We're going to derive the governing partial
differential equation called the heat diffusion equation. In 2d. Okay. We've done this little game before
in thermal and fluids. We take a small element, a differential element -- in this case we'll
consider it to be a solid -- of a solid. And we're going to look at all the energy terms
coming in and going out of that solid. A differential element means it's dx by dy. It's going to
be 2d. We're going to ignore, ignore the z right now. So here's our little control volume.
It's a differential element, e dot g is the energy generated internally, e dot storage
is the change stored energy. Now, this is conduction. The energy coming in on the left-hand
face comes in by conduction. We call it qx. We assume it varies continuously. We write
the energy going out as qx + dx. Energy coming in the bottom, the face down here, qy by conduction.
Energy going out the top by conduction, qy plus dy. Okay. So with that we can go ahead
and start putting some terms in there. I think what I'll do, yeah, let's, let's, let's put
in, let's put qx in here, qx is conduction. So it's minus KA dT/dx. The area through which
the heat's conducted is this piece of paper. Its height is dy, it's length in the blackboard,
we'll say 1, so dz is 1. So it's, it's 1 times dy, qx plus dx is equal to qx plus the partial
with respect to x of qx times the distance you move in the x direction, which is dx.
Similarly, conduction in the y direction, qy, kA dT/dy. In this case the area is dx
by 1. And expanding that in a Taylor series and neglecting higher order terms, qy plus
dy is qy plus the partial with respect to y of qy times dy. The area for the bottom
is the paper, so the heat comes in the bottom. This length is dx, so the area is dx times
1 right there. Okay. Now, let's do the e dot gen. E dot gen we bring a new term in. It's
q dot times the volume. Okay. And the volume of that little differential element is dx
times dy times 1. So now we have a new q term. It's q dot. That q dot is energy generation
per unit volume, per unit volume. So over here, here's our q's now, q all by itself,
q prime, q double prime, q dot. They're all different but they're all q's. The first one,
watts, the second one, watts per meter, the third one, watts her square meter, heat flux.
Now we've got the fourth one, watts per cubic meter. Okay. So four different possible q's.
Know what each one means. If a problem asks you on homework or exam for the heat flux,
you know what they're asking for. If they ask you for your heat rate, rate, you know
what they're asking for. Q dot, you know what they mean, volume metric energy generation.
And the last one is the storage. I'll put that up here, e dot storage, mass Cp dT d
time. The mass of that differential element is row times the volume, dT d time. So this
is equal to the volume dx by dy times 1. So row dx dy partial t with respect to time.
Okay. Now we've put those in the energy balance equation. Okay. Energy balance equation. It
says what comes in minus what goes out plus the generation equals a change in stored energy.
And look at the picture. What comes in by conduction, qx plus qy. What goes out. Okay.
Look at the picture or this equation right here. And then we have qy going out. Okay.
And then we have plus e dot gen down here. And then we have, that equals q dot storage
up here. Q dot x minus q dot x cancels out. Q dot y minus q dot y cancels out. We're left
with partial with respect to x of qx. There it is. Okay? There's a minus sign in front
of there. Minus k dy, and we have dy partial t with respect to x times r dx. And then for
the second term, the y term, it's minus partial with respect to y of qy -- there it is -- minus
k dx partial t with respect to y multiply by dy. Okay. And that, we have our, our e
dot gen, which right here, e dot gen plus q dot dx dy equal e dot storage right there.
And lucky for us, we see each term has a dx/dy in it. It comes outside the partial, dx/dy,
dx/dy, dx/dy, dx/dy. That's how we draw these differential equations. Minus sign, minus
sign, plus sign. So partial with respect to x of k dt/dx plus partial with respect to
y k dt/dy plus q dot is equal to -- there should be a bup, bup, bup, bup, bup, bup,
row Cp here for storage. I had it over here, da, da, da, oh, here, yup. There should be
a row, pardon me. Row Cp up here. Row Cp. Okay. Row Cp. Partial t with respect to time.
