FDST 363 - Heat Exchanger: Fouling

Andreia Bianchini Huebner Author

10/02/2020 Added

20 Plays

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Heat and Mass Transfer

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[00:00:06.110]Good afternoon everybody.
[00:00:07.550]Today, before we get started with our lesson,
[00:00:10.130]I just wanna point out a typo that I made
[00:00:13.600]on our slides that I used on Wednesday.
[00:00:15.780]And I apologize for that.
[00:00:17.460]I just put up this slide once again
[00:00:19.550]so we can double check on that,
[00:00:22.790]this equation that we were using
[00:00:25.530]to calculate the overall heat transfer coefficient
[00:00:30.270]for a heat exchanger system has to match.
[00:00:35.130]So, the outside overall heat transfer coefficient
[00:00:41.400]it's gonna be calculated based on the outside area,
[00:00:43.940]and here on the slide should have said outside.
[00:00:46.530]And I had that as a typo, it said inside.
[00:00:48.920]So make sure that you revisit that if you took notes
[00:00:53.080]and canvas should have now
[00:00:55.780]the correct version of the slide.
[00:00:57.330]So if you wanna download again that would be the best
[00:01:00.200]just to make sure that you don't have any mistakes
[00:01:03.420]on your notes.
[00:01:04.530]And then the following slides is the same thing because here
[00:01:07.750]I had previously outside and it should be inside
[00:01:11.140]because that's the overall heat transfer coefficient
[00:01:14.560]for the inside based on
[00:01:19.952]the inside area of the pipe.
[00:01:22.030]And then, as I was making this correction cause I noticed
[00:01:27.510]as I was reviewing the slides after the lecture,
[00:01:30.490]I noticed that.
[00:01:31.323]And I also thought that maybe
[00:01:32.650]it would be worth adding one more slide
[00:01:36.390]just to make one comment
[00:01:37.920]that I don't think I mentioned during our lecture.
[00:01:41.230]So I added this one.
[00:01:42.890]So what I'm bringing up here is the,
[00:01:45.280]if we look at the equation here in this slide
[00:01:48.170]and the equation on this slide, they're exactly the same.
[00:01:52.090]And this one is associated with the overall heat transfer
[00:01:55.210]on the outside.
[00:01:56.570]This one is associated with the overall heat transfer
[00:02:00.110]for the inside,
[00:02:02.440]but they're essentially the same, right?
[00:02:04.800]The value, the mathematical value is the same.
[00:02:07.540]So how does that come about?
[00:02:09.960]Like how do you reconcile that?
[00:02:12.020]So the overall heat transfer coefficient
[00:02:14.560]is a value that represents
[00:02:16.320]the heat transfer coefficient relative
[00:02:19.060]and that's the word,
[00:02:20.490]relative to the area that you calculate.
[00:02:23.670]So if you use the outside area, you have to tell me
[00:02:28.370]or whoever you are communicating with
[00:02:30.690]that that's the overall heat transfer based on the inside.
[00:02:34.560]If you use the outside, is gonna be based on the outside.
[00:02:38.010]So those two pieces of information always
[00:02:40.210]have to go hand in hand.
[00:02:42.030]If you tell me, this is U
[00:02:44.920]and you don't tell me which area you use to calculate
[00:02:47.760]it's almost missing part of that information.
[00:02:51.530]They always go together.
[00:02:53.590]And then as you know,
[00:02:56.090]the outside area of a pipe
[00:02:57.710]is always gonna be slightly larger, right.
[00:03:00.930]Than the inside area, because the radius is a little bigger,
[00:03:05.650]or the diameter's a little bigger.
[00:03:07.540]So the outside value is gonna be larger.
[00:03:10.550]So the U associated with the outside is gonna be what?
[00:03:14.540]Little smaller.
[00:03:15.580]So this ratio here can be equal to each other,
[00:03:20.520]does that make sense?
[00:03:22.450]So the message is,
[00:03:24.730]if you're calculating overall heat transfer coefficient
[00:03:28.030]you have to communicate if that's based on the outside
[00:03:31.400]or inside area, those two pieces of information
[00:03:34.320]always have to be together, go hand in hand.
[00:03:38.990]And that was it.
[00:03:40.520]So now we're ready to take from where we left on Wednesday,
[00:03:45.240]and we were talking about heat exchangers
[00:03:49.670]and we were talking about calculating
[00:03:53.070]this overall heat transfer coefficient because inside
[00:03:56.850]of that heat exchanger, we have a combination
[00:03:59.260]of conduction and convection that it's going on, right.
[00:04:04.410]So today we're gonna continue to talk about it
[00:04:06.920]but we're gonna add one more piece of information.
[00:04:10.610]When we talked about the types of heat exchangers
[00:04:14.070]that could be used in the food industry.
[00:04:16.370]We talked about fouling,
[00:04:18.240]which is the deposit of materials
[00:04:20.830]on the heat exchanger.
[00:04:22.930]And that's gonna create a layer.
[00:04:24.920]So now if I have a clean,
[00:04:30.190]pipe,
[00:04:31.440]so if I have here, right.
[00:04:35.850]I'm gonna make it fairly thin.
[00:04:38.760]So that's my outside pipe,
[00:04:41.540]my inside pipe.
[00:04:43.440]And I had food traveling here,
[00:04:46.550]and I have steam traveling here.
[00:04:49.500]As food gets the deposited here and creates a little crust.
[00:04:55.590]If I'm thinking about that convection and conduction,
[00:05:00.560]what did I just created?
[00:05:03.500]Another layer that's I need to take into account, right.
