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Gemiola [76]
4 years ago
5

Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle gra

in boundaries.
Engineering
1 answer:
Vlada [557]4 years ago
5 0

Answer:

Explanation:

Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.

Low angle grain boundaries (quasi-coherent) are formed by the dislocation network positioned along the geometric plane with small tilt angle differences between successive peers that is tilt boundary made up edge dislocations therefore it may only divert the slip direction of the incoming gliding dislocation with very little frictional stresses. And on the other hand, a high angle grain boundary region because of their disordered almost liquid like structure which acts as a strong barrier against dislocation slip motion and causes actually formation of dislocations file-up against it by arresting their motion unless that the stress concentration at the leading dislocation becomes high enough to go though the barrier.

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All machines have three fundamental hazards: moving parts, point of operation, and?
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2 years ago
A particular cloud-to-ground lightning strike lasts 500 µµsec and delivers 30 kA across a potential difference of 100 MV. Assu
sasho [114]

Answer:

a) 15 C charge was delivered by the lightening bolt

b) the lightning delivered 3.0 × 10¹² W of power

c)

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is 416666.67 Wh

d)

the residential retail value of the energy delivered by the strike is $ 40.83

e)

a total of 26 lightening strikes would be required to power an average US home for a year.

Explanation:

Given that;

the lighting strike lasted for t ( time ) = 500 μsecs = 500×10⁻⁶ s

Current I = 30 kA = 30×10³ A

voltage V = 100 mV = 100×10⁶ v

a)

we know that; I = Q/t

so Q = I × t

we substitute

Q =  30×10³ × 500×10⁻⁶

Q = 15 C

Therefore 15 C charge was delivered by the lightening bolt

b)

Power P = V × I

we substitute

Power P = 100×10⁶ × 30×10³

P = 3.0 × 10¹² W

Therefore, the lightning delivered 3.0 × 10¹² W of power

c)

we know that; Power = Energy / Time

Energy = Power × Time

we substitute

E = 3.0 × 10¹²  × 500×10⁻⁶

E = 1500 × 10⁶ J

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is;

⇒ 1500×10⁶ / 3600 Wh

= 416666.67 Wh

d)

given that;  1 KWh → $ 0.098

energy delivered by the strike = 416666.67 Wh = 416.66667 KWh

so the residential retail value of the energy delivered by the strike will be;

416.66667 KWh × $ 0.098

= $ 40.83

∴ the residential retail value of the energy delivered by the strike is $ 40.83

e)

Given that; average monthly residential energy consumption is 900 kWh.

for a year; energy consumption = 12 × 900 kWh = 10,800 KWh = 10800000 Wh

Now

1 lightening strike ⇒ 416666.67 Wh

x lightening strike ⇒ 10800000 Wh

x = 10800000 / 416666.67

x = 25.9199 ≈ 26

Therefore; a total of 26 lightening strikes would be required to power an average US home for a year.  

7 0
3 years ago
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