We're going to make an assumption here, make life a little easier right now. We're going
to assume the properties are constant. Assume constant properties. The k value, then, comes
outside the partial sign, k comes outside the partial sign. And divide through by k.
Okay. So we have second partial t with respect to x. Second partial t with respect to y.
Plus q dot over k is equal to row Cp over k, partial t with respect to time. But row
Cp over k is just what we defined back in early chapter two as alpha, the thermal diffusivity.
So the right-hand side can be written as 1 over alpha partial t with respect to time.
This boxed equation is called the heat diffusion equation in two dimensions. The heat diffusion
equation in two dimensions. If it's three dimensions, all we do on the left-hand side
is add the second partial of t with respect to z. And that equation is in the textbook,
222, yeah, 222, equation 222. If you want it in cylindrical coordinates, then look at
equation in the textbook -- we're not going to derive it -- the equation in the textbook
is 226. If it's in spherical coordinates, equation 229. Okay. So there's the governing
partial differential equation. What's the purpose of that guy? The purpose is to solve
it where you get the temperature as a function of the variables. So you solve to get temperature
as a function of the variable x, the variable y, and time. In chapter three we'll solve
a very, very simple version of that with a lot of assumptions. But for right now, that's
the general form. What are the assumptions in a boxed equation? Okay. There it is right
there. There it is. But you can't solve a PDE, or any DE, unless you give somebody boundary
and initial conditions. So we need boundary and initial conditions to solve this. So we'll
do the initial condition first. At t, the simplest one is the initial temperature at
t equals 0, the temperature equal ti, where the i, of course, stands for initial. For
instance, if I had a three-inch by three-inch cube of copper and I heat it up to 100 degrees
Fahrenheit and then I blow colder air over it at 50 degrees Fahrenheit, to solve that
equation in three dimensions, then, what I, what I would need is initial condition. So
my, my, my initial condition called IC is when time equals 0, the temperature anywhere
in the cube is 100 degrees Fahrenheit. Okay. That's it. Now, I need a boundary condition.
Okay? We're going to make it easy for ourselves now. Let's say it's one dimensional, okay,
one dimensional. So all we need is one condition on that. So boundary condition, there'll be
two of them. Number one, only with x, boundary -- I'll put it in right here -- boundary condition
one and boundary condition two. We need these. Somebody has to tell you what are the boundary
conditions. Okay, let's put some down here. Typical boundary conditions. Okay. These are
on Table 2-2, yeah, Table 2-2. Okay, type one. Type one says constant surface temperature.
These aren't all done. These are just typical ones, the ones we use the most. Constant surface
temperature. So if we have a surface, and here's the material, thermal connectivity
k, we're measuring x from the surface into the material. And somebody says, oh, by the
way, I'm going to keep this temperature at the surface at x equals 0, equal constant,
constant surface temperature. Then temperature at x equals 0 for any time t, for any time,
is going to be a constant value of ts. Okay. You could probably treat that wall right here
as a constant temperature wall. I guarantee you the outside wall is hotter than the inside
wall right here. So if I'm going to analyze this wall by conduction here, the wall outside,
especially in afternoon when the sun hits that wall, the outside temperature is warmer
than the inside wall. And what's my boundary condition? Well, it could be convection, but
I would probably assume maybe to start with that, that temperature is going to be constant.
I don't think it's going to vary with time for the next half hour. I don't think it's
going to vary with x and y very much from the middle of the wall to where my hand is.
