a 24volt d.c motor rated at 1/16 of a h.p and 7000 r.p.m.coupled to a 200 to 1 reduction gearbox.what would be the output power.and the r.p.m.
2007-01-14 13:25:06 · 4 answers · asked by tugboat 4 in Science & Mathematics ➔ Engineering
RPM = 7000RPM / 200 (reduction) = 35 RPM
HP = 1/16 of hp / 200 (reduction) = 2.2 HP
2007-01-16 06:33:57 · answer #1 · answered by psychic_hedgehog 2 · 0⤊ 0⤋
Power expended is power expended, no matter what the leverage you expend it through, so you get out what you put in times the efficiency of the gearbox. You use the gearbox to apply the same amount of power to move a heavier load slower. 200:1 means you're turning the output shaft at 35rpm, but you're still putting out not quite 1/16 horsepower.
2007-01-14 23:16:02 · answer #2 · answered by virtualguy92107 7 · 2⤊ 0⤋
35 rpm
2007-01-14 21:34:42 · answer #3 · answered by idhard2find&looking 4 · 0⤊ 0⤋
approx. 2.1875 hp @ 35 rpm with no power loss
2007-01-14 22:13:20 · answer #4 · answered by unpop5 3 · 1⤊ 1⤋