S1: alright we're gonna hand back the exams. i'm gonna tell you the mean first just so you don't get shocked when you see what, <SU-F LAUGH> the grades are. <LAUGH> uh the mean for the exam is forty-three-point-one. okay? so, ye- yep that's right, okay. <P :24> course no one's gonna pick up their exam, um. <P :19> everyone get one? 
SU-M: (xx) 
S1: how did i mix those up? <P :10> hold on Kevin i think i put yours in the wrong, bunch. <P :05> i think (you oughta) lo- a look over it before we start going over it. 
S2: (xx) 
S1: what?
S2: do we have to?
S1: it's your choice.
<P :19> 
SU-M: uhuh
<P :05> 
S1: okay i graded the_ are you gonna look at_ you're not gonna look at it?
S2: i already looked at it, [S1: oh okay ] i don't need to look at it anymore. <LAUGH>
S1: kay well we're gonna go over it so i would suggest to you if you don't wanna look at it to look at, uh i graded the first problem, um problem one A and B which was only worth, twenty points but, it took me seven hours this weekend to grade it. so that's how we had to rearrange the grading so, uh Ali graded the rest of them, uh so, i can answer very detailed questions about the first one as far as the grading, i'm gonna go over the solutions for all of it though. the grading questions in the first one i can answer those, uh, but the other ones i i can i can you help with but Ali will be the one to make the final, decision on that. uh, alright now i'll tell you once i've given you the mean i'll give you, now that i've given you the mean i i'll give you, more detailed information on the statistics. <P :04> okay so, mean is, uh forty-three-point-one and, the standard deviation, is, twenty-point-three. wait no that's not right sorry. the standard deviation is eighteen-point-seven. okay? uh the averages on each of the problems problem one, uh eleven-point-two with a standard deviation of three-point-two. the second problem, eleven-point-six, uh well the first problem was out of twenty this one's out of forty, it's eleven-point-six out of forty, with a, standard deviation of thirteen-point-three. so the standard deviation is, bigger than the s- the average score. and, the third problem, twenty-point-three was the average and the standard deviation was ten-point-two. uh, if you're interested to know our class, average we got all the statistics here we did everything. uh our class, average in this this discussion section, is forty-one with a standard deviation of twenty-f- twenty-four-point-zero. <LAUGH> so there's lots of numbers there you can, do what you want with 'em but at least you have 'em. um, okay um, before i even start going over the, uh exam i wanted to, to say something about, uh, what what should be your approach to taking an exam like this. when you_ when he tells you that, the mean is are are really low like he said something about how the first, exam last year the mean was a seventy something, and, that that was a really high mean so you know that they're normally lower than that, when you have an exam like that your philosophy has to be different than an exam where, you know the mean is like a ninety or ninety-five or something like that or really high, because, you need to go through and like just say i gotta pick up points anywhere. okay and this is gonna apply to_ the reason i'm telling you is not to like say, you know do it differently, or you did it wrong but to say, you know you have a couple more exams ahead of you just to think about it differently uh, when there are when there_ when the mean's so low you wanna try and pick up as many points as possible so what that means is going with the problems first that have the most points, if you don't know how to do a part skipping it and going to do a part you do know how to do. uh a lotta people spent a long time on the first problem there wasn't anyone that didn't do the first problem a lotta people didn't do the second problem. the second problem was wor- f- was wo- was worth forty points the first one was twenty. people spent a long time like giving all kinds of information, explaining things really in-depthly for uh the reasons for simplifying energy balance which is which is good in general but, when you have an exam like this when you, um, don't have a lotta time you can't really, mess with that. so, just keep that in mind uh for future, future e- exams you have, one more regular exam and then a final exam. um <P :06> okay. <P :06> oh wait i didn't give you the highs and the lows i (just) forgot to do that. the high was an eighty-eight, and the low was fifteen. okay? alright now let's, go over this exam. <P :12> (yeah) okay. problem one we're expanding the energy balance like we've done a lotta times. uh we're giving it an energy balance so we're gonna start from that and not, uh something else you know a lotta times we can s- start by, uh, you you know using the book for a liquid or whatever the more simplified equation but here you're told specifically to, start with equation one and, expand it and reduce it so, you need to give reasons for each of the terms, that you're you're