[00:05:08.250]Convection inside, everybody ringing a bell,
[00:05:12.160]conduction through the pipe.
[00:05:13.850]Now I have to consider that the characteristic
[00:05:16.720]of this deposited material
[00:05:19.530]in terms of thermal conductivity and thickness
[00:05:23.640]will play a role.
[00:05:24.640]So I'm gonna have conduction through that.
[00:05:27.660]And then I have convection here,
[00:05:29.770]conduction here and convection outside.
[00:05:32.250]So what I'm doing is
[00:05:33.780]by having this fouling happening, I'm creating
[00:05:37.190]one more layer that I need to take into account.
[00:05:39.370]And that's what we're gonna look at today
[00:05:41.150]and how we go about calculating
[00:05:43.380]and how does that fit into our system.
[00:05:46.300]So we're gonna talk about the types of fouling
[00:05:48.400]and how that affect our U,
[00:05:51.370]which is our overall heat transfer coefficient.
[00:05:54.380]So liquid foods, when they're heated, they may deposit
[00:05:59.370]on the surface of that heat transfer
[00:06:01.710]and that can happen in a circular, heat transfer,
[00:06:07.050]or I should say heat exchanger,
[00:06:08.740]or it could also happen on the heat exchangers
[00:06:10.920]that have the plates.
[00:06:12.790]Either way, the food can deposit there
[00:06:15.870]and that's gonna become one more layer
[00:06:18.760]for our heat to go through.
[00:06:21.510]So,
[00:06:23.020]for the most part that end up hindering our heat exchange.
[00:06:27.410]Because now I have to go through one more layer
[00:06:30.550]and I end up losing energy in that process.
[00:06:34.380]Also, it's gonna influence our fluid flow
[00:06:37.660]because now this food here
[00:06:40.690]doesn't have as much space anymore to flow.
[00:06:45.230]And if this keeps building
[00:06:47.740]and keeps building
[00:06:49.720]my space for my food to flow gets reduced more and more.
[00:06:55.240]So it's also gonna influence other things inside
[00:06:58.850]of that pipe in terms of amount of energy
[00:07:02.210]that I need to pump that fluid,
[00:07:04.550]amount of pressure inside of the pipe.
[00:07:08.150]So, the fluid flowing inside of that pipe will so
[00:07:12.410]experience some restrictions.
[00:07:15.790]And, the same thing is observed when liquids are placed
[00:07:20.270]in contact with subcooled surfaces.
[00:07:22.570]So we can have it in cooling or in heating processes.
[00:07:28.060]So the kinds of fouling that we could observe
[00:07:33.040]are those that are caused
[00:07:35.070]or I should say the types are listed here
[00:07:37.910]and here's the mechanism
[00:07:39.030]or the process that could cause them.
[00:07:41.890]So, you could have maybe precipitation
[00:07:45.300]of minerals that are part of your food
[00:07:49.290]or some chemical reactions that is happening
[00:07:52.740]between the surface
[00:07:55.800]and the components of that food.
[00:08:00.550]For example, protein, sugars and fats,
[00:08:03.710]especially sugars, you know,
[00:08:05.870]they can be heat up and they go through a chemical change
[00:08:10.800]and they become pretty much caramel, right.
[00:08:14.500]Inside of those pipes.
[00:08:16.320]But that's actually a change in the configuration,
[00:08:19.090]chemical confirmation of that sugar.
[00:08:21.310]So that's gonna deposit
[00:08:22.670]and it's gonna be a type of fouling caused
[00:08:25.690]by chemical reaction.
[00:08:26.970]You could have just particles, little pieces of food,
[00:08:29.930]that now deposit.
[00:08:31.510]You could have microorganisms attaching to that surface,
[00:08:35.080]that's another type of fouling,
[00:08:37.410]is biological growth of microorganisms there.
[00:08:41.010]When you are freezing,
[00:08:43.100]the freezing process itself will cause
[00:08:46.130]that to deposit, right.
[00:08:47.410]And, I don't know
[00:08:48.243]if you remember or not, but we talked about
[00:08:49.840]that there's one specific type of heat exchanger
[00:08:52.840]that we use when we are trying to freeze things.
[00:08:55.190]Anybody remember what that was?
[00:08:59.410]We would scrape as we go,
[00:09:01.090]because the name of the heat exchanger was,
[00:09:03.270]its (indistinct) surface, right.
[00:09:05.230]Heat exchanger.
[00:09:06.430]And we can also have deposits that are caused
[00:09:09.160]because of corrosion that it's happening on that equipment.
[00:09:12.800]So just be familiar with those,
[00:09:16.150]because they happen quite a bit in food processing.
[00:09:21.300]And the way that we address this is
[00:09:22.960]by cleaning the equipment and sometimes,
[00:09:26.060]most of the times cleaning in place
[00:09:28.240]and as needed this assembly of the piece of equipment
[00:09:32.800]and scraping and cleaning,
[00:09:34.560]more like a deep clean of that equipment.
[00:09:38.020]So, because now we have that deposit of the food
[00:09:42.070]there are certain things that
[00:09:43.290]we may need to do to compensate.
[00:09:45.640]So either we're gonna have to clean the heat exchanger
[00:09:48.590]on a more frequent basis, which costs money, right.
[00:09:52.980]Or, I'm gonna have to push more heat through the system.
[00:09:59.090]So if I was creating a heat,
[00:10:03.420]like a delta here,
[00:10:05.220]let's say that I was creating a delta of 20.
[00:10:12.520]Let's see, let me think.
[00:10:13.730]Like maybe 70 Celsius degree, that's my delta.