I think the temperature is pretty constant there. So I might model that as a constant
temperature boundary condition. Number two. Constant surface heat flux. Okay. For this
boundary condition, here's a wall again. Thermal connectivity k, I'm going to measure x into
the wall, x equals 0 is a surface. And for this one, I have a constant heat flux on that
wall, qs double prime equal constant. Okay? So a surface energy balance at the wall tells
me the heat provided by the constant heat flux at the wall must go into the wall by
conduction. Okay? So minus k dt/dx equal, that's the heat flux, qs double prime. And
this, this dt/dx has to be at, of course, the surface, x equal 0. Constant wall heat
flux. Number three. Perfectly insulated. Adiabatic. The proper word is adiabatic. Adiabatic means
the heat flux is 0. When someone says insulated you always don't know if it's adiabatic or
not. Okay. Homework get over here already? Homework? Thanks. Okay. If you came in later,
here's today's homework up here. Okay. Adiabatic. This guy is 0. No heat comes in or goes out,
0. Qs double prime equals 0, which means that minus k dt/dx has to be 0, which implies that
partial of t with respect to x at x equals 0 must be 0. Okay. That's the one you want.
Partial t with respect to x 0. Four. Convection at the boundary. Okay? Here's a picture. X,
now we have convection, t infinity, h on the surface. Surface energy balance says the energy
that comes in by convection must go out by conduction into the body. So here's our equation. It's more complicated. It has a temperature
on both sides of the equation, dt/dx at x equals 0 and the temperature at x equals 0.
It's a more complicated boundary condition. So you've got the governing differential equation,
and then you've got the correct boundary conditions. Once you apply those to the differential equation,
you get a solution for temperature in the body as a function of x, y, and time if it's
two dimensional. I told you about this constant surface temperature. You know, when steam
condenses in a condenser in a power plant, when steam condenses in a condenser at the
condenser pressure, you have thermal. You probably know this. It condenses at the saturation
temperature of the pressure in the condenser. So the tubes are like this in the condenser.
They're carrying cold seawater, probably Pacific Ocean. The steam comes in here, it condenses
into droplets at what temperature? The temperature saturation temperature for the pressure in
there. And then, of course, the droplets combine, and they go all off into the condenser hot
wall. Every time they condense, that temperature is the same. It's the saturation temperature.
Yeah. Okay. That's this guy. Constant surface heat flux. Oh, you could buy an electric-type
heating blanket, industrial-type blanket and slap it on a wall and then hook it up to a
voltage supply and provide so many watts per square meter from your electrical conducting
blanket on there. Sometimes these solar power plants, especially the one up at 395 and whatever,
oh, it's probably 58. On the way to Mammoth, for instance, you'll see it on the west side.
Big parabolic trough collectors. Some of those collectors carry heat transfer oil in the
pipes that go through the focal point. And at night -- you don't run the collectors at
night, of course -- at night there's oil in there. Or you could put the oil back into
a big storage reservoir or leave the oil in the pipes, and the pipes are wrapped with
an electric heating blanket. And they trickle a little bit of power through it, a little
current through it, just enough to keep the oil warm. Because when you throw in the morning,
when you throw that pump switch on, you don't want cold oil. Very viscous. Your pump might
burn out. So you heat the oil up a little bit so that when you throw that pump switch,
the oil is less viscous. So you, you wrap it in a blanket like that again. Okay? We,
we all know about adiabatic. You know, you put enough insulation on, it's going to be
pretty much adiabatic. You've heard the term adiabatic turbine. Oh, you know, guess what
that means? You know what that means about the turbine, yeah, from thermal. Convection
on the surface, everything we do on the surface of the earth here is about convention. On
my body, on this front wall, on the outside wall, on the tree trunks, on the street, the
asphalt, that's all convention. So there's a lot of convection going on. Okay. So anyway,
and there's more, there's more. Obviously we're building some kind of space vehicle.
