the particle size class (2 mm for sand, 0.05 mm for silt, and 0.002 mm for clay). The percent of each class for each soil can be found by extending horizontal lines from the points of intersection of the vertical lines with the cumulative fraction lines. % clay (< 0.002 mm) for Field 1 = 29% % silt and clay (< 0.05 mm) for Field 1 = 64% % silt for Field 1 = 64 - 29 = 35% % sand for Field 1 = 100 - 64 = 36% The texture for Field 1 is clay loam. % clay for Field 2 = 16% % silt and clay (< 0.05 mm) for Field 2 = 44% % silt for Field 2 = 44 - 16 = 28% % sand for Field 2 = 100 - 44 = 56% The texture for Field 2 is sandy loam. c. Using the International Soil Science Society particle size classification system, determine the percent sand, silt, and clay in these fields.



Answer: In the International Soil Science Society classification system clay particles are smaller than 0.002 mm (2 µm) in diameter and silt particles are between 0.002 and 0.02 mm (2 to 20 µm) in diameter. Following the procedure used in the previous question we have % clay (< 0.002 mm) for Field 1 = 29% % silt and clay (< 0.02 mm) for Field 1 = 52% % silt for Field 1 = 52 - 29 = 23% % sand for Field 1 = 100 - 52 = 48% % clay for Field 2 = 16% % silt and clay for Field 2 = 34% % silt for Field 2 = 34 - 16 = 18% % sand for Field 2 = 100 - 34 = 66% Note that the International Soil Science Society particle size classification cannot be used to determine the soil textural classes based on the USDA classification system.

Problem 1-4

Air-dried samples of two different soils are submitted to your laboratory for particle size distribution analysis by the pipette method. Twenty grams of each soil sample is treated with hydrogen peroxide for removing organic matter, and 10 cm3 of a 50 g L-1 (0.05 g cm-3) sodium hexameta-phosphate (Na-HMP) reagent solution followed by mechanical disper-sion. The soil suspension for each sample is placed in a 1-L tall cylinder and the volume of suspension is brought to 1000 cm3 with distilled water. Using a plunger, the suspension in each cylinder is mixed thoroughly and then allowed to settle at time t = 0. The temperature of the laboratory where the analysis is taking place is 22 °C and remains constant. a. List the information (e.g., water properties) needed for determining particle size distribution by pipette method. Answer: The concentration of the Na-HMP solution in which particles are suspended - The concentration of the Na-HMP solution can be obtained from the concentration of the original Na-HMP dispersing reagent. However, making a blank solution (in this case mixing 10 cm3 of Na-HMP solution and 990 cm3 deionized water in a cylinder), removing a sample with the same pipette used for sampling soil suspension, and drying the

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sample for determining the amount of Na-HMP in the suspension samples is preferred. The temperature of the soil suspension - Temperature is needed to determine the density and viscosity of the Na-HMP solution in which soil particles are suspended. Although density and viscosity of water at the respective temperature can be used, Gee and Or (2002) offer equations to calculate the density (ρ) and viscosity (η) of the solution based on its Na-HMP concentration: ρ = ρ o (1 + 0.63Cs) η = η o (1 + 4.25Cs) for determining density and viscosity of the solution at temperature T (°C), respectively. In these equations ρo

(g cm-3) and ηo (g cm-1 s-1) are the density and viscosity of pure water at temperature T, respectively, and Cs (g cm-3) is the concentration of Na-HMP in the soil suspension. The particle density (ρs) of soil - In general, the value of 2.65 g cm-3 (density of quartz mineral) is used for the particle density of the soil. If specific information is known, a better estimate of the particle density can be used. b. Assume the temperature of soil suspension in each cylinder is the same as that of the laboratory (22 °C). Using Stokes law, determine the time that a 25 cm3 sample must be collected with a pipette from 10 cm below the level of suspension in the cylinders for determining the silt and clay (particles ≤ 0.05 mm in diameter) content of the soil samples. Also determine the time that a 25 cm3 sample must be collected from 5 cm below the suspension level in each cylinder for determining the clay (particles ≤ 2 µm in diameter) content of the two samples. Answer: Obtain the values for density and viscosity of water from an appropriate reference or the web. From Stokes law, the equation to calculate the time (t, in seconds) for collecting a sample of diameter, d (e.g., 0.005 cm) at depth, h (e.g., 10 cm) below the level of suspension in the cylinder is:

