All categories

3 years ago

Define the difference between elastic and plastic deformation in terms of the effect on the crystal lattice structure.

Engineering

1 answer:

grin007 [14]3 years ago

3 0

Answer:

Elastic Deformation:

Elastic deformation is due to the elasticity of the body i.e., the deformation disappears on removal of external forces and the lattice regain its original structure.
This type of deformation is temporary in nature.
This type of deformation depends mainly on the chemical bonding of the substance.
It occurs due to bending or stretching of chemical bond under applied stress where atoms do not slip pass over each other.
It is a reversible process.

Plastic Deformation:

Plastic deformation is due to the plasticity of the body i,e., the structure can be easily shaped or molded .
This type of deformation is permanent in nature.
This type of deformation occurs mainly due to breakage of chemical bonds(limited in number) between constituent atoms.
Atoms may slip pass each other causing dislocation of atoms , resulting in permanent deformation even after stress is removed.
It is an irreversible process.

You might be interested in

Free brainlist because im new and i just want to but you have t friend me first

Amiraneli [1.4K]

Okay sure.

I’ll 1)chords
2)pulse
3)aerophone
4) the answer is C
5)rhythm

Pretty sure those are the answers

4 0

3 years ago

Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th

Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

$PV=mRT$

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

$T_1 = 38\°C$

$T_2 = 14\°C$

$\eta = 97\%$

$\dot{v} = 510m^3/kg$

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

$PV = mRT$

$\dot{m} = \frac{PV}{RT}$

$\dot{m} = \frac{0.6626*10^{5}*510}{287*311}$

$\dot{m} = 37.85kg/min$

Therfore the mass flow rate at which water condenses, then

$\eta = \frac{\dot{m_v}}{\dot{m}}$

Re-arrange to find $\dot{m_v}$

$\dot{m_v} = \eta*\dot{m}$

$\dot{m_v} = 0.97*37.85$

$\dot{m_v} = 36.72 kg/min$

PART B) Enthalpy is given by definition as,

$H= H_a +H_v$

Where,

$H_a$ = Enthalpy of dry air

$H_v$ = Enthalpy of water vapor

Replacing with our values we have that

$H=m*0.0291(38-25)+2500m_v$

$H = 37.85*0.0291(38-25)-2500*36.72$

$H = 91814.318kJ/min$

In the conversion system 1 ton is equal to 210kJ / min

$H = 91814.318kJ/min(\frac{1ton}{210kJ/min})$

$H = 437.2tons$

The cooling requeriment in tons of cooling is 437.2.

3 0

3 years ago

.provide feedback to the controller about the robot's environment, a limited sense of sight and sound

anyanavicka [17]

Answer:

go to settings and u may figure out

6 0

4 years ago

Read 2 more answers

(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length

EastWind [94]

This question is incomplete, the complete question is;

(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2 $l$ . The flow is laminar and fully developed. The pressure drop for the first pipe is 1.657 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.

D₃ = _____D.

{ the tolerance is +/-3% }

Answer:

the diameter of the second pipe D₃ is 1.13D

Explanation:

Given the data in the question;

Length = 2 $l$

pressure drop in the first pipe is 1.657 times greater than it is for the second pipe.

Now, we know that for Laminar Flow;

V' = πD⁴ΔP / 128μL

where V'₁ = V'₂ and ΔP₁₋₂ = 1.657 ΔP₂₋₃

Hence,

V'₁ = πD⁴ΔP₁₋₂ / 128μL = V'₃ = πD₃⁴ΔP₂₋₃ / 128μL

D₃ = D $($ ΔP₁₋₂ / ΔP₂₋₃ $)^{\frac{1}{4}$

we substitute

D₃ = D $($ 1.657 $)^{\frac{1}{4}$

D₃ = D( 1.134568 )

D₃ = 1.13D

Therefore, the diameter of the second pipe D₃ is 1.13D

8 0

3 years ago

An important fluid property is the kinematic viscosity which determines the viscous, or frictional, forces acting in a flow. The

Aleksandr [31]

Answer:- 400 K - 26.665

600 K - 53.67

Explanation: y−y1=(y2−y1)/(x2−x1)×(x−x1)

Basic usage of the interpolation formula to get the values of the required answers kinematic viscosities at 400 and 600 K

5 0

4 years ago