[00:10:17.060]So inside, maybe I had steam at a 100 Celsius degree,
[00:10:21.980]food traveling at 30 Celsius degree.
[00:10:24.830]And then I had a value of q that was going through,
[00:10:28.960]before fouling.
[00:10:36.310]If I wanna keep the same q after
[00:10:41.160]what I'm gonna have to do?
[00:10:43.490]It's harder now because I have one more layer to go through.
[00:10:47.440]What does it change?
[00:10:49.929]Do you guys remember, from Wednesday when we talked about,
[00:10:53.720]what was one way that we could look at this
[00:10:56.140]and we were adding up, what were we adding up?
[00:11:00.410]Were we adding up the thermal resistance?
[00:11:03.540]So we were adding up the thermal resistance of convection
[00:11:09.360]plus convection plus conduction, do you guys remember that?
[00:11:14.680]And we were adding each little piece of thermal resistance.
[00:11:17.770]So if now I have my fouling happening here,
[00:11:21.660]what did I just do?
[00:11:23.230]I added one term here, right?
[00:11:26.460]So I made this harder to happen,
[00:11:29.290]on increasing the thermal resistance.
[00:11:32.160]So, in order to keep these two equal,
[00:11:35.650]I'm gonna have to increase what?
[00:11:38.390]My delta T,
[00:11:40.750]to compensate for that.
[00:11:42.640]But that costs me money because now
[00:11:45.250]my steam here maybe have to be at 110 Celsius degree.
[00:11:49.910]And that costs me energy, fuel to make happen, okay.
[00:11:55.840]So now let's look at our system
[00:11:59.810]and then let's see how we
[00:12:03.010]would calculate that in our heat exchanger.
[00:12:06.540]So we already know that conduction and convection co-exist
[00:12:11.060]and we know that we're gonna change this U here
[00:12:15.430]when we have fouling because
[00:12:17.910]our thermal resistance also gonna be increased,
[00:12:21.010]so that U is gonna change.
[00:12:24.960]So that's what we're gonna talk about today.
[00:12:26.850]This is just reminding us
[00:12:29.860]of things that we learned and saw so far in this process.
[00:12:34.860]So these are equations that you should already have
[00:12:37.220]on your cheat sheet.
[00:12:38.700]So we have that the overall heat transfer,
[00:12:43.590]it's equal the overall heat transfer coefficient
[00:12:46.210]times the area times delta T,
[00:12:49.030]we have defined our thermal resistance
[00:12:51.300]as the inverse of the overall heat transfer coefficients
[00:12:55.490]and the area.
[00:12:56.850]And we can calculate the thermal resistance
[00:12:59.050]by adding those terms for convection,
[00:13:01.360]conduction and convection,
[00:13:03.250]for a system that it's clean, right?
[00:13:06.920]No fouling, no anything.
[00:13:09.350]We also know that the thermal resistance
[00:13:11.230]is gonna be equal, the two ratios.
[00:13:15.640]And we just discuss about that today
[00:13:19.700]because we brought that piece of information
[00:13:22.550]and the new slide today.
[00:13:24.590]So we have all that already.
[00:13:26.610]What we're gonna add, it's that fouling on the pipe.
[00:13:30.830]So if my clean pipe looks like this,
[00:13:36.070]when fouling happens, I may have an inside layer
[00:13:40.300]caused by deposit of minerals in the water.
[00:13:43.130]That was for example, used to make the steam.
[00:13:45.940]So my steam may have minerals on it
[00:13:49.337]and the minerals are gonna deposit on the inside.
[00:13:52.420]And then I have food sort of baking on the outside.
[00:13:56.610]So potentially I could have even one more layer
[00:13:59.940]to what we have discussed so far.
[00:14:02.560]I would have an internal layer and an outside layer.
[00:14:05.450]And we're gonna consider that conduction
[00:14:08.060]is happening through those,
[00:14:09.240]because we're gonna assume that they are very,
[00:14:11.770]compact and somewhat solid.
[00:14:15.200]So that's how it's gonna look like.
[00:14:18.240]So technically, technically my diameter here for,
[00:14:27.060]I'm gonna draw on the board so I can actually show you,
[00:14:29.510]cause I don't wanna...
[00:14:33.720]So if I have a situation,
[00:14:38.970]that steam is on the inside
[00:14:41.720]this is deposit on the inside.
[00:14:46.430]Then I have the pipe itself.
[00:14:48.640]So this is the pipe,
[00:14:51.010]and here I have another layer of deposit, right.
[00:14:53.500]That's what we are seeing on that slide, deposit again.
[00:15:01.738]So if I'm very precise
[00:15:04.680]and I measure the diameters
[00:15:09.160]and the radius
[00:15:11.940]of each of these layers,
[00:15:14.500]there technically are different.
[00:15:16.230]Would you agree with that?
[00:15:17.540]Because this one is gonna be the shortest
[00:15:20.730]or smallest radius.
[00:15:23.350]So I could say that it's R i
[00:15:26.950]or R one, just for the sake.
[00:15:29.770]Then I would have an R two that goes to here,
[00:15:32.860]and then I would have an R.
[00:15:35.200]Well, actually the one that I would consider is
[00:15:38.350]this one, sorry.
[00:15:40.160]So this here, this distance, is my R one.
[00:15:45.400]This is my R two, and this would be my R three.
[00:15:51.800]So even though does this true
[00:15:53.720]and they are slightly different
[00:15:56.330]for the sake of keeping our calculations
[00:15:59.950]to a manageable level,
[00:16:01.830]we're gonna assume that there all
[00:16:06.090]the R of the pipe.