( )2

18

s

htd g

ηρ ρ

=−

Assume particle density (ρs) = 2.65 g cm-3, density of water (ρo) at 22 °C = 0.998 g cm-3, viscosity of water (ηo) at 22 °C = 0.9548 centipoise = 0.009548 poise or g cm-1 s-1, acceleration due to gravity (g) = 980 cm s-2, and concentration of Na-HMP in soil suspension = (10)(0.05) / 1000 = 0.0005 g cm-3.



From the Gee and Or equations (from part a. above), the density of the soil suspension solution (ρ) at 22 °C = (0.998)[1 + (0.63)(0.0005)] = 0.998 g cm-3, and the viscosity of soil suspension solution (η) at 22 °C = (0.009548)[1 + (4.25)(0.0005)] = 0.009568 g cm-1 s-1. Taking the depth for collecting sample (h) = 10 cm and the diameter of the largest particles to collect (silt- and clay- sized particles) = 0.05 mm = 0.005 cm, the time to collect 25 cm3 sample at 10 cm below the soil suspension level is t = (18)(10)(0.009568) / [(0.005) 2 (980)(2.65 - 0.998)] = 42.6 s The corresponding time for h = 5 cm and clay-sized particles (≤ 2 µm) becomes t = (18)(5)(0.009568) / [(0.0002) 2 (980)(2.65 - 0.998)] = 13297 s = 3 h, 41 min, 37 s c. Instead of collecting a sample from 5-cm depth below the suspension level for determining clay content of the sample, you mistakenly collect a 25 cm3 sample from 10-cm depth below the suspension level. Determine the degree of error in determining the particle size distribution of these soils caused by your mistake. Answer: The diameter (d) of the largest particles collected at depth h below the level of suspension in the cylinder at time t is given by

( )

0.518

s

hdtg

ηρ ρ

=

− Substituting the values into this equation we get d = {(18)(10)(0.009568) / [(13297)(980)(2.65 - 0.998)]}1/2 = 0.000283 cm or 2.8 µm The correct time to collect a sample of clay-sized particles from the 10-cm depth is: (13297 / 5)(10) = 26594 s = 7 h, 23 min, and 14 sec. Degree of error: The correct time to collect a sample from the 10-cm depth, as shown above, is twice the time calculated in part b (i.e., for 5-cm depth). As a result of collecting the sample too early, you have collected very fine silt-sized particles along with clay-sized particles, as evidenced by the average particle size diameter of the sampled suspen-sion shown above (2.8 µm). Consequently, the amount of clay in the sample will be overestimated and the amount of silt will be underesti-

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mated. The importance of this error for textural classification will depend on the location of the sample on the soil textural triangle. d. You mistakenly use the water density and viscosity values for 32 °C instead of 22 °C when calculating the times for collecting samples from 10 and 5 cm depths for determining silt and clay contents of the samples, respectively. Determine the diameter of the largest particles in the samples collected at each step, and discuss the extent of error caused by your mistake. Answer: Density of water (ρo) at 32 °C = 0.995 g cm-3 Viscosity of water (ηo) at 32 °C = 0.7647 centipoise = 0.007647 g cm-1 s-1 Concentration of Na-HMP in soil suspension = 0.0005 g cm-3 From the Gee and Or equations (see part a): Density of soil suspension solution (ρ) at 32 °C = (0.995)[1 + (0.63)(0.0005)] = 0.995 g cm-3 Viscosity of soil suspension solution (η) at 32 °C = (0.007647)[1 + (4.25)(0.0005)] = 0.007663 g cm-1 s-1 Depth for collecting sample (h) = 10 cm For 0.005 cm diameter and a collection depth of 10 cm the calculated time is t = (18)(10)(0.007663) / [(0.005)2 (980)(2.65 - 0.995)] = 34 s The actual diameter of the largest particles collected by the technician at t = 34 s calculated based on the actual temperature of 22 °C is d = {(18)(10)(0.009568) / [(34)(980)(2.65 - 0.998)]}1/2 = 0.0056 cm or 56 µm To collect a sample at 5 cm depth for the clay-sized particles (d = 0.0002 cm) the calculated time is t = (18)(5)(0.007663) / [(0.0002)2 (980)(2.65 - 0.995)] = 10631 s, or 2 h, 57 min, 11 s The actual diameter of the largest particles collected at t = 10631 s, calculated based on the actual temperature of 22 °C, is d = {(18)(5)(0.009568) / [(10640)(980)(2.65 - 0.998)]}1/2 = 0.000224 cm or 2.24 µm Consequently, the amount of sand will be underestimated and the amount of clay will be overestimated, as the size range of the diameter of



