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Calculating Load for a Mini Wind Tunnel? 
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#1
Dec2113, 05:49 PM

P: 8

Hey guys, so I'm building a mini wind tunnel and I would like to calculate the new RPM,HP and Torque since the propeller will act as a load and change the specs of the motor(note I'm using a air wrench as the motor as it's all I had around and have disabled the impact motion) the current specs are 1,000HP 750ftlbs torque and 9500rpm the shaft diameter is 1/2 in and the propellers. Weight is 9lbs and is 45in in diameter I'm looking to find out how the propellers weight will affect HP,Torque and RPM....thanks



#2
Dec2113, 06:28 PM

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The mass of the prop won't affect the HP, Torque, and RPM at all, running at constant speed. More mass might take a bit longer to run up to full speed, but probably not enough to bother about.
More mass would change the rotordynamics of the prop shaft, if you are bothered about the shaft whirling (and prop induced reverse whirl) in your running speed range. (And you should be bothered, if only to be sure they will never happen). But you need a sanity check on your numbers. AT 9500 RPM, the blade tips of a 45 inch diameter prop would be traveling at well over the speed of sound in air. That's not what you want for a "mini" wind tunnel!! And your 1000HP motor would be more than enough to power an 8seater aircraft, let alone a mini wind tunnel. 


#3
Dec2113, 10:27 PM

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It's going to be tricky finding a 1/2" diameter shaft which can handle 750 ftlbs of torque without turning into a pretzel. I didn't realize there were 1000 hp air wrenches available.



#4
Dec2213, 03:24 AM

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Calculating Load for a Mini Wind Tunnel?
750 ftlbs torque  I suspect that's the peak torque in impact mode not continuous rotation.
http://www.milwaukeetool.com/pressr...m18fuelhtiw 


#5
Dec2213, 04:18 AM

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Also, impact wrenches are designed such that they don't deliver continuous torque. They spin up to speed and then engage the chuck for a short duration, releasing and engaging repeatedly. It's the hammering effect which puts the final torque on the bolt or nut when tightening, or releasing the nut or bolt when reversing. If you are building a wind tunnel, you want a prime mover which is capable of delivering continuous torque for an extended duration.
http://en.wikipedia.org/wiki/Impact_wrench 


#6
Dec2213, 10:18 AM

P: 8

Smart guy but I'm using a air powered wrench not electric a little different the electric has peak torque to save energy but the IPM (impacts per minute) are different from RPM so it's true the peak torque but that is continuous because it's separate from IPM in true air powered wrenches 


#7
Dec2213, 10:27 AM

P: 8

Yup I'm planing on using titanium shafts 


#8
Dec2213, 11:00 AM

P: 195

Better check your numbers. If it really is 1000 hp pneumatic, then you need at leat 1000 hp to run the air compressor. 1x400 plus 2x300 hp Diesel engines for example.
For a hand held tool, 1 hp sounds more reasonable. 


#9
Dec2213, 03:11 PM

P: 8

HP=torque x RPM \ 5252 The math says so 750ftlbs torque x 9500rpm /5252= 1,300hp factor In load so about 1000hp so I'm not really sure.... It is pretty strong so I'm not doubting it plus the energy is stored in compressed air tanks so a large engine is not required 


#10
Dec2213, 08:46 PM

P: 1,017

Are you sure it can deliver 750 ftlb while spinning at 9500rpm? If they rate it at 750ftlb peak torque, and 9500rpm max RPM, chances are it can't do both at the same time (and the power level is actually quite a lot lower than that).



#11
Dec2213, 08:47 PM

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Yeah I'm sure it takes a ton of air.....and it does so with a planetary gear system... 


#12
Dec2313, 08:52 AM

P: 1,475




#13
Dec2313, 09:00 AM

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And if the planetary gear is arranged to slow down the speed of the head relative to the the motor, this gear reduction is what multiplies the modest motor torque to the torque of 750 ftlbs available at the head. Neglecting small frictional losses, the power available at the head would still be about 1 HP.



#14
Dec2313, 10:01 AM

P: 8

Yup I just checked the math u r right but I was referencing free spin and trying to factor in the load I only need 150hp(figurative) or 125ftlbs torque and 6000rpm if the 1hp air motor can deliver this at a good CFM rate il use it....I'm guessing the planetary gear reduction won't kill to much of the rpm as I need 6000rpm to achieve my wind tunnels projected airflow...thanks I'm open to any suggestions also 


#15
Dec2313, 11:39 AM

P: 1,475

So whatever hp you air motor produces is not multiplied by gear reduction. At the output, torque and speed of rotation of the head can be varied but not the hp. By the way, have you figured out how much cfm through an air motor will produce 150 hp. Your tank is not going to last a very long time. As someone mentioned previously, the motor for the compressor for the air supply needs a comparable hp. 


#16
Dec2313, 11:55 AM

P: 8

It was just clarified that the hp was a lot less(15hp) but that the torque and rpm are still consistent (I honestly don't know why HP=torque x RPM /5252 must not apply here or something) and the CFM at max free spin Load is 6CFM @90psi my carbon fiber air tanks are just enough to supply this motor for 15min since they are 90CF @4500psi(with psi regulator valve) any suggestions would be great do you think I can achieve 125ftlbs torque at 6000RPM with the air motor since the propeller load is 10lbs the motors "specs" say 750ftlbs torque and 9500rpm no load so any suggestions would be good thanks. 


#17
Dec2313, 01:28 PM

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Power is proportional to the product of torque and the RPM of the shaft delivering the torque. A gear with a ratio of 2:1 input:output means that an input shaft running at 100 RPM will drive an out put shaft running at 50 RPM (2 revolutions of the input shaft produce 1 revolution of the output shaft). At the same time as the shaft speed is reduced by the gear, the torque on the input shaft is multiplied by a factor of 2, so that the torque on the output shaft is now twice the torque which went into the gear.
By looking at the output side of the gear, we see that the output HP is proportional to the product of the output torque and the output shaft speed, which is (1/2)*Input RPM * (2*Input Torque), or proportional to (Input RPM)*(Input Torque), which is where we started. 


#18
Dec2313, 01:42 PM

P: 8

I know a big circle but well worth your thoughts and input...thanks 


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