Respuesta :
Answer:
Among all three tools, the ceramic tool is taking the least time for the production of a batch, however, machining from the HSS tool is taking the highest time.
Explanation:
The optimum cutting speed for the minimum cost
[tex]V_{opt}= \frac{C}{\left[\left(T_c+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]^n}\;\cdots(i)[/tex]
Where,
C,n = Taylor equation parameters
[tex]T_h[/tex] =Tool changing time in minutes
[tex]C_e[/tex]=Cost per grinding per edge
[tex]C_m[/tex]= Machine and operator cost per minute
On comparing with the Taylor equation [tex]VT^n=C[/tex],
Tool life,
[tex]T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)[/tex]
Given that,
Cost of operator and machine time[tex]=\$40/hr=\$0.667/min[/tex]
Batch setting time = 2 hr
Part handling time: [tex]T_h=2.5[/tex] min
Part diameter: [tex]D=73 mm[/tex] [tex]=73\times 10^{-3} m[/tex]
Part length: [tex]l=250 mm=250\times 10^{-3} m[/tex]
Feed: [tex]f=0.30 mm/rev= 0.3\times 10^{-3} m/rev[/tex]
Depth of cut: [tex]d=3.5 mm[/tex]
For the HSS tool:
Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.
So, [tex]C_e=[/tex] [tex]\$20/15+2=\$3.33/edge[/tex]
Tool changing time, [tex]T_t=3[/tex] min.
[tex]C= 80 m/min[/tex]
[tex]n=0.130[/tex]
(a) From equation (i), cutting speed for the minimum cost:
[tex]V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}[/tex]
[tex]\Rightarrow 47.7[/tex] m/min
(b) From equation (ii), the tool life,
[tex]T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}[/tex]
[tex]\Rightarrow T=53.4[/tex] min
(c) Cycle time: [tex]T_c=T_h+T_m+\frac{T_t}{n_p}[/tex]
where,
[tex]T_m=[/tex] Machining time for one part
[tex]n_p=[/tex] Number of pieces cut in one tool life
[tex]T_m= \frac{l}{fN}[/tex] min, where [tex]N=\frac{V_{opt}}{\pi D}[/tex] is the rpm of the spindle.
[tex]\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}[/tex]
[tex]\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc[/tex]
So, the number of parts produced in one tool life
[tex]n_p=\frac {T}{T_m}[/tex]
[tex]\Rightarrow n_p=\frac {53.4}{4.01}=13.3[/tex]
Round it to the lower integer
[tex]\Rightarrow n_p=13[/tex]
So, the cycle time
[tex]T_c=2.5+4.01+\frac{3}{13}=6.74[/tex] min/pc
(d) Cost per production unit:
[tex]C_c= C_mT_c+\frac{C_e}{n_p}[/tex]
[tex]\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc[/tex]
(e) Total time to complete the batch= Sum of setup time and production time for one batch
[tex]=2\times60+ {50\times 6.74}{50}=457 min=7.62 hr[/tex].
(f) The proportion of time spent actually cutting metal
[tex]=\frac{50\times4.01}{457}=0.4387=43.87\%[/tex]
Now, for the cemented carbide tool:
Cost per edge,
[tex]C_e=[/tex] [tex]\$8/6=\$1.33/edge[/tex]
Tool changing time, [tex]T_t=1min[/tex]
[tex]C= 650 m/min[/tex]
[tex]n=0.30[/tex]
(a) Cutting speed for the minimum cost:
[tex]V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min[/tex] [from(i)]
(b) Tool life,
[tex]T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min[/tex] [from(ii)]
(c) Cycle time:
[tex]T_c=T_h+T_m+\frac{T_t}{n_p}[/tex]
[tex]T_m= \frac{\pi D l}{fV_{opt}}[/tex]
[tex]\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 363}=0.53min/pc[/tex]
[tex]n_p=\frac {7}{0.53}=13.2[/tex]
[tex]\Rightarrow n_p=13[/tex] [ nearest lower integer]
So, the cycle time
[tex]T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc[/tex]
(d) Cost per production unit:
[tex]C_c= C_mT_c+\frac{C_e}{n_p}[/tex]
[tex]\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc[/tex]
(e) Total time to complete the batch[tex]=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr[/tex].
(f) The proportion of time spent actually cutting metal
[tex]=\frac{50\times0.53}{275.5}=0.0962=9.62\%[/tex]
Similarly, for the ceramic tool:
[tex]C_e=[/tex] [tex]\$10/6=\$1.67/edge[/tex]
[tex]T_t-1min[/tex]
[tex]C= 3500 m/min[/tex]
[tex]n=0.6[/tex]
(a) Cutting speed:
[tex]V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}[/tex]
[tex]\Rightarrow V_{opt}=2105 m/min[/tex]
(b) Tool life,
[tex]T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min[/tex]
(c) Cycle time:
[tex]T_c=T_h+T_m+\frac{T_t}{n_p}[/tex]
[tex]\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc[/tex]
[tex]n_p=\frac {2.33}{0.091}=25.6[/tex]
[tex]\Rightarrow n_p=25 pc/tool\; life[/tex]
So,
[tex]T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc[/tex]
(d) Cost per production unit:
[tex]C_c= C_mT_c+\frac{C_e}{n_p}[/tex]
[tex]\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc[/tex]
(e) Total time to complete the batch
[tex]=2\times60+ {50\times 2.63}=251.5 min=4.19 hr[/tex].
(f) The proportion of time spent actually cutting metal
[tex]=\frac{50\times0.091}{251.5}=0.0181=1.81\%[/tex]