collected particles extends into the sand category ( > 56 µm). Depending on the actual particle size distribution of the sample, the amount of silt may be over or underestimated. The error in sand content may not be significant if the suspension was passed through a 50 µm opening or 53 µm opening sieve following the standard procedure of the pipette method. The importance of this error for textural classification will depend on the location of the sample on the soil textural triangle. e. The actual temperature of the suspension is 32 °C during the early part of the analysis. It takes approximately 3.5 hours for the soil suspension at 32 °C to reach equilibrium with the laboratory temperature (22 °C). How might you collect the samples for these conditions? Answer: Since the temperature of the suspension does not change rapidly, the time to collect the first sample from 10 cm below the suspension level in the cylinder is 34 s as calculated in d above. If the temperature remained at 32°C, the time to collect the second sample would be 10631 s. After approximately 3.5 hours the temperature of the suspension will reach the laboratory temperature of 22 °C. If the temperature of the suspension had been 22 °C for the entire period, the time to collect the sample would be 13297 s. Assuming the density and viscosity of water change linearly between 32 and 22 °C, the time to collect the second sample would be the average of the above time values, t = (13297 + 10631) / 2 = 11964 s or 3 h, 19, min, 24 s (round it to 3 h, 20 min).

Problem 1-5 A soil sample containing 20% iron oxide (Fe2O3) is being analyzed for particle size distribution by the pipette method in a constant temperature room where the temperature is 21 °C. a. Assume the average particle density for the soil is 2.7 g cm-3 and determine the time that a 25 cm3 sample must be collected from 5 cm below the suspension level to determine the amount of particles with diameters ≤1 µm. Concentration of Na-HMP in soil suspension = 0.0005 g cm-3. Answer: We determine the time that it will take for all particles larger than 1 µm to settle out of the upper 5 cm of the soil suspension in the 1000 cm3 cylinder. Density of water (ρo) at 21 °C = 0.998 g cm-3

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Viscosity of water (ηo) at 21 °C = 0.9779 centipoise = 0.009779 g cm-1 s-1 Concentration of Na-HMP in soil suspension = 0.0005 g cm-3 Acceleration due to gravity (g) = 980 cm s-2 Depth for collecting sample (h) = 5 cm; and Diameter of the largest particles to collect = 0.001 mm = 0.0001 cm From Gee and Or equations (see problem 4): Density of soil suspension solution (ρ) at 21 °C = (0.998)[1 + (0.630)(0.0005)] = 0.998 g cm-3 Viscosity of soil suspension solution (η) at 21 °C = (0.009779)[1 + (4.25)(0.0005)] = 0.00980 g cm-1 s-1 The time to collect 25 cm3 sample of soil suspension at 5 cm below the surface is t = (18)(5)(0.00980) / [(0.0001)2 (980)(2.7 - 0.998)] = 52879 s or 14 h, 41 min, 19 s b. Determine the amount of time that it will take for all the iron oxide particles larger than 2 µm to settle out of the upper 5 cm of the soil suspension in the 1000 cm3 cylinder. Answer: Particle density of iron oxide particles (ρs) = 5.24 g cm-3 Depth of settling (h) = 5 cm Diameter of the largest particles under consideration to collect (clay-sized particles) = 0.002 mm = 0.0002 cm The time for iron oxide particles larger than 2 µm to pass 5 cm depth below the surface of soil suspension is t = (18)(5)(0.00980) / [(0.0002)2 (980)(5.24 - 0.998)] = 5304 s or 1 h, 28 min, 24 s c. Specify the assumptions for determining the times in a and b above, and discuss errors associated with determining particle size distribution of this soil with relatively high iron oxide content by assuming a given particle density value (e.g., 2.7 g cm-3 assumed here). Answer: In calculating the settling time of soil particles in part a, we assumed a soil particle density of 2.7 g cm-3, whereas for part b we used 5.24 g cm-3 for the particle density of iron oxide. The difference in particle densities of soil fractions brings into question whether the Stokes Law assumptions are relevant [namely these particles 1) are spherical, 2) have the same density, and 3) fall independently, at their terminal velocity]. Here 20% of the soil is comprised of iron oxide. In addition to higher particle density,



