An orthogonal cutting operations is being carried out in which uncut thickness is 0.010 mm, cutting soeed is 130 m/min, rake angle is 15° and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is (correct to two decimal places)

The correct answer is 0.44.

Given Data:
Uncut chip thickness, $t=0.010\text{mm}$
Cutting speed, $V=130\text{m/min}$
Rake angle, $\alpha =15°$
Width of cut, $b=6\text{mm}$
Chip thickness, $t_c=0.015\text{mm}$
Cutting force, $F_c=60\text{N}$
Thrust force, $F_t=25\text{N}$

Step 1: Calculate the friction force ($F$) along the rake face
Using Merchant's force circle analysis, the friction force acting along the tool rake face is given by:
$F=F_c\sin \left(\alpha \right)+F_t\cos \left(\alpha \right)$

Substituting the values:
$F=60\sin (15°)+25\cos (15°)$

Evaluating the trigonometric terms:
$\sin (15°)\approx 0.2588$
$\cos (15°)\approx 0.9659$

Therefore:
$F=60\times 0.2588+25\times 0.9659$
$F=15.528+24.148=39.676\text{N}$

Step 2: Determine the chip velocity ($V_c$)
From the continuity equation in orthogonal cutting, the mass flow rate remains constant. The chip thickness ratio ($r$) is related to the velocities and thicknesses as follows:
$r=\frac{t}{t_c}=\frac{V_c}{V}$

Thus, the ratio of chip velocity to cutting velocity is:
$\frac{V_c}{V}=\frac{0.010}{0.015}=\frac{2}{3}\approx 0.6667$

Step 3: Calculate the ratio of friction energy to total energy
The rate of total energy consumption (total cutting power) is:
$P_{\text{total}}=F_c\times V$

The rate of energy consumed in friction on the rake face is:
$P_{\text{friction}}=F\times V_c$

The ratio of friction energy to total energy is:
$\text{Energy Ratio}=\frac{P_{\text{friction}}}{P_{\text{total}}}=\frac{F\times V_c}{F_c\times V}=\left(\frac{F}{F_c}\right)\times \left(\frac{V_c}{V}\right)$

Substituting the values:
$\text{Energy Ratio}=\left(\frac{39.676}{60}\right)\times \left(\frac{0.010}{0.015}\right)$
$\text{Energy Ratio}=0.6613\times 0.6667\approx 0.4408$

Correct to two decimal places, the ratio of friction energy to total energy is 0.44.