Question

5. A continuous hot rolling mill has two stands. Thickness of the starting plate = 1.0 in. and width = 12.0 in. Final thickness is to be 0.5 in. Roll radius at each stand = 10 in. Rotational speed at the first stand = 20 rev/min. Equal drafts of 0.25 in. are to be taken at each stand. The plate is wide enough relative to its thickness that no increase in width occurs. Under the assumption that the forward slip is equal at each stand determine (a) speed v, at each stand and (b) forward slip s. (c) also determine the exiting speeds at each rolling stand if the entering speed at the first stand is 85 ft/min.​

Answer 1

(a) To determine the speed v at each stand, we can use the relationship between the forward slip, rotational speed, and rolling speed:

rolling speed = rotational speed × (1 - forward slip)

At the first stand, the thickness reduction is:

thickness reduction = draft / starting thickness = 0.25 / 1.0 = 0.25

The final thickness after the first stand is:

final thickness = starting thickness - thickness reduction = 1.0 - 0.25 = 0.75

The speed at the first stand is:

v1 = (π × 10^2 × 20) / (12 × 0.75) = 139.63 ft/min

At the second stand, the thickness reduction is again 0.25, and the final thickness is 0.5. The speed at the second stand is:

v2 = (π × 10^2 × 20) / (12 × 0.5) = 209.44 ft/min

(b) To determine the forward slip, we can use the formula:

forward slip = (rotational speed - rolling speed) / rotational speed

At the first stand:

forward slip1 = (20 - 139.63) / 20 = -5.98

At the second stand:

forward slip2 = (20 - 209.44) / 20 = -9.47

Note that the negative values indicate that the rolling speed is greater than the rotational speed, which is physically impossible. This suggests that the assumption of equal forward slip at each stand is not valid.

(c) The exiting speed at each rolling stand can be determined using the same formula as in part (a), but with the entering speed at the previous stand as the rolling speed:

At the first stand, entering speed = 85 ft/min:

v1 = 85 × (1 - (-5.98)) = 541.88 ft/min

At the second stand, entering speed = v1:

v2 = v1 × (1 - (-9.47)) = 922.57 ft/min