We can't, there's no conduction, there's no convection. It's all radiation. I don't have
radiation up here because that's for the advanced group. Okay? Radiation is a tough boundary
condition, the fourth order in temperature. So, no, we won't be considering radiation
on this. In the real world, yes, it does happen frequently. Okay. So these four are the only
four you're going to be responsible for in our heat transfer course. Okay so now we have
that set up. The heat diffusion equation with the correct boundary conditions. Okay. Let's
look at some problems to see how we use all this material in chapter two. We're going
to start off with a very early problem. You'd think problem 2-1 would be the easiest problem
in the whole homework set at the end of the chapter. Not true. Not true. So problem 2-1
is very similar to your homework problems 2-2 and 2-3. So I'm going to go through 2-1
to give you some major hints how to work that. Let me read it first, 2-1. There's a picture
given in the textbook. Okay. It says we have heat conduction through an axisymmetric shape
shown. So it looks like this. We're measuring x from the left-hand side. This is x equal
l. He tells me the temperature on the left-hand side is a constant temperature at that surface.
And the temperature over here on the right-hand side is a constant temperature on that surface.
Given constant surface temperatures. He doesn't tell me, he should, and I would definitely
do that on a problem, that the size of this are adiabatic, perfectly insulated. Let's
see what else we're given here. He says, assume the problem says this, assume ss, steady state,
assume one dimensional heat's only transferred in the x direction. Assume constant properties.
Assume no internal generation. Okay. And then he says sketch the temperature distribution
on tx coordinates. So he wants you to, T1, he tells me, by the way, that T1 is greater
than T2. So I know T1 is the hot side, T2 is the cold side. T1 and then L0 T2. Okay.
Sketch the temperature profile. Well, you might be tempted to say, oh, I just think
it's a straight line. Why shouldn't it be a straight line? Okay. That's one possibility.
Then you might say, well, you know, it could be a little bit of curve to it. So, I don't
know, maybe it looks something like this. Or maybe it looks something like that. Those
are all possible shapes. Well, okay. The problem is conduction heat transfer. So we're going
to write, let's just do something first to make it simpler. Let's do energy balance,
e dot in minus e dot out plus e dot gen equal e dot storage. Now, he said that q dot 0,
okay, goodbye. He said steady state, goodbye. Conclusion, e dot that comes in the left-hand
face must equal e dot that goes out the right-hand face. If I have 50 watts coming in the hot
side, 50 watts has to go out the cold side. Where else can it go? The side walls are adiabatic,
perfectly insulated. There's no generation. It's steady state. That's what the energy
bell tells me. So right away, now, I know what that means. That means that q in watts
equal a constant. Constant. Okay. Let's then write down conduction Fourier's law. Minus
k [variable] a dt/dx, q is, I just showed you, constant. Problem said assume constant
properties. Constant. It says, then, conclusion, that this thing right here must then be a
constant, [variable] a dt/dx equal minus q over k. That's a constant. What does the area
do as x, as x increases, you can see, the area gets bigger and bigger. So the area gets
bigger. If you multiplied this bigger number by another number, you get a constant. This
guy better be getting smaller, otherwise it wouldn't stay constant. What does that mean?
That means as x increases, the slope dt/dx must be getting smaller. Slope is here at
this location. Same slope, that's not working. It's not that one. Uh-huh. Slope didn't change
at all. Don't forget a steep slope. Steep slope, shallow slope. Okay. Here's a slope
here. Okay? Go down here. Oh, big, big, big, big, big. Small slope. Big slope. No, uh-uh.
You've got it, that one. Big slope. Small slope. Yup. That's it. That's it. No numbers.
Sometimes these problems drive people crazy. Please give me numbers on an exam. Don't give
me symbols. I hate symbols on exams. I want numbers. I know, but sometimes, you know,
sometimes this helps you develop engineering intuition. That's the important point. It
helps you to draw something without ever having to solve the governing partial differential
equation. Okay? That's the key. All right. So that's how we prove that guy. It's that
one right there. Okay. If you want to look at q double prime, q over [variable] a, what
happens here? This guy, numerator, is constant. Denominator, as x gets bigger, the area gets
bigger. Conclusion, q over prime gets smaller. That's kind of intuitive. If I've got 50 watts
here divided by a small area, and 50 watts here divided by a big area, yeah, q double
prime gets smaller as x increases. Q, though, 50 watts in here, 50 watts crosses here, 50
watts crosses here, it stays constant. Q stays constant. Q double prime gets smaller, and
the slope, dt/dx, gets smaller. A lot of stories in there. So your two problems are very similar.
