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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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An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are
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This question is incomplete, the complete question is;

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.

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b) The maximum temperature of the cycle is 1893.4 K

c) The cutoff ratio, v₃/v₂ is 1.946

Explanation:

Given the data in the question;

Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K

Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K

Net work per cycle W_{net = 590.1 kJ/kg

Heat transfer input per cycle Qs = 925 kJ/kg

a) compression ratio;

As illustrated in the diagram below, 1 - 2 is adiabatic compression;

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Tγ^{Y-1 = constant { For Air, γ = 1.4 }

hence;

⇒ V₁ / V₂ = ( T₂ / T₁ )^{\frac{1}{Y-1}

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⇒ V₁ / V₂ = (  973 K / 303 K  )^{\frac{1}{1.4-1}

= (  3.21122  )^{\frac{1}{0.4}

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Therefore, The compression ratio is 18.48

b) maximum temperature of the cycle

We know that for Air, Cp = 1.005 kJ/kgK

Now,

Heat transfer input per cycle Qs = Cp( T₃ - T₂ )

we substitute

925 = 1.005( T₃ - 700 )

( T₃ - 700 ) = 925 / 1.005

( T₃ - 700 ) = 920.398

T₃ = 920.398 + 700

T₃ = 1620.398 °C

T₃ = ( 1620.398 + 273 ) K

T₃ = 1893.396 K ≈ 1893.4 K

Therefore, The maximum temperature of the cycle is 1893.4 K

c)  the cutoff ratio, v₃/v₂;

Since pressure is constant, V ∝ T

So,

cutoff ratio S = v₃ / v₂  = T₃ / T₂

we substitute

cutoff ratio S = 1893.396 K / 973 K

cutoff ratio S = 1.9459 ≈ 1.946

Therefore, the cutoff ratio, v₃/v₂ is 1.946

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