Video: Determining the Speed of Airflow from Its Volumetric Flow Rate

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

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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 house’s interior every 15 minutes?

In this problem, we’re looking to solve for the speed of air. And we’ll represent that using the symbol 𝑣 sub 𝑎. And we’re 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. We’re also told the diameter of the main air duct connected to the heater. Using this information along with the cycle time of 15 minutes, let’s solve for 𝑣 sub 𝑎.

To start, we can draw a diagram of the situation. Here is a sketch of what the inside of the house might look like. We’ve been given a height, a width, and a length and told what those values are. And we’re 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.

Let’s 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 we’re told that as air passes through this pipe on its way to the heater, it moves with a velocity we’ll call 𝑣. It moves with a speed that we’ve called 𝑣 sub 𝑎 such that it’s able to empty out the entire volume of the house interior every 15 minutes. Let’s label that time 𝑡 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. We’re told that the diameter is 0.300 metres. We’ll call that diameter 𝑑 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 𝑡 sub cycle every 15 minutes.

Now, as we look to solve for 𝑣 sub 𝑎, the speed of the air in the pipe, let’s look at the pipe in an expanded view. If this is our pipe with air moving through it at a certain speed we’re 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. We’ll call that capital 𝐴.

Now notice this about capital 𝐴, the cross-sectional area of our pipe, and 𝑣 sub 𝑎, if we take their product — that is if we multiply them together — then we would wind up with a value. We’re not sure what that value is quite yet. But whatever the value is, it would have units of volume per time. That’s because capital 𝐴 has units of area and the speed 𝑣 sub 𝑎 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 we’ve 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 𝑡 sub cycle — the time it takes for the entire interior volume of air to be cycled out.

As we look at this equation, we realize we’re able to solve for or are given each one of the variables in this equation other than 𝑣 sub 𝑎. Capital 𝐴, the cross-sectional area of the tube, we can solve for using the tube diameter. The height 𝐻, width 𝑊, and length 𝐿 of the interior of the building are all given and 𝑡 sub cycle is given as well.

So let’s move towards solving for 𝑣 sub 𝑎, the speed of the air in the duct, by replacing capital 𝐴, the cross-sectional area of our tube in terms of values we are given. When we recall that the area of a circle equals 𝜋 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 𝐴 — as we’ve called it — is equal to 𝜋 times the diameter divided by two and that quantity squared. When we square this term 𝑑 over two, we get 𝑑 squared divided by four.

Now, let’s isolate 𝑣 sub 𝑎, 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 𝜋𝑑 squared. Doing this cancels out the four, 𝑑 squared terms and 𝜋 terms on the left side of our equation. So we can write a simplified version of this equation with 𝑣 sub 𝑎 all by itself. 𝑣 sub 𝑎 is equal to the height times the width times the length of the interior of our building times four divided by 𝜋 times the diameter of our duct squared multiplied by the time it takes to cycle the air in and out.

Let’s plug in for the different values in this equation. In the numerator, 𝐻 is 2.75 metres, 𝑊 is 13.0 metres, and 𝐿 is 20.0 metres. Then, in the denominator, 𝑑, the diameter of our duct, is 0.300 metres. 𝑡 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 let’s do a quick conversion.

We’ll create a variable and call it 𝑡 sub 𝑠. That’s the time that this cycling process takes in units of seconds. To convert 𝑡 sub cycle into units of seconds, we’ll need to multiply that number by 60 seconds per minute. So 𝑡 sub 𝑠 is equal to 60 seconds per minute multiplied by 𝑡 sub cycle. Plugging in the value for 𝑡 sub cycle of 15 minutes, when we multiply these two terms together, we find that 𝑡 sub 𝑠, the time it takes to empty all the air in this space, is 900 seconds.

Now, remember this time is the same value as 𝑡 sub cycle, but simply in a different set of units. So we can substitute 𝑡 sub 𝑠 or 900 seconds in for 𝑡 sub cycle in our equation. Now, we’re ready to calculate out what is 𝑣 sub 𝑎. 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 we’ll get a final answer in units of metres per second.

This is encouraging because we’re solving for a speed 𝑣 sub 𝑎 and expected to have units like that. So it seems we’re on right track. Let’s calculate out what this number is. And we find that 𝑣 sub 𝑎 is equal to 11.2 metres per second. That’s 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.