So let's take a look at an example.

So imagine you have a pipeline.

A pipeline where you can say, transport heated water.

Right?

Or yes, heated water.

And inside, you have another pipeline that you have a heater.

So if you heat this inner part, you can imagine the heat will flow in

a radiant direction and make the outer part warm.

So then you can put this under your floor, so

then you can have a very warm apartment say, for instance, in Korea.

So in that case, you may wonder,

what is the temperature distribution and the rate of reaction?

How can I solve this?

And in fact, you can also ask,

what is the rate at which the heat is lost from a length L of the pipe?

So how fast is the heat being transported?

Across this radial direction.

So those questions can be answered in a different way.

And one thing that you may remember

is to use the Gauss law to solve this problem.

And let me ask my TA, Melody, what Gauss law is.

>> So the way we learned about it was

the integral of E.ds is equal to the sum

of charges divided by epsilon not.

>> Exactly, so what welearned was if we have a charge inside a closed volume,

the flux of electric field out of that charge is

exactly the charge divided by epislo non, right?

So we can use the same analogy here, if we use absorption of surface,

which is cylindrical, because we know this is cylindrical symmetry

then the flux of the heat flow vector around the cylindrical surface,

which is Gaussian surface,

will be equal to the heat generated per unit time inside this volume.

So you can say 2 pi rL, which is the area of the cylinder,

times H, which is the heat flow vector Will be the flux,

will be equal to G, which is the heat generated inside the caution surface.

So then h = G over 2 pi r L, and since we know the magnitude

is equal to the h flow of vectrates, so it's equal to -K Gradient of T,

in this case, the gradient is along the radial direction.

So it's dT/dr.

So we can equate these two, and if we equate these two

we wil have this partial differential equations,

where dT/dr is equal to minus G over 2 pi KLr.

So by comparing between the answers for heat flow and

electrostatics you will get a better feeling or

better grasp of how we can use one equation to another phenomenon.

So as you can see if I divide the infinitesimal

increments from dt and dr to both sides.

And do integration for T1 from T2,

which is from the inner pipe to outer pipe temperature.

Then it will be the biggest of inner part pipe to the outer pipe.

So when you do this integration, this is the temperature difference and

on the right side you will have logarithmic term because 1 over r

will be integrated to logarithmic term.

So if I solve for g, then g becomes 2 pi k l over t 1 minus t 2 over log b over a.

And you see this result corresponds.

This result corresponds exactly to the result for

the charge of a cylindrical condenser.

So imagine you have a plus charge and minus charge across this pipe.

So the charge accumulated at one end of the electrode will be depending on

the potential difference between those two as well as the lengths of the pi and

the radius ratio of the pi as well as the permittivity.

So without going further into doing the same kind of calculation,

you can just replace the parameters.

So I'm going to show you a real example in the research that my student,

[INAUDIBLE] Kim Under supervision.

The same kind of calculation for spring type energy harvesters.

Where you coat a PVFTRFE what is a polymer on top of a spring structure.

And then you deposit electrodes on top of it.

So the process is like dip coating.

When you dip coat your ice cream, vanilla ice cream into chocolate sauce,

then you have chocolate layer, and if you it again, you have another layer.

This is how we did.

And for the [INAUDIBLE] we did like barbecue.

So we rotate it, this spring, while we do position.

So in that way we are able to make a cylindrical type of

pipeline of two electrodes.

Which are sandwiched by dialectic.

So if that's the case,

then we can use the equations that we learned from this course.

And that's what he did and you can see how much charge

will be accumulated on the outer electrodes and

we could use the knowledge from this course to get the answer.

So melody, so in your research, do you see you can

use the same type of equations in your for example.

>> Yeah, I think so.

>> Yeah, so she's working on a multi-layer

that could mimic the tectile information.

So for that also we need to know how much charge is accumulated.

On one surface or the other, and

I hope that she can use this equation in our research as well.

>> Perfect.

>> Okay, all right, now let's take a look at our very interesting sample,

Heat source near the ground.

Imagine, somehow, you detonated, say,

TNT Underground, to make a tunnel, right?

If you do that, you want to know, what is the temperature distribution underneath.

Maybe for safety, or maybe for

designing the destruction of the area of interest.

So if that's the case you will ask,

what is the temperature distribution near the surface?

And as you can see air is a very good insulator, and

that's why we use air or vacuum in a tumbler,

to make heat preserved inside the volume of interest.

So you can think is nearly heat conductivity with zero,

kappa is almost zero.

If kappa is almost zero,

it means you cannot have any heat transfer normal to the surface.

So we'll only have parallel component near the surface.

So in that case, you can imagine, how can I solve this?