Chap 15 Solns
ch14 (1)
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- Diffusion in Polymeric Materials
| ρ (g/cm3)
| |
|
crystallinity (%) | |
|
2.144 |
51.3 |
|
2.215 |
74.2 |
(a) Compute the densities of totally crystalline and totally amorphous polytetrafluoroethylene.
(b) Determine the percent crystallinity of a specimen having a density of 2.26 g/cm3.
Solution
(a) We are asked to compute the densities of totally crystalline and totally amorphous polytetrafluoroethylene (c and a from Equation 14.8). From Equation 14.8 let , such that
Rearrangement of this expression leads to
in which c and a are the variables for which solutions are to be found. Since two values of s and C are specified in the problem statement, two equations may be constructed as follows:
In which s1 = 2.144 g/cm3, s2 = 2.215 g/cm3, C1 = 0.513, and C2 = 0.742. Solving the above two equations for a and c leads to
And
(b) Now we are to determine the % crystallinity for s = 2.26 g/cm3. Again, using Equation 14.8
= 87.9%
14.26 The density and associated percent crystallinity for two nylon 6,6 materials are as follows:
|
ρ (g/cm3) |
crystallinity (%) |
|
1.188 |
67.3 |
|
1.152 |
43.7 |
(a) Compute the densities of totally crystalline and totally amorphous nylon 6,6.
(b) Determine the density of a specimen having 55.4% crystallinity.
Solution
(a) We are asked to compute the densities of totally crystalline and totally amorphous nylon 6,6 (c and a from Equation 14.8). From Equation 14.8 let , such that
Rearrangement of this expression leads to
in which c and a are the variables for which solutions are to be found. Since two values of s and C are specified in the problem, two equations may be constructed as follows:
In which s1 = 1.188 g/cm3, s2 = 1.152 g/cm3, C1 = 0.673, and C2 = 0.437. Solving the above two equations for a and c leads to
And
(b) Now we are asked to determine the density of a specimen having 55.4% crystallinity. Solving for s from Equation 14.8 and substitution for a and c which were computed in part (a) yields
= 1.170 g/cm3
Diffusion in Polymeric Materials
14.27 Consider the diffusion of water vapor through a polypropylene (PP) sheet 2 mm thick. The pressures of H2O at the two faces are 1 kPa and 10 kPa, which are maintained constant. Assuming conditions of steady state, what is the diffusion flux [in [(cm3 STP)/cm2-s] at 298 K?
Solution
This is a permeability problem in which we are asked to compute the diffusion flux of water vapor through a 2-mm thick sheet of polypropylene. In order to solve this problem it is necessary to employ Equation 14.9. The permeability coefficient of H2O through PP is given in Table 14.6 as 38 10-13 (cm3 STP)-cm/cm2-s-Pa. Thus, from Equation 14.9
and taking P1 = 1 kPa (1,000 Pa) and P2 = 10 kPa (10,000 Pa) we get
14.28 Argon diffuses through a high density polyethylene (HDPE) sheet 40 mm thick at a rate of 4.0 × 10–7 (cm3 STP)/cm2-s at 325 K. The pressures of argon at the two faces are 5000 kPa and 1500 kPa, which are maintained constant. Assuming conditions of steady state, what is the permeability coefficient at 325 K?
Solution
This problem asks us to compute the permeability coefficient for argon through high density polyethylene at 325 K given a steady-state permeability situation. It is necessary for us to Equation 14.9 in order to solve this problem. Rearranging this expression and solving for the permeability coefficient gives
Taking P1 = 1500 kPa (1,500,000 Pa) and P2 = 5000 kPa (5,000,000 Pa), the permeability coefficient of Ar through HDPE is equal to
14.29 The permeability coefficient of a type of small gas molecule in a polymer is dependent on absolute temperature according to the following equation:
where and Qp are constants for a given gas-polymer pair. Consider the diffusion of hydrogen through a poly(dimethyl siloxane) (PDMSO) sheet 20 mm thick. The hydrogen pressures at the two faces are 10 kPa and 1 kPa, which are maintained constant. Compute the diffusion flux [in (cm3 STP)/cm2 –s] at 350 K. For this diffusion system
Qp = 13.7 kJ/mol
Also, assume a condition of steady state diffusion
Solution
This problem asks that we compute the diffusion flux at 350 K for hydrogen in poly(dimethyl siloxane) (PDMSO). It is first necessary to compute the value of the permeability coefficient at 350 K. The temperature dependence of PM is given in the problem statement, as follows:
And, incorporating values provided for the constants PM0 and Qp, we get
And, using Equation 14.9, the diffusion flux is equal to
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