Respuesta :
Answer:
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
