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seraphim [82]
3 years ago
11

Determine the carburizing time necessary to achieve a carbon concentration of 0.30 wt% at a position 4 mm into an iron–carbon al

loy that initially contains 0.10 wt% C. The surface concentration is to be maintained at 0.90 wt% C, and the treatment is to be conducted at 1100°C. Use the diffusion data for γ-Fe.
Engineering
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer:

idk

Explanation:

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What is one experiment that has been done in the manufacturing
Bingel [31]
Do you have a picture of any sort so I can see if there’s any options ;)
8 0
3 years ago
In casting experiments performed using a certain alloy and type of sand mold, it took 155 sec for a cube-shaped casting to solid
Tems11 [23]

Answer:

A) Cm = 2.232 s/mm²

B) Time taken to solidify = 74.3 seconds

Explanation:

(A) Since a side is 50mm and all sides of a cube are equal, thus, Volume of the cube is;V = 50 x 50 x 50 = 125,000 mm³

There are 6 faces of the cube, thus Surface Area A = 6 x (50 x 50) = 15,000 mm²

So, Volume/Area = (V/A) = 125,000/15,000 = 8.333 mm

Cm is given by the formula; Cm =[Tts] /(V/A)² where Tts is time taken to solidify and it's 155 seconds in the question. Thus;

Cm = 155/(8.333)²= 2.232 s/mm²

(B) For;Cylindrical casting with D = 30 mm and L = 50 mm.;

Volume of cylinder is;

V = (πD²L) /4

So,V = (π x 30² x 50)/4 = 35,343mm³

Surface area of cylinder is;

A = (2πD²)/4 + (πDL)

Thus, A = ((π x 30²)/2) + (π x 30 x 50) = 6126 mm²

Volume/Area is;

V/A = 35,343/6126 = 5.77 mm

Same alloy and mold type was hsed as in a above, thus, Cm is still 2.232 s/mm²

Since Cm =[Tts] /(V/A)²

Making Tts the subject, we have;

Tts =Cm x (V/A)²

Tts = 2.232 x (5.77)² = 74.3 seconds

3 0
3 years ago
The properties of the air in the inlet section with A1 = 0.25ab m2 in a converging-diverging channel are given as U1 = 25a,b m/s
NeX [460]

Answer:

nice cake

Explanation:

3 0
4 years ago
A system consists of a disk rotating on a frictionless axle
kakasveta [241]

The system includes a disk rotating on a frictionless axle and a bit of clay transferring towards it, as proven withinside the determine above.

<h3>What is the angular momentum?</h3>

The angular momentum of the device earlier than and after the clay sticks can be the same.

Conservation of angular momentum the precept of conservation of angular momentum states that the whole angular momentum is usually conserved.

  1. Li = Lf where;
  2. li is the preliminary second of inertia
  3. If is the very last second of inertia
  4. wi is the preliminary angular velocity
  5. wf is the very last angular velocity
  6. Li is the preliminary angular momentum
  7. Lf is the very last angular momentum

Thus, the angular momentum of the device earlier than and after the clay sticks can be the same.

Read more about the frictionless :

brainly.com/question/13539944

#SPJ4

8 0
2 years ago
Electrical strain gauges were applied to a notched specimen to determine the stresses in the notch. The results were εx=0.0019 a
pickupchik [31]

Answer:

\sigma_x=451.8MPa

\sigma_y=265.99MPa

Explanation:

\varepsilon _x=0.0019

\varepsilon _y=0.00072

Poission's ratio for steel =0.28

We know that

\varepsilon _x=\dfrac{\sigma _x}{E}-\mu\dfrac{\sigma _x}{E}

\varepsilon _y=\dfrac{\sigma _y}{E}-\mu\dfrac{\sigma _x}{E}

Now by putting the values

0.0019=\dfrac{\sigma _x}{200\times 1000}-0.28\dfrac{\sigma _y}{200\times 1000}

0.00072=\dfrac{\sigma _y}{200\times 1000}-0.28\dfrac{\sigma _x}{200\times 1000}

So \sigma_x=451.8MPa

\sigma_y=265.99MPa

6 0
4 years ago
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