iron oxide particles are perhaps more uniform than the clay fraction of the soil. Thus, iron oxide particles are expected to fall at a faster rate than the clay mineral particles of the same diameter. As for the implications on measurements of the soil particle size distribution, the presence of iron oxide particles in the suspension at the time of sample collection will influence the results. The time a sample should be collected from 5 cm below the suspension level for determining clay content based on particle density of 2.7 g cm-3 is t = (18)(5)(0.00980) / [(0.0002)2 (980)(2.7 - 0.998)] = 13220 s or 3 h, 40 min, 20 s The diameter of the largest iron oxide particles collected by the pipette from 5 cm depth at t = 13220 s is d = {(18)(5)(0.00980) / [(13220)(980)(5.24 - 0.998)]}1/2 = 0.000127 cm or 1.27 µm For soils containing a high level of iron oxide, assuming a particle density of 2.7 g cm-3 for all soil particles will result in an underestimation of the actual clay-sized (≤ 2 µm) fraction.

Problem 1-6 Two soil samples are collected from 60 cm below the soil surface within a 10-m by 10-m area of an agricultural field. The samples are air dried before a technician uses 20 g of each sample for particle size analysis by the pipette method and determines the water content of a subsample of each sample by the gravimetric method. The amount of organic matter in the samples is negligible. Each sample is mixed with 300 cm3 of deionized water. For chemical dispersion the technician adds 10 cm3 of a 50 g L-1 sodium hexametaphosphate (Na-HMP) reagent to each sample before mechanical dispersion. The technician also prepares a blank solution of Na-HMP by mixing 10 cm3 of Na-HMP reagent solution and 990 cm3 of distilled water in a cylinder (total volume 1000 cm3). After mechanical dispersion the samples are passed through a 53-µm sieve into 1000 cm3 cylinders. After bringing the volume to 1000 cm3 and mechanical agitation the technician collects two 25 cm3 samples from 10 and 5 cm depths in each cylinder at appropriate times for determining the silt and clay contents of the samples, respectively. The data for these analyses are presented below.

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Table 1-2: Data for textural analysis

Sample 1 Sample 2 g g Mass of air-dried soil subsample 125 139 Mass of oven-dried soil subsample 121 133 Mass of air-dried soil for textural analysis 20 20 Mass of oven-dried materials left in the sieve (> 0.053 mm diameter) 3.85 3.24

Mass of oven-dried solids in the first 25 cm3 sample 0.4032 0.3921 Mass of oven-dried solids in the second 25 cm3

sample 0.2417 0.1841

Mass of oven-dried solids in 25 cm3 sample of blank 0.0125 a. Determine the percent of sand-, silt-, and clay-sized particles in each sample. Answer: Gravimetric water content, θm of samples Sample 1: (125 - 121) / 121 = 0.033 g g-1

