>> Midterm will go
through Chapter 4, chapters 1, 2, 3, 4. It won't cover
numerical methods. So the first half
of Chapter 4, you're going to be allowed to bring into class with you one
sheet of paper, eight-and-a-half
by 11, both sides. Anything you want on it, anything at all
that you want. I'm going to pass out
the data package, the tables, the
graphs, the figures. When you turn the exam in, you'll turn the data
package back into me. I'll keep those
for the second, third, midterm,
and final exam. We'll talk more about
that on Wednesday. Okay. Where we left off
last Friday? Yes sir. >> The package that you're going to be passing
already on Blackboard? >> It's on
Blackboard, right. So you know what you've got to resource there. Here's where we
left off last time. We were looking
at table 4.2. We derived one finite difference equation
in conduction. It was for an
interior node. We derived that in class. That equation is
in table 4.2. Table 4.2 also includes other equations, finite
difference equations. For instance, case 2 is the internal corner
node with convection. Case 4 external corner
node with convection. Case 3, a node on a plain wall
with convection. Our last case there, I'm going to erase
this because it's covered the
bottom note. This is a plain wall with a uniform heat flux, q
naught double-prime. So you've got five
possible equations to use to set a problem up with finite differences. Now if this covers
heat flux on a wall, convection on a wall, what about the case where it's an
adiabatic wall? Okay, what you do
there is you set h equal to 0 in
case 2, 3, and 4. Or you set q double prime equal to 0 in case 5. Let's just call that
q double prime. So we'll do that
in just a minute, so an example problem. Okay, so we're going to work an example problem. We're going to look
at this plate. This is a plate whose thermal
conductivity is 25. We don't know
how thick it is, but we do know it's
20 centimeters in this dimension and it's 30 centimeters in
this dimension. Our first goal is
to set up a grid. So we're going to set up a really
coarse grid, not one you want to
do in the real world, but just to show you
how the method works. So we're going to set a grid up of 10
centimeters. So you can see the nodal points,
big black dots. Those are the nodal
points of my grid. The left-hand side is kept at constant
temperature zero degrees C. The
right-hand side is kept at a constant
temperature 100 degrees C. The
top is kept at a constant temperature
100 degrees C. Now I could, like I did last
time we met, number the nodes m and n, m column n row. But that's too complicated for a simple
grid like this. So I'm just going to number them one
through whatever. Well, if I had a 100 nodes, I'd use a double subscript. But for 12 nodes, I'm just call them
one through 12. Node 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Okay,
there they are. The bottom is adiabatic. Question asked is, give me an estimate of the unknown nodal
temperatures. Give me an estimate of
the heat transfer q. Okay. Let's start out with getting the
temperatures. Now some
temperatures I know, by the way, there are 12 total nodes.
12 total nodes. I know T2 equal T3, T6, T9, T12 equal 100 degrees. I know, T4, T7, T10 equal 0 degrees. Unknown, T5, T8, T11. So that's what we're
supposed to find, those guys. Yes, sir. >> [inaudible] T1. >> Oh T1, yeah T1. Here's a game you can play with your
non-engineering friends. So you ask him, Okay now. I want you to give
me your best guess. When x equal two what is y? Half the guy say ten, half the guy say five. Wait a minute, somebody's right and somebody's wrong. No, they're both wrong. That's okay. You can't
ask that question. Illegal. Don't ask me. It was a good question. Thanks for asking because otherwise I wouldn't
have done them. I'm glad you asked. But no illegal
question don't ask me. I don't know when
I can't find out. What's that called in math. Discontinuity,
discontinuous there, infinite jump up there. So anyway, here's what it means in our picture. As I approached the upper left-hand
corner on left-hand wall, the temperature
is 0, 0, 0, 0, 0, 0, 0, 0. Approach from
this side, 100, 100, 100, 100,
100, 100, 100. Get as close as you want, a 100 and 0. If this were a
copper plate, an electrical problem. Where this was zero voltage and this was a 100 volts, that would be called
a short circuit. All the current would flow, to there, nothing
would flow inside. Called a short-circuit.
You can't have that. So it's undetermined. You just say T1
is undetermined. I can't worry about it.
