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Polyvinylpyrrolidone (PVP) is a polymer product used as a binding agent in pharmaceutical applications as well as in personal-care items such as hairspray. In the manufacture of $\mathrm{PVP}$, a spray-drying process is used to collect solid PVP from an aqueous suspension, as shown in the flowchart on the next page. A liquid solution containing 65 wt\% $\mathrm{PVP}$ and the balance water at $25^{\circ} \mathrm{C}$ is pumped through an atomizing nozzle at a rate of $1500 \mathrm{kg} / \mathrm{h}$ into a stream of preheated air flowing at a rate of $1.57 \times 10^{4}$ SCMH. The water evaporates into the stream of hot air and the solid PVP particles are suspended in the humidified air. Downstream, the particles are separated from the air with a filter and collected. The process is designed so that the exiting solid product and humid air are in thermal equilibrium with each other at $110^{\circ} \mathrm{C}$. For convenience, the spray-drying and solid-separation processes are shown as one unit that may be considered adiabatic.
(a) Draw and completely label the process flow diagram and perform a degree-of-freedom analysis. (b) Calculate the required temperature of the inlet air, $T_{0}$, and the volumetric flow rate $\left(\mathrm{m}^{3} / \mathrm{h}\right)$ and relative humidity of the exiting air. Assume that the polymer has a heat capacity per unit mass one third that of liquid water, and only use the first two terms of the polynomial heat-capacity formula for air in Table B.2. (c) Why do you think the polymer solution is put through an atomizing nozzle, which converts it to a mist of tiny droplets, rather than being sprayed through a much less costly nozzle of the type commonly found in showers? (d) Due to a design flaw, the polymer solution does not remain in the dryer long enough for all the water to evaporate, so the solid product emerging from the separator is a wet powder. How will this change the values of the outlet temperatures of the emerging gas and powder and the volumetric flow rate and relative humidity of the emerging gas (increase, decrease, can't tell without doing the calculations)? Explain your answers.

Here is an example of a carnot heat engine cycle using an ideal gas. So a reminder that a carnot engine has to ice A thermal processes. One an expansion want a contraction during the expansion at high temperature heat is absorbed during the contraction heat is expelled to the cold reservoir at low temperature. Then there are two idiomatic processes where heat does not have a chance to exchange with the environment. One is again a contraction one. Again, it's an expansion. Usually when analyzing these cycles, a couple of tools are helpful. One is the ideal gas law provided an ideal gas is used. So I usually like to write that down and then find the pressure and volume and temperature At each of the four vortices shown in the cycle. Mm. Uh The other tool that often is useful is the first law of thermodynamics dealt to you is equal to Q. Heat minus work done by the engine. That could be helpful for processes or cycles that involve a idiomatic processes. It is also good to realize that pressure times volume raised to the idiomatic exponent is equal to a constant. It reminder that the automatic exponent is a ratio of cp two C. B. And depends on the nature of the gas. Here we are, given that this constant is 1.3. So this is usually how I like to start. I usually like to start with a table and fill out the three state variables pressure volume and temperature for each of the vertex is in the cycle Here. There are four that are labeled in the diagram for us. Now this problem is challenging because they don't give us outright numbers but they provide clues as to what's going on with this particular sample to put it mildly. Um And so the first thing they tell us is that there's the ambient Environment that the particle is interacting with at the .4. And they give us the ambient temperature as 123 Calvin. Uh now that's good because we now know the temperature at .1 because of the ice a thermal process. They also tell us that we are at one atmosphere at that point but they don't tell us what one atmosphere is other than yeah, it's ambient there. Mhm. Some other clues at .1 we are told there is something called the trigger volume and that's going to become important Because at .2 we get squeezed to half the trigger volume with the radius given As .08cm cubes. That's kind of no not cubed. Sorry. Yeah, that's kind of a weird way to give a clue. But at .2 they also give us the pressure. Yeah, As 20.3 kilo pascal's. So that's an outright calibrated measurement kilo pascal's at 0.3. They give us the clue that No, it's actually a .4. They give us the clue that the final radius expands from the trigger radius by three And since volume is R. Cubed, that tells us that we have 27 times the trigger volume. So we have some clues and I think these are the the most obvious clues. The rest. We're going to have to use some steps with the gas law and the idiomatic exponent to figure out. So looking at the table, the most information appears to be given. Well we'll start with two because there's a nice calibrated point right there. And so at point to what we know is the volume Is 4/3 pi r cubed. And that's going to unravel a lot. So we have 2.14 Times 10 to the -3 m3 figuring that out. I won't show all the math there and then what happens is we can now find the trigger volume By just multiplying it by two and that will give us the Entry for the volume at .1 and it will also give us then The volume at .4 Which is 27 times that. Yeah, 0.116 m3. Yeah. Yes. And we can fill in that entry into