Sample 2: (139 - 133) / 133 = 0.045 g g-1 Oven-dried mass of soil in 20 g samples used for particle size analysis Sample 1: 20 / (1 + 0.033) = 19.36 g Sample 2: 20 / (1 + 0.045) = 19.14 g Mass of oven-dried silt and clay in the first 25 cm3 samples Sample 1: 0.4032 - 0.0125 = 0.3907 g Sample 2: 0.3921 - 0.0125 = 0.3796 g Mass of oven-dried clay in the second 25 cm3 samples Sample 1: 0.2417 - 0.0125 = 0.2292 g Sample 2: 0.1841 - 0.0125 = 0.1716 g Mass of clay particles in the original 20 g air-dry sample Sample 1: (0.2292)(1000) / 25 = 9.17 g Sample 2: (0.1716)(1000) / 25 = 6.86 g Mass of silt particles in the original 20 g air-dried sample Sample 1: (0.3907 - 0.2292)(1000) / 25 = 6.46 g Sample 2: (0.3796 - 0.1716)(1000) / 25 = 8.32 g Total mass of particles recovered Sample 1: 3.85 + 6.46 + 9.17 = 19.48 g Sample 2: 3.24 + 8.32 + 6.86 = 18.42 g Difference between mass of particles recovered and oven-dried mass of soil in the 20 g soil sample. Sample 1: 19.36 - 19.48 = -0.12 g Sample 2: 19.14 - 18.42 = 0.71 g



We assume the organic matter content to be negligible. The errors for samples 1 and 2 are -0.6% [(-0.12)(100) / 19.36] and 3.7% [(0.71)(100) / 19.14], respectively. One way to resolve the discrepancy between the total oven-dried mass of the soil and the mass of sand, silt and clay particles recovered is to assume the discrepancy is mainly due the presence of very fine sand-sized particles that passed through the sieve (mesh opening of 0.053 mm) and settled past the 10 cm depth when the first 25 cm3 sample was collected. Accordingly, we can resolve the discrepancy between observa-tions and expected results by adding the difference to the sand-sized fraction for Sample 2 (sand content = 3.24 + 0.71 = 3.95). However, for Sample 1, the discrepancy between the mass of oven-dried soil in the first sample and the amount of sand, silt, and clay particles recovered is < 1%, which is within measurement error. Thus, we can subtract the difference from the sand fraction or distribute it among the three particle sizes. Here, for Sample 1, we use the total mass recovered to determine the percent of the sand- , silt-, and clay-sized particles. For Sample 2, we use the oven-dried mass of soil and recalculated mass of sand to determine the particle size distribution. Table 1-3: Results for textural analysis

Mass(g) (%) Sand Silt Clay Total Sand Silt Clay Sample 1 3.85 6.46 9.17 19.48 20 33 47 Sample 2 3.95 8.32 6.86 19.13 21 43 36

b. Using the USDA textural triangle (Figure 1-1), determine the textural class of each sample. Answer: Sample 1: clay (c) Sample 2: clay loam (cl) c. Explain why there is a difference between the particle size distribution results for the two samples. Answer: The difference between the results for the two samples is due to spatial heterogeneity in the field and/or analysis error. Natural heterogeneity: Soils are variable by nature. The difference between the sand content of the two samples is not significant. However, there is a substantial decrease in clay content from Sample 1 to Sample

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2. Information about the soil profile at the site can be very helpful in determining the cause of discrepancy. For example, the soil may be shallow, and Sample 1 may be from the middle of the B horizon while Sample 2 may have come from the boundary between the B and C horizons in which C materials with substantially lower clay content are mixed with the clayey B horizon. Another possibility is that the depth of the boundary between different B horizons (B1, B2, B3, etc.) varies, so the samples may be from different B horizons. Since the sand contents are approximately the same, there does not seem to be a chance for the A and/or E horizon be mixed with the B horizon by deep plowing. Measurement Errors: Assuming that the 25 cm3 samples were collected at correct time and depth for each of the two soil samples, one possibility for error is that the original samples were not thoroughly mixed when the 20 g subsamples were collected, or the soil suspensions were not adequately agitated to disperse all the particles uniformly in the cylinders. Another possibility is that the samples may have contained some organic materials (OM) that cause soil particles to flocculate. Since the OM was not removed, there is a possibility that the clay particles, particularly in Sample 2, did not disperse completely. Yet, another possibility is that the amount of dispersing agent added to disperse the soil particles was not adequate to disperse the clay particles completely. In addition, the amount of original air-dried sample used for this analysis may have been too high. Generally, for the pipette method, a 10 g sample is recommend-ed for a clayey textured soil. In these analyses, a 20 g sample was used.

