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    

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