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Mass Transport Processes

  Processes which move pollutants and other compounds through the air, surface water, or subsurface environment or through engineered systems (for example, treatment reactors) are of particular interest to environmental engineers and scientists. Pollutant transport acts to move pollutants from the location at which they are generated, resulting in impacts which can be distant from the pollution source. On the other hand some pollutants, such as sewage sludge, can be degraded in the environment if they are sufficiently dilute. For these pollutants, slow transport---slow dilution---can result in excessively high pollutant concentrations, with resulting increased adverse impacts.

In this section, we discuss the processes which distribute pollutants in the environment. The goals of this discussion are twofold: to provide an understanding of the processes which cause pollutant transport, and to present and apply the mathematical formulas used to calculate pollutant fluxes.

Advection and Diffusion

Transport processes in the environment may be divided into two categories: advection and diffusion. Advection refers to transport with the mean fluid flow. For example, if the wind is blowing toward the east, advection will carry any pollutants present in the atmosphere toward the east. Similarly, if a bag of dye is emptied into the center of a river, advection will carry the resulting spot of dye downstream. In contrast, diffusion refers to the transport of compounds through the action of random motions. Diffusion works to eliminate sharp discontinuities in concentration and results in smoother, flatter concentration profiles. Advective and diffusive processes can usually be considered independently. In the example of a spot of dye in a river, while advection moves the center of mass of the dye downstream, diffusion spreads out the concentrated spot of dye to a larger, less concentrated region.

Definition of the Mass Flux Density

We used the term mass flux (, with units of mass/time) earlier when we calculated the rates at which mass was transported into and out of a control volume in a mass balances. Because mass balance calculations are always made with reference to a specific control volume, it was clear that this value referred to the rate at which mass was transported across the boundary of the control volume. In our calculations of advective and diffusive fluxes, however, we will not restrict ourselves to a specific, well-defined control volume. Instead, we will calculate the flux density across an imaginary plane oriented perpendicular to the direction of mass transfer. The resulting mass flux density is defined as the rate of mass transferred across the plane per unit time per unit area, with units of (mass)/. We will use the symbol J to represent the flux density.

J represents the mass flux density, expressed as the rate per unit area at which mass is transported across an imaginary plane. J has units of [M]/[LT].

The total mass flux across a boundary () can be calculated from the flux density simply by multiplying J by the area of the boundary:


In the following sections, we will consider the flux density which results from advection and from diffusion. The symbol J will be used to represent the flux density in each case, whether the flux is a result of advection, diffusion, or a combination of both processes.

Calculation of the Advective Flux

The advective flux refers to the movement of a compound along with flowing air or water. The advective flux density depends simply on concentration and flow velocity.


The fluid velocity, v, is a vector quantity---it has both magnitude and direction, and the flux J refers to the movement of pollutant mass in the same direction as the fluid flow. In this course, we will generally define our coordinate system so that the x-axis is oriented in the direction of fluid flow. In this case, the flux J will reflect a flux in the x-direction, and we will generally ignore the fact that J is really a vector.

  Example .. Calculation of the advective flux density. Calculate the average flux density J of phosphorus downstream of the sewage pipe of example 2.1. The cross-sectional area of the river is 30 m.
Solution: In example 2.1, we found the following conditions downstream of the spot where a sewage pipe added effluent to a river: volumetric flow rate and as phosphorus. The average river velocity is . Using the definition of flux density (equation 32), we find:


Diffusion results from random motions of two types: the random motion of molecules in a fluid, and the random eddies which arise in turbulent flow. Diffusion from the random molecular motion is termed molecular diffusion; diffusion which results from turbulent eddies is called turbulent diffusion or eddy diffusion. We will compare these two types of diffusion below. First, however, we consider why diffusion occurs.

