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.

Calculate:

a) lenghtb) exit velocity of the final slab

L_f = 26.025 ft

v_f = 51.77 ft/min

Explanation:

Given:-

- The thickness of the slab initially, t_o = 2 in

- The width of the slab initially, w_o = 10 in

- The Length of the slab initially, L_o = 12.0 ft

- The reduction in thickness in each of three steps, r = 75%

- The widening of the slab in each of three steps, m = 3%

- The entry speed vi = 40 ft/min

- The roll speed remains the same

Find:-

a) length

b) exit velocity

Of the final slab

Solution:-

- The final thickness (t_f) after three passes is as follows:

t_f = (r / 100) ^n * t_o

Where, n = number of passes.

t_f = (75 / 100) ^3 * (2.0)

t_f = 0.844 in

- The final width (w_f) after three passes is as follows:

w_f = (m / 100 + 1) ^n * w_o

Where, n = number of passes.

w_f = (3 / 100 + 1) ^3 * (10.0)

w_f = 10.927 in

- Assuming the Volume of the slab remains the same. Zero material Loss. The final length of slab can be determined:

t_o*w_o*L_o = t_f*w_f*L_f

L_f = (2 * 10 * 12) / (0.844 * 10.927)

L_f = 26.025 ft

- We can use the volume rate equation as the roll speed remains constant i. e change in rate of volume is zero. Hence, we can write the before and after the 3rd step formulation:

t_i*w_i*v_i = t_f*w_f*v_f

Where, v_i : The entry step speed

v_f : Third step exit speed.

(0.75) ^2 * 2 * (1.03) ^2 * 10 * 40 = (0.844) * (10.927) * v_f

v_f = 51.77 ft/min