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storchak [24]
3 years ago
11

What types of degradations that can take place in polymers and ceramics?

Engineering
1 answer:
Luda [366]3 years ago
5 0

Explanation:

Polymer degradation is a change in the properties of the polymer – such as tensile strength, color, shape, molecular weight – or of a polymer-based product under the influence of one or more environmental factors, such as heat, light, chemicals, or any other applied force.

  • <em>Photoinduced degradation:</em>  Electromagnetic waves with the energy of visible light or higher, such as ultraviolet light, X-rays and gamma rays are usually involved in such reactions.
  • <em>Thermal degradation:</em> Chain-growth polymers can be degraded by thermolysis at high temperatures to give monomers, oils, gases and water. This kind of degradation can happen by pyrolysis, hydrogenation or gasification:
  • <em>Chemical degradation:</em> it can happen through many mechanism such as oxidation, ozone cracking, galvanic actior or Chlorine-induced cracking
  • <em>Biological degradation:</em> by microorganisms to give lower molecular weight molecules  

In the case of ceramics, environmental factors are the major cause. There are several ways in which ceramics break down physically and chemically

  • <em>Physical degrdation:</em> from mishandling, impacts and abrasion, frost and mold growth
  • <em>Chemical degradation:</em> because of water and soluble salts

I hope you find this information useful and interesting! Good luck!

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7 0
3 years ago
Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a
Arada [10]

Answer:

a) h_c = 0.1599 W/m^2-K

b) H_{loss} = 5.02 W

c) T_s = 302 K

d) \dot{Q} = 25.125 W

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, T_a = 25^0 C = 298 K

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600}  = 1.25 kg/sec

Rate of heat transfer,

\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W

a) To calculate the convection coefficient relationship for heat transfer by convection:

\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, k_c = 0.8

\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K

b) Heat loss from the pipe to the environment:

H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W

d) The required fan control power is 25.125 W as calculated earlier above

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