Okay. Fine. It's out. Undetermined. Okay. Is that it then you kind
of get the idea. Yeah. Oh, I'll say, in the real-world,
can this happened? No, you can't do that. What you do is
you might put a little thermal
insulation, right there. Little piece of
insulation right there, so they don't
touch each other. That's the whole point. Don't let them
touch each other. That's what you do
in the real world. Okay, back to here. Three unknown temperatures, which means I need three equations for three unknowns
temperatures. So let's take node 5. Node 5 is an interior node. So here's what I
found out last time. For an interior node,
here's what you do. You add the temperatures of the four surrounding
neighbors and subtract off four
times the temperature of the center node. That should be equal to 0. Center node T5,
surrounding nodes 2, 4, 6, 8 minus four times
the temperature of the center node
must equal 0. Node 8, interior node. Add the four surrounding
temperatures of the center
node, 5, 7, 9, 11 minus four times
the temperature of the center node equal 0. Node 11 adiabatic boundary. >> I'll just say it's
case 3 with h equals 0. Two times T_8, plus T_10, plus T_12, minus 4 times T_11 equal 0. You can get that from case 3 by setting
h equal to 0. Plug in known
temperatures to get, okay, let's see
where you got it. Adiabatic wall behaves like there was a T_8 here and a T_8 down here. So now add your four
neighbors together, T_8 plus T_10 plus T_12
plus T_8, two T_8s. It's symmetrical about the adiabatic
wall, that's why. You just solve
it, I don't know. XL matrix inversion,
you solve it, T_I with a solver
function, or whatever. So it's pretty simplistic, there are three
equations, algebraic three unknowns,
solve for them. When I do that,
I get T_5 63.5, I get T_8 53.8, and I get T_11 51.9. So that's my
first estimate on the interior
temperatures and the adiabatic wall
temperature at the center. I know those are
bad answers, I mean the grid is
way too coarse. If I want better answers, I'll make that grid
a lot smaller. Say okay, let's make this as 10 centimeter grid, let's make it five
centimeter grid. There, there, there,
there, there, there, there,
there, there, five. I cut the grid
size down by half, unknown temperatures, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 14, 15, 16, 17, 18. Now you've got 18
equations, 18 unknowns. Those answers now will be a lot closer to the
real answers, but I really want
accurate answers. Okay, make the grid
size one centimeter, hundreds of nodes now. You're solving a
150 by 150 matrix, or on your TI, you
wait overnight for 12 hours until it
spits the answers out. Yeah, so what's a
penalty you pay? More computing time.
What's the good news? Better accuracy.
Is it worth it? That's your call,
you're the engineer, you figured out,
what do you want? What if I have
the grid size? The temperature
don't change by more than 110 degrees
C. Okay fine, that's your criteria, keep going until
you get there. So that's the game
you play as far as, what accuracy do you want? Now this is, ME 4150, the title is heat transfer. So same old game. With conduction, once
we get to temperatures, the next thing to do is, to find out how
much heat is transferred. From
where to where? From the top and the
right-hand side, out the left-hand side, 100 degrees goes out
the zero degree phase. So I'm going to find out how much heat we have. First, we'll look
at q that comes into the zero degree phase. Here is how we do it. Remember, we first talked about numerical
methods in the grid, and we said, you lump
the mass at the nodes, and the nodes are connected by heat conducting rods. The only way for heat
to be conducted, is through the heat
conducting rods. Here's a node on the
zero-degree phase. Heat comes in from
the right-hand side, heat goes from five
to four through this heat conducting rod right there, q_5 to four. Heat goes from node 8 to node 7 through the
heat conducting rod. >> Heat goes from 11-10 through the
heat conducting rod. But don't get confused
and say yeah, but that's an
adiabatic wall. I know, but remember what adiabatic wall means? Heat will not
be transferred normal to the wall.
I understand that. Can heat be transferred
parallel to the wall? Sure it can. What's this
temperature? A hundred. What's that temperature?
Zero. Of course, there's going to be heat
transfer along that. This is a heat
conducting rod. Heat can't go this way, but it can go this way.
