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3 years ago

The smallest crystal lattice defects is a) cracks b) point defects c) planar defects d) dislocations.

Engineering

1 answer:

ss7ja [257]3 years ago

8 0

Answer:b) Point defects

Explanation: The point defect is the tiny defect that occurs in the lattice. It usually occurs when there is the atoms or ions missing in the lattice structure that creates a irregularity in the structure.The name point defect itself describes that the occurring defect is having a size of point thus is the smallest defect. Therefore option(b) is the correct option.

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djverab [1.8K]

Answer:

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2 years ago

Steam enters a turbine at 120 bar, 508oC. At the exit, the pressure and quality are 50 kPa and 0.912, respectively. Determine th

Archy [21]

Answer:

The turbine produces <u>955.53 KW</u> power.

Explanation:

Taking the turbine as a system. Applying Law of Conservation of Energy:

m(h₁ - h₂) - Heat Loss = P

where,

m = mass flow rate of steam = 1.31 kg/s

h₁ = enthalpy at state 1 (120 bar and 508°C)

h₂ = enthalpy at state 2 (50 KPa and x = 0.912)

Heat Loss = 225 KW

P = Power generated by turbine = ?

First, we find h₁ from super steam tables:

At,

T = 508°C

P = 120 bar = 12000 KPa = 12 MPa

we find that steam is in super-heated state with enthalpy:

Due to unavailibility of values in chart we approximate the state to 500° C and 12.5 MPa:

h₁ = 3343.6 KJ/kg

Now, for state 2, we have:

P = 50 KPa and x = 0.912

From saturated steam table:

h₂ = hf₂ + x(hfg₂) = 340.54 KJ/kg + (0.912)(2304.7 KJ/kg)

h₂ = 2442.4 KJ/kg

Now, using values in the conservation equation:

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3 years ago

The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,

babunello [35]

Answer:

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × $10^6$ Pa

viscosity n = 100 Pas

angle = 18°

so Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 $\times 10^{-3}$ )²× 9 $\times 10^{-3}$ × sin18 × cos18

Qd = 94.305 × $10^{-6}$ m³/s

and

Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2

Qb = 11 × $10^{6}$ × π × 85 $\times 10^{-3}$ × ( 9 $\times 10^{-3}$ )³ × sin²18 ÷ 12 × 100 × 2

Qb = 85.2 × $10^{-6}$ m³/s

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volume flow rate Qx = Qd - Qb ..............3

Qx = 94.305 × $10^{-6}$ - 85.2 × $10^{-6}$

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

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