[00:16:09.290]So the area
[00:16:12.330]of the inside deposit,
[00:16:14.610]the area of the outside deposit
[00:16:17.067]and the area of my clean pipe,
[00:16:19.880]I'm gonna consider the same.
[00:16:21.800]That's what this slide is telling us here.
[00:16:25.740]The area of the fouled,
[00:16:28.720]the pipe with the fouling inside,
[00:16:31.860]it's equal the area of that pipe when it's clean.
[00:16:36.380]And the same thing for the outside.
[00:16:39.090]And the reason is because that film,
[00:16:41.200]it's so thin, so small,
[00:16:44.040]that you would make our calculations a lot more difficult
[00:16:47.840]and not add more to the story.
[00:16:51.100]But what adds to the story is what that film is made of.
[00:16:55.880]So we will have a certain amount of thermal conductivity
[00:17:04.150]that is associated with that,
[00:17:07.182]and that will take into account.
[00:17:09.560]So we still gonna count,
[00:17:11.580]but we're just not gonna be as precise as to the level
[00:17:16.680]of saying that the area of the pipe changed.
[00:17:20.010]We're just gonna consider that area is the same.
[00:17:22.520]But we will have some on the outside and some on the inside
[00:17:27.450]that's gonna hinder that transfer.
[00:17:30.780]So because of it, we're adding those two terms here,
[00:17:39.399]the amount of energy that it's gonna be affected
[00:17:43.010]because of that fouling on the inside
[00:17:46.240]and because of the fouling on the outside of that pipe.
[00:17:49.880]But look, when I'm calculating the area
[00:17:52.260]I'm calculating the area of the inside,
[00:17:54.910]like if it was clean.
[00:17:56.760]And the area of the outside, like if it was clean.
[00:18:00.290]And that R f value for this class
[00:18:04.030]it's always gonna be given.
[00:18:06.090]So all that you need to remember
[00:18:07.840]is that if your pipe went through the process of fouling
[00:18:12.020]you have to take that into account and add one more term.
[00:18:16.640]But those values for this class are gonna be provided
[00:18:20.240]because they're really difficult to estimate and calculate.
[00:18:23.440]So you're gonna have that information
[00:18:25.470]and you're just gonna add it to your equation.
[00:18:29.790]So as you can see now, our thermal resistance
[00:18:33.760]it's a little bit larger, right.
[00:18:36.570]Because I have the convection inside going
[00:18:40.950]through that first layer of junk,
[00:18:44.310]the conduction through the pipe itself,
[00:18:47.480]second layer of junk,
[00:18:49.610]and then convection of my food.
[00:18:52.500]So the equation just became a little longer.
[00:19:00.570]And we can calculate that based on the...
[00:19:07.180]If we're looking at the overall now,
[00:19:09.800]overall heat transfer coefficient,
[00:19:12.070]it can once again,
[00:19:13.120]be calculated based on the outside area of the pipes
[00:19:17.090]or the inside area of the pipe.
[00:19:20.700]If we consider the outside area,
[00:19:23.110]then we're gonna have an overall heat transfer coefficient
[00:19:26.060]for the fouled pipe based on the outside area.
[00:19:29.730]And again, like I mentioned to you,
[00:19:31.510]we have to provide both information together.
[00:19:34.860]Are you calculating based on the outside area of the system
[00:19:38.590]or the insight area of the system.
[00:19:41.240]And that's how we go about.
[00:19:44.750]Any questions so far?
[00:19:46.270]Because that's part of like what we build on.
[00:19:49.940]It's putting that little piece on our equation
[00:19:53.070]and now we can kinda look and see how that plays
[00:19:55.800]on a problem.
[00:19:57.220]But do you have any questions about the theory,
[00:19:59.450]how we got here?
[00:20:04.380]Cause otherwise then we're gonna just apply to this problem,
[00:20:07.350]which you will see that
[00:20:09.810]it's exactly the same problem that we worked on,
[00:20:12.650]on Wednesday.
[00:20:14.120]And I posted online.
[00:20:16.210]The only thing that we're gonna do is
[00:20:17.630]we're gonna take that system
[00:20:19.220]and we're gonna now add the layers for fouling.
[00:20:24.820]So I'm just gonna get everything.
[00:20:28.630]See if I have it here.
[00:20:45.829]Yes.
[00:20:49.843](faintly speaking)
[00:20:57.055]Oh, I should have left this picture actually,
[00:21:00.530]cause we could have used just add our layers there
[00:21:02.970]in our information from the problem.
[00:21:10.970]Let's read it together.
[00:21:13.260]So in a double-pipe heat exchanger, made of stainless steel.
[00:21:19.690]So how (indistinct),
[00:21:20.870]is I'm gonna just draw on top of what I have.
[00:21:26.070]So we have our pipe here,
[00:21:30.810]where food is gonna be navigating through.
[00:21:35.020]Then we have the inside,
[00:21:38.329]oh, that's...
[00:21:44.900]And that's my pipe for my steam.
[00:21:49.620]So I'm gonna have steam on the inside
[00:21:52.420]or how does it go?
[00:21:55.070]Does it say, which one is which?
[00:21:57.300]So in a double-pipe heat exchanger made of stainless steel,
[00:22:00.810]the inside pipe has an inner diameter of two centimeters.
[00:22:05.820]So I'm gonna save that to say that later.
[00:22:08.420]So an inner diameter of two centimeters,
[00:22:15.900]an outside diameter of 2.5.
[00:22:18.940]So outside diameter of 2.5,
[00:22:25.940]the outer shell has an inner diameter of four.
[00:22:33.909](indistinct) different color.
[00:22:36.680]So this one here,
[00:22:39.740]diameter of four.
