# Thermal 7 Part 4: R Values

Okay now, this is not something that you're going to be standing on in the physics course that it is important for engineers to understand. So this is how this theorem is applied to engineering. So often when you buy a material to build something from its cousin are very much are is equal to our length terms, divided by our conductivity term, and it's referred to as an r-value.

So the formula for the rate of flow is keeps with our values is P is equal to the surface area. The hot sides, the difference. In temperatures between the hot side in the colored side over the sides, we are valued for whatever materials you're building it from. So this just says that if you have a composite material to get the total hourly, you just need to add the R values from each part. This table presents the bodies for some substances that are commonly used in building this tables actually taken from the textbook, which is an American textbook. And so it has very strange units.

You can see the unit for our bodies here. A seat where's, her degrees Fahrenheit per hour, curve British thermal unit. So in our more normal units, the feet squared becomes meters squared degrees, Fahrenheit becomes degrees Celsius. And this is time divided by energy units. British thermal units are related to joules.

You get from one to the other units multiplied by 1055. But the time divided by the energy is equal to the inverse of what so we have per watt here. Instead of hb2 you normally to get from these very weird, American units to our unit. You just divide by a number, which is approximately six. These are just presented here. So that if you ever see these units that you know what it's talking about, but in physics will not been using these units, we will be using assign so here's, a question for us to try to calculate, the total r-value for a wall constructed as shown. So we've got bricks.

We've then got some shading. We've got an S face. And finally, we've got some plasterboard. Okay.

So let's just draw a little diagram. We've got bricks here. Those 10 centimeters of bricks, we've then got one point two centimeters of sheathing. We've got two centimeters of air and one point two centimeters of plaster board, and we know that T is equal to a TH, minus TC over the sums of the are values. And this is equal to K, a delta T over L. And so we know that the sums of the eyes is equal to L on K. And so all we need to do is add the L on K for each of these different materials. So we'll need to look that up in the table, Kate is equal to zero point. Seven.

Zero, what per meter per Kelvin K for cheesing is equal to two point, zero watts per meter per Kelvin K for and equal to zero point. Zero, two, three, four watts per meter Kelvin. And finally, we've got past the board. So k possible is equal to zero point.

One, seven, four watts per meter per Kelvin. So our R is equal to zero point. One zero me just over 0.70. Plus. We've got the sheeting for that 0.01 2 meters over 2.0, plus the air, which is a 0.02 over the 0.023 4, plus the pass the board, which is the 0.01 to.

Over 0.174 so solving this on the calculator. We end up with 1 point, 0 7, 3 meters squared Kelvin per watt and so that's, the R value for this wall.