Question over there? >> That means an isotherm? >> What? >> The heat conducting rod. >> No. I'll give
you an example. That temperature is 100, 90, 80, 70, 60, 50, 40, 30, 20, 10, 0, is not the
same temperature. I'm glad you mentioned
that, thank you. I'm so glad you
mentioned that because what I
meant to do first was to draw the heat
flux plot just to give you an engineering
interpretation of this. Heat flux plot. Here is our adiabatic surface
at the bottom. I'll say T1 is here, T2 is here, T2 is here. Geometric symmetry, right
through the middle, vertical and
horizontal lines. Thermal symmetry,
vertical line, T1, T2? No horizontal line through the middle adiabatic T2? No. There's no lines
with thermal symmetry. Conclusion, I
can't use that. But here's something
we did twice already. We said, where two corners come together at
different temperatures, the heat pretty much flows around the closer you get, almost like a
quarter circle. At the bottom,
this is adiabatic. Let's assume T1 is
hot, T2 is cold. He comes in near the bottom of a
hot surface, T1. It can't get
out the bottom, so it pretty much
travels along here, somewhat horizontal
like that. Okay. There's an adiabatic, that's an
isothermal corner. I know an adiabat comes out of that
corner and it bisects that corner and then
it has to go out down here or vice versa,
fill in the rest. There they are, these
are the adiabats. Now we'll draw
the isotherms, we did it twice before. The isotherms that come
out of a corner where two different
temperatures are there at the corner, T1, T2, they come out in a
fan pattern like that. When the isotherms get to an adiabatic boundary, it says they come
in at right angles. Everywhere inside
the boundaries it says they must cross
at right angles. Okay, draw a few
of these in. These guys are
the isotherms. So that gives you
a little bit of engineering intuitive feel for what's going on. If you want to get
more accurate, you can construct
curvilinear squares. He's real good,
not bad at all. You can do that and
that'll give you an estimate of the heat
flow MK Delta T over n. It'll give you
an estimate of the temperatures
in there if you construct reasonably
curvilinear squares. But we're doing it now
by numerical methods. But that's the physical picture of what's going on, which gives you some
feel for the problem. Numbers on a spreadsheet do not give you a feel
for a problem. Graphical things
like that give you a feel for a problem. Why do you think mohr
circle is so important? You can do it by
equations and they're accurate to 15
place its accuracy. Mohr circle tells
a great story, a phenomenal story, but
you see it visually. It gives you
that engineering intuitive feel
for a problem. Same thing here.
Okay, let's do it. Q5 to 4. Now, this heat conducting rod replaces that much
material five to four. Here's the heat flow. What's the area? My hand. What's that?
Delta y times 1. The area is
Delta y times 1. We're using Chapter 1. K, the area Delta y times 1 divided by the distance the heat traveled, Delta x. Eight to seven, same thing, K. Put the temperatures
in there T5 minus T4 plus K area Delta
y times 1 Delta x T8 minus T7 plus, now we've got
to be careful. Let's do eight
to seven first. There's eight to seven, but now 11 to 10, the area is half of y. If we make the grid
square Delta x equals Delta y
makes life easier. Cancel, cancel, cancel,
cancel, cancel. Not k, Delta y, Delta x. I know K 25, I know all the temperatures put him in there and Q is equal to 360 watts. You say yeah but look
at this picture. The zero-degree
face is covered. We've covered
that much of it, but we've ignored
that much of it. We didn't account for
that right there. I know, that's why my
grid is so coarse. If I want to
get a better Q, I'm going to put
more nodes in here and I'm going
to reduce this now to this much
area that's unknown by having
the grid size. If I have the grid size, this will be ignored, this will be accounted for. But I want a better answer. Okay, make more nodes, have it, have the spacing. Now, you're ignoring
this little piece. You get better and
better estimate of Q as you make the grid
finer and finer, at the cost of what?
Computing time. Now, what comes out of a zero-degree face should have come in the top
and right-hand sides, what comes in here in steady-shade should
go out here. Let's try that then, Q out of 100-degree face. Look at the heat conducting rods from five to two, from five to six. Let's do q5 to
2 plus q5 to 6, plus q8 8 to 9. Sorry, should be six to
five, nine to eight. We've got five there, so the hot temperatures, two to five, six to five. >> Then we have 9-8, and then we have 12-11. I'm not going to
go through that. Do it the same way
I did it here, but it comes out to be 356. Yeah, I know they're not equal because my grid
sizes are so coarse. I'm surprised it's
that close, actually. I'm very surprised it's
that close, actually. So what's my answer? Add them together and
divide by two, 358. That's my answer
of what Q is. That's my
approximate answer. The finer your grid size, the closer they are
going to agree. But again, at the
expense of what? Computing time. Is it worth it? That's your call. The finer you make
the grid size, the more accurate your
temperatures are, and the closer Q is
to the actual Q. These guys here, no, that nine means nothing. It might be anywhere from 50-51-53,
that temperature. Any question on this
then? Yes, sir. >> Why do we start in
between losses again? >> Let's take 6-5. This rod replaces how much material?