[00:22:45.610]The convective heat transfer coefficient
[00:22:47.730]on the inside surface of the inner pipe, it's 550.
[00:23:01.960]And the outside surface of the inner pipe is 900.
[00:23:12.450]Over continuous use of the heat exchanger,
[00:23:15.180]fouling which is depositing of solids from the liquids
[00:23:18.530]on the pipe surfaces
[00:23:20.770]caused additional resistance to heat transfer.
[00:23:25.490]It is determined that the resistance
[00:23:27.910]to heat transfer due to fouling on the inside surface
[00:23:32.470]of the inner pipe.
[00:23:34.290]So I'm gonna have deposit,
[00:23:36.360]on the inner surface of my inner pipe.
[00:23:42.140]And this value is going to be,
[00:23:46.330]it's the R f inner,
[00:23:54.254]0.00038 square meter.
[00:24:01.830]Oops!
[00:24:04.862]Per watts.
[00:24:08.090]And on the outside surface of the inner pipe is 0.02.
[00:24:11.920]So I'm also gonna have deposit,
[00:24:15.490]on the outside surface in (indistinct).
[00:24:19.150]It's on the, it's my R for fouling.
[00:24:22.910]It's my thermal resistance
[00:24:24.240]because of fouling on the outside,
[00:24:26.320]that's what the letters mean.
[00:24:29.440]It's 0.002.
[00:24:31.330]0.0002.
[00:24:33.380]Did I get?
[00:24:34.880]Yes.
[00:24:38.940]What else did they tell us?
[00:24:40.930]Okay.
[00:24:41.950]We also know that it's,
[00:24:44.360]this pipe is made of stainless steel,
[00:24:46.470]and has a K of
[00:24:51.490]515 watts.
[00:24:58.240]That's another piece of information.
[00:25:04.620]Calculate the total thermal resistance of the heat exchanger
[00:25:09.380]per unit length.
[00:25:11.260]Calculate the total thermal of resistance.
[00:25:14.677]So, total thermal resistance of
[00:25:20.800]the fouled pipe.
[00:25:21.930]So that's what we're gonna try to do.
[00:25:25.030]Per unit length.
[00:25:26.460]So the length here would be one meter.
[00:25:32.480]So with the information that has been provided to us,
[00:25:37.060]we're trying to calculate the total thermal resistance.
[00:25:40.940]So what type of thermal resistance
[00:25:44.680]should I be considering here?
[00:25:46.960]That one associated first with what?
[00:25:51.630]From inside out, what comes first?
[00:25:55.200]The convection, right.
[00:25:57.670]And, I just have way too many markers in my hand,
[00:26:04.290]and I need this second one.
[00:26:10.010]That's not what I want.
[00:26:14.760]It's not the one I want.
[00:26:18.448]We gonna look at this one first.
[00:26:20.820]It's not...
[00:26:23.040]Here, so we're gonna look at that.
[00:26:24.500]So you guys can help me.
[00:26:25.610]The first one is the convection
[00:26:28.140]and its gonna be one over my area on the inside
[00:26:36.640]times my edge, on the inside.
[00:26:41.010]And then after that, what will come?
[00:26:43.550]If I go in order, from inside out,
[00:26:46.010]what do I encounter next?
[00:26:49.920]The fouling, right?
[00:26:51.150]Which is the R f for the inside
[00:26:55.600]divided by the area of the inside.
[00:26:59.360]Then again, moving from inside out,
[00:27:02.010]my next thermal resistance
[00:27:03.570]is gonna be associated with what?
[00:27:05.960]The pipe itself, which is the,
[00:27:15.810]two pi K L,
[00:27:18.600]there we go.
[00:27:20.130]And then, moving yet from inside out,
[00:27:24.460]I come across the second layer of fouling,
[00:27:28.160]R f o, the area on outside.
[00:27:32.270]And then I have one more thermal resistance associated
[00:27:35.680]with the convection on the outside,
[00:27:37.360]which is gonna be one over,
[00:27:40.100]outside and outside.
[00:27:44.640]Do I have information to calculate all
[00:27:46.620]of this different things.
[00:27:52.010]I should have, right?
[00:27:53.230]It's just a matter of finding them all.
[00:27:54.900]So what is my area associated with my inside
[00:27:59.890]of the inside of this pipe?
[00:28:01.440]We're focusing on this pipe here.
[00:28:04.440]So focusing on that pipe, it would be
[00:28:07.190]pi D L,
[00:28:10.610]and you guys may have to help me cause
[00:28:13.500]I don't have all my numbers with me.
[00:28:17.830]So area on the inside pi D,
[00:28:23.480]for the inside, 0.02 meters,
[00:28:28.030]length, one meter.
[00:28:30.270]Is that correct?
[00:28:32.370]So my area on the inside, this one,
[00:28:35.340]I think I have it.
[00:28:36.550]You guys double check my math,
[00:28:38.170]0.063 square meter.
[00:28:44.510]And then we have an area for the outside.
[00:28:47.510]I'm just gonna keep moving in,
[00:28:48.960]somebody please double check the numbers.
[00:28:52.870]For the outside would be pi D,
[00:28:56.790]for the outside now I'm talking about 2.5
[00:29:03.950]and I used the wrong number here.
[00:29:05.800]So I'm definitely gonna need help.
[00:29:08.500]0.025 times one meter.
[00:29:14.050]And this meter, one meter.
[00:29:20.330]I do not have this number.
[00:29:24.780]0.078.
[00:29:28.040]Thanks Jeffard.
[00:29:29.910]Yup, thank you much.
[00:29:32.390]Okay, so now we have this value,
[00:29:35.250]we have this value.
