1
KE-100.3410 Polymer properties
Exercise 4:
Viscoelasticity and rheology
Effect of molecular weight on viscosity:
Above the critical molecular weight the zero viscosity of polymer can be calculated using
equation:
log 0    log M w  B
or
0  kM w
Below the critical molecular weight the constant  is 1.0, above the critical molecular weight
 = 3.4.
Effect of temperature on viscosity:
There is no one equation applicable to all polymers. For amorphous polymers above the glass
transition, WLF equation can be applied:
lg
  C1  T  Ts 

s
C 2  T  Ts 
When T ranges Ts ± 50, universal constants C1 = 8.86 and C2 = 101.6 are applicable a wide
variety of polymers. If, alternatively, the reference temperature is chosen as glass transition
temperature of the polymer, Tg, then the constant values are different: C1 = 17.44 and C2 =
51.6.
Many partly crystalline polymers have the Arrhenius type correlation between viscosity and
temperature.
 E 

 RT 
  k  exp 
Tensile stress
 
F
A
2
Strain
Shear rate
Viscosity
Creep compliance
 t  
l
l0
dy
dt



 t 
J t  
 

Exercise 4.1
The viscosity of amorphous PVC was measured to be 3.9105 Pas at temperature 122 oC.
For processing, the viscosity should be below 2104 Pas, but at least 5000 Pas. At what
temperature should the processing be done?
Solution 4.1
Amorphous PVC follows Williams-Landel-Ferry equation in the temperature range T = Ts50
°C.
lg
  8.86  T  Ts 

 s 101.6  T  Ts 
Solve for the temperature at which the viscosity is at most 2104 Pas :

s
T 
 Ts

8,86  lg
s
 101,6  lg
Where by substituting the values:
2  10 4 Pa  s
3.9  10 5 Pa  s  395 K  412.3K  139 o C
T
2  10 4 Pa  s
8.86  lg
3.9  10 5 Pa  s
 101.6  lg
(1)
3
When the lowest acceptable viscosity is 5000 Pas, the temperature is:
5000 Pa  s
3.9  10 5 Pa  s  395K  422.6 K  149 o C
T
5000 Pa  s
8.86  lg
3.9  10 5 Pa  s
 101.6  lg
So within the temperature range 139 - 149 oC.
Exercise 4.2
Zero viscosity of a linear polyethylene was determined to be 676000 Pas at 190°C. For
polyethylene the constants for comparison of Mw and zero viscosity are k = 3.410-15 Pas and
 = 3.5. The temperature dependence of the viscosity of PE in melt can be estimated with
Arrhenius-type equation and activation energy for HDPE is 27 kJ/mol.
a) What is the molecular weight Mw of PE?
b) How much should the temperature be altered in order to reduce the viscosity by half?
Solution 4.2
a) The weight average molecular weight for the polymer from equation 0
Mw  
0
 3.5
k
676000 Pas g
g

630000
3.4 *10 15 Pas mol
mol
b) Temperature dependence for the viscosity by Arrhenius:
 E 

 RT 
  k  exp 
Activation energy for HDPE is 27 kJ/mol, the temperature can be solved:

 kM w :
4
 E 

k  exp 
 E  1 1 
RT1 
1 2

 
 exp     
2 1
 E 
 R  T1 T2  

k  exp 
 RT2 
8.314 J
1
1 E
1
1
molK ln 2
  ln 2 


J
T2 T1 R
T2 463K 27000
mol
=1/0.001946
T2=513.8K  241°C
The temperature should be increased by 51 °C in order to reduce the viscosity by half.
Exercise 4.3
Polypropylene PP rod attached to the ceiling (length 200 mm, width 25.0 mm, thickness 3.0
mm) is loaded with 30 kg´s. How much will the polymer creep in two minutes when the creep
compliance J(t) follows the equation (t is time in minutes)?
J(t) = 1.5 - exp(-t/6min) GPa-1
Solution 4.3
Stress imposed on the cross section of the polymer rod is:
m
F
N
s2
 
 3.9  10 6 2
A 0.0030m  0.0250m
m
30kg  9.81
Creep  at the moment t is obtained from the Strain:
 t   J t  
 2 min 
1
Where J t   1.5  exp  
GPa = 0.783 GPa-1
6
min