taking you're taking out that you're getting rid of, uh you can't just, say it's a liquid and then, write a more simplified form. uh <WRITING ON BOARD NEXT 12:00 OF UTTERANCE> so let's start with, this form, the same thing for, part A and part B. okay. so what we have here is we have a liquid, film, that's falling, along a wall, so the l- it wasn't clear from the picture this is the wall here, on the left side. with, uh T-S is the temperature there and, you're given the uh velocity profile, and you guys have seen these before, in fluids and what that is. um, giving the velocity, for this plane right here when it is going down, um in a Z-direction. and (y-) you can give it a coordinate system so some people, or at least one person i saw used radial coordinates, um which, shouldn't because you have the information right there. okay so liquid film is falling down the surface of a cool wall, the other side of the liquid film is exposed to ambient air at temperature T-infinity. okay? so it's T-infinity right here, T-S. before i get into actually simplifying it, let's talk about physically what's going on cuz this is how you would approach it you'd say h- you know before i can s- cross off terms i have to know, what's going on in my situ- in in my situation. so, it's T-S here right along this wall and T-infinity anything in contact with this alright? as far as what's going on here you hafta think about it alright it's not given for you. now you have to measure thi- this fluid comes in at some temperature okay? T-naught let's say, you know when it's, when it just comes in from wherever it's starting so it's all uniform temperature which is a pretty normal thing to assume. and uh, i'll draw this for the case, uh wh- what the temperature profile would look like if T-naught was in between, uh T-infinity and T-S and T-S is higher. when it first came out, like the temperature profile would be uniform right there, it's all T-naught alright? and if T-S is higher then after some time it'll look like, this, okay? this is lower here, some heat is lost out here, comes in here, and then eventually this looks more and more like this until, it's looking like this, looking like that, and eventually when you're down low enough you'll have a straight line here, okay? uh, it's kind of an intuitive, for me it's kind of intuitive like how to draw this but you could also, i- once you simplify the final equation, i'll show you why this y- you'll know that this is the temperature profile. but just, if you know at this temperature here starting off it's not necessarily gonna be the same as either of these, something like this is gonna happen where there's gonna be, heat transfer, in the Z in the uh the Y direction okay? alright so, expanding this for this case, uh and ask me any questions if you have 'em anytime please. okay. the first thing is we know this is steady state because no information is given, about it being at start-up or anything else okay? that's kind of the clue and a lot of people did a good job on that writing because no one_ no information is given we can assume steady state, so, that's steady state. uh, this term here we'll expand in a second but just, seeing what we can eliminate in this first one, the K has been moved out because it's uh, we're assuming constant uh connectivity for, a, uh, for the situation, it's not a function of temperature, and then you can do that it's sc- scou- it's how it's been for all of our problems. uh this fluid is incompressible, okay? it's a liquid so it's incompressible, which means this term here, is is zero... okay? there was a lotta confusion on this term here for both parts okay so i should explain what this means. this term right here is zero when you have an incompressible fluid so for a solid or a liquid this term is zero, alright? this term right here, is, almost zero for a liquid or a solid okay? so but_ so we have a liquid here so this is almost zero it's very small this is zero, the reason why we take this term off is cuz this is zero not because this is zero okay? because people that did that then made, a mistake on the second part, alright you need to understand that. for a gas this is significant and this is significant. so for a gas both of these remain, alright? but for a liquid, or a solid, they leave. right? this right here, is a thermodynamic property. this is not a this is not anything having to do with our system, for the second problem people talk about the pressure being constant, things like that, you should recognize this from, from thermo this is saying when the, this is the molar volume or the density that's the inverse of the density, is fixed, how does the pressure change with temperature? okay so it's not anything to do with the system but just, the fluid that's in the system or the s- solid whatever it is in the system, how does it behave when you fix its, density and then you change the temperature? how does the pressure change? if the pressure doesn't change that term is zero, if it does, that's what it is. you can write explicit, uh equ- you can put an