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3 years ago

Explain in detail the difference between the microstructures of a cold worked sample and Recrystallized sample ?

Engineering

1 answer:

MrRa [10]3 years ago

7 0

Answer:

A recrystallization process differs mainly from a restoration process (microestructures of a cold worked sample) in that the first, the microstructure of the final product consists mainly of high angle borders.

Explanation:

The microstructure of a cold worked material has a high energy stored in dislocations and subgrains. During a heat treatment, this microstructure evolves in order to achieve a more stable state by reducing its energy. The complex microscopic mechanisms that take place during this process have been traditionally encompassed in three categories: restoration, recrystallization and grain growth, which will be discussed later.

These processes generally lead to the total or partial recovery of the original properties of the material (ductility, resistance). They can take place under dynamic conditions, that is, during thermomechanical processing or when the material is subsequently deformed at high temperature, and under static conditions, that is, during a heat treatment after thermomechanical processing.

The term restoration encompasses the following succession of micromechanism: formation of dislocation cells, annihilation of dislocations within them, formation of subgrains and their growth.

A recrystallization process differs mainly from a restoration process in that in the first one the microstructure of the final product is mainly formed by high-angle borders.

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3 years ago

To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at TH and Tc, would you incr

alukav5142 [94]

$\ T_{c}$ has greater effect.

Explanation:

$\eta_{\max }=1-\frac{T_{c}}{T_{A}}$

$T_{c}\\$ = Temperature of cold reservoir

$T_{H}$ = Temperature of hot reservoir

when $T_{c}$ is decreased by 't',

$\eta_{\text {incre }}$ = $1-\frac{\left(\tau_{c}-t\right)}{T_{H}}$

$=n \ + \frac{t}{T_{n}}$ $-(i)$

when ${T_{H}}$ is increased by 'T'

$\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)$

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3 years ago

Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the br

vagabundo [1.1K]

Answer:

Explanation:

Using the kinematics equation $v = v_o + a_ct$ to determine the velocity of car B.

where;

$v_o =$ initial velocity

$a_c$ = constant deceleration

Assuming the constant deceleration is = -12 ft/s^2

Also, the kinematic equation that relates to the distance with the time is:

$S = d + v_ot + \dfrac{1}{2}at^2$

Then:

$v_B = 60-12t$

The distance traveled by car B in the given time (t) is expressed as:

$S_B = d + 60 t - \dfrac{1}{2}(12t^2)$

For car A, the needed time (t) to come to rest is:

$v_A = 60 - 18(t-0.75)$

Also, the distance traveled by car A in the given time (t) is expressed as:

$S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

Relating both velocities:

$v_B = v_A$

$60-12t = 60 - 18(t-0.75)$

$60-12t =73.5 - 18t$

$60- 73.5 = - 18t+ 12t$

$-13.5 =-6t$

t = 2.25 s

At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars

i.e.

$S_B = S_A$

$d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

$d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2$

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft

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3 years ago

A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 12

nydimaria [60]

Answer:

The value of Modulus of elasticity E = 85.33 × $10^{6}$ $\frac{lbm}{in^{2} }$

Beam deflection is = 0.15 in

Explanation:

Given data

width = 5 in

Length = 60 in

Mass of the person = 125 lb

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We know that moment of inertia is given as

$I = \frac{bt^{3} }{12}$

$I = \frac{5 (1.5^{3} )}{12}$

I = 1.40625 $in^{4}$

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We know that deflection of the beam in this case is given as

Δ = $\frac{PL^{3} }{48EI}$

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3 years ago

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Answer:

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3 years ago

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