Fick's Law

In this section, we will derive Fick's Law, the equation which is used to calculate the diffusive flux density, by analyzing the results of random motion of a hypothetical box of molecules. The purpose of this derivation is to provide a qualitative and intuitive understanding of the reason that diffusion occurs, and the derivation itself is useful only for that purpose. In problems where it is necessary to actually calculate the diffusive flux, we will normally start at the end of this derivation---that is with the Fick's Law equation (equation 40). (This derivation is based closely on one presented in Mixing in Inland and Coastal Waters, H. B. Fischer, E. J. List, J. Imberger, and N. H. Brooks, Academic Press, New York, 1979.)

Consider a box which is initially divided into two parts, as shown in Figure 8. Each side of the box has a height and depth of 1 unit, and a width of length . Initially, the left portion of the box is filled with 10 molecules of gas x and the right side is filled with 20 molecules of gas y, as shown in the top half of Figure 8. We now remove the divider, and observe what happens.

As we know, molecules are never stationary. All of the molecules in our box are constantly moving around, and at any moment they have some probability of crossing the imaginary line at the center of the box. We will check the box and count the molecules on each side every seconds; we will call the probability that a molecule crosses the central line during the period between observations k. Let's say for now that . So the first time we check the box, after a period , 20% of the molecules that were originally on the left will have moved to the right, and 20% of the molecules that were originally on the right will have moved to the left. When we count the molecules on each side, then, we will find the situation shown in the bottom of Figure 8, with 8 ``x'' molecules remaining on the left and 2 on the right, and 16 ``y'' molecules remaining on the right, 4 having moved to the left. If we assume that the mass of each molecule is 1 unit, we can calculate concentrations in units of mass/volume, and find that the concentration differences between the two boxes have been reduced from 10 to 6 for molecule ``x'', and from 20 to 16 for molecule ``y.'' This is a fundamental result of diffusion---that concentration differences are reduced. We also observe that the movement of molecule ``x'' is essentially independent from that of ``y'', so the diffusion of each molecule can be considered as a separate problem. That is, we don't have to worry about molecule ``x'' when we are calculating the diffusion of molecule ``y'', or vice versa.

So diffusion moves mass from regions of higher concentration to regions of lower concentration, and if left to continue indefinitely, it would result in equal concentrations on both sides of the box. We now need to find the flux density J which diffusion causes. For this calculation, we will again use the situation shown in Figure 8, with a probability of any molecule crossing the central boundary during a period \ equal to k. Since each molecule can be considered independently, we will analyze the movement of a single molecule type, say molecule ``y.''

Let be the total mass of molecule ``y'' in the left half of the box, and equal the mass in the right half. Since our box has unit height and depth, the area perpendicular to the direction of diffusion is one square unit. Thus, the flux density---the flux per unit area---is just equal to the rate of mass transfer across the boundary. The amount of mass transferred from the left to right in a single time step is equal to , while the amount transferred from right to left during the same period is . Thus, the rate of mass flux from left to right across the boundary is equal to divided by , or

Since it is more convenient to work with concentrations than with total mass values, we need to convert this equation to concentration units. The concentration in each half of the box is given by

Thus, substituting for the mass in each half of the box, the flux density is equal to

Finally, we note that as , , and therefore


(Note that the negative sign in this equation is simply a result of the convention that flux is positive when it flows from left to right, but the derivative is positive when concentration increases toward the right.) This equation states that the flux of mass across an imaginary boundary is proportional to the concentration gradient at the boundary. Since the resulting flux cannot depend on arbitrary values of or , the product must be constant. This product is the value we call the Diffusion Coefficient, D. Thus, we obtain Fick's Law:


The units of the diffusion coefficient are clear from an analysis of the units of equation 40 or from the units of the parameters in equation 39; the diffusion coefficient has the same units as . Since k is a probability, and thus has no units, the units of D must be (length)/(time). Diffusion coefficients are commonly reported in cm/s.

Before we go on, note the form of equation 40:

This form of equation will also appear later when we discuss Darcy's Law, which governs the rate at which groundwater flows through soil pores. The same equation also governs heat transfer, if we replace the concentration gradient with a temperature gradient.

