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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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pochemuha

Answer:

laminar flow

Explanation:

In fluid dynamics, laminar flow is characterized by fluid particles following smooth paths in layers, with each layer moving smoothly past the adjacent layers with little or no mixing.

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3 years ago
If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less
liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19}  } =0.93ohm*cm

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2}  } , if n_{i}=1.5x10^{10}cm^{-3}  \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15}  }{(1.5x10^{10})^{2}  } )=0.83V

w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15}  } } =3.35x10^{-5} cm=0.335um

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm

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3 years ago
96/64 reduced to its lowest term
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Answer:

3/2

Explanation:

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Steam enters an adiabatic turbine at 10MPa and 500 C and leaves at 10 kPa with a quality of 90%. Neglecting the changes in kinet
Amanda [17]

Answer:

flow ( m ) = 4.852 kg/s

Explanation:

Given:

- Inlet of Turbine

        P_1 = 10 MPa

        T_1 = 500 C

- Outlet of Turbine

        P_2 = 10 KPa

        x = 0.9

- Power output of Turbine W_out = 5 MW

Find:

Determine the mass ow rate required

Solution:

- Use steam Table A.4 to determine specific enthalpy for inlet conditions:

          P_1 = 10 MPa

          T_1 = 500 C            ---------- > h_1 = 3375.1 KJ/kg

- Use steam Table A.6 to determine specific enthalpy for outlet conditions:

          P_2 = 10 KPa       -------------> h_f = 191.81 KJ/kg

          x = 0.9                  -------------> h_fg = 2392.1 KJ/kg

          h_2 = h_f + x*h_fg

          h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg

- The work produced by the turbine W_out is given by first Law of thermodynamics:

          W_out = flow(m) * ( h_1 - h_2 )

          flow ( m ) = W_out / ( h_1 - h_2 )

- Plug in values:

          flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )

          flow ( m ) = 4.852 kg/s

3 0
3 years ago
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