Question

A 2.0-in-thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each step will reduce the slab to 75% of its previous thickness. It is expected that for this metal and reduction, the slab will widen by 3% in each step. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps, determine: (a) length and (b) exit velocity of the slab after the final reduction

Answer 1

Answer:

26.02 ft

86.7690 ft/min

Explanation:

After 3 steps

0.75³(2.0 thickness)

T = 0.84375

W = (1+0.03)³10

= 10.92727 inches

A To get length

2.0 x 10 x 12 x12 = 0.84375 x 10.92727x lf

= 2880 = 9.21988Lf

Lf = 2880/9.21988

= 312.368 inches

Convert to feet

322.368 x 0.0833

= 26.02 ft

B.

= 2 x 10 x 40 = 0.84375 x 10.92727 x vf

800 = 9.21988vf

Vf = 800/9.21988

Vf = 86.7690 ft/min