Molecular Diffusion

The molecules-in-a-box analysis used above is essentially an analysis of molecular diffusion. Purely molecular diffusion is relatively slow. Typical values of the diffusion coefficient D are approximately 10--10 cm/s for gases, and much lower, around cm/s for liquids. The difference in diffusion coefficient between gases and liquids is understandable if we consider that gas molecules are free to move much greater distances before being stopped by ``bumping'' into another molecule. The diffusion coefficient also varies with temperature and the molecular weight of the diffusing molecule. This is because the average speed of the random molecular motions is dependent on the kinetic energy of the molecules. As heat is added to a material and temperature increases, the thermal energy is converted to random kinetic energy of the molecules, and the molecules move faster. This results in an increase in the diffusion coefficient with increasing temperature. However, if we compare molecules of differing molecular weight, we find that at a given temperature a heavier molecule moves more slowly, and thus the diffusion coefficient decreases with increasing molecular weight.

  Example .. Molecular Diffusiongif Gasoline-contaminated groundwater has been transported under a house from a nearby gas station. Two meters below the dirt floor of the house's basement, the concentration of hydrocarbon vapors in the airspace within the soil is g/cm. Estimate the flux density of gasoline vapor transported into the basement by molecular diffusion. The diffusion coefficient for gasoline vapor in the air space within the soil is equal to cm/s. Assume that the basement is well-ventilated, so that the concentration of gasoline in the basement is very small in comparison to the concentration in the soil.
Solution: To calculate the flux density, we use Fick's Law (equation 40), assuming that the gradient of concentration with height is linear over the 2 m depth. We will not be careful about the sign in the equation, however, since we know that the diffusive flux will be from the ground, where concentration is higher, into the basement, where the concentration is lower.

The calculated flux density of can be used to calculate the total mass flux of gasoline into the basement using equation 31. If the basement floor has an area of 100 m, then the total flux into the basement is

Turbulent Diffusion

In turbulent diffusion, mass is transferred through the mixing of turbulent eddies within the fluid. This is fundamentally different from the processes which determine molecular diffusion---in turbulent diffusion, it is the random motion of the fluid that does the mixing, while in molecular diffusion it is the random motion of the pollutant molecules that is important. Earlier in this section, we used an example in which a spot of dye was dropped into the center of a river. If we follow the center of the dye spot down the river, we would see that spreading out of the spot by molecular diffusion would occur (slowly) as a result of the random motion of dye molecules across the edges of the spot. Turbulent diffusion would occur (much more quickly), as a result of eddies in the river mixing clean water from outside the spot with dye-colored water within the spot.

To indicate this difference in causation, the diffusion coefficient for turbulent diffusion is often referred to as the eddy diffusion coefficient. The value of the eddy diffusion coefficient depends on the properties of the fluid flow, rather than on the properties of the pollutant molecule we are interested in. Most important is the flow velocity---turbulence is only present at flow velocities above a critical level, and the degree of turbulence is correlated with velocity. (More precisely, the presence or absence of turbulence depends on the Reynolds Number, a non-dimensional number which depends on velocity, width of the river or pipe, and the viscosity of the fluid.) In addition, the degree of turbulence depends on the material over which the flow is occurring, so that flow over bumpy surfaces will be more turbulent than flow over a smooth surface, and the increased turbulence will cause more rapid mixing.

Finally, the value of the eddy diffusion coefficient depends to some extent on the size scale of the problem we are considering. This is best illustrated by an example. Figure 9 shows three examples of the mixing of an isolated puff of pollutant (or spot of dye) in a turbulent flow. In Figure 9 a, the size of the pollutant puff is large compared to the size of the turbulent eddies. The result is that turbulent diffusion is slow. The opposite extreme is shown in Figure 9 b, where the size of the puff is very small compared to the turbulent eddies. In this case, the entire puff is moved along with the fluid eddies. The result is not diffusion at all---we would normally call this advection, since the puff is being ``blown'' along intact. The third example (Figure 9 c) shows the intermediate situation. Here, the size of the puff is comparable to the size of the turbulent eddies, and the puff is rapidly stretched out and mixed with the surrounding fluid. In this case, the eddy diffusion coefficient would be rather large. In the real world, of course, turbulent eddies of all sizes are present simultaneously. Therefore, any given case of turbulent diffusion will be a mixture of the three situations shown in Figure 9, and only a single eddy diffusion coefficient would be used.