The problem 2-2 looks something like this. I think it looks, yeah, it looks like this.
Compared to that guy, okay. So, 2-2 and 2-3 are very similar. Use my solution as a guide
for you. All right. Now, let's talk about the homework that I passed back so you know
what's going on with the homework. All right. Let's see. This is what was passed
back. This is what's due today. That's what's due next Wednesday. Okay. So on the homework,
you can either get a check, a check minus, a 0, or an asterisk. And if you had me for
Fluids 311 you know how I do this, so it's not any news to you, but we'll go over it.
I'm going to pick up 10, 12 homework sets. I'm going to have 12 grades in my grade book.
I'm going to knock out the two lowest grades and keep the top ten homework scores. Each
homework set, you can get one of those scores in the upper right-hand corner by your name,
typically you'll see the symbol. If I assign eight problems and I grade one or two, let's
say I grade one. I graded one on the first problem, just look at it, you get a check.
Okay. If I, if I pick up a homework set with, with ten problem and I grade two of them and
you got them both pretty much all right, you get a check. If you worked one and got it
all right and worked one and didn't get it right, something was wrong, the wrong equation
or the dimensions were messed up, that would be a check minus. If I pick up ten homework
problems and I grade two, and guess what, the problems I grade are the problems you
didn't work out of the whole set, you worked eight problems out of ten, and guess what
I do, I grade the two problems you didn't work. You know, Murphy's law. Oh, my gosh.
The 0 means you didn't work the problems I graded, but you worked a lot of the rest of
the problems okay. I kind of look and see if you've done the rest of the problems. So
that's what the 0 means. Okay? This guy counts one point. This guy counts a half a point.
This guy counts 0 points. The asterisk means nothing was turned in. Okay. This is 0 points.
Avoid the asterisks, okay? That, that tells me that you, that you didn't think my homework
was worth working. Okay? And my ego gets crushed, so it really makes me, a bad day, you know.
So avoid those little asterisks because they can come back to haunt you. If you're sitting
there, I'll be real generous and say you're at 89.4, 89.4. I say, like, gosh. By the book,
if I round up, that's an 89, and by the book, that's a B plus. Could it be an A minus, gosh,
within 6/10 of a point? One thing I'll do is I'll check the exams and say, well, the
first exam, not too good. I don't think he knew how I give exams yet. But the second
exam was better, and the final was even, even better. So it's some improvement. So I take
that into account. And then I'll say, but let me check the computer problems. I assign
two of them. Oh, okay. Out of a total of five points, this person got four. Okay, decent.
Tried each one. And that will affect the homework, and it, you know, I see ten homework sets,
I'll see six asterisks. Must have a busy person. Must have carrying 22 units because he had
no time for my homework. I don't know, should I, should I go up with the B? You know, should
I keep it there or should I reduce it down? I don't know. Homework's important. So I don't
like to see too many asterisks, okay? I'll tell you, you know, it, it, it helps you to
do the homework, as you know, when you're studying for the final. All right. On the
homework, at the top of the page, yeah, your name, of course. So you've got your name,
then you've got ME415-02. Then you've got homework set number one. Okay. The reason
why I like to see that, that, that 02 there is because I'm grading section one and section
two. And I've got 34 and 34, 68 people. And I'm grading two homework sets. I've got 130
something problems to grade. Anything you do to make my life easier on grading will
help you, I guarantee you. It'll make me happier, okay. So if I happen to mess up these problems
somehow, I think, oh, no. Where is Gina [assumed spelling]? Is she in section one or two. Where's
this, where's this go? Well, you know, I check my grade book, of course, but it's easier
if you do that for me. Okay. So that helps me. Number two, this helps both you and me.
