Lowering the viscosity of a polymer

In processing a batch of polycarbonate at 453 K to make a certain part, it is found that the polycarbonate’s viscosity is twice that desired. Determine the temperature at which this polymer should be processed.

Question

Lowering the viscosity of a polymer

In processing a batch of polycarbonate at 453 K to make a certain part, it is found that the polycarbonate’s viscosity is twice that desired. Determine the temperature at which this polymer should be processed.

Accepted Answer

From Table 10.2,

Glass-Transition and Melting Temperatures of Selected Polymers
Material	[latex]T_{g}[/latex] (K)	[latex]T_{m}[/latex] (K)
Nylon 6,6	330	538
Polycarbonate	423	538
Polyester	346	538
Polyethylene
high density	183	410
low density	163	388
Polymethylmethacrylate	378	—
Polypropylene	259	449
Polystyrene	373	512
Polytetrafluoroethylene (Teflon)	183	600
Polyvinyl chloride	360	485
Rubber	200	—

the [latex] T_{g}[/latex] for polycarbonate is 423 K. Its viscosity at 453 K is determined by Eq. (10.8):

[latex]\log \eta =12-\frac{17.5 \ \Delta T }{52+\Delta T}.[/latex] (10.8)

[latex]\log \eta =12-\frac{17.5(30)}{52+30}=5.6[/latex]

Hence, [latex]\eta =0.396[/latex] MPa-s, and because this magnitude is 2.2 times what is desired, the new viscosity should be 0.18 MPa-s. We substitute this new value to find the new temperature:

[latex]\log (0.18 imes 10^6) =12-\frac{17.5(\Delta T)}{52+\Delta T}.[/latex]

Therefore, [latex]\Delta T = 32.6 [/latex], or the new temperature is 423 + 32.6 = 455.6 K Note that in this solution the numbers for temperature have been rounded, and that viscosity is very sensitive to temperature.

Question 10.2: Lowering the viscosity of a polymer In processing a batch of...

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