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3 years ago

What time ----–- the train arrve?

Engineering

1 answer:

Tcecarenko [31]3 years ago

4 0

I think it’s TIME FOR U TO GET A WATCH !!! lol I’m jk jk I believe it arrives at 12:33

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What should be your strongest tool be for gulding your ethical decisions making process

valkas [14]

Answer:

Recognize that there is a moral dilemma.

Determine the actor. ...

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Test for right versus wrong issues. ...

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7 0

2 years ago

Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be compared for the same turning operation on a ba

Tpy6a [65]

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

$V_{opt}= \frac{C}{\left[\left(T_c+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]^n}\;\cdots(i)$

Where,

C,n = Taylor equation parameters

$T_h$ =Tool changing time in minutes

$C_e$ =Cost per grinding per edge

$C_m$ = Machine and operator cost per minute

On comparing with the Taylor equation $VT^n=C$ ,

Tool life,

$T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)$

Given that,

Cost of operator and machine time $=\$40/hr=\$0.667/min$

Batch setting time = 2 hr

Part handling time: $T_h=2.5$ min

Part diameter: $D=73 mm$ $=73\times 10^{-3} m$

Part length: $l=250 mm=250\times 10^{-3} m$

Feed: $f=0.30 mm/rev= 0.3\times 10^{-3} m/rev$

Depth of cut: $d=3.5 mm$

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So, $C_e=$ $\$20/15+2=\$3.33/edge$

Tool changing time, $T_t=3$ min.

$C= 80 m/min$

$n=0.130$

(a) From equation (i), cutting speed for the minimum cost:

$V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}$

$\Rightarrow 47.7$ m/min

(b) From equation (ii), the tool life,

$T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}$

$\Rightarrow T=53.4$ min

(c) Cycle time: $T_c=T_h+T_m+\frac{T_t}{n_p}$

where,

$T_m=$ Machining time for one part

$n_p=$ Number of pieces cut in one tool life

$T_m= \frac{l}{fN}$ min, where $N=\frac{V_{opt}}{\pi D}$ is the rpm of the spindle.

$\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc$

So, the number of parts produced in one tool life

$n_p=\frac {T}{T_m}$

$\Rightarrow n_p=\frac {53.4}{4.01}=13.3$

Round it to the lower integer

$\Rightarrow n_p=13$

So, the cycle time

$T_c=2.5+4.01+\frac{3}{13}=6.74$ min/pc

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc$

(e) Total time to complete the batch= Sum of setup time and production time for one batch

$=2\times60+ {50\times 6.74}{50}=457 min=7.62 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times4.01}{457}=0.4387=43.87\%$

Now, for the cemented carbide tool:

Cost per edge,

$C_e=$ $\$8/6=\$1.33/edge$

Tool changing time, $T_t=1min$

$C= 650 m/min$

$n=0.30$

(a) Cutting speed for the minimum cost:

$V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min$ [from(i)]

(b) Tool life,

$T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min$ [from(ii)]

(c) Cycle time:

$T_c=T_h+T_m+\frac{T_t}{n_p}$

$T_m= \frac{\pi D l}{fV_{opt}}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 363}=0.53min/pc$

$n_p=\frac {7}{0.53}=13.2$

$\Rightarrow n_p=13$ [ nearest lower integer]

So, the cycle time

$T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc$

(e) Total time to complete the batch $=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.53}{275.5}=0.0962=9.62\%$

Similarly, for the ceramic tool:

$C_e=$ $\$10/6=\$1.67/edge$

$T_t-1min$

$C= 3500 m/min$

$n=0.6$

(a) Cutting speed:

$V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}$

$\Rightarrow V_{opt}=2105 m/min$

(b) Tool life,

$T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min$

(c) Cycle time:

$T_c=T_h+T_m+\frac{T_t}{n_p}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc$

$n_p=\frac {2.33}{0.091}=25.6$

$\Rightarrow n_p=25 pc/tool\; life$

So,

$T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc$

(e) Total time to complete the batch

$=2\times60+ {50\times 2.63}=251.5 min=4.19 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.091}{251.5}=0.0181=1.81\%$

3 0

4 years ago

Technician A says that a way to prevent galvanic corrosion is to duplicate the original installation method. Technician B says t

sertanlavr [38]

Answer:

Technician A

Explanation:

Galvanic corrosion is not on only one metal alone but caused when two metals are interacting. Thus, Duplicating the original installation method is a better option because re-using a coated bolt doesn't prevent galvanic corrosion because both materials must be coated and not just the bolt and in technician B's case he is coating just the bolt. Thus, technician B's method will not achieve prevention of galvanic corrosion but technician A's method will achieve it.

8 0

3 years ago

The inner surface of a hollow cylinder is subjected to tangential and axial stresses of 40,000 and 24,000 psi, respectively. Det

Furkat [3]

Answer:

15,000 psi

Explanation:

The solution / solving is attach below.

5 0

3 years ago

1. Examine the following circuit. Find RT, I3, R1, R2, R3, V1, V2 and V3. Show all of your work clearly below.

Mkey [24]

Explanation:

Ohm's law is used here. V = IR, and variations. The voltage across all elements is the same in this parallel circuit. (V1 =V2 =V3)

The total supply current is the sum of the currents in each of the branches. (It = I1 +I2 +I3)

Rt = (8 V)/(8 A) = 1 Ω . . . . supply voltage divided by supply current

I3 = 8A -3A -4A = 1 A . . . . supply current not flowing through other branches

R1 = (8 V)/(3 A) = 8/3 Ω

R2 = (8 V)/(4 A) = 2 Ω

R3 = (8 V)/(I3) = (8 V)/(1 A) = 8 Ω

V1 = V2 = V3 = 8 V

6 0

3 years ago

What time ----–- the train arrve? ​​

What time ----–- the train arrve?