Good idea? Bad idea? Cheap Blower intake.
01-16-2008, 03:51 AM
#24
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the reason I say this is they have taken the time to look back 3 or 4 years they should have read every thread that WOULD COVER THE BASICS like what lift and what tires what MPG should I get for a 4x4 the infamous clunk or creek from the front of my truck when I turn the steering wheel full right or left... and the list go's on and on
01-16-2008, 04:38 AM
#25
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There are videos of people doing this all over the 'net
http://youtube.com/watch?v=xwkSr60fk0g
http://youtube.com/watch?v=xwkSr60fk0g
01-16-2008, 02:55 PM
#26
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with at least 4 cylinders, you'll always have at least one cylinder in the intake stroke (of course this assume equal spacing in the firing). if the air was massless, then you may be able to notice different speeds because the cylinder will suck air in at different rates in the inake stroke (very slow at the ends of the stroke, and fastest in the middle). but since the air has mass, it will be flowing at mostly the same speed all the time. it's interesting to note, that on the 22re all four injectors are fired at the same time and the fuel just 'sits' in the intake until the valves open. (in the 3vze there the 3 injectors on each side are fired at the same time). so this leads me to beleive, that even on the scale of one cylinder, the air in the intake will be flowing at a constant rate (otherwise the fuel would probably fall due to gravity).
here's a quick and dirty calculation to show that the air most likely is moving at a constant speed.
assumptions: 2.4L 22re at 3000 rpm (numbers will be the same order of magnitude for just about any engine running at a normal speed)
2.4L @ 3000 rpm means that the engine is sucking in 120L/s of air.
120L/s of air is 120,000 cm^3 of air. The cross section of the intake is about 100cm^2 so that means that the air is flowing in the intake at about 1200cm/s (27mph!)
lets make the assumption that the air is stopped and then needs to be accelerated to 1200cm/s in the time of 1 intake stroke and calculate the forces required to do that.
3000rpm=50hz or only .02sec per revolution. however, only 1/4 of the time is intake so that's realy .005sec for the whole intake time.
so we need to calculate how much force is required to accelerate 120L of air to 1200cm/s in .005sec
120L of air is about 80g (.080kg)
the acceleration is 1200cm/s/.005s or 24000cm/s^2 or 240m/s^2
F=ma = .080gk*240m/s^2 = 192N or about 43 pounds
what all of this means is that if the air needed to be accelerated all the time that the engine would have apply 43 pounds of force on the air (and then when the air stopped it would put 43 pounds of force on the engine). in my 22re the intake is mounted across the front of the engine, that means that there would be 43 pounds of force going back and forth sideways on the truck. i think you could imagine the kind of violent shaking that would occur if you pushed sideways on the truck with 43 pounds of force 3000 times per minute.
i think that this shows that it's very likely that the air is at a constant flow.
here's a quick and dirty calculation to show that the air most likely is moving at a constant speed.
assumptions: 2.4L 22re at 3000 rpm (numbers will be the same order of magnitude for just about any engine running at a normal speed)
2.4L @ 3000 rpm means that the engine is sucking in 120L/s of air.
120L/s of air is 120,000 cm^3 of air. The cross section of the intake is about 100cm^2 so that means that the air is flowing in the intake at about 1200cm/s (27mph!)
lets make the assumption that the air is stopped and then needs to be accelerated to 1200cm/s in the time of 1 intake stroke and calculate the forces required to do that.
3000rpm=50hz or only .02sec per revolution. however, only 1/4 of the time is intake so that's realy .005sec for the whole intake time.
so we need to calculate how much force is required to accelerate 120L of air to 1200cm/s in .005sec
120L of air is about 80g (.080kg)
the acceleration is 1200cm/s/.005s or 24000cm/s^2 or 240m/s^2
F=ma = .080gk*240m/s^2 = 192N or about 43 pounds
what all of this means is that if the air needed to be accelerated all the time that the engine would have apply 43 pounds of force on the air (and then when the air stopped it would put 43 pounds of force on the engine). in my 22re the intake is mounted across the front of the engine, that means that there would be 43 pounds of force going back and forth sideways on the truck. i think you could imagine the kind of violent shaking that would occur if you pushed sideways on the truck with 43 pounds of force 3000 times per minute.
i think that this shows that it's very likely that the air is at a constant flow.
Ok maybe that was overly critical.
01-16-2008, 04:04 PM
#28
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with at least 4 cylinders, you'll always have at least one cylinder in the intake stroke (of course this assume equal spacing in the firing). if the air was massless, then you may be able to notice different speeds because the cylinder will suck air in at different rates in the inake stroke (very slow at the ends of the stroke, and fastest in the middle). but since the air has mass, it will be flowing at mostly the same speed all the time. it's interesting to note, that on the 22re all four injectors are fired at the same time and the fuel just 'sits' in the intake until the valves open. (in the 3vze there the 3 injectors on each side are fired at the same time). so this leads me to beleive, that even on the scale of one cylinder, the air in the intake will be flowing at a constant rate (otherwise the fuel would probably fall due to gravity).
here's a quick and dirty calculation to show that the air most likely is moving at a constant speed.
assumptions: 2.4L 22re at 3000 rpm (numbers will be the same order of magnitude for just about any engine running at a normal speed)
2.4L @ 3000 rpm means that the engine is sucking in 120L/s of air.