That much material. All the heat from here to here is called Q 6-5. It's replaced by a symbol, node 6 to node 5. How high is this? Delta Y. How far
heat travel? Delta X. That's
how you do it. >> But are you in between
2-5 and then 1-4? That's arbitrary
and [inaudible]. >> Wait, 2-5 and what now? >> They're in
between two and five and then one and four. >> Right. >> That's just
arbitrary because it's just going to
cancel out, right? >> There's one here. >> How high up? As
long as they are equal to each another, it doesn't really matter. >> What's equal
to one another? >> How you put it
into the equation. Where the Delta
Ys and Delta Xs ended up canceling each other as long as you stay consistent with
that spacing. >> Yeah. Otherwise,
things would change. These would change. You want to make life easy, so you try and
make Delta X equal Delta Y. Yes, sir. >> So for finding
the Q out of a base, you just use the adjacent nodes to that base, right? >> Here's what you have
to do. You say, well, how about heat going
that way? I say, no. Illegal. Heat only is transferred along
heat-conducting rods. Is there a rod from 5-3? No. Is there a
rod from 5-6? Yes. From 5-2? Yes. All the
heat's conducted along those guys there. When you draw this picture, these are the
heat-conducting rods. Is there something else?
You had a question? >> Yeah. If you're using
smaller nodes, say, you had five
going this way, so it would be
1, 2, 3, 4, 5. Would you just use 5-4? >> Along here? >> Yeah. >> Like the top? >> Yeah. >> Okay. Four to nine, 3-8, 2-7, 10-9, blah,
blah, blah. >> So how come
we ignore 2-1? >> Okay. I got it. >> I'm not sure if you
made that clear also. >> That's a good question. What's the
temperature here? >> Two or 100. >> Whats the
temperature there? What's the
temperature right there? A hundred. >> Even though the zero is coming from inside. >> The problem said,
the top surface is at 100 except
the corner. That's what the
problem says. It's got to say
that. Over here, the left-hand face is zero. Where? Except right
there. Illegal. >> I thought it
was an assumption, >> No. It's
going to say it. It's going to say
it in the problem. Was there another
question? Yes, sir. >> If you draw the
box for the area that it replaces, for 6-5. >> Six to five.
The original one? >> Yes. Then if you
draw the box for 2-5, they overlap slightly. >> Six to five.
This was five? >> Yeah. >> Okay. There's
6-5. They overlap. Yeah, you got it. In a way. How about this area here? Conclusion. I ignored it. So I double count
here, ignore that one. You know there's errors
built in this thing. But guess what? As you keep reducing the grid size,
they go away. The areas get
smaller and smaller. The problem areas
get smaller and smaller, that's right. >> You only have
another corner. >> This area gets smaller, now you're missing
that much. Now you're missing
that much. Now you're missing
that much. At what cost? More
computing time. That ends our chapter
on 2D heat conduction. Now I'm going to go over some problems for homework. Did it get over you
guys yet, the homework? >> Yeah. >> Okay. Hand it
to me, please. Thank you. Did
anybody come in late that that didn't get
their homework back? Just so you know the
numbers on this. Three, 15, main problems on that was getting the
correct K values, pretty much for that
wall with studs, insulation, wallboard,
and siding. So the hard wood siding, K equal 0.094. Gypsum, 0.038. Insulation, 0.038. This guy here is 0.17. Gypsum, 0.17. Area of the total wall, two-and-a-half meters high by six-and-a-half
meters long. Area of studs; there
were 10 studs. Each stud is four centimeters and
two-and-a-half high. Each stud four centimeters by two-and-a-half meters. Area of insulation; area of the wall minus
area of the studs. You need those areas, of course, for
your resistance. KA over L, there's
your areas. If you didn't get
these K values, see me after class and I'll point out exactly in the tables where
the correct value is for you so you know. Three, 27; the correct answer for the
chip temperature, 48.1 degrees C. I think I had down
46 point something degrees C. The
correct answer is 48.1 degrees C. Three, 111 is a solar
collector panel which behaves like a fin. There's two tubes carrying a fluid, generally water. X measured from
here, L is halfway. There's convection
coming out to the air. There is radiation
coming in from the sun. The way you start
