|Chapter 2||Particle characterisation
In Table 2.2 the Malvern is stated as providing and area based diameter; surely it gives a volume diameter?
As stated in the sentence above the table, this provides the equivalent spherical diameter measured by the device. The Malvern uses Fraunhofer laser diffraction which is more closely related to the area of the particles than the volume. However, the results are then converted within the software, and with the assistance of a calibration and appropriate model, into a volume diameter. Hopefully, the inclusion of Section 2.7 helps to avoid any confusion about how to interpret the Malvern's results and the on-line example printout.
In question 7, part (i), I obtain 1/100, a constant, what do I do next?
You are correct that n3(x) is 1/100, which can be taken out of the integral. This leaves us integrating dx/x which gives a natural log. The answer should be (where the 6 comes from the equation at the top of page 19):
6/100 [ln(x)] where the lower limit is 1 and the upper 101 microns.
|Chapter 3||Fluid flow through porous media
In question 2, part (viii), what is meant by static pressure?
This comes form the well-known equation: depth x density x gravity = pressure - which actually comes from Bernoulli's equation just considering the potential energy and pressure energy terms.
So, 2 x 1100 x 9.81 gives 21.6 kPa. This additional pressure would need to be considered if specifying a pump as the fluid needs to be raised this height, together with the pressure drop due to friction when pumping through the bed; i.e. dynamic pressure loss.
|Chapter 4||Liquid filtration
In question 3, what causes the pressure drops?
The main principle behind the question is to demonstrate the two components to pressure drop: that over the cake and the cloth. In constant rate, the pressure over the cloth must remain constant - a consequence of Darcy's law and the constant rate. The cloth, or medium, resistance is revisited again in the next question when the filtration starts as one of constant rate and then swaps over to constant pressure. This needs to treat the previously deposited layers of cake, and the original medium resistance, as an effective new medium resistance which is obviously much greater than the original.
So how do I interpret the empirical equation in Q.3?
The pressure rise equation is: pressure = 168 Pa per seconds plus 6670 The first equation in the 'cake filtration' box on page 43, ignoring the pressure addition equation, is valid for both constant pressure and constant rate filtrations. You have a constant rate filtration so dV/dt is equal to V/t; i.e. a straight line through the origin on a graph of V against t. In order to provide this, the pressure must go up to overcome the increasing cake depth during filtration - hence the empirical pressure rise equation. Thus, comparing the cake filtration equation in this box and the pressure equation above it gives us the pressure drop over the medium at 6670 Pa, and this remains the same for all the constant rate period (Darcy's law and the same filtrate rate again), but the pressure over the cake increases. The pressure rise term (168 Pa per second) must equal the increasing pressure required to force filtrate through the cake part; i.e. 168 Pa per second = 168 multiplied by t = mu alpha c over A2 multiplied by (V over t) multiplied by V Hence, rearranging gives: (V over t)2 = (168 A2 ) / (mu alpha c) take a square root to get the filtrate rate (V/t).
It all hinges on dV/dt = V/t, which for constant rate filtration will be the case! This type of filtration occurs when a positive displacement pump feeds a filter press - for example. Such pumps are the progressing cavity type, e.g. Mono pumps. Once you have the rate, the time taken to filter 2.3 litres can be deduced.
In Question 4, I obtain a much greater filtration pressure than those given. Where have I gone wrong?
The filtrate during the constant rate period should be 0.0023 m3, and at a rate of 4.6E-6 [Q.3 pt (ii))] this takes 500 seconds. The pressure after 500 seconds will be: 6670 + 500 * 168 = 90670 Pa.
In Question 4, I obtain a much greater filtration time than those given. Where have I gone wrong?
At constant rate we need 500 seconds, using a constant pressure of 90670 Pa and assuming dry cake mass per unit filtrate volume and cake specific resistance remains the same as before, then putting these values in to equation (4.19), rearranged for time, should give 731 seconds. Adding the two separate times for the different types of filtration gives 1230 seconds.
|Chapter 5||Dilute systems
The Heywood Table technique is complicated, Stokes is simple and the Archimedes is ok. Can't I just use Stokes?
The answer to your question on which technique to use is yes, and no; as the tables technique is much more complicated it is best to use Stokes' law first, then calculate the Particle Reynolds number. If this suggests minimal turbulence in the fluid, Re< 0.2 - approximately, then you have confidence in the result. However, for Re> 0.2 then it may be better to use the Tables. For Re >> 0.2 then the Tables, or some other correlation is required. The problem with the other correlations (e.g. Archimedes') is that they are only valid over a restricted range of Re; whereas the Heywood Tables are valid for all Re.
|Chapter 6||Hindered systems and rheology
In Question 3, how do I obtain the batch flux data?
You really need a series of tangents to the batch settling curve: this gives you a series of settling velocities. In order to calculate the concentration of these velocities you need to see where they intercept the height axis and read off the value. Applying this value in the equation: CoHo=C1H1 rearranged to give: C1=CoHo/H1 will provide the value of C1, which is the concentration that you would have started with, from the height H1, in order to give the tangential line you graphically constructed above. So, Co=0.037 and H0=0.456 m and H1 is read off as above. Now, knowing the velocity and concentration you have sufficient information for a set of data for a batch settling curve: flux (CU) plotted against concentration (C).
OK, so I've now used equation (6.8) rearranged to give me a value fo TCu, now what do I do?