(1)
5
Creep at two minutes
 t   0.783  109
1
 3.9  106 Pa  0.0031
Pa
and thus the PP rod has strained during the two minutes time
l  200mm  0.0031  0.62mm
Polymer viscosity
William-Landel-Ferry (WLF) –equation:
lg
  C1  T  Tg 

g
C 2  T  Tg 
where Tg is the glass transition temperature of the polymer, Tg is viscosity at glass transition
temperature, C1 and C2 constants for specific polymer. Values C1 = 17.44 and C2 = 51.6 can
be used for linear amorphous polymers when T is above Tg:
lg
  17.44  T  Tg 

g
51.6  T  Tg 
Important parameters regarding the correlation of molecular weight with viscosity are the
number average degree of polymerization X n and the number of atoms in the polymer
backbone (Z). For styrene, acrylate and vinyl polymers, the value is Z  2 X n . When the
polymer chain are long enough to form stable entanglements, longer than the critical chain
length Zw > Zc,w, the polymer viscosity  and chain length Zw can be connected by (Sperling,
Introduction to physical polymer science, 4th ed., Wiley & Sons, 2006, p. 533-537.):
 0  KZ w3.4
where K is a constant.
6
Exercise 4.4
The usual processing temperature of polystyrene cups is 160oC and the melt viscosity is then
1.5102 Pa  s, provided that the mainchain length of PS is Zw=800. The quality of the
polymer however varies and one day the Zw=950. Processing is tuned for a particular viscosity
range so how should the processing temperature be altered so that the melt viscosity would
still be 1.5102 Pa  s? Glass transition temperature of PS is Tg=100oC.
Solution 4.4
Viscosity is increased when the molecular weight increases. By increasing the temperature the
viscosity can be kept lower. Solving the constant K first:
 0  KZ w3.4  K 
 0,1
Z w3.,41

1.5  10 2 Pa  s
 2.02  10 8 Pa  s
800 3.4
Viscosity of the novel polymer grade at 160oC:
0,2  KZ w3.,42  2.02 10 8 Pa  s  9503.4  2.69 10 2 Pa  s
The viscosity of this polymer at the glass transition temperature can be obtained using WLF
equation:
lg
  17.44  T  Tg 

 g 
g
51.6  T  Tg 

10
2.69 102 Pa  s

 6.40 1012 Pa  s
17.44T Tg 
17.44( 433K  373K )
51.6  T Tg 
10 51.6  433K 373K 
The new processing temperature T2 can be solved from WLF equation:
7
lg
  17.44T2  Tg 

 g 51.6  T2  Tg 
 lg

T2  Tg   17.44  T2  Tg   51.6  lg 
g
g
lg
 T2 
 T2 
lg

Tg  51.6  17.44Tg
g

lg
 17.44
g
1.5  10 2 Pa  s
373K  51.6  17.44  373K
6.40  1012 Pa  s
 436.8 K  163.6 o C
2
1.5  10 Pa  s
lg
 17.44
6.40  1012 Pa  s
So the processing temperature should be about 4oC higher so that the viscosity would remain
the same.
Exercise 4.5*
There is a novel polymer available for the production line with the following properties: melt
viscosity at 140
o
C is 1105 Pas, glass transition temperature 110
o
C but some
o
decomposition starts at 160 C. The production line is tailored for polymer viscosity 2102
Pas running at 160 oC. What would the processing temperature have to be for the novel
polymer grade in order to have viscosity in the range appropriate for the production line?
How could the decomposition temperature of the polymer be altered (increased)?
Solution 4.5*
Viscosity of the novel polymer at glass transition temperature can be obtained using WLF
equation:
8
lg
  17.44  T  Tg 

g
51.6  T  Tg 
 g 

1.5 105 Pa  s
11

  17.44( 413K 383K )  2.58 10 Pa  s
51.6  T Tg 
10 51.6  413K 383K 
10
17.44 T Tg
and at 160oC temperature:
lg
2  17.44  T2  Tg 

g
51.6  T2  Tg 


17.44 T Tg
  g   g  10
51.6  T Tg


 2.58 10 Pa  s  10
11
17.44( 413K  383K )
51.6   413K  383K 
 675Pa  s
So the viscosity of the melt at 160oC is clearly higher than the extruder optimal 200 Pas .
Increasing the temperature would lower the viscosity. Required temperature would be:
lg
3  17.44T3  Tg 

 g 51.6  T3  Tg 
 lg
3
T3  Tg   17.44  T3  Tg   51.6  lg 3
g
g
lg
 T3 
 T3 
lg
3
T  51.6  17.44Tg
g g

lg 3  17.44
g
2 102 Pa  s
383K  51.6  17.44  383K
2.58 1011 Pa  s
 439.6 K  166oC
2
2 10 Pa  s
lg
 17.44
2.58 1011 Pa  s
So the melt stability of the novel polymer would require improvement so that it could be
processed at this higher temperature. Improvement could be achieved by adding some
stabilizers. On the other hand the viscosity of the novel polymer could be lowered by reducing
the average molecular weight (this might on the other hand lower the melt stability and thus
all changes in the polymer properties would be very much affected).