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45 For this (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. Thus, there is the equivalence of 2 atoms associated with this FCC (100) plane. 1)]; and, thus, the area of this square is just 2 R 2 )2 = 8R2. Hence, the planar density for this (100) plane is just PD100 = number of atoms centered on (100) plane area of (100) plane = 2 atoms 8 R2 = 1 4R 2 That portion of an FCC (111) plane contained within a unit cell is shown below.

Thus, the direction is represented as [112 3] . (b) This portion of the problem asks for the same conversion of the (010) and (101) planes. A plane for hexagonal is represented by (hkil) where i = - (h + k), and h, k, and l are the same for both systems. For the (010) plane, h = 0, k = 1, l = 0, and i = - (0 + 1) = -1 Thus, the plane is now represented as (hkil) = (01 1 0) . For the (101) plane, i = - (1 + 0) = -1, and (hkil) = (10 1 1) . 36 For plane A we will leave the origin at the unit cell as shown.

The density of an alloy ρave is just the total alloy mass M divided by its volume V ρ ave = Or, in terms of the component elements 1 and 2 63 M V m1 + m2 ρ ave = V + V 1 2 Here it is assumed that the total alloy volume is equal to the separate volumes of the individual components, which is only an approximation; normally V will not be exactly equal to (V1 + V2). 11b) for Aave is requested. The alloy average molecular weight is just the ratio of total alloy mass in grams M' and the total number of moles in the alloy Nm.

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Airbus A320 SOP 06Cruise

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