Simply draw a line from TCu that is a tangent to the batch settling curve and see where it cuts the solid concentration axis. Hopefully, this gives you something that will convert to 19% by mass, see Section 3.6 for conversion between mass and volume.
One advantage of this flux technique is that it demonstrates how a change of feed condition (F or Co) will have an influence on the underflow concentration and that influence can be numerically predicted. Hence, it is useful for understanding how a thickener performs rather than just a design method such as Coe and Clevenger.
Can I obtain all settling fluxes from a single batch sedimentation, at a lower concentration than I need?
Yes, the solids settling velocities, and hence the settling fluxes, come from the interface settling plot. Taking the example of the column at 0.039 v/v solids, Question 2, you need to draw a tangent to the settling curve that goes through the point 21.5 cm on the y-axis intercept. This is the settling curve (line) for a suspension starting at 3.9% v/v solids, from an initial height of 21.5 cm. It is slower than the original 3% line - as would be expected. The justification for this construction is in equation 6.5 and Figure 6.7. Remember, you have not added, or taken away, any solids - simply redistributed what solids are present in a lower volume of water.
My settling flux, Question 3, increases with concentration shouldn't it decrease with concentration?
Knowing the concentration and velocity from the above you can now calculate the flux and plot it on the figure. This process needs to be repeated for all the other concentrations and, as you expected, the flux should reduce - not increase. However, there is no visible maximum for the batch settling flux with this data. This simply means that it is too close to the y-axis to be plotted. Be very careful calculating your tangent values; this has caused some major problems with students in the past.
|Chapter 8||Centrifugal separation
In question 3, part (i), I don't get any of the answers!
The correct answer is 190 radians per second, and a radian is dimensionless therefore 1/s. Reason: 1 revolution is 2 pi but rpm is revs per minute so you need to divide by 60.
In question 5, I don't know how to start!
You have to think circular here! The question asks for volumes, but that simply means that you need to multiply the length of the centrifuge by the cross-sectional area. Now, it is this area that is circular. The entire cross-section will be: pi (ro^2 - rL^2); allowing for a liquid level, or radius, removed from the centre of rotation - see Figure 8.2. The cross-sectional area of the start radius will be: pi (ro^2 - rs^2) - note that rs replaces rL - again refer to Fig 8.2. Hence, the volume fraction between the start radius and the outer radius (ro) will be the latter divided by the former expression. Now, what does this represent? Imagine particles of our size of interest (critical size) entering the centrifuge between rs and rL, they will not be removed from the flow because they will not make it to the outer wall before leaving the centrifuge. Particles entering between rs and ro will be removed. Hence, this gives us the fraction of particles removed for this critical size for given operating conditions of the centrifuge; i.e. rotation speed, volume flow rate, etc. The rest of this question is all about converting the imaginary start radius (rs) into actual particle size (related via settling velocity). If this procedure is repeated for different start radii (particle sizes) it provides an expression for collection efficiency for all particles in the centrifuge. I know this sounds complicated, but try the mathematics and see if it makes sense when you have finished!
I've tried the rearrangement and I get an equation with start radius on both sides and I can't get rid of this imaginary term.
You should have the equation p = (r outer squared - r start radius squared) / (r outer squared - r inner squared) for the proportion of the volume within the machine that is processed. Now, if you equate the two sigma expressions in the box on page 89 AND to make things clearer substitute for the volume of the machine: i.e. remove pi (r outer squared - r inner squared) L by, let's use the term, Vol: to represent the volume of the machine. This leaves you with an equation with a single r start (rs) term that can be rearranged to give: rs = r outer / exp[Vol omega squared Ut / Q and g] Now, substitute this equation into the one that says p = ... and rearrange things. This is an equation that tells you how the grade efficiency of the centrifuge varies with particles size (i.e. via terminal settling velocity), centrifuge properties and volume flow rate. The problem is a literal translation of 'substitute rs for rL' used in the text. This really should refer to changing the start radius from rL to rs for the particle trajectory BUT to leave the volume of the centrifuge term alone: i.e. pi (ro ^2 - rL ^2) L. If you really want to master this question, try creating a spreadsheet to predict the grade efficiency with particle size for a centrifuge using this theory. Obviously, rs has to be restricted to be between ro and rL, but this should give you a good idea of what is going on and how the theory can be used to rationalise/explain practical observations. If you do this, use Stokes' law for Ut - as this provides an easy relation between particle diameter and settling velocity.
What's happened to questions 6 and 7?
Sorry, they are missing from all the books!
In question 8, I'm convinced the volume ratio should have an exponent of 3 in it, this is right isn't it?
The correct answer is (R/Ro)^2 not to the power 3. The reason being that the volume is achieved by multiplying the cross-sectional area (hence R^2) by the length. The length will cancel out, leaving R^2. You have to think circular again for this question. Of course, the final answer is dimensionless as the exponent '2' is both on the top and bottom of the equation.
In question 3, I obtain a negative velocity - this can't be correct!
You need to multiply the mean particle by mass to the power 0.3 by 0.0638 and the solids density to the power 0.5. There has been some misunderstandings using this equation. The main trouble is that this is an empirical equation - so it is not easy to check what makes sense. However, values for solid velocity must lie between the gas velocity and zero! If the solids velocity is close to the gas velocity either the mean size must be very small or the solids density very low.
|Chapter 10||Powder flow and storage
|Chapter 11||Crushing and classification
|Chapter 12||Solid/solid mixing
|Chapter 13||Colloids and agglomeration
|Chapter 14||Gas cleaning
|Chapter 15||Powder hazards
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