equation of state in there for a, uh for a gas which i'll show you for the, the second problem but, uh for this one you don't need to it's incompressible. this term right here there's no generation given, so no generation. i was pretty lenient on, people not giving really, full answers but just crossing off the right terms. except if they gave, uh there're certain ones i needed to look for the exact, thing which was, with th- this but i was leaning on the first problem part A but not part B but this one right here, the viscosity times uh, the viscous dissipation function. uh, you actually don't really know if in this in this situation if it's gonna be, uh negligible or not, okay? it's gonna depend on what the fluid is. so i accepted both answers but only if you gave good answers for or a good reasoning for why that's there so, if you said, the uh, the viscosity's small for this i'll assume, <ENTERS S4> okay i'll give you your test. mm that was (a) forty-three. [S4: mkay ] uh, so if you said the viscosity i assume here is is small, therefore, uh, this term is is negligible that that was a good answer. uh also if you said i don't know anything about the viscosity, so i can leave it in that's a good answer also. cuz both of them are kinda true. uh, but if you, if you crossed if you just crossed it off blindly, uh not recognizing that it could be significant you lost points for that okay? you had to recognize that i- it depended on situation and, he says clearly describe all assumptions, uh so that would be an assumption you had to make about that. so even if you kept the term you hafta give a reason for keeping it. alright? um <P :08> oh, and the the the correct thing to do for this that wasn't even done in the solution, uh some people did is expanding this function here, and then crossing off the velocity terms. and then you're actually just left with one velocity, uh V-Z as a function uh D-V-Z-D-Y which you actually have an expression for Z- V-Z so, that was, the best way to solve it cuz you could explicitly put that in. alright? but, and we didn't_ you didn't hafta do that that i accepted, uh just crossing it off if you gave a good reason. um, and so for this we'll cross it off and say, um, um you know it's not beca- it doesn't have to do with the velocity profile okay? it's not, that there's no high-velocity gradage you don't really know how that'll look i mean you have a a picture but, you don't really know the magnitude of that, okay? so it's the viscosity that's the key here. okay? and so expanding with what we have left okay, if we did it correctly, that's what should remain. alright from this diagram here, you see that, the only velocity we have is going in the Z direction right? it's going down. there's no veloc- velocity in the Y direction, there's no velocity, in the X direction because it's not even given in the problem it's going back into the page so, right away you can cross out V-X and V-Y and you just say that they're zero that that's fine, uh you don't have to give much of a reason for that, and because of that because at least the Y- the X direction, it's into the plane of the board, by symmetry you know that this, uh, temperature doesn't vary with X okay so you can cross that off. uh, and the only remaining thing to say, is, to be completely, correct uh and Professor Thompson talked about this in, in lecture, was comparing the convection term and the conduction term, okay? saying that, this term here, the conduction is negligible, compared to, uh, convection something like that. you could leave it in but, uh, we didn't take off if you left it in but the most correct thing and and and actually only one person, did this that i saw, is saying that, this term was negligible compared to this, okay? i hope that_ i make i- i hope that makes sense that you can't say that, there's no temperature variation in the Z direction because if you look at how i drew the temperature profile you can see that there is a variation in the Z direction, okay? cuz as it moves down, it changes, at any at any given point if you, if you go down it changes, so you can't say that_ you can't cross out both of these and say the temperature doesn't vary in the Z direction it's only crossing off this and one saying it's negligible compared to this term, not that there's good_ there still will be a temperature variation in the Z direction okay? so conduction is negligible, compared to convection, and that's the final form, (xx) P-sub-V V-Z (xx) and then you're able to... and like i said you'd have another term here for this viscosity if you left that in for D-V-Z D-Y but, you're just left with V-Z and you can plug it in, uh, you can plug in the V that's given there but, we didn't require that of people, uh, cuz then you'd actually have everything, in terms of Y and you could really s- well, i dunno if you'd be able to solve (a) differential equation since we didn't require it i haven't looked into it but, that should be your final answer. um and it's kinda interesting to note, if you if you divide through and just make this some