### Video Transcript

The inside volume of a house is equivalent to that of a rectangular solid 13.0 metres wide by 20.0 metres long by 2.75 metres high. The house is heated by a forced air gas heater. The main uptake air duct of the heater is 0.300 metres in diameter. What is the average speed of air in the duct if it carries a volume equal to that of the houseโs interior every 15 minutes?

In this problem, weโre looking to solve for the speed of air. And weโll represent that using the symbol ๐ฃ sub ๐. And weโre told that that speed matches a certain condition in that it can replace the volume of air in this house, whose dimensions were given, every 15 minutes. Weโre also told the diameter of the main air duct connected to the heater. Using this information along with the cycle time of 15 minutes, letโs solve for ๐ฃ sub ๐.

To start, we can draw a diagram of the situation. Here is a sketch of what the inside of the house might look like. Weโve been given a height, a width, and a length and told what those values are. And weโre also told that inside this house and the interior, there is a main uptake valve through which the air in the interior of the house passes between the house and the heater.

Letโs say this dotted yellow circle here is a cross section of this intake valve. And that this valve or this pipe passes from the outside of that space to the inside. And weโre told that as air passes through this pipe on its way to the heater, it moves with a velocity weโll call ๐ฃ. It moves with a speed that weโve called ๐ฃ sub ๐ such that itโs able to empty out the entire volume of the house interior every 15 minutes. Letโs label that time ๐ก sub cycle for the time it takes to cycle the air through the interior of the house. And again, that time is 15 minutes.

Another piece of information we were given in the problem statement is the diameter of this duct. Weโre told that the diameter is 0.300 metres. Weโll call that diameter ๐ and record its value down below. So with that as setup, what we want to solve for here is again that speed of the air as it moves through this duct such that it empties out the volume of the space in time ๐ก sub cycle every 15 minutes.

Now, as we look to solve for ๐ฃ sub ๐, the speed of the air in the pipe, letโs look at the pipe in an expanded view. If this is our pipe with air moving through it at a certain speed weโre looking for and a diameter of 0.300 metres, then that diameter means that we can solve for the cross-sectional area of the pipe. Weโll call that capital ๐ด.

Now notice this about capital ๐ด, the cross-sectional area of our pipe, and ๐ฃ sub ๐, if we take their product โ that is if we multiply them together โ then we would wind up with a value. Weโre not sure what that value is quite yet. But whatever the value is, it would have units of volume per time. Thatโs because capital ๐ด has units of area and the speed ๐ฃ sub ๐ has units of length per time. So when we take the product of these two terms, we get a volume per time.

Now based on what weโve been told in the problem, we indeed have a volume and we have a time. The volume is the product of the height and the width and the length of our interior. And the time is ๐ก sub cycle โ the time it takes for the entire interior volume of air to be cycled out.

As we look at this equation, we realize weโre able to solve for or are given each one of the variables in this equation other than ๐ฃ sub ๐. Capital ๐ด, the cross-sectional area of the tube, we can solve for using the tube diameter. The height ๐ป, width ๐, and length ๐ฟ of the interior of the building are all given and ๐ก sub cycle is given as well.

So letโs move towards solving for ๐ฃ sub ๐, the speed of the air in the duct, by replacing capital ๐ด, the cross-sectional area of our tube in terms of values we are given. When we recall that the area of a circle equals ๐ times the radius of the circle squared, we see that we can use this equation to solve for the area of our duct. That area capital ๐ด โ as weโve called it โ is equal to ๐ times the diameter divided by two and that quantity squared. When we square this term ๐ over two, we get ๐ squared divided by four.

Now, letโs isolate ๐ฃ sub ๐, the speed of the air in the duct, on the left side of the equation by multiplying both sides of our equation by four divided by ๐๐ squared. Doing this cancels out the four, ๐ squared terms and ๐ terms on the left side of our equation. So we can write a simplified version of this equation with ๐ฃ sub ๐ all by itself. ๐ฃ sub ๐ is equal to the height times the width times the length of the interior of our building times four divided by ๐ times the diameter of our duct squared multiplied by the time it takes to cycle the air in and out.

Letโs plug in for the different values in this equation. In the numerator, ๐ป is 2.75 metres, ๐ is 13.0 metres, and ๐ฟ is 20.0 metres. Then, in the denominator, ๐, the diameter of our duct, is 0.300 metres. ๐ก cycle, the time it takes to cycle the air in and out of this volume is a time that was given in minutes, but we would like this time to be in seconds. So letโs do a quick conversion.

Weโll create a variable and call it ๐ก sub ๐ . Thatโs the time that this cycling process takes in units of seconds. To convert ๐ก sub cycle into units of seconds, weโll need to multiply that number by 60 seconds per minute. So ๐ก sub ๐ is equal to 60 seconds per minute multiplied by ๐ก sub cycle. Plugging in the value for ๐ก sub cycle of 15 minutes, when we multiply these two terms together, we find that ๐ก sub ๐ , the time it takes to empty all the air in this space, is 900 seconds.

Now, remember this time is the same value as ๐ก sub cycle, but simply in a different set of units. So we can substitute ๐ก sub ๐ or 900 seconds in for ๐ก sub cycle in our equation. Now, weโre ready to calculate out what is ๐ฃ sub ๐. But before we do that, take a moment to look at the units of our numerator and denominator. In the numerator, we have a combined set of units of metres cubed. And in our denominator, we have a metre squared term and a time term in seconds. We can see as a final result that two of the metre terms in the numerator will cancel out with the metre squared term in the denominator. And weโll get a final answer in units of metres per second.

This is encouraging because weโre solving for a speed ๐ฃ sub ๐ and expected to have units like that. So it seems weโre on right track. Letโs calculate out what this number is. And we find that ๐ฃ sub ๐ is equal to 11.2 metres per second. Thatโs the speed that the air would move through this duct in order to clear out the entire interior volume in a time of 15 minutes.