Re: Putting a fan on top of air intake?
Crufty Dusty said:
Hmm... I would disagree because you're assuming that the fluid is incompressible, fluid velocity is constant throughout and that the fan blades have no inertia.
Moving greater mass at pressure through the same pipe, you are compressing the air. Obviously air is compressible. Primary school kids learn that in science class, so why would I assume that it isn't?
Fluid velocity is contant from inches downstream of the fan, to inches before the TB. Of course when the blades kick on or off, there is spring effect as the compression starts or decompression starts but this is momentary. I am not sure why you are trying to obfuscate these simple things which make extremely little difference in how much is sent back out the MAF. The raw number shown has already proved this. The small wave effects within the air column when it is abruptly trapped, are irrelevant at these velocities, pressures. There is no 100 CFM / CF to be backed out through the MAF. I have shown so.
Let's say the fan is 10cm (0.1m) in diameter. Circumference = 0.1*pi m. It is spinning at 4600 RPM or 76 revs/sec. Thus the angular velocity is 0.1 * pi * 76 = 23.8716 m/sec. Let's say a split second is 0.1 sec. Deceleration = 23.8716/0.1 = 238.716 m/sec^2 or 24G's. I'm not sure if the fan can withstand such repeated deceleration.
G is half the picture. To complete the picture and find actual loads, you need component mass. 24G is nothing to a low mass fanblade. 24 G applied to large mass amounts to great loads, but the fanblade is not a large mass. Also note I said in earlier post 0.1-0.5 seconds. You are taking the extreme short end. Also, your estimate of 10cm diameter is high. Bilge blowers are not that large in diameter. Computer fans I am not interested in involving, as they have been shown to be lower pressure, lower volumetric flow units, vs bilge blowers.
24G applied to 2gram reed valve is a 48gram load.
24G applied to 30gram impeller is a 720gram load.
24G applied to 500gram piston is a 12,000gram load.
The shaft in a blower is not going to have to support the rough 700 gram load radially since balanced rotation cancels out, but there will be a moment on the shaft. Moment will be reduced somewhat for mass distribution reasons listed above. Even assuming the huge 10cm dia you started out with, 700grams at 10cm makes for a very low moment. approx 0.515 lb-ft torque. A toothpick can take that kind of torque. Even this 0.515 lb-ft figure is on the high side, since mass is distributed not just on the very circumference of the blade, but also all the way along it to the center. To keep things simple, we have assumed that all mass is concentrated on the circumference, artificially raising the torque the shaft has to deal with. Still the force is tiny.
On top of this, shaft speed does not need to reach zero in order for it to stop pressurizing. There is a threshold shaft speed below which there is no meaningful amount of air moved. This is the same for non-positive-displacement blowers whether they be radial, axial, or anything in between. This is demonstrated clearly in turbochargers and centrifugal superchargers that require very high shaft speeds. The first few thousand RPM in those systems produce insignificant amounts or no boost.
So take decel time at 0.25 seconds to reach say 15% of max shaft speed, at a realistic 3.0 inch impeller diameter, consider mass distribution from center of rotation.......and the above mentioned 0.515 lb-ft torque, being miniscule to begin with, is reduced to near nothing.