kind of, you can, recognize what's the constants are let's just call it beta for now just, this is the unsteady state equation uh except for this is a Z okay? we're used to saying D-T-D, T is equal to s- alpha, and then the second derivative of temperature with respect to a direction, and so that's that's kind of exactly what we, we see here like if w- if this was a flat plate and instead of this being Z this was time, this is how we'd expect it to look over time, it would eventually reach steady state where, the temperature gradient look like that, okay? so that's problem, one A which we spent way too much time on any questions? <P :04> okay <P :04> part B. alright you have, well lemme tell you the common, mistakes on the first one was saying D-T-Z, or D-T-D-Z is zero which isn't true, people didn't recognize that V-Y was equal to zero and they didn't recognize it was steady state. okay? this problem here we have, <WRITING ON BOARD NEXT 6:00 OF UTTERANCE> glass, and the glass is right here, this is air in the center, T, say T-outside and and T-inside for now, okay? and the system as he says is the air that's between the two glass plates, and these are at different temperatures, and there's no reason to assume that they're the same, let's say that the T-inside is the, uh, is the hotter one so T-I is greater than T-naught. what it'll look like inside is, it'll be convection in here. okay? due to a density gradients. this is hotter over here, so it's gonna rise along this wall here, it's gonna come down on the cooler side, and this is what we have this is what we talk about when we have natural convection, and it'll be like this too in here, alright? s- maybe seventy percent of the people didn't recognize that there's natural convection here. they they treated this as if it was a solid and they ended up getting the, the form that we'd expect for, a a solid which is the (replauchian) of the temperature, is equal to, zero. which isn't tru- correct for, for air. okay? uh, so starting with that same one up there, and i guess we're writing, this here, the same coordinates as the other one, uh i'll just go through each of these terms, this is also steady state, okay? there's no indication that this is in start-up, so steady state. this term like i explained is not zero for this case alright this is a gas and this is not zero, it's a gas so this is not zero this needs to stay. you can't say it leaves because, it's incompressible or_ cuz it's definitely not an incompressible gases aren't incompressible. but also you can't say anything about the pressure in the situa- in the situation not changing (xx) there's a thermodynamic property that pertains to air. okay so, this is not, correct for, this case so that stays. this one, low mu for uh, for air most people were able to pick that up, and so this term is is negligible, and there's no generation system, okay? so... i'll just rewrite it and be lazy... (xx) okay okay so if you keep that term you'll be left with that over there, the steady state takes it off those two terms, drop off, and then... because there's, uh natural convection happening, uh there wasn't a coordinate system given but, i i accepted it i had to look at everyone_ this problem it took me so long cuz i was looking at everyone's, coordinate system to see what they they put down, so it's independent of which coordinate system you have but assuming that Z is this way Y is up and X is into the board, you know that V-X is zero because it's symmetrical in that direction there's not gonna be anything moving over there, uh, so something like symmetry. you didn't hafta give an answer i_ or you can say that D-T-D-X is zero because the temperature doesn't depend on X so either of those things, and the same thing here the temperature doesn't depend on X so, this is gone by symmetry. the V-Y and V-Z are present, just based on a diagram that you would know for natural convection. the temperature gradients are, also present because the temperatures are different, it's changing as it's going up it's changing as it's going over to the left and to the right. uh, these terms are the same thing they're both, they're all present. uh this right here, this here we haven't done anything yet with this but this is the idea of taking the concepts they use and actually applying them you you have expression for, the the gradient, of, a vector and this is velocity, and so you can rewrite this this is what it is, D-V-X-D-X plus D-V-Y-D-Y plus D-V-Z-D-Z. and because of symmetry again this this term is, uh there's no velocity in the X direction. so that's zero, these things, remain and so that's what you put in here and, that's your final form uh an optional thing that he had in the solution but no one did was, saying if this is a, an ideal gas you can actually write out what this is, cuz you know, P-V equals N-R-T, uh, so you can, D D-P-D-T and you can see that that would be, um D-P-D-T, N-R over V, right? and then, you're multiplying that by temperature, and so then it becomes T uh, N-R over V which is equal to pressure. okay? so