Fick's Law applies to turbulent diffusion just as it does to molecular diffusion. Thus, flux density calculations are the same for both processes; only the magnitude of the diffusion coefficient is different.

Mechanical Dispersion

The final diffusion-like process that we will consider is similar to turbulence, in that it is a result of variations in the movement of the water (or air) which carries our pollutant. In mechanical dispersion, these variations are the result of (a) variations in the flow pathways taken by different fluid parcels that originate in the nearby locations near one another, or (b) variations in the speed at which fluid travels in different regions.

Dispersion in groundwater flow provides a good example of the first process. Figure 10 shows a magnified depiction of the pores within a soil sample, through which groundwater flows. (Note that, as shown in Figure 10, groundwater movement is not a result of underground rivers or creeks, but rather is caused by the flow of water through the pores of the soil, sand, or other material underground.) Because transport through the soil is limited to the pores between soil particles, each fluid particle takes a convoluted path through the soil and, as it is transported horizontally with the mean flow, it is displaced vertically a distance that depends on the exact flow path it took. The great variety of possible flow paths results in a random displacement in the directions perpendicular to the mean flow path. Thus, a spot of dye introduced into the groundwater flow between points B and C in Figure 10 would be spread out, or dispersed into the region between points B and C as it flows through the soil.

The second type of mechanical dispersion results from differences in flow speed, and is only important in non-steady state problems, such as the accidental, sudden release of a pollutant into a flowing stream. Anywhere that a flowing fluid contacts a stationary object, the speed at which the fluid moves will be slower near the object. For example, the speed of water flowing down a river is fastest in the center of a river, and can be very slow near the edges. Thus, if a line of dye were somehow laid across the river at one point, it would be stretched out as it flowed down the river, with the center part of the line moving faster than the edges. This type of dispersion spreads things out in the longitudinal direction---in the direction of flow---in contrast to diffusion and dispersion in groundwater, which spread things out in the direction perpendicular to the direction of mean flow.

The Movement of a Particle in a Fluid---Stoke's Law

In this section, we will analyze the forces which determine the movement of a particle suspended in a fluid. Suspensions of particles must be analyzed in a variety of applications, from the cleaning of particulate pollutants from coal-fired power plants and the settling out of suspended particles in wastewater treatment plants, to the dense suspension of particles in fresh cement. In each of these examples, it is useful to understand the movement of particles within the air or water fluid.

The movement of a particle in a fluid can be determined by a balance of the viscous drag forces resisting the particle movement with gravitational or other forces which cause the movement. The solution to this balance of forces in the specific problem of particle settling under gravity is known as Stoke's Law. This law is derived and applied in this section.

Gravitational Settling

Consider the settling particle shown in Figure 11. The particle is settling in a fluid, which may be air, water, or any other fluid. The particle moves in response to the gravitational force, , which is equal to the mass of the particle times the gravitational constant, . An upward buoyancy force is also present, and is equal to the mass of the displaced fluid times g. Let us assume that the particle is a sphere of radius and density \ (), and that the fluid density is . Then the volume of the particle is , and we have

The only remaining force to determine is the drag force, . The drag force is the result of frictional resistance to the flow of fluid past the surface of the particle. This resistance depends on the speed at which the particle is falling through the fluid, the size of the particle, and the viscosity, or resistance to shear of the fluid. (Viscosity is essentially what one would qualitatively call the ``thickness'' of the fluid---honey has a high viscosity, while water has a relatively low viscosity and the viscosity of air is much lower yet.) Over a wide range of conditions, the friction force can be correlated with the Reynolds number, which is discussed in CE361. However, for most situations we are interested in, we can use Stoke's Law, which states that


where is the fluid viscosity (units of ) and is the velocity of the particle relative to the fluid.

