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vovangra [49]
3 years ago
12

The lattice constant of a simple cubic lattice is a0.

Engineering
1 answer:
Oksi-84 [34.3K]3 years ago
7 0

Answer:

A)The sketches for the required planes were drawn in the first attachment.

B)The sketches for the required directions were drawn in the second attachment.

To draw a plane in a simple cubic lattice, you have to follow these instructions:

1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment)

2- The coordinates of that plane are written as: π:(1/a₀ 1/b₀ 1/c₀) (if one of the coordinates is 0, for example (1 1 0), c₀ is ∞, therefore that plane never cross the direction c).

3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.

4- Join the points.

To draw a direction in a simple cubic lattice, you have to follow these instructions:

1- Identify the points a₀, b₀, and c₀ in the cubic cell.

2- Draw the direction as a vector-like (a₀ b₀ c₀).

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Answer:

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Explanation:

The efficiency of the pump = 78%

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The height of the pool above the underground water, h = 30 m

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The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

\eta_t = The efficiency of the pump = 78%

\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}

Q_{max} = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump Q_{max} ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

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