the table. Okay, so now we've got to do a little bit more work. Um we see that there's quite a bit of information given in steps one and 2 and they are connected through an idiomatic process. So what we can do between one and 2 because it is idiomatic, We can set the pressure one Times The Volume one raised to 1.3 is equal to the pressure 0.2 times the volume, it too Raised to the 1.3. And let's see that is going to give us of the unknowns. It will give us the pressure at 0.1. So we can solve for that. And we get uh huh 8.24 Killer Pascal's and bingo. Now we've got everything inside appoint one. Uh, we've got, yeah, definitely a bingo type situation. Everything is filled in so we can now use one that everything at that point to solve for the number of moles are, is just the gas constant 8.31 jules per bowl kelvin. And we could solve for that high temperature, which is a very important quantity. Um, once we find the number of moles. Okay, but actually we're using that point to saw for the number of moles. So let me be clear about that. Oh, okay. So we can solve for the number of moles And that's definitely important quantity 3.45 Times 10 to the -2 moles, which I'll just write off to the side because that is an important clue that we can then use in 0.2 to find the high temperature. It's like a mystery that unravels and doing that again, I won't show all the plugging and chugging, But we get a temperature of 151 0.5 kelvin. And that's nice because we now know both the low temperature and the high temperature and that is definitely important things for the operation of a carnot cycle. But while we're at it we have a few more things to fill out so we might as well go ahead and fill those out. There's a lot of information missing at .3. Um and as well as point for But what we see is .3 is connected 2.4 through a uh a dramatic so let's stop with one more step. What about .3? It's connected to four through idiomatic and it's connected to through ice a thermal. So what we can say is the following two relationships the unknown pressure at three Times the unknown volume at three raised to 1.3. That idiomatic exponent is equal to pressure it for volume it for Raised to the 1.3. Okay. Um well we don't know pressure it for, do we? But we can use the fact that four is connected to one through an ice a therm. And so we can say that pressure for volume for is equal to pressure one volume one. And we definitely know both of those at a .1 and we also know the volume at point for so we can solve for pressure for and again, I'll spare the details but we basically get mhm 301 pascal's Okay. Now what do we know? We don't know either pressure three or volume three. So we have to connect it to point to through the ISIS are okay. So we know both Pressure volume at .4. Same with two. We have two equations into unknowns and we can definitely then solve For both pressure three and volume three. And what are the steps to that? Well, I would probably take the bottom relationship And solve it for pressure three and put that back into the top equation. Yeah, 1.3 is the exponent And solving that we get volume three is equal to Okay. Uh huh. Uh huh. 5.84 Times 10 to the -2 cubic meters. And then we can solve for pressure three. Okay. So yeah, we've got the table full at this point. Yeah. So I'll put a little faced by that to show that that one is the most challenging and it's probably the least interesting too on top of it. So at this point, we have all the information about this cycle um we can do some other things with the cycle, basically knowing what the carnot efficiency is. I will use the expression without the percentage because certainly getting the fraction is good enough. But that efficiency is basically the difference between the high and the low temperature divided by the high temperature. So we know both of these temperatures and we can figure out that efficiency, It's about 18.8%. But .188 in terms of a fraction and a reminder that any engine has another definition of efficiency is the work that you get out divided by the heat absorbed at high temperature during the high temperature is a thermal expansion. Yeah. So yes, while the gas is hitting up, it is expanding and pushing against something um whether it's the wheels, the shaft of your car with a piston that's pushing or whether it's this little particle um doing something expanding. Anyway, what we're told is that the work out is 60 killer jewels per hour and that's really a rate at which work is being done, 60 killer jewels per hour Is the same as one killer jewel per minute. Just thinking about How hours translate two minutes. Um and then we can figure out the heat that is absorbed at least the rate. Just turning those into rates does not change the fraction there. And so qh dot the rate at which he dis absorbed is simply uh yeah 5.32 killer jewels per minute. And then we want to use the first law to figure out the difference between um the heat absorbed minus the heat expelled exhaust heat is equal to the output work. And really what you're doing with that is using the first law With Delta U. equals to zero. And solving for qc dot it's the heat absorbed minus the output work, energy must be conserved. And so this is four 32, kill the jewels permit it

