Question

A strip of 120mm width and 8 mm thickness is rolled between two 300 mm diameter rolls to get a strip of 120 mm width and 7.2 mm thickness The speed of the strip at the exit is 30m/mm There is no front or back tension. Assuming uniform roll pressure of 200 MPa in the roll bite and 100% mechanical efficiency, the minimum total power (in kW) required to drive the two rolls is

Answer 1

Method -I:



Projected length,

$L_P=Rsin\alpha =\sqrt{R\Delta h}$

= $\sqrt{150\times (8-7.2)}=10.954$

Projected area,

$A_p=L_p\times b=10.954\times 120$

= 1314.48 $m{m}^2$

Roll separating Force,

F = ${\sigma }_0\times L_P\times bN$

$=200\times 1314.48N=262.9$

$[{\sigma }_{0}inN/m{m}^{2}i.e]$

Arm length (a in mm) = 0.5 $L_P$ for hot roll

= 0.5 $\times$ 10.954 mm = 5.477

= 0.45 $L_P$ for cold rolling

Torque per roller,

T = $F\times \frac{a}{1000}Nm$

= 262.9 kN $\times$ 5.477 mm

= 1439.9 Nm

Total power for two roller,

$P=2T\omega =2T\times \frac{2\pi N}{60}$

For the calculation of N we need velocity of neutral plane

Continuity equation

$h_0{b}_0{v}_0=h_f{b}_f{v}_f$

$8\times 120\times V_0=7.2\times 120\times 30$

or $V_0=27m/min$

Assuming velocity of neutral plane

average velocity,

$V=\frac{V_0+V_f}{2}=\frac{27+30}{2}$

= 28.5 m/min

Now, $V=\pi DN$

or 28.5 = $\pi \times 0.300\times N$

or N = 30.24 rpm

$\therefore$ Total power for two roller,

$P=2T\times \frac{2\pi N}{60}$

= $2\times 1439.9\times \frac{2\times \pi \times 30.24}{60}W$

= 9.1195 kW $\approx$ 9.12 kW

Method-II :

Velocity at the neutral plane = ?

$H_n{V}_n=H_2{V}_2\left(Applyingcontinuity\right)$

$H_n=H_2+0.25\Delta H$ (Standard relation)

= 7.2 + 0.25 $\times$ 0.8 = 7.4 mm

$V_n=\frac{H_2{V}_2}{H_n}=29.19m/min$

N = $\frac{29.19\times 1000}{\pi \times 300}=30.98rpm$

Total power = $2T\omega$

= $2\times 1439.9\times 2\pi \frac{30.98}{60}$

= 9.34 kW