Question

A specimen is being grinded on a surface grinder with grinding wheel speed of 1600 m/min, and workpiece speed of 25 mm/min, grinding wheel diameter of 200 mm, grinding depth of cut of 0.025 mm. Take transverse feed of workpiece = 4.4 mm/pass and specific cutting energy = $2200J/m{m}^3$. Assume that pitch of grits at periphery of grinding wheel is 12 mm, the factor for negative rake angle of abrasive grains as 5. Mean grinding power in above operation will be___kW

• A. 0.504

Answer 1

The correct option is A 0.504
Mean grinding power $=\frac{(Velocityofworkpiece\times depthofcut\times transversefeed\times \times factorfornegativerake\times specificcuttingenergy)}{60\times 1000}$

$=\frac{25\times 0.025\times 4.4\times 5\times 2200}{60\times 1000}=0.504166kW$