Using this relationship, we now have an equation for all of the forces acting on the particle. The net downward force acting on the particle is equal to:


The particle will respond to this force according to Newton's second law, which states that . Thus,


This differential equation can be solved to determine the time-varying velocity of a particle which is initially at rest. The solution indicates that, in almost all cases of environmental interest, the period of time required before the particle reaches its final settling velocity is very short. For this reason, we will consider here only the steady-state situation in which the velocity is constant, and thus the right-hand side of equation 49 is equal to zero.

In this case, . Setting equal to zero and noting that is equal to the settling velocity at steady state, we can rearrange equation 48 to obtain


This is the fundamental equation which is used to calculate terminal settling velocities of particles in both air and watergif. Because it is based on the Stoke's Law for the drag force, this equation is also often referred to as Stoke's Law, and the settling velocity is often call the Stokes velocity. A fundamental result of this equation is that the settling velocity increases as the square of the particle diameter, so that larger particles settle much faster than do smaller particles.

  Example .. Settling in water. A tank called a grit chamber is to be used to remove suspended sand particles from water by allowing them to settle out. The main purpose of this chamber is to prevent the sand from wearing out pipes and pumps within the wastewater treatment plant. To design the tank in a sufficient size, we must first determine the settling velocity of the sand particles. If the sand particles are spheres with diameter m and density , what would be their settling velocity? (Note that the viscosity of water is and the density of water is .)
Solution: We will use our steady-state solution based on Stoke's Law, equation 52.

(Note that .) The settling velocity is relatively slow. For very small particles, the drag force is significant in comparison to the gravitational force acting on their small mass.

  Example .. Settling in air. Calculate the gravitational settling velocity for two atmospheric particles having diameters of 0.1 and 100. , respectively, and with density . Based on the result, determine whether removal of these particles in a settling chamber of 1 m depth would be a realistic proposition. The viscosity of air is . The density of air is extremely small, and thus the buoyancy force may be neglected.
Solution: Again, we simply need to plug the parameters of the problem into equation 52. For the 0.1 particle, we obtain

and, for the 100. particle,

The 0.1 micron particle would take about 900 hours to settle a distance of 1 m. However, because the settling velocity is proportional to the square of the particle size, the 100 m particle settles 10 times faster, and would settle a distance of 1 m in only 3 s. It would clearly be realistic to construct a chamber with a residence time of >3 s. However, a much larger chamber would be required to reach a residence time of 9 hr, and such a chamber would likely not be economical.

Note that the 100 particle in this example settles with a velocity that is much greater than that of the 100 particle settling through water in example 3.3. This is due to the much lower viscosity of air relative to water.

  Example .. Calculation of the minimum diameter removed by a settling chamber. A settling chamber is used to remove sand particles from the sewage flow through a wastewater treatment plant. The chamber is 2 m deep and the residence time (retention time) of water in the chamber is 4.4 hr. What is the minimum size particle which would be completely removed by settling to the bottom of the chamber during passage through the chamber? The density of sand particles is 2.65 g/cm, and the viscosity of water is 0.01185 g cm s. Assume that any particle that settles to the bottom of the chamber is removed.
Solution: Since only those particles which reach the bottom of the chamber are removed, 100% removal requires that the distance settled during passage through the chamber is equal to the chamber depth. This results in a minimum settling velocity:

Plugging in our equation for settling velocity,

Thus, the minimum size particle removed entirely has a diameter of 13.