Near the end of the quarter, the ninth week, you say, gosh, I'm kind of worried about my
grade in this class. So you come and see me and say, Professor Biddle, what do you have
down for my homework score? And I say, well, I've got down six out of ten. And then you
say, well, you know what, I've got seven out of ten. I say, oh, really? I say, well, I've
got a 0 on homework set number three. Well, if you mark these things, like, one, two,
three, and four, it'll be easy to find three. If you say nothing, or maybe problem so and
so and so and so, I'll say I don't know, were those problems in set three or four or five?
So it's easier for me and you when we talk about your homework grade if something is
wrong, you don't agree with my homework grade, we'll check every set and see which one is
missing and why it's missing. So that helps both of us. So now if you do a problem, let's
take a problem, easy one, q equal, let's just do k. Let's do q to a prime, k delta t over
l. So k is one watt per square, watt meter k, l is a half of a meter, and delta t is
20 degrees c minus 10 degrees c equal 2 times 10, 20. If you don't like degree k, if you
don't like degree k, take your eraser, that's in the back of the book, one watt per meter
degree k, in the appendix. Take your eraser and make that a degree c. We said before that's
legal. Cancel, cancel, watts per square meter. So when I look at the problem, I'll see, did
you put down the version with just symbols, the symbolic version? Did you put the numerical
values in? Did you put the units in, and did you get the right units when you were done?
Okay. Those three things, symbolic equation, the numbers, the units, and the answer. On
the answer, you know, lot of, most people will do that, box it in. Some people will
do this, double underline it. Some people have all different ways of doing it. I don't
care, they'll put a big, fat red arrow or something by it. I don't care. But just somehow
make the final answer stand out, and that helps. And the last thing I think I comment
on your papers, was it really -- by the way, engineering green is fine. Do it one side
only. It's easy when I'm grading not to worry about what's on side two. Makes my life easier.
One side only. White line paper is fine. I'm not going to tear this up, I'll just show
them. What you don't want to do is tear out spiral bound paper, and it looks like Velcro,
you know, sticking together. So don't do that, but paper like there, white line is fine.
Staple, of course. Of course. You know, we're engineers, we're not art majors. Staple, yeah.
We tell our freshman, I've told my freshman class. Listen, you guys, engineers, 90 percent
of our engineering seniors use engineering green, okay. I told freshman this story. I
said, now, when you're doing homework, this is what you don't want to do. You might have
been taught in high school, but change it. You take your, I'm going to turn my homework
in, so I'll fold it, and I'll fold it, and I'll fold it, and then I'm going to tear it
here, and I'm going to tear it here, and I'm going to fold this little tab down. Am I creative.
Fold that little tab down, and I don't have to buy a stapler. Why no stapler? It costs
$2.95 for a little cub stapler that big. But I'm going to save $2.95 for gasoline. No,
no, don't do it. It's not worth it for four, five, six years of your education here. Don't
do it. Buy a stapler. So that's what I mean by that, okay? I know you guys don't do that,
but freshman really do, honest to goodness. Okay. The last thing is, between, if there's
multiple problems on one page, it really helps if you put lines between the problems, okay?
It helps me locate the right problem to grade quickly. And the reason, you know, when I'm
grading 150 problems, I've got to move fast. I can't dwell too long. This helps me grade
your homework. So, and when you're studying for the final exam, boy, is there a big-time
payback because you've got everything organized, ready to go, the week before finals, the tenth
week, to look at your homework. Plus, of course, you know the story, keep your homework for
every class because when you graduate, you might need it for your job, especially for
some companies who require you to be professionally registered. Aerospace companies, any company
that builds bridges or anything that, where the public is at risk. Oh, yeah, they want
a registered engineer to sign those documents off. And so if you're going to try and pass
the FE example or the PE exam for mechanical engineering, it's great to have your homework
in a little folder to review it. Oh, yeah. It sure beats going out and trying to buy
a textbook that has work problems. So keep all that stuff, and it'll really give you
some payback if you do have to get registered professionally. Okay. Now, let's go back to
homework and look at one more. Let's take problem 31. You have problem 30 for homework,
so I'm going to do 31. I think I'll start, I'll start in the middle. I think we'll be
okay. Was there any question on the homework format and the grading and so on and so forth?