[00:29:36.930]We have, we have, we have, we have,
[00:29:40.410]should have it all.
[00:29:42.160]Okay, let's plug it in and see what we get.
[00:29:44.970]The thermal resistance for this four pi
[00:29:48.450]would be one over 550 watts square meter Celsius times
[00:29:57.440]inside area, 0.063 square meter.
[00:30:03.450]Plus R f one,
[00:30:06.442]0.00038 square meter Celsius per watts
[00:30:15.870]times the area of the inside,
[00:30:18.120]0.0623 square meter, plus,
[00:30:25.640]when you guys do this math, can you write down each term
[00:30:30.040]so we can double check ourselves.
[00:30:32.890]So if you can write it down
[00:30:33.900]and then we can compare notes
[00:30:35.780]to make sure that we're calculating correctly.
[00:30:38.750]What is my R o,
[00:30:41.970]my R o here.
[00:30:43.930]It's going to be 0.0125, right?
[00:30:49.730]And my R i inside,
[00:30:54.660]it's going to be 0.01 meter.
[00:30:58.730]Everybody agree with that.
[00:31:00.400]So my log here should be R o, 0.0125 meter
[00:31:07.610]over 0.01 meter divided by two pi K.
[00:31:14.700]It was given, 515 watts per square
[00:31:24.205]per meter Celsius.
[00:31:26.543]Let me write it down,
[00:31:31.510]meter Celsius.
[00:31:33.900]This meter goes away with this meter.
[00:31:37.090]What else I have?
[00:31:37.923]Plus R f o.
[00:31:41.910]R f o, 0.002.
[00:31:46.561]How many zeros?
[00:31:47.570]Three.
[00:31:52.880]Area of the outside,
[00:31:55.350]0.078 square meter,
[00:31:59.850]plus one over
[00:32:03.380]outside convective heat transfer coefficient, 900 watts
[00:32:11.880]times outside area
[00:32:14.189]0.078 square meter.
[00:32:20.570]Square meter goes away with square meter.
[00:32:23.560]What is my R t f?
[00:32:24.970]The first value here,
[00:32:29.610]0.029.
[00:32:33.010]See if you all agree, cause I don't have the numbers
[00:32:36.700]and that would be Celsius per watts.
[00:32:39.580]Plus second term, this meters go away.
[00:32:47.330]I think I missed one zero Jeffard?
[00:32:49.281]No.
[00:32:50.550]It's right, and then three,
[00:32:53.100]zero three, thank you.
[00:32:55.290]Zero three.
[00:32:56.570]And my units, Celsius per for watts.
[00:32:59.550]This is a great way to check if our units,
[00:33:02.133]were are where they belong.
[00:33:04.330]And if all our terms are ending up
[00:33:06.400]with the same units at the end.
[00:33:08.760]So now what is this term?
[00:33:10.940]The calculation, the value.
[00:33:17.020]And I forgot one meter here.
[00:33:19.700]And I noticed that because I wanted to cross this meter
[00:33:22.460]and I didn't have a partner for it,
[00:33:25.430]but I just added because our length is one.
[00:33:28.910]So now I'm all good with my units.
[00:33:30.850]I'm gonna end up with Celsius per watts again.
[00:33:34.290]So, anybody has it for me.
[00:33:42.260]Did I get all the zeros?
[00:33:43.620]Cause it's a lot of zeros.
[00:33:46.700]All good?
[00:33:48.730]Plus this terminal,
[00:34:00.568]0.0025,
[00:34:04.960]units, Celsius per watts plus.
[00:34:12.190]You got it.
[00:34:15.060]The last one, 0.014 Celsius per watts.
[00:34:20.640]And I finally found my page two,
[00:34:22.810]and let's see if we're gonna match all the way to the end.
[00:34:27.540]So, the thermal resistance total is,
[00:34:32.250]when we add it all up?
[00:34:49.863]0.05,
[00:34:53.280]are we good?
[00:34:54.450]That's what I got.
[00:34:55.650]Do you guys got that too?
[00:34:57.690]Celsius per watts.
[00:35:00.550]Okay, in addition to the busy math work
[00:35:03.170]because that was just busy math,
[00:35:05.910]as long as we understand
[00:35:08.040]that the thermal resistance is the sum
[00:35:11.410]of every little bit that contributes to it,
[00:35:15.380]we can get to that number, right?
[00:35:17.780]There was no,
[00:35:19.400]nothing else to it other than plugging
[00:35:21.610]the numbers on the right place.
[00:35:24.624]What I want you to consider is which of this,
[00:35:28.710]and again, this is just like for us to understand
[00:35:31.560]what's going on in our system.
[00:35:33.410]Which of the layers are the ones that contribute the most
[00:35:38.460]to my thermal resistance.
[00:35:43.340]If I tell you that I wanna improve my heat transfer,
[00:35:48.560]which layer or which piece I should work on?
[00:35:53.360]Let's look at the pipe.
[00:35:55.370]Is the pipe a big number or small number?
[00:35:57.830]So my pipe is good, right.
[00:35:59.350]Using stainless steel was a good idea.
[00:36:02.330]It's not impeding my transfer.
[00:36:06.850]How about comparing mineral deposits
[00:36:12.660]on the inside versus the food on the outside,
[00:36:15.290]which one is giving me more trouble,
[00:36:17.960]the fouling on the inside or the fouling on the outside?
[00:36:21.880]Maybe I should filter my water.
[00:36:23.900]You see where the practical applications come into place.
[00:36:27.780]The math is just the busy work.
[00:36:29.610]What I need from you
[00:36:31.060]it's to think about and tell me what this means at the end.
