Question

In an external cylindrical grinding operation on a hardened steel tube whose outside radius = 42.5 mm, the grinding wheel diameter = 125 mm and wheel width = 20 mm. The work piece rotates at a surface speed of 25 m/min, the wheel rotates at 1800 rev/min, infeed (depth of cut) = 0.05 mm, and traverse feed = 0.50 mm/rev. There are 50 active grits/cm2 of wheel surface, and the operation is performed dry.

Determine the (a) volume rate of metal removed, (b) number of chips formed per unit time, and (c) average volume per chip. (d) If the tangential cutting force on the work = 45 N, compute the specific energy in this operation.

Answer 1

a

The volume rate of metal removed

`V_R= 625mm^2/mm`

b

The number of chips formed per unit time

`N_C = 176,715 chips/min`

c

The average volume per chip

`V= 0.003537 mm^3 / chip`

d

The the specific energy in this operation

`Q= 50.8938 N \cdot m / mm^3`

Step-by-step explanation:

From the question we are told that

The outside radius is
`R = 42.5mm`

The grinding wheel diameter is
`D = 125 \ mm`

The grinding wheel width is
`w = 20 mm`

The surface speed of the work piece rotation of
`v = 25 m/min =25 * (1000mm)/(m) = 25 *10^3 mm/min`

The speed of rotation of the wheel is
`N = 1800 \ rev/min`

The depth of cut is
`D = 0.05mm`

The transverse feed is
`T_r = 0.50 mm/ rev`

The number of
`grit /cm^2` of wheel surface is
`C = 50\ grits/cm^2 = 50 * [(10^(-2) cm )/(mm^2) ] = 50 *10^(-2) grits/mm^2`

The cutting force is F = 45 N

The volume rate of the metal removed is mathematically represented as

`V_R = v * D* T`

Substitution value

`V_R = 25*10^3 * 0.05 *0.5`

`V_R= 625mm^2/mm`

The speed of the wheel is mathematically represented

`v = N \pi D`

`= 1800 * \pi * 125`

`= 706,858mm/min`

The number of chips formed per unit time is mathematically represented as

`N_C = 706858 * 0.5 * 50 *10^(-2)`

`= 176,715 grits/min`

`N_C = 176,715 chips/min`

The average volume per is mathematically represented as

`V = (V_R)/(N_C)`

`= (625 )/(176,715)`

`V= 0.003537 mm^3 / chip`

The specific energy is the operation mathematically

`Q = (F v )/(V_R)`

`= ((45)[ (706,858.3) * (1m)/(1000mmm) ])/(625)`

`Q= 50.8938 N \cdot m / mm^3`