I want to build a hydraulic log splitter. The only requirements I have are it has to have a tonnage of 45,000tons. I want to be able to understand how in other splitters they use only a pump capable of 28gpm and max psi of 3,000psi to create the same force. I've done the math. I don't know if I did it wrong or something else is adding tonnage. Any help?
You probably are thinking of 45 tons, not 45,000. The latter is 90 million pounds force. The F-1 engine from the Saturn V booster (the most powerful single chamber rocket engine of all time) creates only 1.5 milllion lbs force. You don't, um, need 90 million pounds force to split a log.
So, 45 tons is (45 x 2000 = ) 90,000 lbs force. You have a compressor that is 3000 psi. That's pounds force per square inch. Force = pressure x area. So area = Force/pressure. That is,
area = Force/pressure = 90,000 lbs force/(3000lbs force/square inch) = 30 square inches.
You need a hydraulic cylinder with a ram that is 30 square inches in cross section. Using pi * r squared, that works out to a little more than a 6 inch diameter ram.
If you truly have 28 gpm at 3000 psi (that's about a 20 horsepower pump), you have 28 gpm * 239 cubic inches/gallon, or 6692 cubic inches/minute. That gives you
a nominal ram speed of 6692 cubic inches / minute /30 square inches, or 232 inches per minute, or 3.7 inches per second. May be faster than you need.
But... A 40 ton log splitter with a 25 gpm pump and a 14 hp Kohler engine can be had for 2 grand from Tractor Supply. Used double acting cylinders in that rod diameter range, from Surplus Center, are alone going for more than 2 grand. And you need the frame, suspension, wheels, engine, valves, hosing and fittings, and wedges to boot.
Is there some magic about 45 tons? Are you splitting 3 foot diameter ebony?