Problem 1-7 A retention pond collects runoff from a construction site. After a heavy rain, the retention pond is almost full. At time to, runoff into the retention pond ends, and sediments start settling. The temperature of the water in the pond is 33 °C. a. Twenty-four hours after runoff to the pond stops, water in the upper 30 cm of the pond is discharged into a creek. Assume the pond temperature remains constant and determine the equivalent diameter of the largest particles discharged from the retention pond into the creek for this rain event. Specify your assumptions. Answer: Time for settling = (24)(3600) = 86400 s Height of column of water discharged after 24 hours = 30 cm Acceleration due to gravity, g = 980 cm s-2 Particle density of sediment (ρs) = 2.65 g cm-3



Viscosity of water (η) at 33 °C = 0.007491 poise or g cm-1 s-1 Density of water (ρw) at 33 °C = 0.995 g cm-3 Diameter of the largest particles discharged is d = {(18)(30)(0.007491) / [(86400)(980)(2.65 - 0.995)]}1/2 = 0.00017 cm or 1.7 µm b. Under local regulations particles larger than 5 µm should not be dis-charged into the stream receiving runoff water from this retention pond. Determine the time when the gate valve for the pond overflow system can be opened to allow water from the upper 30 cm of the pond to be discharged into the stream. Assume temperature remains constant. Answer: The diameter of the largest particles that can be discharged is the same as the diameter of the smallest particles that can settle 30 cm. We need to determine the time that all particles larger than 5 µm settle 30 cm or more. t = (18)(30)(0.007491) / [(0.0005)2 (980)(2.65 - 0.995)] = 9976 s or 2 h, 46 min, 16 s c. A skimmer is a floating device that is used to drain water from the upper part of a retention pond. A skimmer that drains water from the top 5 cm of the ponded water is used to drain water into the stream. The flow rate through the skimmer can be adjusted to attain a given rate of drop in the water level in the pond. No soil particle larger than 5 µm in diameter can be discharged into the stream. Determine the earliest time after runoff into the pond stops that the skimmer can be opened to discharge water into the stream. Determine the rate at which water level in the pond can be dropped to prevent discharge of particles larger than 5 µm. Answer: Diameter of the largest particles that can be discharged = 5 µm or 0.0005 cm Because the skimmer opening is 5 cm in depth, the earliest time that the skimmer can be opened is when all particles larger than 5 µm settle at least 5 cm. The time when all particles larger than 5 µm settle 5 cm is t = (18)(5)(0.007491) / [(0.0005)2 (980)(2.65 - 0.995)] = 1663 s or 27 min, 43 s The rate of drop in water level cannot exceed the rate at which a 5 µm diameter particle can settle. To obtain a general equation for the rate of

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drop of soil particles in the pond we rewrite the equation for falling particles in soil suspension (i.e., Stokes law equation) in terms of h:

( )2

18std g

hρ ρη

−=

The rate of fall is found by differentiating h with respect to time t: ( )2

18sd gdh

dtρ ρη

−=

For 5 µm diameter particles we get dh/dt =(0.0005)2(980)(2.65 - 0.995) /[(18)(0.007491)] = 0.003007 cm s-1 or 10.83 cm h-1 The rate of drop for 5 µm diameter particles can also be obtained by dividing the settling distance (5 cm) by the time it takes the particles to settle that distance, 5 / 1663 = 0.0030 cm s-1 or 10.8 cm h-1. d. The level of water in the pond can be dropped 100 cm. Six hours after runoff into the pond stops, the skimmer valve is opened. Determine the rate at which the level of water in the pond can be dropped to prevent any particle larger than 5 µm to be discharged into the stream. Specify your assumptions. Answer: The distance that a 5 µm particle settles in 6 hours (21600 s) is h = (21600)(0.0005)2 (980)(2.65 - 0.995) / [(18)(0.007491)] = 64.95 cm This distance can also be calculated from the rate of drop we found in section c, h = (10.83)(6) = 64.95 cm. To lower the water level 100 cm, the bottom of the skimmer opening will be at 105 cm depth. To meet the requirement, all particles larger than 5 µm must settle past the 105 cm depth. The time all particles larger than 5 µm settle through the upper 105 cm of the pond is t = (18)(105)(0.007491) / [(0.0005)2(980)(2.65 - 0.995)] = 34917 s or 9 h, 41 min, 57 s The skimmer can be opened after 6 hours or 21600 s and can drain down to the 65-cm depth, but the pond cannot be drained to 100 cm depth before 34917 s. Thus, the skimmer can be open for 13317 s (34917 - 21600). Based on this, the maximum rate of drop in water level in the pond is (100)(3600) / 13317 = 27.0 cm h-1.



