Tom King Posted January 23, 2019 Report Share Posted January 23, 2019 I'm going to change the motor on my 10hp 3 phase compressor to a 7-1/2 hp single phase. I bought a VFD for it, but after running it some, I'm pretty sure that neither the VFD, nor the motor will have a long life. I'll just keep the VFD as a spare for the big bandsaw, or some future use. The 7-1/2 will provide plenty of air for anything I'll use it for. 10 hp is just too complicated to do electrical for at different jobsites, and the service I have to the building where it's used now will be asked to do more pretty soon for an addition. I'm having a hard time figuring out how to downsize the pulley on the motor. Should the difference be figured simply by the difference in diameter of the pulley? I'm probably just too tired to think about it clearly, but if anyone knows right off, without guessing, it would be greatly appreciated. Quote Link to comment Share on other sites More sharing options...
Chestnut Posted January 23, 2019 Report Share Posted January 23, 2019 Yes Diameter radius or circumference they are all directly related so the ratio between all of them will be the same. Quote Link to comment Share on other sites More sharing options...
Tom King Posted January 23, 2019 Author Report Share Posted January 23, 2019 So it's just as simple as going to 3/4 the size of the pulley on the 10 hp? Quote Link to comment Share on other sites More sharing options...
wtnhighlander Posted January 23, 2019 Report Share Posted January 23, 2019 Drew, correct me if I'm out of line, but going from 10 to 7.5 hp is 25% reduction. So Tom should increase the ratio of the pulleys by 25%, not just reduce the drive pulley, correct? Quote Link to comment Share on other sites More sharing options...
Chestnut Posted January 23, 2019 Report Share Posted January 23, 2019 Oh ahh that might be a bit more complicated.... Give me a few min need to think about that Quote Link to comment Share on other sites More sharing options...
Chestnut Posted January 23, 2019 Report Share Posted January 23, 2019 I get a 33% increaise in motor pully..... trying to figure out why... Quote Link to comment Share on other sites More sharing options...
Tom King Posted January 23, 2019 Author Report Share Posted January 23, 2019 I knew I was too tired to think clearly about it tonight. Quote Link to comment Share on other sites More sharing options...
Tom King Posted January 23, 2019 Author Report Share Posted January 23, 2019 Google says this pump runs a 6-3/8" pulley with a 10hp motor, and 8-1/8" with a 15 hp. I did the same thing with a 5hp down to a 3hp, for similar reasons, but that was over 40 years ago. Quote Link to comment Share on other sites More sharing options...
Chestnut Posted January 23, 2019 Report Share Posted January 23, 2019 OK i got it. You have 75% the power. The inverse of 75% is 1.3333 which is how much bigger the smaller pulley needs to be so the torque on the compressor is the same but the rotation is slower. Quote Link to comment Share on other sites More sharing options...
Chestnut Posted January 23, 2019 Report Share Posted January 23, 2019 crap i have this backwards don't i? Quote Link to comment Share on other sites More sharing options...
Tom King Posted January 23, 2019 Author Report Share Posted January 23, 2019 I wasn't too tired to see that. Quote Link to comment Share on other sites More sharing options...
Tom King Posted January 23, 2019 Author Report Share Posted January 23, 2019 I get 4.78 inches, if indeed the one on the 10hp is 6-3/8". I haven't even measured it yet. I just decided to do this tonight. Quote Link to comment Share on other sites More sharing options...
Chestnut Posted January 23, 2019 Report Share Posted January 23, 2019 The difference is 33% i know that for sure. Quote Link to comment Share on other sites More sharing options...
Tom King Posted January 23, 2019 Author Report Share Posted January 23, 2019 Probably this one: https://www.zoro.com/tb-woods-v-belt-pulley-1-18fixed-425od-iron-2bk45118/i/G2977737/?q= Thanks for your help. I'll think about it another day. Quote Link to comment Share on other sites More sharing options...
wtnhighlander Posted January 23, 2019 Report Share Posted January 23, 2019 1 hour ago, Chestnut said: The difference is 33% i know that for sure. 33% of what? Reduction ratio, small pully diameter, large pully diameter? None of the above? In any case, reducing the motor pully to allow the 7.5 hp motor torque to match the load of the compressor will reduce the rpm of the compressor, and the tank will take longer to fill. Assuming the synchronous speed of the 7.5 hp 1 phase motor is the same as the 10 hp 3 phase... Quote Link to comment Share on other sites More sharing options...
