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nirvana33 [79]
3 years ago
5

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.
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A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k
goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = \frac{ln\frac{r2}{r1}}{2 \pi *k * L}   .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

so

heat loss is change in temperature divide thermal resistance

Q = \frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}

Q = \frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

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4 years ago
Meaning of temporary stitches​
algol13

Answer:

They are use to hold garment or fabric pieces together before pernament stitches are made

6 0
3 years ago
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An aggregate blend is composed of 55% aggregate A (Sp. Gr. 2.631), 25% aggregate B (Sp. Gr. 2.331) and 20% sand (Sp. Gr. 2.609).
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Answer:

The right choice would be Option b (2.545).

Explanation:

The given values are:

The aggregate blend will be:

W_A = 55%

G_ A = 2.631

W_ B = 25%

G_B = 2.331

W_C = 20%

G_C = 2.609

Now,

On applying the formula, we get

⇒ G_{BA}=\frac{W_A+W_B+W_C}{\frac{W_A}{G_A} +\frac{W_B}{G_B} +\frac{W_C}{G_C}}

On substituting the values, we get

⇒          =\frac{55+25+20}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}

⇒          = \frac{100}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}

⇒          =2.545

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