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Vanyuwa [196]
3 years ago
15

In terms of the atomic radius, R, determine the distance between the centers of adjacent atoms for the FCC crystal structure alo

ng the [100] direction.

Engineering
1 answer:
timama [110]3 years ago
6 0

Answer:

The distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2

Explanation:

From the image uploaded, a Face centered cubic structure (100) plane, there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell.

In terms of the atomic radius, R, we determine the distance between the centers of adjacent atoms.

Let this distance = AC

the two adjacent sides = AB and BC

AB = a = 2R

BC = a = 2R

Using Pythagoras theorem

AC² = AB² + BC²

AC² = a² + a²

AC² = 2a²

AC = √2a²

AC = a√2

But a = 2R

AC = 2R√2

Therefore,  the distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2

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Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

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Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

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B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

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At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

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