e. Polyacrylamide (known as PAM) is sometimes used to reduce the amount of sediment leaving a retention pond by enhancing settling of particles through flocculation. During a storm event, PAM is applied to the runoff on its way into the retention pond. Thirty minutes after the runoff into the pond stops, a large portion of the soil particles entering the pond settle out of the upper 40 cm. Determine the equivalent diameter of the smallest sized soil aggregates that settle out of the upper 40 cm of the retention pond 30 minutes after the runoff stops. Answer: Assume the soil aggregates that are formed are spherical with a particle density of 2.65 g cm-3, and that we can apply Stokes law to calculate the rate of settling of the aggregates. The diameter of the smallest aggregates settling through the upper 40 cm of the pond in 30 minutes is d = {(18)(40)(0.007491) / [(30)(60)(980)(2.65 - 0.995)]}1/2 = 0.00136 cm or 13.6 µm

Problem 1-8

A retention pond, 24 m long, 19 m wide and 120 cm deep, is constructed for catching runoff from a two-ha construction site. [NOTE: Here we assume the retention pond is rectangular with vertical walls. In reality, a retention pond constructed at a site is not rectangular and has sides that are sloped in order for the pond to be stable]. The top soil in the area is removed and the Bt horizon is exposed. The average dry bulk density (ρb) for the Bt horizon at this site is 1.35 g cm-3, and its average sand and silt contents are 41 and 36%, respectively. Also, the average percentage of particles less than 1 µm is 13%. One day after constructing the pond, a 26-mm rain occurs and 80% of the rainfall enters the retention pond as runoff. The average concentration of the soil materials eroded from the site in the runoff is 53 g L-1 (53 kg m-3). Assume the particle size distribution for particles suspended in the runoff water is the same as the Bt horizon. Water containing suspended soil materials from the upper 40 cm of the pond is released into a stream 24 hours after the runoff into the pond stops. Assume the average temperature of water in the pond is 20 °C, soil particles in the incoming runoff are dispersed completely, and that their particle density is 2.7 g cm-3. a. Plot the cumulative particle size distribution of the soil particles dispersed in the runoff water (same as the particle size distribution for the B horizon) (Figure 1-3)

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Answer: Fig. 1-3: Cumulative particle size distribution for the Bt horizon at the site b. Estimate the amount and distribution of sediment that accumulates at the bottom of the pond 24 hours after runoff into pond stops. Answer: Amount of rain water falling over the 2-ha field = (2)(10000)(26) / 1000 = 520 m3, Amount of runoff entering the pond = (520)(0.8) = 416 m3 Surface area of the pond = (24)(19) = 456 m2, Depth of water in the pond after the rain = 416 / 456 = 0.91 m or 91 cm, Mass of particles transported to the pond = (416)(53) = 22048 kg Particles settle in the pond at different rates based on their sizes. As particles settle, their concentration in the pond decreases from top to bottom. To determine the amount of sediments at the bottom, we divide the depth of water in the pond into different sections, and determine the amount of particles remaining in suspension in each section after 24 hours. The amount of particles settling at the bottom is then equal to the total amount entering the pond minus the amount remaining suspended. For example, we divide the column of water in the pond into 9 sections, each 10 cm deep, except for the bottom layer, which will be 11 cm. Then, we estimate the average diameter of the particles that settle out of each layer. For this, we choose the average depths of settling for the 9 sections, from top to bottom, to be 5, 15, 25, 35, 45, 55, 65, 75, and 85.5 cm and assume that the average diameter of the particles settling through each section is directly related to the diameter of the largest particles settling the corresponding average depth for that section. Assuming the density and viscosity of water in the pond at 20 °C to be 0.998 g cm-3 and 0.01002 g cm-1 s-1, respectively, we calculate the diameter of the largest particles settling 5, 15, 25, 35, 45, 55, 65, 75, and 85.5 cm after 24 hours (84600 s) by