Other applications of force balances on a particle

Analyses similar to the one we used to calculate the gravitational settling velocity can be used in a number of other applications. Essentially, in any situation where a particle moves in response to an applied force, the applied force can be balanced against the drag force to determine the particle's terminal velocity. In air pollution control, devices called electrostatic precipitators are used to remove smaller particles by applying an electric force to the particles, which causes them to move out of the airstream and onto charged collection plates. Other air pollution control devices, called cyclones, are based on the use of inertial forces to remove particles. To determine the particle removal efficiency in these devices, it is necessary to balance the drag force against a centrifugal force.

Flow of water through a porous medium---Darcy's Law for groundwater flow

The final problem we will consider in Part I of the course is the calculation of the rate at which water flows through a porous medium, that is, the velocity of groundwater flow. This problem is similar in some respects to the particle settling problem we just analyzed. In both situations, a fluid and a particle or particles are moving relative to one another, and the relative velocity results from a balance of the drag forces with an applied force. For a settling particle, the applied force is the downward gravitational force acting on the particle, and the drag force is due to the movement of a solid particle through a stationary fluid. In groundwater flow, the drag force is due to the movement of water (the fluid) past stationary pore surfaces. The applied force which causes groundwater movement is a pressure force. gif

Groundwater is water which fills the cracks and pore spaces of underground soil and rock. It constitutes the world's largest freshwater resource, containing more than twice as much freshwater as the world's glaciers and about 40 times the freshwater in all the world's lakes and rivers. Groundwater is used extensively as a water source for agricultural and industrial use, and about half of the U.S. population uses groundwater resources for drinking water. However, groundwater flow is extremely slow in comparison to surface water flow speeds. In combination with the large volume of groundwater reservoirs, this results in slow pollutant transport but very long residence times. For this reason, contaminated groundwater moves quite slowly, but once groundwater is contaminated it can be very difficult to clean up. In this section, we will present and apply the equations that govern the rate at which groundwater moves through the subsurface environment. Before we begin, however, some definitions are in order.

Hydraulic Gradient

The hydraulic gradient is the gradient of the height of the water table. For example, if two wells are drilled a distance of m apart and the height of the standing water within each well (called the head of water at that location) is measured, the hydraulic gradient between the two wells is given by

Just like any fluid above or below the ground, groundwater flows from regions of higher head to regions of lower head.


Porosity is defined as the fraction of the total volume of soil or rock that is empty pore space capable of containing water or air. Thus, porosity, with the symbol , is defined as

Since the units cancel, porosity is a unitless value. Typical values of porosity range from 15--20% for sandstone, sand, and gravel, to 45% for clay.

Darcy's Law

We are now ready to consider the rate at which water flows through the subsurface environment. The force which drives this flow is proportional to the hydraulic gradient. Based on the drag force we used for particles (equation 45), we might expect that the drag force is proportional to the velocity of groundwater flow. Thus, since the driving force must be equal to the drag force, we would expect the velocity of groundwater flow to be proportional to the hydraulic gradient. Thus, we would expect

where Q is the volumetric flow rate of groundwater, and A is the cross-sectional area through which the flow is occurring. This result was observed in the nineteenth century by a French civil engineer named D'Arcy. The equation he used to describe this relationship is termed Darcy's Law:


The constant K in equation 58 is called the hydraulic conductivity. The hydraulic conductivity depends on properties of the soil or rock medium and on properties of the fluid. For example, it depends on the smoothness or roughness of the pore surfaces and on how wide or narrow or how straight or tortuous the pores are. It also depends on the viscosity of the fluid (just as the drag force in Stoke's Law depends on the fluid viscosity). Representative values of hydraulic conductivity K are .2--.5 cm/s for gravel (which has big pores that are easy for water to flow through); -- cm/s for sand; and cm/s for clay (which is composed of very fine particles and thus has very small, tight pores that are difficult to force water through).

