I'm working with a gearbox manufacturer to upgrade a customer's machine. All the machine does is spin a reel of wire. The diameter of the reel is 3ft and the weight is 1000lbs when full and the diameter is 1ft and the weight is ~150lbs when empty. It needs to pay off wire at 1000 feet per minute. I calculated 300 max rpm of the gearbox, which will occur when the reel is almost empty.
My customer prefers that the gearbox manufacturer handle the selection of the motor because they want it to mate to the gearbox, out of the box. So this gearbox manufacturer is coming up with out-to-lunch (in my opinion) horsepower figures and trying to convince my customer that they need an expensive motor and drive when they don't. I spoke up about it and tried to back myself up with some math, but I'm not 100% confident that I figured it right. Can someone please look over this and let me know if I got it wrong.
From the gearbox people:
My customer prefers that the gearbox manufacturer handle the selection of the motor because they want it to mate to the gearbox, out of the box. So this gearbox manufacturer is coming up with out-to-lunch (in my opinion) horsepower figures and trying to convince my customer that they need an expensive motor and drive when they don't. I spoke up about it and tried to back myself up with some math, but I'm not 100% confident that I figured it right. Can someone please look over this and let me know if I got it wrong.
From the gearbox people:
My reply:Based on the info you gave me (300 RPM, 1000 FPM, 1 ft Drum diameter, and up to 1000 lbs of load) your motor needs to be 30.3 HP, 30 is close but gives you no service factor, 40 HP would be best Are you sure you want to go that big? Confirm this info please and I will get a quote on the gearbox for you.
What acceleration time are you using to figure horsepower? Using a value of 5 seconds (sounds reasonable to me) to get from 0 to 300RPM gives me 11HP, and 30HP would work out to around 2 sec accel time. So we are on the same page, here is how Im figuring it:
ACCELERATING TORQUE =
\(\frac{WK^{2}N}{308(t)}\)
Where:
N = Change in RPM
W = Weight in Lbs.
K = Radius of gyration
t = Time of acceleration (secs.)
WK2 = Equivalent Inertia
308 = Constant of proportionality
ACCELERATING TORQUE = \(\frac{(1000lbs)(1ft^{2})(300rpm)}{(308)(5sec)}\)
Accelerating torque = 194FT*LBs
HP = \(\frac{Torque x rpm}{5250}\)
HP = \(\frac{194ft*lbs x 300rpm}{5250}\)
HP = 11.13
30 or 40 HP seems like way overkill to me. I think that 10HP would be sufficient, as that is the size motor I have seen on similarly sized payoffs. 15HP would be a safer bet since the math doesnt reflect mechanical losses and such. Consider that the extremes of the parameters never coincide. When the reel is fully loaded, the diameter will be 3ft, so the payoff will only need to run ~100 RPM, not 300RPM. We will never need to accelerate 1000lbs to 300rpm in 2 or 5 seconds. Maybe 150lbs to 300rpm in 5 seconds or 1000lbs to 100rpm in 5 seconds, but not both.