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MSE 230 Solutions for Assignment 12 Fall 2014

Problem 1*

a)

The specific surface area decreases rapidly with increasing particle size. Note that the log scalemakes it much easier to follow the trend.

This plot suggests that smaller particles have a much greater driving force for sintering because

reducing surface energy/area is the driving force for sintering and the smaller particles have a lotmore surface area to lose in contrast to the larger particles.

b)

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The main point is that smaller particles with larger specific surface areas have a larger drivingforce for sintering. Therefore, all other things being equal (e.g., packing density, temperature,

etc.), smaller particles will sinter/densify at a faster rate than larger particles.

2. a) For FCC packing we are simply looking at the volume of the spherical particles making up

the unit cell divided by the volume of the cubic. It is completely analogous to FCC packing of

atoms except the space not taken up by the particles is taken up by air (not sure what you find in

between atoms of metal).

Volume of particles (4 particles per unit cell) = 4!43"r

3

Volume of unit cell = a3

Relation between a and r: a = 2 2r

Packing density =4!4

3"r

3

2 2r( )3 = 0.74 or 74% of theoretical density.

b) For a cylinder, V = !r2h. The volume of the powder compact is

mass

theoretical density( ) packing fraction( ) =

10 g

3.97g cm3( ) 0.74( )= 3.40cm3

h = V

!r2 = 3.40cm

3

!" 1.25cm( )2 = 0.69 cm

c) If you assume that the proportions of the cylinder (e.g., height to diameter ratio) stay the same

than you can calculate the height after the powder compact has reached full density.

h

r=

069cm

1.25 cm= 0.552 h = 0.552r

V = !r2h = !r2 0.552r( )

mass

theoretical density( ) packing fraction( ) =

10 g

3.97g cm3( ) 1( )= 2.52cm3

2.52 cm3 = !r2 0.552r( ) r =1.13 cm, h = 0.625 cm

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3.* Given a crystalline and an amorphous polymer, would you expect the cycle time formanufacturing a given component out of the crystalline polymer to be greater, smaller or the

same. Assume that the thermal conductivity of the two polymers is identical. Briefly explain

your answer.

The cycle time for the crystalline polymer will typically be greater because of the thermal

arrest (e.g., the temperature is stuck at the melting temperature until crystallization is

complete) that occurs during crystallization. An amorphous polymer does not have a

melting temperature and therefore can cool continuously as it solidifies.

4. You are asked to purchase HDPE resins for parts to be processed by extrusion and injection

molding. Describe how the desired properties of the resins will differ depending on processing

method and explain why.

The main difference is that the resins for injection molding have a higher melt index

(lower viscosity) than resins for extrusion. Molding filling in small spaces is an issue ininjection molding therefore a higher melt index is desirable. In extrusion, a higherviscosity is helpful because it helps a component maintain its shape after it leaves the

extruder. The magnitude of the viscosity difference may be seen in the polymer melt

viscosity slide. Note the difference in viscosity between injection molding and extrusion

grade nylon.

5. You are developing a new product using polymers. Recycling the polymer after use is an

important consideration. Would you prefer to use thermoplastic or thermosetting polymers foryour product? Please explain your answer based on the contrasting molecular structures of

thermosets and thermoplastics.

Thermosets form a three-dimensional interconnected network of polymers chains uponcuring. Once formed, heating causes decomposition of the thermoset rather than

remelting as in thermoplastics. Therefore, one would prefer to use thermoplastics where

having a recyclable component is important.

6. a) Name the following polymer(s) that would be suitable for the fabrication of cups to contain

hot coffee: polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate.Why?

b) Of those polymers listed in Table 15.2, which polymer(s) would be best suited for use as icecube trays? Why?

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a) This question asks us to name which, of several polymers, would be suitable for thefabrication of cups to contain hot coffee. At its glass transition temperature, an amorphous

polymer begins to soften. The maximum temperature of hot coffee is probably slightly below

100C (212F). Of the polymers listed, only polystyrene and polycarbonate have glass transition

temperatures of 100C or above (Table 15.2), and would be suitable for this application.

b) In order for a polymer to be suited for use as an ice cube tray it must have a glass-transition

temperature below 0C. Of those polymers listed in Table 15.2 only low-density and high-density polyethylene, PTFE, and polypropylene satisfy this criterion.

7.* Briefly explain how each of the following influences the tensile modulus of a semicrystalline

polymer and why:(a) Degree of crystallinity, (b) Deformation by drawing, (c)Annealing of a drawn material

(a) Tensile modulus increases with increasing degree of crystallinity for semicrystallinepolymers. This is due to enhanced secondary interchain bonding which results from adjacent

aligned chain segments as percent crystallinity increases. This enhanced interchain bondinginhibits relative interchain motion.

(b) Deformation by drawing also increases the tensile modulus. The reason for this is thatdrawing produces a highly oriented molecular structure, and a relatively high degree of

interchain secondary bonding.

(c) A drawn semicrystalline polymer that is annealed experiences a decrease in tensile modulus

as a result of a reduction in chain-induced crystallinity, and a reduction in interchain bondingforces.