Labour cost per hour = Rs. 12.00

Machine overhead per hour = Rs. 40.00

Grinding cost per hour = Rs. 15.00

Grinding machine overhead per hour = Rs. 50.00

ldle time = 5 minutes

Taylor's tool life equation is given by $\mathrm{VT}^{0.22}=475$

The operation can be carried out using tungsten carbide tools either as brazed tools or throwaway tools.

For brazed tools,

Initial cost = Rs. 60

Grinding time = 5 minutes/edge

Tool changing time = 2 minutes

9, Grinding per tool before salvage

For throw away tips,

Initial cost = Rs. 40

Tool change time = 1.5 minutes

Total cutting edges = 8.

Find the optimum cutting speed, tool life and the cost of the operation for both the brazed tip and throwaway tips using the criteria of minimum production cost, and maximum production rate. $\mathrm{MU}-$ May $2013,15$ Marks

Solution:

$\mathrm{L}=600 \mathrm{mm}, \mathrm{d}=150 \mathrm{mm}, \mathrm{f}=0.25 \mathrm{mm} / \mathrm{rev}$

$x=\frac{12+40}{60}=R s .0 .867 / \min$

$\mathrm{VT}^{0.22}=475$

For Brazed Tools

$\mathrm{C}_{\mathrm{t}}=\mathrm{Rs} 60$

Grinding time / edge $=5 \mathrm{min}$

$C_{g}=5\left(\frac{15+50}{60}\right)=\operatorname{Rs} 5.42$

$y=\frac{C_{f}+r C_{g}}{r+1} ; \quad r=9$

$y=\frac{60+9 \times 5.42}{10}=R s / 0.88$

$\mathrm{T}_{\mathrm{d}}=2 \mathrm{mm}$

$\mathrm{R}=\mathrm{t}_{\mathrm{d}}+\frac{\mathrm{y}}{\mathrm{x}}=2+\frac{10.88}{0.867}=14.55$

$x=0.22$

$\mathrm{T}_{\mathrm{o}}=\mathrm{R}\left[\frac{1}{\mathrm{n}}-1\right]=14.55\left[\frac{1}{0.22}-1\right]=51.59 \mathrm{min}$

$\mathrm{V}_{\mathrm{o}}=\frac{475}{51.59^{0.22}}=199.5 \mathrm{m} / \mathrm{min}$

$\mathrm{T}_{\mathrm{co.}}=\frac{600}{0.25 \times \frac{199.5 \times 1000}{\pi \times 150}}=5.67 \mathrm{min}$

Optimum machinery cost:

$\mathbf{x}=\left[\mathbf{T}_{\mathbf{c}}+\mathbf{t}_{\mathbf{c o}}\left(1+\frac{\mathbf{R}}{\mathbf{T}_{\mathbf{e}}}\right)\right]$

$=0.867\left[5+5.67\left(1+\frac{14.55}{51.59}\right)\right]$

$=\mathbf{R} s .10 .64$

For maximum production:

$\mathrm{T}_{d}=\mathrm{t}_{4}\left[\frac{1}{\mathrm{n}}-1\right]$

$=2\left[\frac{1}{0.22}-1\right]=7.09 \mathrm{min}$

$\mathrm{V}_{\mathrm{o}}=\frac{475}{7.09^{0.22}}=308.71 \mathrm{m} / \mathrm{min}$

$t_{c o}=\frac{600}{0.25 \times \frac{308.71 \times 1000}{\pi \times 150}}=3.66 \mathrm{min}$

$\mathrm{C}=0.867\left[5+3.66\left(1+\frac{14.55}{7.09}\right)\right]=\operatorname{Rs} 14.02$

2. For Throw away tool:

$Y=\frac{\text { Cost }}{\text { No. of cutting edges }}=\frac{60}{8}=7.5$

$t_{d}=1.5$

For minimum cost:

$\mathrm{R}=\mathrm{t}_{\mathrm{d}}+\frac{\mathrm{y}}{\mathrm{x}}=1.5+\frac{7.5}{0.867}=10.15$

$\mathrm{n}=0.22$

$\mathrm{T}_{\mathrm{o}}=\mathrm{R}\left[\frac{1}{\mathrm{n}}-1\right]=10.15\left[\frac{1}{0.22}-1\right]=36 \mathrm{min}$

$\mathrm{V}_{\mathrm{o}}=\frac{475}{36^{0.22}}=215.93 \mathrm{m} / \mathrm{min}$

$t_{\mathrm{co}}=\frac{600}{0.25 \times \frac{215.93 \times 1000}{\pi \times 150}}=5.24 \mathrm{mm}$

Optimum cost $=\times\left[\mathrm{T}_{\mathrm{e}}+\mathrm{T}_{\mathrm{c}}\left(1+\frac{\mathrm{R}}{\mathrm{T}_{\mathrm{o}}}\right)\right]$

For maximum products:

$\mathrm{T}_{\mathrm{o}}=\mathrm{t}_{\mathrm{d}}\left(\frac{1}{\mathrm{n}}-1\right)$

$=1.5\left(\frac{1}{0.22}-1\right)=5.32 \mathrm{min}$

$\mathrm{V}_{\mathrm{o}}=\frac{475}{5.32^{0.22}}=328.85 \mathrm{m} / \mathrm{min}$

$t_{\mathrm{co}}=\frac{600}{0.25 \times \frac{328.85 \times 1000}{\pi \times 150}}=3.44 \mathrm{min}$

cost $=0.867\left[5+3.44\left(1+\frac{10.15}{5.32}\right)\right]$

$=\mathbf{R} s 13.0$ $\dots \boldsymbol{A} \boldsymbol{n} \boldsymbol{s}$