Help me help some kids approximate the work done to move air through a tube
March 9, 2012 4:52 AM Subscribe
I need to approximate, very simply and avoiding any mention of fluid dynamics because this is for schoolkids aged 14-16, the forces acting on a volume of air that is being pulled through a tube by a fan. My physics is failing me. Help!
This is for an activity to approximate the power required to move a certain volume of air per second through a large tube. There will be explanatory notes for teachers pointing out that this isn't really how you do it.
My thinking is that considering two forces will do:
If we think of the air moved in one second as a cylinder of air within the tube, the first force is the frictional forces between the surface of this cylinder of air and the inner lining of the tube.
The second force is the resistance of the air behind this cylinder, which we can think of like the head of a piston, but in reverse: there's a slight lowering of pressure and so this acts over the area of the tube to resist the movement of the cylinder of air I'm talking about.
Together these make up the overall force that the fan needs to balance in order to maintain constant motion for the cylinder of air. In which case if that volume of air moves a distance d in one second, the power of the fan is the work done per second, or d x the sum of these forces.
Or would I be better thinking about the kinetic energy of the cylinder of air? And what about the rest of the air moving through the tunnel?
I'm in a dreadful muddle. Help! I don't need a real-world answer, just a simple but not completely ridiculous way of thinking about this. I can't change the scenario.
This is for an activity to approximate the power required to move a certain volume of air per second through a large tube. There will be explanatory notes for teachers pointing out that this isn't really how you do it.
My thinking is that considering two forces will do:
If we think of the air moved in one second as a cylinder of air within the tube, the first force is the frictional forces between the surface of this cylinder of air and the inner lining of the tube.
The second force is the resistance of the air behind this cylinder, which we can think of like the head of a piston, but in reverse: there's a slight lowering of pressure and so this acts over the area of the tube to resist the movement of the cylinder of air I'm talking about.
Together these make up the overall force that the fan needs to balance in order to maintain constant motion for the cylinder of air. In which case if that volume of air moves a distance d in one second, the power of the fan is the work done per second, or d x the sum of these forces.
Or would I be better thinking about the kinetic energy of the cylinder of air? And what about the rest of the air moving through the tunnel?
I'm in a dreadful muddle. Help! I don't need a real-world answer, just a simple but not completely ridiculous way of thinking about this. I can't change the scenario.
Best answer: Very, very simply, there are only two things to worry about here - friction forces of the air against the wall of the tube and the inertia that wants to keep it stationary. If you go any more complicated than that, I think you're going to be in trouble.
Going the kinetic energy route is probably your best bet. Consider the conservation law of energy (cannot be created or destroyed, etc.). The total energy in your plug of air upstream of the impeller can be said to be comprised of its internal energy (i.e., its temperature) and its velocity (kinetic energy). Downstream of the impeller, its total energy is also equal to its internal energy (for the sake of argument let's say it's equivalent to the upstream) and its downstream velocity, which is now faster than upstream. That energy had to come from somewhere, and that somewhere is the impeller.
If you want to make it a little more life-like, you can add some efficiencies (motor is only 75% efficient, impeller is 90% efficient, therefore size of the motor must be larger than the total energy required to speed up the air).
Do you actually need to discuss forces over energy? If this is the case, you only need to worry about the drag from the cylinder walls. Your second "force" doesn't need to be considered, for two reasons. First, assume the whole system is incompressible (it's a good assumption at low airspeeds that you would find in a situation like this). Second, you have an infinite supply of air behind your "air plug" so these vacuum forces aren't going to exist. It's less like a reverse piston and more like a long train of discrete air units all being pulled together. And while we're at it, let's also assume a steady state condition such that the fan isn't really accelerating the upstream air from a dead stop, but rather keeping all the air moving at a constant velocity.
Drag calculation is very simple - D = (0.5)(rho)(U^2)(S)(Cd)
D - total drag force
rho - density of the air
U - airspeed
S - planform area; for this exercise, pick a sufficiently long pipe length (100 ft, say) and determine the surface area of the inside of the pipe. "Sufficiently long" so that your area of interest is all uniform and you're not worrying about inlet and outlet effects
Cd - drag coefficient. You've got a nice smooth pipe without many aerodynamic effects... call it 0.02.
That's really all you have to worry about for this exercise. Is this in any way accurate? Not really, but for a sophomore physics class it'll do.
In summary: if you want to look at work, use kinetic energy. If you actually want the force balance, use the drag calculation.
posted by backseatpilot at 5:30 AM on March 9, 2012 [1 favorite]
Going the kinetic energy route is probably your best bet. Consider the conservation law of energy (cannot be created or destroyed, etc.). The total energy in your plug of air upstream of the impeller can be said to be comprised of its internal energy (i.e., its temperature) and its velocity (kinetic energy). Downstream of the impeller, its total energy is also equal to its internal energy (for the sake of argument let's say it's equivalent to the upstream) and its downstream velocity, which is now faster than upstream. That energy had to come from somewhere, and that somewhere is the impeller.
If you want to make it a little more life-like, you can add some efficiencies (motor is only 75% efficient, impeller is 90% efficient, therefore size of the motor must be larger than the total energy required to speed up the air).
