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Anit [1.1K]
3 years ago
12

The end of a large tubular workpart is to be faced on a NC vertical boring mill. The part has an outside diameter of 38.0 in and

inside diameter of 24.0 in. If the facing operation is performed at a rotational speed = 40 rev/min, feed = 0.015 in/rev, and depth = 0.180 in, determine: (a) the cutting time to complete the facing operation, (b) the cutting speeds and metal removal rates at the beginning and end of the cut.

Engineering
1 answer:
nata0808 [166]3 years ago
5 0

Answer:

(a) the cutting time to complete the facing operation = 11.667mins

b) the cutting speeds and metal removal rates at the beginning= 12.89in³/min and end of the cut. = 8.143in³/min

Explanation:

check attached files below for answer.

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A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar
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Answer:

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Detention time 2.34 hr

weir loading  114.06 m^3/d/m

Explanation:

calculation for single clarifier

sewag\  flow Q = \frac{12900}{2} = 6450 m^2/d

surface\  area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2

surface area = 314.16 m^2

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                             =314.16\times 2 = 628.32m^3

Length\ of\  weir = \pi \times diameter of weir

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overflow rate =v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2

Detention timet_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr

weir loading= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m

6 0
3 years ago
The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100),
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Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.

Therefore, a particular unit cell consist only 1/8th part of an atom.

The lattice constant of a simple cubic primitive cell is 5.28 Å

We know formula of distance,

d = \frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}

a)(100)

a=5.28 Å

Distance = \frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}=5.28 Å

b)(110)

Distance = \frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}} = 3.73 Å

c)(111)

Distance= \frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}= 3.048 Å

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2 years ago
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Answer:

2 ohms

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