Power required in rolling

A 250 mm-wide annealed brass 70-30 strip is rolled from a thickness of 20 mm to 12 mm. For a roll radius of 300 mm and roll rpm of 100, estimate the total power required for this operation.

Question

Power required in rolling

A 250 mm-wide annealed brass 70-30 strip is rolled from a thickness of 20 mm to 12 mm. For a roll radius of 300 mm and roll rpm of 100, estimate the total power required for this operation.

Accepted Answer

The power needed for a set of two rolls is given by Eq. (6.43) as

Power[latex]=\frac{\pi FLN}{60,000} [/latex]kW,

where F is obtained from Eq. (6.39) and L from Eq. (6.38). Hence,

[latex]F=Lw\overline{Y^{′}} [/latex] and L=[latex]\sqrt{R\Delta h} . [/latex]

Therefore,

L[latex]=\sqrt{(0.30)(20-12)/1000} = 0.0489 [/latex] m and [latex]w=0.25[/latex] m

For brass 70-30, we have K = [latex]895[/latex] MPa and [latex]n =0.49[/latex] (from Table 2.3).

Table 2.3,

Typical Values for K and n in Eq. (2.11)([latex]\sigma =K\epsilon ^n[/latex] )at Room Temperature
Material	K (MPa)	n
Aluminum, 1100-O	180	0.2
2024-T4	690	0.16
5052-O	210	0.13
6061-O	205	0.2
6061-T6	410	0.05
7075-O	400	0.17
Brass, 7030, annealed	895	0.49
85–15, cold rolled	580	0.34
Bronze (phosphor), annealed	720	0.46
Cobalt-base alloy, heat treated	2070	0.5
Copper, annealed	315	0.54
Molybdenum, annealed	725	0.13
Steel, low carbon, annealed	530	0.26
1045 hot rolled	965	0.14
1112 annealed	760	0.19
1112 cold rolled	760	0.08
4135 annealed	1015	0.17
4135 cold rolled	1100	0.14
4340 annealed	640	0.15
17–4 P-H, annealed	1200	0.05
52100, annealed	1450	0.07
304 stainless, annealed	1275	0.45
410 stainless, annealed	960	0.10

The true strain in this operation is

[latex]\epsilon _{1}=\ln \left\lgroup\frac{20}{12} ight group=0. 5108 [/latex]

Thus, from Eq. (6.10),

[latex]\overline{Y} =\frac{K\int_{0}^{\epsilon _{1}}{\epsilon ^{n}d\epsilon } }{\epsilon _{1}} =\frac{K\epsilon ^{n}_{1} }{n+1}. [/latex] (6.10)

we have

[latex]\overline{Y} =\frac{(895)(0.5108)^{0.49}}{1.49} =432 [/latex] MPa

and

[latex]\overline{Y^{′}}=\overline{Y} \left[1+\frac{\mu L}{2h_{av}} ight] =432\left[1+\frac{0.1 imes 0. 0489}{2 imes 16 imes 10^{-3}} ight],\left[h_{av}=\frac{20+12}{2}=16 \ mm = 16×10^{-3} ight] [/latex]

[latex]\overline{Y^{′}}= 497[/latex] MPa

Therefore,

F =[latex](0.0489)(0.25)(497 × 10^{6}) = 6.07 × 10^6[/latex] N

and for two rolls

Power = [latex]\frac{2\pi (6 .07 imes 10^6)(0 .0489)(100)}{60,000} =3111 [/latex] k W

Question 6.4: Power required in rolling A 250 mm-wide annealed brass 70-30...

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