Answer:
The roll force is 1.59 MN
The power required in this operation is 644.96 kW
Explanation:
Given;
width of the annealed copper, w = 228 m
thickness of the copper, h₀ = 25 mm
final thickness, hf = 20 mm
roll radius, R = 300 mm
The roll force is given by;
![F = LwY_{avg}](https://tex.z-dn.net/?f=F%20%3D%20LwY_%7Bavg%7D)
where;
w is the width of the annealed copper
is average true stress of the strip in the roll gap
L is length of arc in contact, and for frictionless situation it is given as;
Now, determine the average true stress,
, for the annealed copper;
The absolute value of the true strain, ε = ln(25/20)
ε = 0.223
from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.
Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa
Finally determine the roll force;
![F = LwY_{avg}](https://tex.z-dn.net/?f=F%20%3D%20LwY_%7Bavg%7D)
![F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F = 1.59 \ MN](https://tex.z-dn.net/?f=F%20%3D%20%28%5Cfrac%7B38.73%20%7D%7B1000%7D%29%28%5Cfrac%7B228%7D%7B1000%7D%29%2A180%20%5C%20MPa%5C%5C%5C%5CF%20%3D%20%20%201.59%20%5C%20MN)
The power required in this operation is given by;
![P = \frac{2\pi FLN}{60}\\\\P = \frac{2\pi (1.59*10^6)(0.03873)(100)}{60}\\\\P = 644955.2 \ W\\\\P = 644.96 \ kW](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B2%5Cpi%20FLN%7D%7B60%7D%5C%5C%5C%5CP%20%3D%20%20%5Cfrac%7B2%5Cpi%20%281.59%2A10%5E6%29%280.03873%29%28100%29%7D%7B60%7D%5C%5C%5C%5CP%20%3D%20644955.2%20%5C%20W%5C%5C%5C%5CP%20%3D%20644.96%20%5C%20kW)