Hey guys! Leads to problem 50. The first question is way. Need to calculate the volumetric flow rate it is given that they put the problem. Is that a stripy? Therefore one to normal of guests. Apple pies 22.4 Major Cube and the number off most of gasses. The number off Muncey's that is getting in is the volume that's getting in, which is 3.74 given in the problem. Divided by 22.4 m too far. Guillemin therefore the number of most gasses 0.1 67 Kinnaman Fine minute. It means that every minute each minute 0.167 Philomel gas is getting in. Given the feed cast contents 22 more percentage of D. C. D. C s and 60% of the disease has reacted. Therefore, the most of these refugees That is second to the 60% age what is reacted, which is which means 0.6 multiplied by the mole fraction of D. C s and the total number of monster was seated and therefore 0.6 multiplied by 0.2 too much. That is the man fractional diseases and you can buy the total number of phones, which is your 0.167 Guillemin a minute. And from here we get to them most of disease strategies 0.0 to to to the moon. Four minutes since two months off nitrogen nitrous oxide reacts with one matloff diseases. Therefore the number off more off my dress up sides the number of most of nitrous excited that one which reacted is to multiple ratified a number of most of DCs that reacted. Which means that two multiplied by zero one has zero to do or 0.44 to the moment. One minute. Yeah, therefore, the amount of unrecognized Chris oxide. Let us calculate that that is the kind of the moon fraction off nitrous oxide and deployed with the number of moons that was needed minus the number off moves off nitrous oxide which reacted. It doesn't serve develops. Common fraction is 0.78 The total months. Caesar advanced 67 months kilometers per minute, minus the recommended 044 cumin for a minute. From here, we get the amount of underactive nitrous oxide ease 0.863 Kill them all per minute. Similarly, the amount of contracted DCs is the percentage 1% of D. C. S multiplied by the total number off moves minus the amount of disease that reacted from here We get the amount of 100 diseases Eupen 0147 Hilleman per minute. Since each more lost nitrous oxide reacted, it creates one mole of nitrogen and hydrochloric acid. One more off nitrous oxide produces one more off nitrogen on one more off hydrochloric acid. Therefore, the number off on love nitrogen and hydrochloric acid at one day at the same as the number of most off. I just excited one day and a number of months of hydrochloric acid at containing is is a country the number off off nitrous oxide which reacted, which is 0.44 kill the more for a minute. Now we can calculate the total number off. Yes, this will do it. They put the total number of most of us that one day. Yeah, yeah, yeah. It is the submission off down most of all guesses that pointing the number of months off DCs at 20 plants the number of months off nitrous oxide at 20 the last number of months off nitrogen, that pointy. That's the number of Monthsof hydrochloric acid at point E. Therefore, we can do the summation here. Therefore, the total number of months of gas at 20 is 0.189 Antiquated writing in verse, three small per minute. We can't kill them. One. Let's converted two months more permanent. Therefore, we can calculate the volume off gas as NRT, divided by P number of most 0.189 multiplied waiting in verse. Three more permanent. This constant is 62.36 multiplied waiting in verse three. Meet a cute door moral universe. One killed the Universe one, and the temperature given is 900. Did you sell shares, or 100% to 73 Children? Pressure given here is 600 for Milito or 600 for multiple written in verse. Three off. When we do this calculation, we get 2.2 900 goodbye. Tend to the bar for meter you for a minute as the volumetric for it. The next problem says what is the rate off organization? Let's calculate the rate off deposition. The partial pressure of D C s Can we calculate it? The partial impression off D. C S s. You can do the moon fraction off. Do you see us at point A. Multiplied by the total pressure and the moon fraction can be found by dividing them Number of months off the CS that 0.8 divided by the number of most off. Guess at one day want to graduate the total pressure The number of months of this is that 20 is 0.147 killing more than we need and total I want to go gas at one days 7.189 for the one for my neck The pressure is 604 Really drunk. Therefore, the passion pressure off DCs is 47 0 new little similarly, the partial pressure off nitrous oxide. Therefore, the function pressure of nitrous oxide is 276 mil It'll now the rate off deposition The relation is provided for innovative deposition. The question is are is the culture of 3.16 multiplied waiting universe For the partial pressure off D. C s multiplied by the partial impression off I just offside to depart their 0.65 So we have all these values here. Latest in some deck. The partial pressure on this is this. 47 in the passion pressure off 90 subsidies to 76. Once we do that calculation, we get the rate of deposition is 5.73 months of graduating Universe five. And finally, how thick is that? Sit on drugs. Sadly, questions see. From the rate of deposition, we can calculate the most soft silicon dioxide deposited after two hours. From the rate of deposition, we can calculate the malls off silicon dioxide deposited after drawers. Therefore, the thickness can we calculated as the Mueller Mass off silicon dioxide is 60 k g in 60 multiplied Veteran anniversary. As you 600.1, the specific gravity is given 2.25 and the density of water is Thank you. KC 1 may help you. Once we do this calculation, we will get 1.1 multiple waitin universe 5 m that's converted toe and strong on 1.11 to put right into the bar. Fine, I am strong. Is the thickness off the silicon dioxide there and this thickness of the ticker? The thickness would be taken towards the entrance off director because the partial pressure off disease and then I just oxide would be higher, something in a higher value of the deposition rate. That's right it down, resulting in a hang a lot of deposition red.

