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Andrews [41]
2 years ago
5

Chemical milling is used in an aircraft plant to create pockets in wing sections made of an aluminum alloy. The starting thickne

ss of one workpart of interest is 20 mm. A series of rectangular-shaped pockets 12 mm deep are to be etched with dimensions 200 mm by 400 mm. The corners of each rectangle are radiused to 15 mm. The part is an aluminum alloy and the etchant is NaOH. The penetration rate for this combination is 0.024 mm/min and the etch factor is 1.75.
Determine
(a) metal removal rate in mm3/min,
(b) time required to etch to the specified depth.
Engineering
1 answer:
Lelu [443]2 years ago
3 0

Answer:

a) metal removal rate is 1915.37 mm³/min

b) the time required to etch to the specified depth is 500 min or 8.333 hrs

Explanation:

Given the data in the question;

starting thickness of one work part of interest = 20 mm

depth of series of rectangular-shaped pockets = 12 mm

dimension of pocket = 200 mm by 400 mm

radius of corners of each rectangle = 15 mm

penetration rate = 0.024 mm/minute

etch factor = 1.75

a)

To get the metal removal rate MRR;

The initial area will be smaller compare to the given dimensions of 200mm by 400mm and the metal removal rate would increase during the cut as area is increased. so'

A = 200 × 400 - ( 30 × 30 - ( π × 15² ) )

= 80000 - ( 900 - 707 )      

= 80000 - 193

A = 79807 mm²

Hence, metal removal rate MRR = penetration rate × A

MRR = 0.024 mm/minute × 79807 mm²

MRR = 1915.37 mm³/min

Therefore, metal removal rate is 1915.37 mm³/min

b) To get the time required to etch to the specified depth;

Time to machine ( etch ) =  depth of series of rectangular-shaped pockets / penetration rate

we substitute

Time to machine ( etch ) = 12 mm / 0.024 mm/minute

Time to machine ( etch ) = 500 min or 8.333 hrs

Therefore, the time required to etch to the specified depth is 500 min or 8.333 hrs

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5 0
2 years ago
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Andrei [34K]

Answer:

(a) Work out put=692.83\frac{KJ}{Kg}

(b) Change in specific entropy=0.0044\frac{KJ}{Kg-K}

Explanation:

Properties of steam at 1 MPa and 200°C

        h_1=2827.4\frac{KJ}{Kg},s_1=6.69\frac{KJ}{Kg-K}

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of turbine is 0.83 and temperature T=40°C.So from steam table we can find pressure corresponding to saturation temperature 40°C.

Properties of saturated steam at 40°C

      h_f= 167.5\frac{KJ}{Kg} ,h_g= 2537.4\frac{KJ}{Kg}

 s_f= 0.57\frac{KJ}{Kg-K} ,s_g= 8.25\frac{KJ}{Kg-K}

So the enthalpy of steam at the exit of turbine  

h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}

h_2=167.5+0.83(2537.4-167.5)\frac{KJ}{Kg}

h_2=2134.57\frac{KJ}{Kg}

s_2=s_f+x(s_g-s_f)\frac{KJ}{Kg-K}

s_2=0.57+0.83(8.25-0.57)\frac{KJ}{Kg-K}

s_2=6.6944\frac{KJ}{Kg-K}

(a)

Work out put =h_1-h_2

                      =2827.4-2134.57 \frac{KJ}{Kg}

Work out put =692.83 \frac{KJ}{Kg}

(b) Change in specific entropy

     s_2-s_1=6.6944-6.69\frac{KJ}{Kg-K}

Change in specific entropy =0.0044\frac{KJ}{Kg-K}

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2 years ago
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jonny [76]

Answer:

ALL CAREFULLY ANSWERED CORRECTLY

Explanation:

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4) Technology has boosted every sector of our lives and it has the capacity to do more. Restricting it's importance to entertainment alone would be an underusing of its potentials. Engineering students infact should not need any drive to be encouraged about maximizing all it can do in shaping our world.

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~Never ceases to leave us in amazement

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4 0
2 years ago
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goblinko [34]

Answer:

<em>0.0386 hr</em>

<em></em>

Explanation:

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<em>There are 24 hrs in a day,</em> therefore rate in hrs will be

220/24 = 9.17 mole/m^2-hr

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therefore,

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<em>time that will be taken will be for water to pass = mole flux/mole flux rate</em>

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7 0
2 years ago
Your task is to fill in the missing parts of the C code to get a program equivalent to the generated assembly code. Recall that
Rudik [331]

Answer:

See Explaination

Explanation:

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long loop (long x, long n)

{

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long mask;

//For loop

for(mask = 1; mask != 0; mask = mask << (n & 0xFF))

{

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result | = (x&mask);

}

//Return result

return result;

}

6 0
2 years ago
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