this term right here this whole thing is equal to pressure for an ideal gas. does that make sense? does it make sense? we're just doing this derivative here and for an ideal gas this is true, so, if you take the derivative of this, if you divide this it looks like this, so D-P-D-T is N-R over V, okay? then you_ that you stick that in here, and you multiply that by T, and then you're left with_ then you get this, which according to the ideal gas equation is equal to P, so this whole thing becomes P, and so, you can just put a P there but you don't need to do that okay but you need_ you did need to recognize that this thing can expand and one of the terms drop off. uh... a lotta people said, common mistakes were not realizing that convection is present like i said about pre- maybe seventy percent of the students, didn't do that, didn't recognize that and then saying this term is zero for, a whole slew of reasons like i said. uh that's it's incompressible the pressure doesn't change, remember it's a thermodynamic property. okay? any questions on that one? <P :07> mkay <P :04> alright <P :07> alright problem three, probably do this, on the overhead. <P :25> alright. this problem the key thing for the second problem, sh- show you that again, is recognizing that this is a fin, okay? it's a fin, in that it it the exact physical situa- the physica- the physical situation is exactly the same as what we know for a fin, except that it's not told to you it's a fin. so it's you having to recognize, what it was the fin, beh- how a fin behaved, and seeing this as exactly the same. and so everything that we derived for a fin applies to this situation. alright we even uh, gave you, oh uh if this was unclear this first part, uh i thought that it might be unclear thinking of this as an unsteady state problem, uh cuz it surprised me how long (it would) take time but we put up the hint thing that said uh, that we're talking about L for length not time here okay so there could there, there shouldn'ta been any confusion here on that but a lotta people took this to be, an unsteady state problem. <WRITING ON OVERHEAD NEXT 2:45 OF UTTERANCE> it's a steady state problem, uh you have a long square copper rod, uh just like that, this end right here is maintained at a hundred in the same way that, the fan is maintained at T-naught at that end, and it's exposed to a fluid at twenty-five degrees Celsius, you're given the heat transfer coefficient so you have the temperature which is T-infinity, the K of the copper and the the H of the fluid. okay? you have those in two places, okay? so once you recognize it's a fin, that's why th- there was standard, such a large standard deviation on this f- is because you almost, got it and then you did really well and got a lotta points or you didn't realize it and you didn't get anything or you got a very small amount. so, once you recognize that it it is a fin, i- then nothing nothing was difficult about the problem, uh after you recognize that that was the h- the hardest step i guess. uh, hold on okay so yeah we're h- we're solving for the length. okay? and you're asked to, find the minimum temperature approximately how long does the rod have to be for the minimum temperature? if you know it's a fin you know the minimum temperature is at the end of fin, right there, okay? so we have, the temperature profile for, a fin, this right here is the, uninsulated case uh, (we decided) this cuz this is so small you can assume it's unin- uninsulated but if you wanted to use, his equation and put L-sub-C in well he u- he did use L-sub-C, in the solution but, you didn't need to have the uh the exact case to a solution that Professor Thompson had derived. remember it's not really necessary for the things that we're dealing with to do that, you can always just modify this case-one solution with, L-sub-C. uh let's see, yeah so remember this is T we're talking about T-minimum occurs at X is equal to L, and so you plug in X is equal to, to L here this is zero this is zero the cosi- um yeah hyperbolic cosine of zero is one, and then you're left with uh, this right here. the hyperbolic cosine of of M-L, and then you, change this to T-minimum, and... you're left with that the cosine of M-L is equal to, ten, and then you solve for, for M for this case right here, H you're given K you're given, area is the cross-sectional area as we talked about it comes under the derivation, and the perimeter, you just , you can, so you can figure out (it's) the thickness of this is is twenty-five and there's four sides around. or point point-zero-two-five meters, and then you multiply that by four that's the perimeter you get, <WRITING ON BOARD NEXT 1:30 OF UTTERANCE> so that's eight inverse meters, um if you you do the in- to get this from this step to this step you do the uh, inverse hyperbolic cosine of ten and you get this and you're left with M-L... and, plugging in L is equal to this, right here just plugging in the numbers. so y- in the solution did_ y- (you can) use L you could use L-sub-C, just the same either of those would've