A strip of 1120mm width and 8mm8mm thickness is rolled between two 300 mm-diameter rolls to get a strip of 120mm width and 7.2mm thickness. The speed of the strip at the exit is 30m/min. There is no front or back tension. Assuming uniform roll pressure of 200MPa in the roll bite and 100% mechanical efficiency. Calculate the minimum total power (in kW) required to drive the two rolls.

please find the below solution.

Step-by-step explanation

b= 120 mm ; σ_o = 200 MPa

Dia of roll = 300 mm => R = 150 mm ; ?h = (8 -7.2 ) = 0.8mm

Projected length, Lp =Rsinα≈ square root ( R?h ) = Square root ( 150*0.8 ) = 10.954 mm

Projected area = A_p = L_p*b = 10.954 *120 = 1314.48 mm²

Roll separating force = F=σ_o×L_p×b

F = 200*10.954*120 = 262896 N

F = 262.896 kN

Arm length ( a in mm) a= 0.5L_p for hot rolling and a= 0.5 L_p for cold rolling

For maximum La = 0.5* 10.954 = 5.477 mm

Torque per roller , T = F*(a/1000) = 262.896 kN*(5.477/1000) m = 1.43988 kN-m

Total power for two roller

P = 2T*ω = 2T*(2πN/60)

• Exit velocity is 30 m/min that is the maximum value.
• So, the maximum value of resistive power is obtained when we consider maximum velocity.

V=πDN

N= (V/πD) = ( 30/π*0.3) where D = 300 mm = 0.3m

N = 31.833 RPM

Therefore, ω = (2πN/60) = 3.333 radians /sec

P = 2T*ω = 2T*(2πN/60)

P = 2*1.43988 kN-m*3.333 radians /sec

P = 9.598 kN-m/sec = 9.598 kJ/sec = 9.598 kW

Therefore with 100 % mechanical efficiency

Required minimum power = P = 9.6 kW ( Approximated to first decimal place)

P = 9.6 kW