Chestnut Posted January 23, 2019 Report Share Posted January 23, 2019 8 hours ago, wtnhighlander said: 33% of what? Reduction ratio, small pully diameter, large pully diameter? None of the above? In any case, reducing the motor pully to allow the 7.5 hp motor torque to match the load of the compressor will reduce the rpm of the compressor, and the tank will take longer to fill. Assuming the synchronous speed of the 7.5 hp 1 phase motor is the same as the 10 hp 3 phase... I think i said above either the pully needs to be 33% bigger or smaller. They are measured in diameter right? I just had it backwards..... Quote Link to comment Share on other sites More sharing options...
Tom King Posted January 23, 2019 Author Report Share Posted January 23, 2019 Thinking more clearly this morning, after a good nights sleep, it's much more complicated than a simple ratio. The power required to spin the pump goes down, non-linear, as the pump rpm's go down. With the given sizes of drive pulleys for 10, and 15 hp, it will help a lot. There is a 22% drop in pulley size from 15hp, down to 10hp. I expect the required power drop, for the 7-1/2hp motor, will be less than the difference for higher hp motors. This pump is pressure lubricated, so frictional differences in rpm's might not be as much as with a regular pump. So it looks like its a Calculus problem, rather than simple Algebra. My best friend, an Astrophysicist, is also one of the world's best Mathematicians. This is the kind of problem he does for a living, and fun. I'll turn it over to him. Once figured, pulley can be tried the next step up from calculated size, which will probably not land exactly on available pulley sizes, and tried to see how fast the motor gets warm. If it doesn't, it should be good. If it does, I'll go down in pulley size until it runs cool. Quote Link to comment Share on other sites More sharing options...
Tom King Posted January 23, 2019 Author Report Share Posted January 23, 2019 18 hours ago, wtnhighlander said: 33% of what? Reduction ratio, small pully diameter, large pully diameter? None of the above? In any case, reducing the motor pully to allow the 7.5 hp motor torque to match the load of the compressor will reduce the rpm of the compressor, and the tank will take longer to fill. Assuming the synchronous speed of the 7.5 hp 1 phase motor is the same as the 10 hp 3 phase... It will absolutely produce less air volume, but there is no way around that. It still should be a Lot of air. With 10hp it's 38 cfm at 175 psi. I know I'll have to wait for it using a rock drill, but should be plenty to run a 3/4" air wrench fast enough for me. The refrigerated dryer can handle 58 cfm at 100 psi, so that was oversized to start with. I should still have all the dry air I need almost all of the time. I didn't find a comparison chart for pressure lubricated Speedaires, but here's one for splash lubricated IR's: 5 hp 14.0 cfm 60.0 gal Vertical 76 in 48 in 40 in Ingersoll Rand $1,840.61 5 hp 14.0 cfm 80.0 gal Vertical 76 in 48 in 40 in Ingersoll Rand $2,259.82 7 1⁄2 hp 24.0 cfm 80.0 gal Vertical 76 in 48 in 40 in Ingersoll Rand $3,351.62 10 hp 34.1 cfm 120.0 gal Horizontal 54 in 69 in 25 in Gamut Approved $3,305.00 10 hp 35.0 cfm 120.0 gal Horizontal 65 in 83 in 36 in Ingersoll Rand $3,790.40 10 hp 35.0 cfm 120.0 gal Vertical 83 in 51 in 46 in Ingersoll Rand $4,121.03 10 hp 34.1 cfm 90.0 gal Vertical 52 in 69 in 24 in Gamut Approved $3,797.98 15 hp 50.0 cfm 120.0 gal Horizontal 54 in 69 in 25 in Gamut Approved $5,659.84 1 Quote Link to comment Share on other sites More sharing options...
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