Do you actually need to discuss forces over energy? If this is the case, you only need to worry about the drag from the cylinder walls. Your second "force" doesn't need to be considered, for two reasons. First, assume the whole system is incompressible (it's a good assumption at low airspeeds that you would find in a situation like this). Second, you have an infinite supply of air behind your "air plug" so these vacuum forces aren't going to exist. It's less like a reverse piston and more like a long train of discrete air units all being pulled together. And while we're at it, let's also assume a steady state condition such that the fan isn't really accelerating the upstream air from a dead stop, but rather keeping all the air moving at a constant velocity.
Drag calculation is very simple - D = (0.5)(rho)(U^2)(S)(Cd)
D - total drag force
rho - density of the air
U - airspeed
S - planform area; for this exercise, pick a sufficiently long pipe length (100 ft, say) and determine the surface area of the inside of the pipe. "Sufficiently long" so that your area of interest is all uniform and you're not worrying about inlet and outlet effects
Cd - drag coefficient. You've got a nice smooth pipe without many aerodynamic effects... call it 0.02.
That's really all you have to worry about for this exercise. Is this in any way accurate? Not really, but for a sophomore physics class it'll do.
In summary: if you want to look at work, use kinetic energy. If you actually want the force balance, use the drag calculation.
posted by backseatpilot at 5:30 AM on March 9, 2012 [1 favorite]
On rereading what I just wrote, I think I threw a red herring in there about inertia in the first sentence. I would ignore that.
(I am not your fluid dynamicist.)
posted by backseatpilot at 5:31 AM on March 9, 2012
(I am not your fluid dynamicist.)
posted by backseatpilot at 5:31 AM on March 9, 2012
HVAC people have a bunch of empirical formulas for this — yea much drag (or pressure loss) per foot for air at a certain speed in a certain size duct; yea much loss for each kind of opening, baffle, expansion or reduction. You could use those and add a brief explanation of the lower-level physical forces that give rise to each kind of drag.
posted by hattifattener at 12:33 PM on March 9, 2012
posted by hattifattener at 12:33 PM on March 9, 2012
Well, you have frictional drag acting from the inner wall of the tube, you have a pressure gradient from inside and outside of the tube proportional to the radius of the tube acting as a normal force onto the air, and you have to consider that air is a compressible fluid. Assuming steady state, you should have two components within the tube: an entrance zone and a fully developed zone. Look up laminar Couette flow for a simplified 2D model, and laminar Poiseuille flow for a 3D example of what you're talking about. There should be some good diagrams from university courses, just try and think through all the forces, and if they make sense to you, then you can explain it! Also you might have luck searching for some of these flow models on YouTube, MIT has some REALLY fantastic fluid dynamics videos from the 50s hosted, and I've watched them in a few grad level fluid dynamics courses. Lots of visualization is key, kids get pictures and videos. Hope that makes sense coming from my little iPod touch!
posted by oceanjesse at 12:59 PM on March 9, 2012
posted by oceanjesse at 12:59 PM on March 9, 2012
Best answer: I think the way you present this depends on what tools you're trying to give them. If you're interested in them gaining an understanding of "friction" inside a duct, then I'd head in the direction of the HVAC approach which will lead to you calculating the static pressure in the duct due to the fan/air draw. In order for this to have real meaning you'd have to be looking at a situation with some combination of strong fan/long duct/narrow duct that actually led to some appreciable choking, so they could see the real effect. I'm not sure you can pull that off with e.g. a desk fan and a paper towel tube but I might be wrong.
On the other hand, you could approach this from the "power in flow" direction, which would lead to the opposite set of assumptions (resistance of tube is negligible for the dimensions involved, but does serve to define the flow area) but would give them the tools to understand wind turbines, hydro-power, etc. That approach would lead them to the equation for power in a (incompressible, ideal, blah blah blah) fluid flow:
P = 1/2 * ρ * A * v3
Where ρ is the fluid density, A is the area of flow, and v is the fluid velocity. There are several caveats with this approach too (eg, you can only be looking at situations in which you can safely call air "incompressible," which is somewhat limiting), but it at least gives a first-crack approximation of a very difficult calculation, using no-calculus physics concepts that sophomore and juniors in high school are probably ready for.
(To derive this, you start with Kinetic Energy = 1/2 mv2, extrapolate to Kinetic Energy / volume = 1/2 ρv2, and work from there to figure out how much KE/sec (aka power) is flowing, relying on the fact that in constant situations the volume/second is equal to cross-section area (A) * velocity.)
posted by range at 7:13 PM on March 9, 2012
On the other hand, you could approach this from the "power in flow" direction, which would lead to the opposite set of assumptions (resistance of tube is negligible for the dimensions involved, but does serve to define the flow area) but would give them the tools to understand wind turbines, hydro-power, etc. That approach would lead them to the equation for power in a (incompressible, ideal, blah blah blah) fluid flow:
P = 1/2 * ρ * A * v3
Where ρ is the fluid density, A is the area of flow, and v is the fluid velocity. There are several caveats with this approach too (eg, you can only be looking at situations in which you can safely call air "incompressible," which is somewhat limiting), but it at least gives a first-crack approximation of a very difficult calculation, using no-calculus physics concepts that sophomore and juniors in high school are probably ready for.
(To derive this, you start with Kinetic Energy = 1/2 mv2, extrapolate to Kinetic Energy / volume = 1/2 ρv2, and work from there to figure out how much KE/sec (aka power) is flowing, relying on the fact that in constant situations the volume/second is equal to cross-section area (A) * velocity.)
posted by range at 7:13 PM on March 9, 2012
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posted by Slap*Happy at 5:16 AM on March 9, 2012