Answer:
L_f = 26.025 ft
v_f = 51.77 ft/min
Explanation:
Given:
- Initial slab thickness, t_o = 2 in
- Initial slab width, w_o = 10 in
- Initial slab length, L_o = 12.0 ft
- Thickness reduction each step, r = 75%
- Width increase each step, m = 3%
- Initial entry speed vi = 40 ft/min
- Roll speed remains constant
Required:
a) Length
b) Exit velocity
For the final slab
Solution:
- After three passes, the final thickness (t_f) is determined as:
t_f = ( r / 100 )^n * t_o
Where n signifies the number of passes.
t_f = ( 75 / 100 ) ^3 * ( 2.0 )
t_f = 0.844 in
- The final width after three passes is calculated as:
w_f = ( m / 100 + 1 )^n * w_o
Where n denotes the number of passes.
w_f = ( 3 / 100 + 1 ) ^3 * ( 10.0 )
w_f = 10.927 in
- Assuming there is no loss in material, the final length of the slab can be obtained:
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
- The equation for volume rate can be applied since the roll speed stays the same. Thus, we equate the conditions before and after the third step:
t_i*w_i*v_i = t_f*w_f*v_f
Where v_i is the initial speed at entry:
v_f signifies the exit speed after the third step.
(0.75)^2 * 2 * (1.03)^2 * 10 * 40 = (0.844)*(10.927)*v_f
v_f = 51.77 ft/min