120L/s of air is 120,000 cm^3 of air. The cross section of the intake is about 100cm^2 so that means that the air is flowing in the intake at about 1200cm/s (27mph!)
lets make the assumption that the air is stopped and then needs to be accelerated to 1200cm/s in the time of 1 intake stroke and calculate the forces required to do that.
3000rpm=50hz or only .02sec per revolution. however, only 1/4 of the time is intake so that's realy .005sec for the whole intake time.
so we need to calculate how much force is required to accelerate 120L of air to 1200cm/s in .005sec
120L of air is about 80g (.080kg)
the acceleration is 1200cm/s/.005s or 24000cm/s^2 or 240m/s^2
F=ma = .080gk*240m/s^2 = 192N or about 43 pounds
what all of this means is that if the air needed to be accelerated all the time that the engine would have apply 43 pounds of force on the air (and then when the air stopped it would put 43 pounds of force on the engine). in my 22re the intake is mounted across the front of the engine, that means that there would be 43 pounds of force going back and forth sideways on the truck. i think you could imagine the kind of violent shaking that would occur if you pushed sideways on the truck with 43 pounds of force 3000 times per minute.
i think that this shows that it's very likely that the air is at a constant flow.
here's a quick and dirty calculation to show that the air most likely is moving at a constant speed.
assumptions: 2.4L 22re at 3000 rpm (numbers will be the same order of magnitude for just about any engine running at a normal speed)
2.4L @ 3000 rpm means that the engine is sucking in 120L/s of air.
120L/s of air is 120,000 cm^3 of air. The cross section of the intake is about 100cm^2 so that means that the air is flowing in the intake at about 1200cm/s (27mph!)
lets make the assumption that the air is stopped and then needs to be accelerated to 1200cm/s in the time of 1 intake stroke and calculate the forces required to do that.
3000rpm=50hz or only .02sec per revolution. however, only 1/4 of the time is intake so that's realy .005sec for the whole intake time.
so we need to calculate how much force is required to accelerate 120L of air to 1200cm/s in .005sec
120L of air is about 80g (.080kg)
the acceleration is 1200cm/s/.005s or 24000cm/s^2 or 240m/s^2
F=ma = .080gk*240m/s^2 = 192N or about 43 pounds
what all of this means is that if the air needed to be accelerated all the time that the engine would have apply 43 pounds of force on the air (and then when the air stopped it would put 43 pounds of force on the engine). in my 22re the intake is mounted across the front of the engine, that means that there would be 43 pounds of force going back and forth sideways on the truck. i think you could imagine the kind of violent shaking that would occur if you pushed sideways on the truck with 43 pounds of force 3000 times per minute.
i think that this shows that it's very likely that the air is at a constant flow.
Youre assuming the intake volume is the total displacement on each stroke.. But the displacement is the sum of all four cylinders. So on each stroke, only one cylinder fills with air (2.4l/4=0.6l).. So on each stroke, the engine intakes 0.6l.
And rpm is revolution of the crank, so the frequency of intake isnt 1/4 of the time of revolution, but 1/2 (one up/down movement of the piston) still only 100/sec at 3000 rpm.
I always remember; suck (half revolution; stroke down), squish (half revolution; up), bang (half revolution; stroke down), blow (half revolution; stroke up). Each one of four of the cylinders is doing one action at one time.
So the correct volume calculation would be;
V/s=(2.4l/4)[(3000rpm/60)*2]
V/s=60l/s
Using your estimated values for the mass of air and intake length..
m=60l*0.08kg/120l
m=0.040kg
a=(600cm/s)/(0.01s)
a=60m/s^2
F=ma
F=(0.040kg)*(60m/s^2)
F=24N [or about 5.4 pounds]
...and honestly, there's no these calculations even with the hypothetical estimations would even come reasonably close to the actual force or intake volume due to variables like volumetric efficieny, haha.
Last edited by ornery; 01-16-2008 at 04:08 PM.
01-16-2008, 07:24 PM
#30
Registered User
That would be a good Mythbusters experiment whether it would actually work or not. Here is the link to the forum for Mythbusters:
http://community.discovery.com/eve/f...m/f/2991937776
The leaf blower turbo myth has already been posted more than once, but has never made the show to my knowledge.
By the way, if anyone is thinking about trying this on their daily driver, I would recommend against it. Most leaf blowers have a plastic fan, and airflow from a vehicle is likely to be way more than the leaf blower puts out, which will be enough to spin the fan beyond it's limits and either two things will happen:
1. It will break the fan.
2. Vacuum will spin the fan against the crankshaft of the blower, which will apply torque to the bolt opposite of the way it was designed and loosen it.
Either way, parts are going to get sucked in your intake, not to mention all of the dust from it being used as a leaf blower.
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