1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Andrews [41]
3 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]3 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

You might be interested in
Which phrase best describes a safety-critical system? A. a system that faces a very high risk of failure B. a system isolated fr
KIM [24]

Answer:

B.

Explanation:

A safety-critical system (SCS) or life-critical system is a system whose failure or malfunction may result in one (or more) of the following outcomes: death or serious injury to people. loss or severe damage to equipment/property.

7 0
3 years ago
4.116 The lid of a roof scuttle weighs 75 lb. It is hinged at corners A and B and maintained in the desired position by a rod CD
babunello [35]

Answer:

(a) The magnitude of force is 116.6 lb, as exerted by the rod CD

(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.

Explanation:

Step by step working is shown in the images attached herewith.

For this given system, the coordinates are the following:

A(0, 0, 0)

B(26, 0, 0)

And the value of angle alpha is 20.95°

Hope that answers the question, have a great day!

5 0
3 years ago
I need help with this question
Ad libitum [116K]

Answer:

LOL where is the question, that u need help with?

Explanation:

5 0
3 years ago
According to the video, what are examples of systems that Stationary Engineers oversee? Check all that apply. electrical systems
garik1379 [7]

Answer:

electrial systems

fire systems

heating systems

air systems

Explanation:

3 0
3 years ago
Read 2 more answers
soccer is also called association football" A soccer ball is a sphere, with circumference of 70 centimeters. in developing a new
timama [110]

Answer: Weight on Mars = 0.02593N

Explanation:

Given; Circumference C of Sphere = 70cm = 0.7m,

Specific Gravity S. G. of material = 1.21,

acceleration due to gravity in the Mars gm = 3.7m/s^2

We know that Weight W = mass m × acceleration due to gravity.

Let the Weight in on the Mars be Wm.

Wm = m × gm

Since we are given gm, we need to calculate for m. (Note that mass m is the same everywhere)

But mass = specific gravity × volume

Since we know the specific gravity, let's go ahead to calculate for the volume of the ball.

We know that Volume of a Sphere V = (4/3)πr^3

To get r, we know that C = 2πr

Therefore, r = C/(2π) = 0.7/(2π) = (7/10)/2π = 7/20π (in meters)

V = (4/3)*π×(7/20π)^3 = 343/6000π^2 (in meter^3)

m = 343/6000π^2 × 1.21 = 7.01×10^(-3)kg

Wm = 7.01×10^(-3) × 3.7 = 0.02593N

8 0
3 years ago
Other questions:
  • The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8 What is the probability of observing
    6·1 answer
  • Consider the following pulley system. A block of mass m is connected to a translational spring of stiffness k through a cable, w
    10·1 answer
  • What are the two safety precautions taken before driving a car​
    12·1 answer
  • For problems 1 and 2, six luminaires, similar to Style E used in the Commercial Building, are to be installed in a room that is
    13·1 answer
  • Discuss the chemical and physical properties of crude oil​
    6·1 answer
  • A 220-V electric heater has two heating coils that can be switched such that either coil can be used independently or the two ca
    15·1 answer
  • This app, I'm done, bye... I can't, bye
    11·1 answer
  • A jackhammer uses pressurize gas to change it forced to the hammer what type of mechanical system is it a jackhammer it uses in
    5·1 answer
  • A hammer can be used to see how a mineral breaks. If you observe square chunks of the mineral when broken, what can you conclude
    15·1 answer
  • Explain wet and dry compression tests​
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!