Everyone, welcome to the next session of Answers Workbench Tutorials.
In the session we are solving another numerical on steady state thermal analysis.
I will first double click on steady state thermal.
Now you're solving for a plain wall.
So I'll just rename the file as plain wall.
Next, I'll go to geometry and double click on it.
Now let's go to the problem.
The inner surface of a plane brick wall is that 60 degrees Celsius and the outer surface is that 35
degree Celsius.
Calculate the rate of heat transfer o meter square of surface area of the wall which is to 20 mm thick.
The thermal conductivity of the brick is 0.51 watt meter degree Celsius.
So I will go to design modeler, I'll go to units, I'll choose that as mm x y at Now we know that this
numerical that we are solving, had we been solving it using the numerical method, theoretical one,
then we would have used for a lot of heat conduction.
There is one particular assumption of four year law that the cross section area is not important.
But the thickness over which you are measuring your heat transfer from one end of the wall to the other
is going to be important and that thickness is already given as.
220 mm.
So rest of the cross section is immaterial.
So I'll go to a rectangle.
I'll draw a rectangle, I'll give it some dimensions.
These dimensions are random.
You can choose any value.
I'm taking 101 hundred.
Next I'll go to Extrude, so I'll go to create Extrude.
I'll select the sketch that I've just made.
Sketch one apply.
Now, here, this depth that I'm giving is very important.
It is to 20 and I choose generate.
So this is the geometry made.
Now, this is important because here you can see the direction in which my arrow is moving in this direction.
I will have to check the rate of heat transfer which is taking place and all these two walls.
I'm going to apply the temperature.
So this is done.
Next, I'll go to model and double click on it.
Now I'll go to Mesh.
I'll select the sizing.
As fine.
An update.
Next I go to steady state thermal.
I'm going to insert temperature.
I will choose this option face.
I'll select your apply.
And one end was 60 as given.
And the other 35.
Next I'll choose this side.
Right click insert temperature and this value is 35.
Apart from temperature, no other data is given.
But here it is, given that the material of the wall is brick.
So I will have to apply a new material because if you select this geometry or solid over here, you
can see the assignment of material is by default, structural steel.
I need to change it, so I'll click your new material.
Okay.
This is taking me to engineering data.
I'm going to add a new material here.
Let me rename it as Brick.
I need to give it some property so I'll go to thermal isotropic thermal conductivity.
The values given in the question as .51 what per meter degree Celsius?
You can check the unit here.
So this is done.
Next I'll go to project in the model and right click.
Update.
So the material now gets added in my model.
Or you can also update project over here.
Either way, you can do it.
Next I'll go to model and when I see the solid now if I click on this button, I can see brick and this
brick has now been applied to this one.
Next I'll go to solution.
Right click insert thermal temperature and the question asks for calculating total heat flux.
So I'm going to click on it.
Next, I'll go to Solution Solve.
Now we click on temperature so I can see a temperature variation on this wall.
Also we can check the total heat flux.
The value that we obtain over here is 57.955.
That is a constant value over this entire body.
We will go to this question and here you can see this is a solution which I have obtained by using numerical
calculation.
So I have just marked the value here for your reference and the value that I obtained by 30 and by answer
solution is exactly the same.
It is 57.95.
So this proves that the answer, which I have got to answer, is correct.
If you want, you can solve this question by using a very simple formula that is four years low.
That is Q is equal to minus K, a D by the x.
So this is the formula for calculating Q.
Now, we always take a negative sign because that shows the temperature which is final.
That is over here on the other end is 35 less than the initial value 60.
You can just subtract over here at 60 -35 and not take the negative sign.
Or if you take the negative sign, then just see to it that you are taking.
It has 35 -60 and the x is the length that is the thickness to 20.
It changes from 0 to 220.
K values also give it an area since it is not given it is four meters squared.
So you can take it as one.
When you use this formula, you will get this answer.
That's how I have solved it, and I've also proved the answer by using answers.
So this is how you solve it by 30 and compare it by using a software.
So with this, I end the session.
I hope you have understood the procedure for solving this kind of a question.
If you have any doubts, please write to me the comments section.
See you in the next session.
Thank you.