been, just fine, alright? questions on that part? <P :11> okay. and if the rod is point-five meters long how much heat is lost? and so you need the efficiency, for a fin and when we talk about the heat that's lost we always talk about that. so you can calculate the efficiency of this, uh using, uh the equation, that we have for efficiency and plugging in the numbers that we know, and getting that efficiency, um you can also do it with the charts and i'll show you that method in a second but then, then we know that, that equation there, heat is lost through a fin and we just have a fin now we don't have a pipe so we can deconsider an- any of the parts of the pipe, it's just the standard equation the efficiency of the fin, H times the area of the fin, times the temperature difference, and we know all of those things, now, we're able to solve, for the number a hundred-and-forty-point-five watts, uh, questions on that? <P :05> no questions? alright, alright. and then you can also do this from the charts, and the charts use L-sub-C so look up L-sub-C uh, you calculate L-sub-C you calculate these quantities that the chart's looking for for, this fin, and you get uh two-point-zero-five for that quantity this, uh would be a longitudinal fin okay not a circular fin which looks different. and, you can see that that, the efficiency it gives point-three-five which is is different than what you had up here, here we got point-two-five, imagine this is looking it up and i think this this would be closer to this if you had plugged in L-sub-C here, cuz you (need) L-sub-C in this part. either way you'd done it would be fine. that's the Q that you get using, that method, okay? questions, on that? <P :04> yeah you need a_ a key thing was recognizing the area of the fin, at four times, uh the, length which is point-five meters, and, point-zero-two-five was, the thickness of each of the sides. you didn't need to include this end because it is pretty small compared to the area here, but you definitely need to include all four sides in the book the equation for the area of the fin they, they're talking about a very thin fin and so, they say like just the top and the bottom they account for and the sides they ignore they use two-L-sub-C-W, but in this case the the thickness of the fin is significant compared to, the fin that we've seen before so you need to make sure you include it in. okay? so as long as you, you could include this or not that looks fine but, you just need to account for all four sides, that were exposed to the fluid. kay...? i'm guessing there's no question. cuz there hasn't been so far. no questions? okay. <P :06> problem three <P :04> okay <P :09> alright, we have discs here that are being heated in a furnace, give you the thickness and the diameter, and they're initially_ you get the initial temperature at twenty degrees Celsius, and air at five hundred degrees Celsius is flowing around them so they're gonna heat up you have the velocity, of that uh air, and you're given this thing here th- the characterizing Nusselt number, okay? uh the important to note is that the (char) convention is_ characteristic dimension is the disc thickness, which is five centimeters, that goes in the Reynolds number okay? people used eight centimeters for some reason, e- when it says specifically to use the disc thickness you need to use that. and with the Reynolds number, it just needs to have some kinda characteristic dimension for it okay? and, it's kinda arbitrary what you choose it's just that when you define what's what Reynolds number are for laminate and turbular (sic) regimes you would take into account whatever, whatever length you use but, for this case you use the the uh the disc thickness so you need to make sure you use that, okay? uh, no so this is an unsteady state problem, we can all recognize that right? it's it's a lot clearer than the second one which might've been confusing when you had to recognize that it was it was a fin it was unsteady state, this is unsteady state and you're talking about, uh, the temperature, eventually it asks for, for pretty much the the define the temperature at a certain time, so, what we do when we see these problems is to see what the (B-O) number is remember? to compare the uh the the resistances in the inside it, the solid and in the, the gas and this is ceramic so you can, you could guess right away that, the (B-O) number is gonna be large and it's you're not gonna be able to treat this as a lump sum he- heat capacity where you can, uh, treat the temperature inside it as the exact same all the way through, so if you would've done that and just said, alright this is ceramic, (B-O) number's gonna be large i can just go right into the unsteady state looking at the charts or just, like i said before you can always go into the chart for unsteady state problems it's always gonna be more accurate than the lump ti- sum heat-capacity method, but it just might be more work, so but if you did that that's fine but, one of the things that you can do is, uh solve for the (B-O) number, but, uh you