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sergij07 [2.7K]
3 years ago
6

An annealed copper strip of 228 mm wide and 25 mm thick being rolled to a thickness of 20 mm, in one pass. The roll radius is 30

0 mm, and the rolls rotate at 100 rpm. Calculate the roll force and the power required in this operation
Engineering
1 answer:
sdas [7]3 years ago
5 0

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}

where;

w is the width of the annealed copper

Y_{avg} 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;

L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm

Now, determine the average true stress, Y_{avg}, 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}

F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F =   1.59 \ MN

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

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