[00:36:36.130]All good?
[00:36:38.220]Awesome, that's what matters.
[00:36:40.540]So let's go back and see,
[00:36:41.590]cause I think there's one more thing.
[00:36:43.780]The overall heat transfer.
[00:36:46.910]So, it's also asking us to calculate
[00:36:49.040]the overall heat transfer coefficient based
[00:36:52.040]on the outside area.
[00:36:54.160]We have that equation that tells us that
[00:36:56.410]the total thermal resistance,
[00:36:59.890]it's one over U, A,
[00:37:03.600]who are they asking you to calculate based on?
[00:37:07.194]Outside, so I plug like that.
[00:37:11.220]0.05, now it's busy math work again.
[00:37:17.690]One over overall heat transfer coefficient,
[00:37:22.060]area on the outside, 0.078 square meter.
[00:37:27.520]So what is my U o here?
[00:37:30.550]See if we agree.
[00:37:36.580]252 each.
[00:37:40.220]What do you got?
[00:37:42.102]256.
[00:37:43.595]I got 252, but it might be just approximation.
[00:37:47.320]Units, this one goes here.
[00:37:51.000]Watts, go on top.
[00:37:53.320]Celsius go on the bottom
[00:37:55.290]and square meter goes on the bottom.
[00:37:58.990]Any questions?
[00:37:59.830]Again, just the importance here,
[00:38:02.920]it's understanding that we can't forget any little layer
[00:38:06.570]that we go through.
[00:38:08.160]And also then looking back and understanding
[00:38:11.190]who is doing what for our system,
[00:38:13.190]who's helping, and who's hindering.
[00:38:18.890]So the question then becomes,
[00:38:21.780]we know that in a clean pipe,
[00:38:24.960]I have that equation, right.
[00:38:26.890]The overall heat transfer times the area times the delta T,
[00:38:31.284]and the c over there is just to remind us
[00:38:34.070]that it's the clean pipe.
[00:38:36.580]For a fouled pipe,
[00:38:38.700]it would be a value of U f because the,
[00:38:41.950]U now will change, right?
[00:38:43.770]The clean pipe didn't have this two pieces.
[00:38:49.460]And the fouled pipe, which is the second equation now
[00:38:52.370]has this two pieces.
[00:38:55.040]I know that this is true for the fouled pipe.
[00:38:59.100]What if I wanna my heat transfer
[00:39:01.540]to be exactly the same in both situations.
[00:39:05.600]Practical application, I have a fouled pipe
[00:39:09.410]and I wanted to keep heating my food to the same extent
[00:39:14.330]as my clean pipe.
[00:39:16.430]Looking at the equations, what can you do?
[00:39:20.430]And the answer is already here,
[00:39:21.730]but I want you to look at it and realize yourself as well,
[00:39:25.070]that if we want this value to continue to be the same
[00:39:28.560]we're gonna either have to increase this value,
[00:39:33.540]if this number is smaller
[00:39:36.350]or we're gonna have to change the area
[00:39:39.830]of the piece of equipment.
[00:39:41.880]I can make it longer, right?
[00:39:44.830]Then I can achieve the same effect.
[00:39:50.670]Any questions?
[00:39:53.080]Because I think it's gonna ask us to calculate that now,
[00:39:57.390]in this problem.
[00:39:58.223]So it's exactly the same problem.
[00:40:00.320]Double-pipe heat exchanger made of stainless steel.
[00:40:05.080]The inside pipe has an inner diameter of two,
[00:40:08.070]outside of two and a half.
[00:40:10.340]Outer shell has an inner diameter of four.
[00:40:12.920]It's exactly the same problem.
[00:40:15.050]Convective heat transfer on the inside it's 550,
[00:40:18.920]on the outside 900.
[00:40:21.050]Over continuous use, there was fouling
[00:40:24.500]that caused additional resistance to heat transfer.
[00:40:27.930]It's determined that the heat resistance
[00:40:30.170]is 0.0038 on the inside,
[00:40:33.042]0.002 on the outside.
[00:40:36.264]How much extra area would I need to have,
[00:40:41.550]if I want both pipes to show me the same heat transfer,
[00:40:48.640]how would we go about that?
[00:40:53.000]What I'm gonna erase?
[00:40:53.860]I'm gonna erase this side here,
[00:40:55.160]cause I wanna save some of this.
[00:40:57.270]Cause we might wanna look at it.
[00:40:59.550]But we already have the values for areas,
[00:41:01.670]so hopefully, we won't miss this one.
[00:41:06.820]So, according to the equation that we just talked about,
[00:41:11.330]we want the U,
[00:41:15.743]pull it up just for you to see.
[00:41:18.070]We want the U off the fouled pipe,
[00:41:23.990]times the area of the foul of pipe,
[00:41:27.140]T one minus T two of the fouled situation, be equal,
[00:41:34.200]the U of the clean, area of the clean,
[00:41:38.550]T one minus T two of the clean pipe.
[00:41:41.890]So we want that q for the fouled pipe
[00:41:45.640]to be equal to q of the clean pipe.
[00:41:50.850]So we want to maintain the same temperature gradient,
[00:41:57.380]so we can calculate this based on our U's.
[00:42:04.670]So we're gonna be looking at U's, just like we did here.
[00:42:08.690]We want the temperature gradient to be the same.
[00:42:12.970]That's what our problem tells us.
[00:42:15.560]Let's look at the problem.
[00:42:17.040]It says, calculate the percent of extra area
[00:42:20.310]that would need to be added
[00:42:22.080]to ensure the heat transfer
[00:42:24.200]is the same as with the clean pipe.