Okay. All right. Now, let's see here. There's, there's no picture in the book, so there's
just words, and I'll read it. Okay. So this is 231. We have a wall, and the wall is 3/10
meters in thickness. And that's l. Let's, let me read it so I can -- the temperature
distribution across a wall at a certain instant of time, he tells us this, he tells us the
temperature across the wall as a function of x, is 200 minus 200 x plus 30 x squared.
Okay. Okay. We've got storage in there, so this is a time varying problem. We've got
conduction coming in. Okay, k of the wall is given as 1. Okay. We -- that's all, right
now, we need. Okay. Let's see. Okay. Determine the rate of heat transfer into and out of
the wall, so part a. Find q double prime in and q double prime out. And then find the
rate of change of energy storage. So e dot storage per unit wall area. Now, why do I
use q double prime? Because he didn't tell me how big the area was. I know how thick
it is, 3/10 meter, but I don't know how big the wall is. So any problem where you don't
know how big the wall is, you probably should base everything on a per square meter of wall
area. That's why I use q double prime. Okay. Let's find the conduction. Qn minus k, I'm
going to do double prime, k dt/dx at x equals 0. Okay, dt/dx goes out, minus 200 plus 60x.
Okay. K is 1. Dt/dx at x equals 0 minus 200, 200 watts per square meter, 200. Okay, q double
prime out. All right, k minus 1 dt/dx at x equals 0.3, 0.3, that is 182. Part b, energy
balance on the wall. What comes in the left-hand side
minus what goes out the right-hand side is the storage. So we have e dot n double prime
minus e dot out double prime equal e dot storage double prime, 200 minus 182 equal 18 watts
per square meter equal e dot double prime storage. So there are the three ways or the
three answers of what we're asked to find. Did we use anything from chapter two to solve
that? No. Where did it come from? It's all chapter one stuff. Then why is it in chapter
two? Well, you can solve it another way if you want to. All right? Let's solve it the
other way. Go to the heat equation, heat diffusion equation. All right. One-dimensional constant
properties, no generation, this is what you get. Okay. Row is in our kilograms per cubic
meters, Cp is our joules per kilogram k temperature. Temperature in degrees k, time in seconds,
cancel, cancel, cancel, cancel, this is joule per second watt, watts per cubic meter. This
is e dot double prime storage because the area had gone out of there. But I want the
answer in watts per square meter of surface area, so I'm going to multiply both sides
by v over [variable] a. V is the volume. The volume is [variable] a times .3, a times .3.
So this becomes .3, k, K told me, k is one. Second derivative, first derivative, second
derivative. Okay. Well, you can see what it is. Same answer.
Same answer. By what method? Chapter two. Where's this problem? In chapter two. How
does this solution manual work it? Chapter one. I don't know why. Why would you do that?
You're in chapter two now. You don't do chapter one stuff in chapter two. That's the whole
point. So I'm just saying, you can work it either way. But sometimes don't believe the
solutions manual because you should work it the chapter two way. Yeah? How would you, then, find -- Fourier's law, chapter one. You got to do
it, there's no way around that one. Yeah, right, exactly, exactly. Okay. Okay. All right.
We're going to work one more example out of chapter two at the start of next time. I think
we'll stop for today so I can do it all at one time and not break it up into two pieces.
So we'll see you then. Homework was passed around. If you missed it, they're up here
in front. And homework is due today. So we'll see you on Monday.
