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Shkiper50 [21]
3 years ago
6

Briefly explain how each of the following influences the tensile modulus of a semicrystalline polymer and why:(a) molecular weig

ht(b) degree of crystallinity(:1 deformation by drawing(d) annealing ofan undefonned material(e) annealing of a drawn material
Engineering
1 answer:
marin [14]3 years ago
5 0

Answer:

(a) Increases

(b) Increases

(c) Increases

(d) Increases

(e) Decreases

Explanation:

The tensile modulus of a semi-crystalline polymer depends on the given factors as:

(a) Molecular Weight:

It increases with the increase in the molecular weight of the polymer.

(b) Degree of crystallinity:

Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.

(c) Deformation by drawing:

The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.

(d) Annealing of an undeformed material:

This also results in an increase in the tensile strength of the material.

(e) Annealing of  a drawn material:

A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.

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Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

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tensile stress = 345 MPa

Y = 1.04

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solution

we find here length of critical interior flaw from formula that is

α  =  \frac{1}{\pi} (\frac{K}{\sigma Y})^2     ....................1

put here value we get

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α  = 15.043 mm

so minimum length of a surface crack is 15.043 mm

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Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
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Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

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p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

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ΔU =3030 J

ΔU =3.03 kJ

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ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

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