that problem is to go back to fins at the derivation of the fin equation
for temperature, which is equation 3,61. >> You have qx
minus qx plus dx minus q
convection equals 0. That was equation 361. We did that in class. We took a small
differential element of length dx out the fin. We wrote that equation,
energy balance on it. From that, we got a differential
equation relating temperature as a
function of x. In words, heat coming in the left side of the element by
conduction qx minus heat going out the right side
of the element by conduction qx plus dx minus the heat loss by convection to the
fluid to infinity. Now, what's a
difference here? You see the picture. The difference is,
I've got a throw in qSUN as an input
is coming in. Coming in means positive. So this becomes
plus qSUN equals 0. That's a constant,
it's just a constant, qSUN is given
as a constant. You then follow what
I did in class or in the textbook on
that page until you end up with an equation
similar to the 366, which is the governing
differential equation. So it's a pretty
straightforward solution. All you do is throw
a constant in and then follow
line by line, five lines what he does
in the textbook to get something similar
to equation 366, which is the governing
differential equation. 3.149. This was the circular tube separated by there's
baffles here. There's baffles here. There's fluid flowing. This tube comes out the blackboard,
these are baffles. So the fluids flow in
these four quadrants. So I have q equal q from the fins plus
q from the wall. These guys behave
like fins. This is T base, T base, T base, T base. In the middle, no heat
crosses horizontally, no heat goes up and
down vertically. Conclusion that this point, these fins behave as if they have an adiabatic tip. Case two, Table
3.4 Find qf. Got it. How many fins? Four. Got it. Table 3.4, case b. Plus q from the walls. What's the total wall area? The circumference times the length; circumference, pi D. But if there's
a subtract out 4 times T because
that's not bare wall. The fin's touching
the wall there, times the length, which is one would do it on
a per meter basis. So that's how you
get that guy, 3.158. Any questions on
him before I go on? [inaudible]. Yeah, that's right.
I would go to that. Because like you say, is not touching
the fluid area. 3.158 is this picture. Circumferential fins. By the way, on
this problem here, you don't need to find
the fin efficiency. That's a waste of
time on a timed exam. You can, if you're asked for it, you sure can. But this is the easy way out because I know a qf is capital M hyperbolic
tangent, little ml. I get that really quick. But if you want to
find the efficiency, you can do it that way. It just takes more time. So on this exam, time is valuable, real valuable. Come up Friday. Let's check how many fins there are on
that thing 158. The first part says find the heat loss by
one fin. Got it. That's part A. Heat
loss per fin, 200 fins. So I'm going to
use figure 3.20 to get ta fin and then
q for one fin is equal to eta area of the fin times h T base
minus T infinity. The area of the fin
there's top and bottom. So area of the fin is two because of
top and bottom times pi r squared minus the
inside radius squared. There's your
area of the fin. For 200 fins q total equal 200 q_f plus q from the bare wall. So there's the bare wall. Now I'll show you
the bare wall. That's the bare wall.
So how many spaces are there between the fins? One hundred and
ninety-nine. If you want to do
it that way, it's fine, but be careful. He says the fins are spaced at five-millimeter
increments. Well, I told you in
this chapter on fins, there's so many
different areas. There's a_c, there's
a_p, there's a_f. You got to know
that most people on fins make their
mistakes on areas, which should be the
simplest thing of all, but they make their
mistake on the areas. So the spacing of the fins five-millimeter increments. Wall in your house,
somebody tells you the studs are 60 inches,
spaces 16 inches. Does that mean
there's 16 inches between this face and
that face of the stud? No. What's the correct
way to say that? Center to center, right? Yeah. Center to center. So what does this mean? Five millimeters
center to center? No, you got it wrong. Your area is wrong. It's from the middle
to the middle. People make mistakes like that, which seem small. But when you multiply
it by 199 spaces, it becomes rather large. So get the areas correct on problems
like that.