( )

0.5

s

18hdtg

ηρ ρ

=

−

0.001 0.01 0.1 1 2

100

80

60

40

20

0

Perc

ent P

artic

les

Particle Diameter, mm



Table 1-4: Results for various average settling depths

Average Settling Depth Diameter of particles Settling cm cm µm 5 7.9105E-05 0.8

15 0.0001370 1.4 25 0.0001769 1.8 35 0.0002093 2.1 45 0.0002373 2.4 55 0.0002624 2.6 65 0.0002853 2.9 75 0.0003064 3.1

85.5 0.0003271 3.3 With these calculations we have determined the average diameter of the largest particles remaining in a column of suspension within each 10-cm section. For example, after 24 h, on the average, all particles larger than 0.8 µm settle out of the upper 10 cm. Alternatively, we can say that the average diameter of particles remaining in the top 10 cm of the pond is < 0.8 µm. As another example, we can say that all particles greater than 3.3 µm settle to the bottom of the pond after 24 hours. Assume all particles are dispersed uniformly at the beginning, and that particles settle uniformly based on their size and density. The total mass of particles initially suspended in the pond is (22048)(10) / 91 = 2423 kg in each 10 cm section, and (22048)(11) / 91 = 2665 kg in the lower 11 cm of the pond. Using the above particle size distribution graph, we determine the percent of each size group between 2 mm and 1 µm by interpolation, and extrapolate for the percent of ≤ 0.8 µm size particles. Alternatively, a regression equation can be developed for estimating the cumulative percent of particles. For example, using SLOPE and INTERCEPT func-tions of EXCEL, the following equation is obtained relating the logarithm of particle sizes to their cumulative percentage: % particles = 11.341ln(Particle Diameter in µm) + 14.143 Graphically or by using the regression equation, we determine the amount of particles that remain in each 10 cm layer of the pond. For example, as indicated before, the average diameter of particle remaining in suspension in the upper 10 cm of the pond is 0.8 µm. From the graph, we estimate that approximately 10% of the particles in the runoff are smaller than 0.8 µm. Using the regression equation, the percentage of particles comes out as 11.341ln(0.8) +14.143 = 11.6%

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Using the particle percentage figured from the regression equation, the amount of particles present in the upper 10 cm of the retention pond after 24 hours is (2423)(0.116) = 281 kg. The amount of particles settling out of this layer is 2423 - 281 = 2142 kg. In a similar manner, on the average, the diameter of particles settling out of the upper 20 cm of the pond is 1.4 µm. The percent of particles that are smaller than 1.4 µm is 11.341ln(1.4) + 14.143 = 18% The amount of particles present in the upper 10 to 20 cm of the pond after 24 hours is (2,423)(0.18) = 436 kg. The amount of particles settling out of this layer is 2423 - 436 = 1987 kg. We continue the calculation and determine that the amount of particles in suspension in the lowest 11 cm of the column of water in the pond after 24 hours to be (2665){[11.341ln(3.3) + 14.143] / 100} = 738 kg. Table 1-5: Total mass in suspension

Settling Depth Diameter of

Largest Particles Settling

Particles Smaller

Mass in Suspension

cm cm µm % kg 0-10 5 0.8 11.6 281 10-20 15 1.4 18 436 20-30 25 1.8 20.8 504 30-40 35 2.1 22.6 548 40-50 45 2.4 24.1 584 50-60 55 2.6 25 606 60-70 65 2.9 26.2 635 70-80 75 3.1 27 654 80-91 85.5 3.3 27.7 738

Total in Suspension 4986

Total Settled = 22048 - 4986 = 17062 kg c. Determine the concentration of particles released in the outflow from the upper 40 cm of the pond after 24 hours. Answer: From above table, the amount of particles suspended in the upper 40 cm of the pond is 281 + 436 + 504 + 548 = 1769 kg Volume of water released from the upper 40 cm of the pond is (24)(19)(0.4) = 182.4 m3. Concentration of suspended particles in outflow is 1769 / 182.4 = 9.7 kg m-3

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