can't really, well the first thing, b- before you need that i guess that's for part B. to do part A heat transfer coefficient, okay? you just need to calculate these these things knowing which, the H comes from here, according to this relationship here H (is equal) Nusselt number times K over L, and you have this expression here so you need to solve for the Reynolds number using, the disc thickness and the Prandtl number, and here you go this is the density you need to decide what things they apply to, the Reynolds number are the properties of the, uh the ga- the gas air in this case, plugging in the viscosity for that. uh fifteen-eighty is the Reynolds number you have a lotta people said this is laminar, uh, there's no reason to assu- to say it's laminar or turbulent because you're given the equation for finding H already, and also you don't know if it's laminar or turbulent because of what i said, s- s- it, the d- the uh disc thickness is the characteristic dimension depending on how you change that it would change the regime for, laminar or turbulent flow, so, that that's incorrect. Ali marked that but didn't take off points for that. so you have fifteen-eighty for that, the Prandtl number, uh mu-C-sub-P and mu is for air, uh the heat capacity, for air also and the K is for air, okay? remember that the Prandtl number is a property of the fluid you guys looked it up for air, i mean you you couldn't in this problem cuz you were given the numbers, but since you know you can look it up on a chart just for air you should know that all the properties that go into it, are the ones for air, not for the ceramic, okay? and you get point-six-four-one, and you're able to calculate the Nusselt number with those two numbers, plugging into that equation above, point-five-eight-four is the constant in front you get twenty, and then using this equation right here, you can solve for H, which actually turns out to be the same number he said to assume in part B so i'd assume it's sixteen usually, if he says that he'll give you a new number to start with just so you have something to go with in part B if you couldn't get part A, it happens to be the exact same number so that was kind of a check for you for some reason you got sixteen you saw that he was saying sixteen you probably knew you were, correct on that one. <WRITING ON BOARD NEXT 3:00 OF UTTERANCE> um, okay and then for part B this is where you would need to check the (B-O) number, okay? it's kinda hard to see there uh, let me write it on board. <P :07> (B-O) number is equal to H, how you do know that? the volume over the area times K, uh you couldn't just use_ now this is equal to X-one according to the book and then they give you different things for X-one, uh a lotta people took the X-one for a long, cylinder this isn't a long cylinder this is a short cylinder okay? a long cylinder needs infinitely long so there's really no, uh there's no end effects there's no, area off the end. but th- in this case we do, and so you need to account for that, so you, figure out the, the volume of this is, pi times point-zero-four-squared point-zero-five uh which is equal to, point-zero-zero-zero-three meters-squared, okay? so it's pi-R-squared times the, the height to get the, the volume, and then A is the surface area, two-pi times point-zero-four-squared, and that's accounting for the stuff off the top, cuz this is, uh pi-R-squared times two of them okay that's the area for the the circles on top of this thing, plus, pi times the diameter times the height to get the surface area up the sides, okay? point-zero-five, okay? which is equal to, point-zero-two-three meters squared, and V over A, with a point-zero-one-three-three, and now you know this you already have H, and then, K is gonna be the K for the (sow) this time because of the (B-O) number compares, and we're talking about what's going on in the solid compared to what's going on in the fluid, we know that K is gonna be, that for, the ceramic which is point-zero-eight there, and if you, plug those numbers in, uh, you can, kind of see right there the (B-O) number is equal to, sixteen times this divided by point-zero-eight, and this is two-point-six-five, okay? so two-point-six-five is the value for the (B-O) number, which is greater than point-one which is probably what you would've expected knowing it's a ceramic, so you do need to use the charts you can't assume lumped heat capacity, um, and you're asked to find the minimum temperature, in the disc after, two thousand two hundred and fifty seconds, assuming this heat-transfer coefficient so again if you weren't able to get this, you could start off with it so you were able to do part B, remember if you couldn't do A try and do B. pick up whatever points you can when you know that the mean's gonna be low. uh, is it sufficient to increase the minimum temperature to four-twenty-five? the minimum temperature inside disc that's flowing with uh fluid on the outside is gonna be in the center okay? lowest temperature cuz it starts off there it's gonna be the farthest from the