A strip of 120 mm 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 30 m/min. 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 ________.

In the range: 8.5 - 10

Concept:

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

T = F × R

where P = Power, F = Rolling Force, R = roll radius, W = angular velocity and N = RPM

Calculation:

Given:



In this question, the velocity of exit is given. From the velocity of exit, we may assume the velocity of the neutral plane. But the correct answer will be slightly different.

$ProjectedLength,L_p=Rsin\alpha =\sqrt{R\mathrm{\Delta }h}=\sqrt{150\times \left(8-7.2\right)}=10.954mm$

Projected area, Aₚ = Lₚ × b = 10.954 × 120 = 1314.48 mm²

Roll separating force, F = σ₀ × Lₚ × b = 200 × 1314.48 = 262.9 kN

Arm length = 0.5 Lₚ for hot rolling = 0.5 × 10.954 mm = 5.477 mm

$TorqueperrollerT=F\times \frac{a}{1000}=262.9\times 5.477mm=1439.9Nm$

Total power for two rollers,

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

Now, for finding N, applying the continuity equation

Hₒb${}_{oV}$ o = h${}_f$ b${}_f$ V${}_f$

8 × 120 × V${}_O$ = 7.2 × 120 × 30

V${}_O$ = 27 m/min

Assuming the velocity of the neutral plane

$averagevelocity,V=\frac{V_o+V_f}{2}=\frac{27+30}{2}=28.5\frac{m}{min}$

V = πDN

28.5 = π × 0.300 × N

⇒ N = 30.24 rpm

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

$=2\times 1439.9\times 2\pi \times \frac{30.24}{60}=9.12kW$