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almond37 [142]
4 years ago
12

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of wid

th 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm
Engineering
2 answers:
dmitriy555 [2]4 years ago
3 0

Complete Question:

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.

Answer:

T_{min} = 26 mins 40 secs

Explanation:

Reduction in depth, Δd = 20 mm

Depth of cut, d_c = 0.5 mm

Number of passes necessary for this reduction, n = \frac{\triangle d}{d_c}

n = 20/0.5

n = 40 passes

Tool width, w = 5 mm

Width of metal plate, W = 200 mm

For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times

Speed of tool, v = 100 mm/s

Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec

minimum time required to reduce the depth of the plate by 20 mm:

T_{min} = number of passes * Time/pass

T_{min} = n * Time/pass

T_{min} = 40 * 40

T_{min} =  1600 = 26 mins 40 secs

Degger [83]4 years ago
3 0

Answer:

the minimum time required to reduce the depth of the plate by 20 mm is 26 minutes 40 seconds

Explanation:

From the given information;

Assuming the tool moves 100 mm/sec

The number of passes required to reduce the depth from 30 mm to 20 mm can be calculated as:

Number of passes = \dfrac{30-20}{0.5}

Number of passes = 20

We know that the width of the tool is 5 mm; therefore, to reduce the depth per pass; the tool have to travel 20 times

However; the time per passes is;

Time/pass = \dfrac{20*L}{velocity \ of \ the \ feed}

where;

length L = 400mm

velocity of the feed is assumed as 100

Time/pass  =\dfrac{20*400}{100}

Time/pass = 80 sec

Thus; the minimum time required to reduce the depth of the plate by 20 mm can be estimated as:

T_{min} = Time/pass *number of passes

T_{min} = 20*80

T_{min} = 1600 \ sec

T_{min} = 26 minutes 40 seconds

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F = 155.41 N - 0.4 × 351.82 N

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Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

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Substituting the values of the variables into the equation, we have

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x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

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x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

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Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

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Where \Delta t is the heat transfer time, measured in hours.

If we know that \dot Q = 417.492\,\frac{BTU}{h} and \Delta t = \frac{1}{3}\,h, then the net energy transfer is:

Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)

Q = 139.164\,BTU

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

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