So just drop of some contextual information. I'll just draw up on the screen. The equations of interest to have log he started the bay for pressure is equal to a minus. B over T f C ABC The coefficients t is all temperature we have pieced are a since Y a multiplied by P. P start is a partial pressure. Why is the mole fraction of Pete is the total pressure? So the full form of L. F Ellis, the lower flammability limit. So what we have is a flashpoint of us over a 15 flashpoint Brussels based 75 degrees Celsius and in the ambient temperature of our lab is 20 to 25 degrees Celsius. So the temperature of solving a is 15 degrees Celsius, which is lower than the ambient temperature in the lab so soften a can ignite easily have exposed exposed to a flame. However, if we take a look at solving B, it's 75 degrees Celsius, which is higher than the ambient temperature in the lab. To solve and be cannot ignite is easily if exposed to a flame, so solvent be is safer to keep in the lab two. Therefore, it must be monitored the temperature of the solvent. They does not rise above 15 degrees Celsius streets, which it must be refrigerated. However, there is no need to put solvent be in the fridge. It's moving on, Susie next part here. So the LF l of methanol in air is about six miles to sound, and the total pressure is one atmosphere. So one atmosphere we know is equal to 100 and 1.3 to 5. Killer Paschal's more fraction of methanol is equals the ratio one component in the total mixture. So why, for methanol is equal to 9.6 fried by 100 we get not quite, not six to the mole. Fraction of math does not point North. Six. We can calculate the partial pressure of methanol. You start. Um, it's equal to not point not six. We'll swap by one a 1.3 to 5. That gives us 6.795 kill of our skulls. So the part of pressure methane is 6.0 76 95 Killer pass girls. I was just written on the screen there. So now that we know the partial pressure of methanol, methanol coefficients. We have listed as a B and C weaken. Substitute them in on solve our equation. Hands we have logged 6.795 seek to 16.5785 Take away 3638.27 Divided by T. R. 239.5 We sold the T. We got a value of 6.768 degrees Celsius, so the flashpoint of methanol and the temperature are closer to each other. Therefore, the values of flashpoint were taken around 7 to 8 degrees Celsius. Eso, the flashpoint of methanol, is a lot less than the ambient temperature of the environment. So if methanol is exposed to an open flame, it will easily evaporate. Entertain? LF l and if it's exposed to ignition source, it can easily catch fire. So the common ignition sources we have our sparks heart plates, Bunsen burners in the lab

Hello, everyone. I hope all is well today. I'll be helping you with the 37th problem of the chapter 29 problem set. And 37 is a multiple step problem. So we'll go through all the steps. So we have a which is says, ah, pose questions about the fitness of various biological solutions to the problem of capturing strong and using free energy. So one of my questions would be can cork really store energy, then release it right? So how much could actually released? And I said, for releasing energy with the best material used be wood or enamel because you have these types of things, so how would you rate the scale? So then we move on to be which says refined this representation by identifying an important property of a bio material that is missing and explain its importance in term of free energy acquisition in use. So So you have this type of chart and you have toughness here, and then you have one over elasticity, right? And it and then I would also like to ad that personally for me, right? I would also include some sort of way like, um, some size. So how would you include size? Because size is huge because, um, if something is the need for a small space to conduct energy but, um but it's actually huge. You can't really fit it because sometimes when you break up objects, they're not going to be the same. So now we move on to see you. Correct? And I says, Explain how these how these data indicates that the properties of bio polymers will lead to better bio Ah, biomimetic design of the humanoid robot that isn't imagined in see Threepio. Ah, Star Wars fame, right? So I said this will lead to a better REB state representation in a humanoid form because these materials will be able to make all the impacts and workings of of a human be more realistic on how it interacts with outside conditions. So I hope you found the cell phone, and I hope you have a great


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