fluid. so the minimum temperature's at the center, uh now this right this this shape that we have here, it's a s- it's a cylinder, or uh yeah. it's a cylinder, but as far as what, how to treat it from the charts we need to treat it, using the tubes remember how we talked about you n- be, to treat like a tube for this very same case here, you have, a cylinder, we have the case for a long cylinder where there's no end, (remember) the case for two flat plates and you know how to relate those two, i'll show you how to do that but you need to recognize that. okay? you need to get, M and, X, fro- for the charts okay for both of those charts that that (we) were gonna get and this is the equation that relates them the Y of the disc for this whole thing is the Y of the cylinder times Y of a, of a plate, and so you know you need to get_ you're gonna be looking at two different charts to get that information, cuz we wanna get the M for, cylinder, uh same K and M for a plate the same K for both of them from there, uh H is the same at sixteen the axis is gonna be different it just depends on how the the chart defined X-one as the characteristic dimension here it's, um <P :07> point-zero-four so it's half of this this is the dimension here for these, uh for the cylinder and then for the flat plate half of this distance from the center up, and so that's, uh, two-point-five, where is that at? right there, point-zero-two-five so you'll be able to calculate the M for both of those cases, the alpha is the same for for both of them. alright? you'll need to calculate one for, for each. this is the uh, thermal, uh diffusivity, for the solid, okay? so, you need to plug in the properties of the solid, you get that number there, and from that, looking at the charts you're able to, well you have to_ i'm sorry you have calculate X first, there's the equation for that, right there alpha-T over X-one-squared, that's why you need to calculate alpha, so you can plug in alpha here, and the time you're given (equal to) two thousand two hundred fifty seconds, X-one is the characteristic dimension depending on which s- system you have, here it's point-zero-four like this and here you use point-zero-two-five the math isn't shown but, it comes out to one-point-zero-one for the plate, and the cylinder is point-three-nine-four, but now that you have, the M and an X for the both the cylinder and the plate you're able to, to look on the chart and find the Y, so the Y of the cylinder, point-four for this figure, the Y of the plate is point-two from that figure <P :04> and then you multiply 'em together, with two Ys just like we see up here, multiply them together, get the Y for the disc, and then you have the equation for for Y, it's not shown here so i'll write it on the board... <WRITING ON BOARD NEXT :40 OF UTTERANCE> Y is equal to, T-one minus T over T-one minus, T-naught. okay? and we're looking for this, temperature here, T-naught is the initial temperature of the solid and T-one is i'm sorry yeah T-naught is the initial temperature of the solid T-one is the temperature of the fluid going outside, (and so) just multiplying this off solving for T, with with Y in there that you got, uh plugging in five hundred as the outside temperature, the initial temperature of the solid is twenty, and so you get four-sixty-two degrees Celsius, which is in fact um, greater than four-twenty-five so the answer is, is yes. <P :10> okay, and that's everything for that problem. uh any questions on that one or, questions in general about the test or anything in the last couple minutes that we have? you gotta have some questions. 
SU-F: i think that you added wrong 
S1: you have a what? 
S3: added wrong
S1: you added wrong? 
S3: uh no my points are added wrong.
S1: oh okay bring that to me afterwards. questions on the on the test?
S4: can we have more time on the next one? [S1: uh ] cuz that fin problem i don't know, i figured it out at the end but i wasn't [S1: okay ] done and, didn't [S1: yeah ] have the time.
S1: again like, we don- we don't write the exams exams Professor Thompson's the one so you'd have to talk to [S4: right ] him, but that's just kinda how he's been doing it for many years and, it seems like that's how most of the engineering exams are in chemical engineering it's that you're given a lot to do and a lot of times the mean's really low so, uh that's why a good question ask professors beforehand before an exam is, you know what's the typical mean in your exam? and, if it's lower you know you should have a different way of approaching it. i mean if it's a high, if it's a high mean you wanna go through and and and do every part perfectly and you'll probably have enough time to do it because most people that do it have enough time to get a high mean, uh otherwise you wanna just go around looking for points if there's a low mean. okay? other questions? <P :04> nothing. okay, that's it. see you guys later. you can come talk to me individually if there's anything oh i should give the forms back to you
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