I have a super old DeVilbiss air compressor. It's about 6 feet tall, has a motor rated at 1.5 HP (but probably weights 100 pounds!) and has a gigantic 2-stage compressor (probably 100 pounds, too!).

The tank is 3 feet tall (plus the spherical caps) and is 64 inches around. Based on my math, that would put it at around 80 cubic feet volume. Check me on that.

I have read that you convert cubic feet to gallons by dividing by 7.48 (I admit that doesn't make any sense to me--maybe it's multiply, but that doesn't make sense either). That gives around 11 gallons. The tank obviously holds way more than 11 gallons (of water, for example). A similar size compressor at Home Depot is labeled as holding more like 60 gallons. Neither way of moving from cubic feet to gallons gives a number that makes any sense.

Where's the disconnect?

Thanks,
Jay

Using the equation I came up with at this thread:

viewtopic.php?f=3&t=15196&start=0

0.0003445 * 64 * 64 * 36 = 51 gallons, plus the end cap half spheres.

You have a 60 gallon tank there

Using your approach: circumference of 64" means radius of 10.2". So 0.85 feet.
Height of 3 feet, so volume of that cylinder is 6.81 cubic feet. Then you multiply by 7.48 (not divide) -- so you get 51 gallons. Add the end cap half-sphere volumes and you will end up with 60 gallons.

The motor worries me. Size/weight means it is built well but we need to go by the spec plate as stated HP rating. A 2-stage pump matched with the correct motor will give you maybe a max of 4 CFM per HP (theoretical limit), so at 1.5HP it will be underpowered for auto body work, unfortunately.

So the limitation of a compressor is really based on how quickly the compressor component can replenish the tank? If I'm not doing large swaths of work at one time and the tank can be replenished in between, would it work okay in that instance?

I found some article that talks you through figuring out the actual CFM of your rig. I may try doing that, just for fun (and to see if I can paint!).

And thanks for the help with the math!

Jay

Sort of - the limitation of the compressor is how much CFM the pump can deliver at the required pressure. The tank is meant to relieve the duty-cycle of the motor and provide a collection point for water that condenses as the air cools coming out of the pump head.

A 60 gallon tank is 8 cubic feet. You won't be able to use all that volume since it won't deliver the entire tank at the required PSI. So figure half... 4 cubic feet. That might be around 30 seconds of time with a spray gun.

If I test my compressor and discover that it doesn't deliver the requisite CFM for my needs, isn't it really a matter not of more HP, but of spinning the compressor faster? That may be splitting hairs, but you get the point. In other words, if the motor had the oomph, wouldn't putting a larger pulley on the motor solve the problem? Of course, this is assuming the motor has the HP to do that. How do I know what RPM the compressor component can handle? Like I said, it's a jumbo, two-stage unit, and also has a jumbo oil-filled crankcase (I don't know if that helps you answer the question). If I recall, the motor spins at 1740 RPM. I'm guessing the current reduction is 3:1 or 4:1. Worst case, I guess I could just stick a new motor on it. Then I should be in business, right? (assuming I'm right about the RPM part of it and the compressor can handle the greater RPM)

Thanks for bearing with my attempt to understand this.

Jay

By the way, for anyone who is interested in what was wrong with my math when trying to figure out the capacity of my tank, here it is.

• I measured the height of the tank as 3 feet (not including end caps).
• I measured around the tank and got 64 inches.
• I divided that by pi to get diameter = 20 inches.
• I squared the radius (10 inches) and multiplied by pi to get area of the cross section of the tank = 314 square inches.
• I multiplied that by 36 to get volume of the tank (sans end caps) = 11,304 cubic inches.
• I divided by 144 (12 squared) to convert to cubic feet. D'oh!
• I needed to divide by 1,728 (12*12*12) to convert from cubic inches to cubic feet. That gives 6.54.
• Then multiply by 7.48 to convert from cubic feet to gallons (I still have no idea how 1 cubic foot = seven-and-a-half gallons—a gallon of milk looks like about a cubic foot to me! Probably something to do with pressure.)

And there you have it. Three cheers for math!

Jay

I know, the idea that you can fit seven gallon containers of milk in a one cubic foot box just seems wrong...but it's not. A gallon of milk is around 9 inches high by just under 6 inches wide. Also, it is tapered at the top.

Anyway, regarding your motor question. It is both RPM and horsepower (torque, really).

As the air in the tank is compressed, the pump needs to work harder. When the pump needs to work harder it's going to put more of a load on the compressor.

Instead of a larger pulley on the motor, think about a smaller pulley on the pump - it'll give the same result but make it easier to show you my point. If I asked you to turn a bolt with a socket wrench, and that bolt offered 100 ft/lbs of resistance, and I gave you a choice of a 2' long wrench or a 1' long wrench, which would you take? Right - the longer one. The less mechanical leverage you have, the harder you need to work. A pulley is nothing more than a lever, to the motor.

So you could find that the 1.5 HP motor will simply stall at a certain PSI threshold. Or just overheat and burn up.

Regarding the pump, it may or may not tolerate a 2x speed increase. Also you'd need to know the displacement of the pump (piston surface area times travel distance) to figure out how much air volume it can move with each rotation of the pump. If that math works then maybe a 5HP motor along with a pulley change would do it.

That's a lot of rope to pull in and I think you are dealing with too many unknowns.

I got the compressor for free (just had to haul it away), so I may do some experimenting. I'll let you know what happens.

Thanks for your help, Chris.

Jay