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.
1 answer:
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:
110 m or 11,000 cm
Explanation:
- let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
- let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
- let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively
M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s
C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c
<u>Using effectiveness-NUT method</u>
- <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>
C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c
C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c
From the result above cold fluid heat capacity rate is smaller
Dimensionless heat capacity rate, C = minimum capacity/maximum capacity
C= C<em>min</em>/C<em>max</em>
C = 5.016/8.62 = 0.582
.<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>
Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)
Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW
.<em>3 Third, we determine the actual heat transfer rate, Q</em>
Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)
Q = (5.016 kW/°c)(80 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW
.<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε
ε<em> </em>= Q/Qmax
ε <em>= </em>303.66 kW/702.24 kW
ε = 0.432
.<em>5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>
NTU =
NTU =
NTU = 0.661
<em>.6 sixth, we determine Heat Exchanger surface area, As</em>
From the question, the overall heat transfer coefficient U = 640 W/m²
As =
As =
As = 5.18 m²
<em>.7 Finally, we determine the length of the heat exchanger, L</em>
L =
L =
L= 109.91 m
L ≅ 110 m = 11,000 cm