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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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Air is compressed by a 40-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process a
AlexFokin [52]

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS_{air = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS_{air =  -40 kW / 298.15 K

ΔS_{air =  -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

7 0
3 years ago
The worst time you have had with a mechanical issue
Paul [167]

Answer:

When I was taking my sat exam online and my phone battery died

8 0
3 years ago
Read 2 more answers
A section of highway has a free-flow speed of 55 mph and a capacity of 3300 veh/hr. In a given hour, 2100 vehicles were counted
alexira [117]

Answer:

Space mean speed = 44 mi/h

Explanation:

Using Greenshield's linear model

q = Uf ( D - D^{2}/Dj )

qcap = capacity flow that gives Dcap

Dcap = Dj/2

qcap = Uf. Dj/4

Where

U = space mean speed

Uf =  free flow speed

D = density

Dj = jam density

now,

Dj = 4 × 3300/55

    = 240v/h

q = Dj ( U - U^{2}/Uf)

2100 = 240 ( U - U^{2}/55)

Solve for U

U = 44m/h

5 0
3 years ago
The pattern in a simple____ truss is a zigzag
liberstina [14]

Answer:

Bridge

Explanation:

A common, simply bridge truss is the zigzag.

6 0
3 years ago
A room is filled with saturated moist air at 25°C and a total pressure of 100 kPa. If the mass of dry air in the room is 100 kg,
ANTONII [103]

Answer:the mass of water vapor is=d. 2.04 kg

Explanation:

The humidity ratio by mass is given as

x = mw / ma                                  (1)

where

mw = mass of water vapor

ma = mass of dry air

And also humidity ratio by pressure by  Ideal Gas Law is given as

x = 0.62198 pw / (pa - pw)                            (2)

where

pw = partial pressure of water vapor in moist air

pa = atmospheric pressure of moist air

Equating both equation we have that

Humidity ratio =  0.62198 pw / (pa - pw)= mw / ma      (3)

From the from the water table at 25°C,

water vapor partial pressure = 23.8 torr

 1 atm =760 torr.

1 atm = 101.325 kPa,

So, 760 torr =101.325 kPa,

Therefore 23.8 torr =3.173kPa

Humidity ratio =  0.62198 pw / (pa - pw)= mw / ma  

0.62198 x 3.173 / (100-3.173)= mw / 100

1.9736 /96.827=mw/100

0.0204=mw/100

mw=2.04kg

6 0
3 years ago
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