The Darcy Velocity versus the True Velocity

Darcy's Law relates the volumetric flow rate of groundwater, Q\ (units of volume water per unit time) to the cross-sectional area A\ of the soil or rock through which the flow occurs. The ratio of these two quantities gives the Darcy velocity:


However, this velocity does not reflect the true speed at which groundwater moves through the subsurface environment, because the cross-sectional area of the pores through which groundwater flows is smaller than the total soil or rock cross-sectional area. The ratio of pore cross-sectional area to total cross-sectional area is equal to the porosity, . Thus, the true velocity is given by


  Example .. Transport time of groundwater between two wells An underground storage tank has been discovered to be leaking diesel fuel into groundwater. A drinking water well is located 200 m from the fuel spill. To ensure the safety of the drinking water supply, a monitoring well is drilled halfway between the drinking water well and the fuel spill. The difference in hydraulic head between the drinking water well and the monitoring well is 40 cm (with the head in the monitoring well higher). If the porosity is 39 percent and hydraulic conductivity is 45 m/day, how long after it reaches the monitoring well would the contaminated water reach the drinking water well?
Solution: To calculate this period, we need to determine the true velocity of the groundwater between the two wells. The time for travel between the two wells will then be . The hydraulic gradient is equal to . The Darcy velocity is given by (equation 59)

The true velocity is equal to this value divided by the porosity (equation 61):

Thus, the period for flow from the monitoring well to the drinking water well is days. This result is typical of groundwater flow speeds---groundwater transport is usually very slow.


  .. The concentration of of a pollutant along a quiescent water-containing tube is shown in the figure below. The diffusion coefficient for this pollutant in water is equal to cm/s.

What is the initial pollutant flux density in the x-direction at the following locations: , 1.5, 2.5, 3.5, and 4.5?
If the diameter of the tube is 3 cm, what is the initial flux of pollutant mass in the x-direction at the same locations?
As time passes, this diffusive flux will change the shape of the concentration profile. Draw a sketch of concentration in the tube versus x-axis location showing what the shape a later time might look like. (It is not necessary to do any calculations to draw this sketch.)

answer: (a) At , .
(b) At , .

  .. The tube in problem 1 is connected to a source of flowing water, an water is passed through the tube at a rate of 100 cm/s. If the pollutant concentration in the water is constant at 2 mg/l, what is

The mass flux density of the pollutant through the tube due to advection?
The total mass flux through the tube due to advection?

answer: (a) .
(b) .

  .. A settling chamber is used to remove particles from an air flow. If the settling chamber height is 3 m and the residence time of air in the settling chamber is 10 s, what is the minimum size particle that the chamber would remove? Assume that the particles settle according to Stoke's Law, and that, to be removed, a particle must settle the full 3 m during the period it is in the chamber. The particle density is 1 g/cm, air viscosity is , and the air density is small enough that the buoyance force may be ignored. Express your answer as particle diameter in .

answer: 96

  .. A wastewater treatment plant uses a grit chamber to settle out sand particles. Water flows at a rate of 100 gallons per minute through the chamber, which is 2 m in height and has a volume of 100 m. Calculate the minimum size sand particle that would be removed in the chamber. To be removed, the particles must settle the depth of 2 m during the period the wastewater spends in the chamber (). (Sand density is 2.65 g/cm, water density is 1.0 g/cm, and the viscosity of water is .

answer: 12.9

  .. Two groundwater wells are located 100 m apart in permeable sand and gravel. The water level in well one is 50 m below the surface, and in well two the water level is 75 m below the surface. The hydraulic conductivity is 1 m/day, and porosity is 0.60. What is:

The Darcy velocity, .
The true velocity of the groundwater flowing between the wells.
The period it takes water to travel between the two wells, in days.

answer: (a) 0.25 m/day; (b) 0.42 m/day; (c) 238 days.

  .. The hydraulic gradient of groundwater in a certain location is 2 ft/100 ft. Groundwater in this location flows through sand, with a high hydraulic conductivity equal to 40 m/day and a porosity of 0.5. An oil spill has caused the pollution of the groundwater in a small region beneath an industrial site. How long would it take the polluted water from that location to reach a drinking water well located 100 m downgradient?

answer: 62.5 days

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