[00:42:27.330]So we want these two values to be the same.
[00:42:31.020]We're not talking about changing temperature.
[00:42:33.990]So the gradient temperature is gonna be the same.
[00:42:37.410]I'm not gonna change the temperature of my heating medium.
[00:42:41.890]I'm not gonna change the temperature of the food.
[00:42:43.710]So I can almost cross those two values out, right.
[00:42:47.320]Because the problem is asking us to fix this,
[00:42:52.227]fix the situation that we have at hand based on the area.
[00:42:56.650]So that's what we gonna do.
[00:42:58.870]Do we have the U of the f?
[00:43:02.750]We have based on the outside area, remember?
[00:43:05.290]So I can put outside, outside.
[00:43:08.020]I can put here outside of the clean,
[00:43:10.330]area outside of the clean.
[00:43:12.720]So I can now, I have this value.
[00:43:16.610]I know my area.
[00:43:19.489]What would be my U have the clean, do I have that?
[00:43:27.143]Or,
[00:43:32.500]I'm sorry, I'll take this back.
[00:43:34.660]Because we have the overall heat exchange coefficient
[00:43:39.810]of my fouled pipe.
[00:43:41.750]That's what it was right here and I erased.
[00:43:44.250]The area of the foul pipe it's what I'm looking for.
[00:43:47.900]This would be my new area,
[00:43:52.900]this area of the clean I have is the area that was,
[00:43:56.140]we were given.
[00:43:57.780]Do we know this yet?
[00:44:02.620]Can we calculate?
[00:44:03.690]I bet we calculated on Wednesday,
[00:44:06.730]on that problem that we solved
[00:44:08.330]that I put the solution online.
[00:44:10.390]I bet we calculated there
[00:44:11.740]because we were talking about the clean system there.
[00:44:15.010]Even if you didn't look at that problem at home,
[00:44:19.050]how would you calculate that?
[00:44:21.030]You would go back to here, and what would you do?
[00:44:25.690]Take this away,
[00:44:27.490]take this away.
[00:44:29.140]So we would add up what?
[00:44:31.090]Somebody with a calculator really quick.
[00:44:33.160]This term plus this plus this term,
[00:44:38.560]what would that be?
[00:44:39.640]The R t of the clean pipe.
[00:44:55.610]Did I get all the zeros?
[00:44:58.400]So now my R t,
[00:45:02.800]it's one over U clean o, A o.
[00:45:08.980]So R t is 0.044,
[00:45:17.130]one over, overall clean.
[00:45:21.360]It's what I'm looking for times the area of the outside.
[00:45:27.240]Area outside,
[00:45:28.580]area outside, 0.078.
[00:45:33.713]0.078, did I get it right?
[00:45:37.150]So what is my overall of the clean pipe
[00:45:42.610]based on the outside area?
[00:45:49.221]291.4,
[00:45:54.780]my units would be watts per Celsius square meter.
[00:46:00.640]Now I have that one too, right?
[00:46:03.130]So now I have this value
[00:46:07.810]and I can finally figure out my new area.
[00:46:12.470]Let's try and replace those numbers here,
[00:46:15.560]see if we can fit.
[00:46:18.430]U f o,
[00:46:22.510]what was that?
[00:46:25.700]252.
[00:46:26.970]Now, you said what Jeffard?
[00:46:28.430]Cause we had slightly different,
[00:46:30.367]it was 254 or six?
[00:46:35.450]Was 256.
[00:46:37.600]Anybody took note of that?
[00:46:40.060]Was 256?
[00:46:42.110]Watts Celsius square meter.
[00:46:46.950]Area of the foul pipe, it's what I'm looking for.
[00:46:53.287]291.4
[00:47:01.330]times clean.
[00:47:02.670]What was the area of the outside clean?
[00:47:05.120]0.
[00:47:09.580]area outside clean, 0.078.
[00:47:17.272]Is that right?
[00:47:20.040]While somebody does this math for us
[00:47:22.870]so we can figure it out what is the area of the fouled pipe.
[00:47:28.720]While does that for us.
[00:47:33.560]0.89 square meter.
[00:47:37.630]So let's look again.
[00:47:40.140]What does it mean?
[00:47:41.770]Because everything else is just busy math.
[00:47:44.650]What does it mean?
[00:47:45.840]If I look at my overall heat transfer of the foul pipe,
[00:47:50.730]heat transfer, is it bigger or smaller, clean versus fouled.
[00:47:59.580]The fouled pipe, this one here transfer less heat
[00:48:06.700]then my clean pipe, does that make sense?
[00:48:09.950]Totally, right?
[00:48:12.290]Now the area of my fouled pipe is it bigger or smaller
[00:48:17.320]of my clean pipe?
[00:48:19.770]Bigger if I wanna do the same job, does that make sense?
[00:48:23.660]Totally again.
[00:48:25.480]That's what matters for us here.
[00:48:28.280]Where does this fit and does it make sense.
[00:48:31.390]If I wanna calculate as far as percentage,
[00:48:34.110]how much more this is compared to that?
[00:48:37.550]So, if 0.078 square meter would represent 100%
[00:48:43.970]of the clean pipe.
[00:48:47.780]0.089 square meter would be, what?
[00:48:57.313]114%, so I would have to increase my heat exchanger
[00:49:05.350]by 14% more to achieve the same heat transfer.
[00:49:10.300]Now, because of this fouling that I'm observing it.
[00:49:17.510]That's the implication.
[00:49:19.470]That's what we care about.
[00:49:24.560]Any questions?

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FDST 363 - Heat Exchanger: Fouling

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