Calculating Load for a Mini Wind Tunnel?

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Discussion Overview

The discussion revolves around calculating the effects of a propeller load on the RPM, horsepower (HP), and torque of a mini wind tunnel powered by an air wrench motor. Participants explore the implications of using a specific motor and propeller configuration, addressing theoretical and practical aspects of the setup.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks to understand how the weight of the propeller affects the motor's HP, torque, and RPM.
  • Another participant argues that the mass of the propeller does not affect HP, torque, and RPM at constant speed but may influence the time taken to reach full speed.
  • Concerns are raised about the feasibility of using a 1/2" diameter shaft to handle 750 ft-lbs of torque without failure.
  • Some participants suggest that the torque rating provided for the air wrench may refer to peak torque in impact mode rather than continuous torque during operation.
  • There is a discussion about the relationship between torque, RPM, and horsepower, with some participants providing calculations to illustrate their points.
  • Participants question whether the air wrench can deliver the specified torque at the maximum RPM simultaneously, suggesting that the actual power output may be lower than claimed.
  • One participant mentions the need for a significant air supply to maintain the desired horsepower and torque, indicating that the compressor's power requirements must be considered.
  • There is a mention of gear reduction effects, where torque may be multiplied while RPM is reduced, but horsepower remains constant.
  • Another participant expresses uncertainty about the application of the horsepower formula in this context, suggesting that the motor's specifications may not align with expected performance.

Areas of Agreement / Disagreement

Participants express differing views on the torque and horsepower capabilities of the air wrench motor, with no consensus reached on the actual performance metrics. The discussion remains unresolved regarding the feasibility of achieving the desired torque and RPM with the proposed setup.

Contextual Notes

Participants highlight potential limitations in the specifications of the air wrench, including the distinction between peak and continuous torque, and the implications of gear reduction on performance. There are also concerns about the air supply's capacity to sustain the required power levels.

questionguy
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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
 
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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 8-seater aircraft, let alone a mini wind tunnel.
 
It's going to be tricky finding a 1/2" diameter shaft which can handle 750 ft-lbs of torque without turning into a pretzel. I didn't realize there were 1000 hp air wrenches available.
 
750 ftlbs torque - I suspect that's the peak torque in impact mode not continuous rotation.

http://www.milwaukeetool.com/press-releases/cordless-m18-fuel-htiw
 
Last edited by a moderator:
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
 
CWatters said:
750 ftlbs torque - I suspect that's the peak torque in impact mode not continuous rotation.

http://www.milwaukeetool.com/press-releases/cordless-m18-fuel-htiw
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
 
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SteamKing said:
It's going to be tricky finding a 1/2" diameter shaft which can handle 750 ft-lbs of torque without turning into a pretzel. I didn't realize there were 1000 hp air wrenches available.
Yup I'm planing on using titanium shafts
 
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.
 
anorlunda said:
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.
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
Are you sure it can deliver 750 ft-lb while spinning at 9500rpm? If they rate it at 750ft-lb 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
cjl said:
Are you sure it can deliver 750 ft-lb while spinning at 9500rpm? If they rate it at 750ft-lb 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).
Yeah I'm sure it takes a ton of air...and it does so with a planetary gear system...
 
  • #12
questionguy said:
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

I would think the 750 ft-lbs is at the tool head which does not spin at 9500 rpm. If the tool head spins at 9.5 rpm then your horsepower is more like 1.3 hp.
 
  • #13
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 ft-lbs available at the head. Neglecting small frictional losses, the power available at the head would still be about 1 HP.
 
  • #14
SteamKing said:
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 ft-lbs available at the head. Neglecting small frictional losses, the power available at the head would still be about 1 HP.
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
questionguy said:
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

Power in = Power out
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
256bits said:
Power in = Power out

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.
It was just clarified that the hp was
a lot less(1-5hp) 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
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
SteamKing said:
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.
I know a big circle but well worth your thoughts and input...thanks
 

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