Polymer Properties
Exercise 4
Viscoelasticity and rheology
Effect of molecular weight on
viscosity
Effect of temperature on
viscosity
• Above the critical molecular
weight the zero viscosity of
polymer can be calculated using
equation:
• For amorphous polymers above
the glass transition, WLF equation
can be applied:

0  kM w
• Below the critical molecular
weight the constant  is 1.0,
above the critical molecular
weight  = 3.4.
• Zero viscosity is determined by
rheology measurement: dynamic
viscosity at zero frequency.
  C1  T  Ts 
lg

s
C 2  T  Ts 
Reference temperature Ts
C1 = 8.86 and C2 = 101.6
Reference temperature Ts = Tg
C1 = 17.44 and C2 = 51.6
1. Viscosity
• Viscosity of an 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?
1)
• Amorphous PVC follows WLF equation in the temperature
range T = Ts  50 °C.
  8.86  T  Ts 
lg

 s 101.6  T  Ts 
• Solve for the temperature at which the viscosity is at most
2104 Pas :
2  10 4 Pa  s

 101,6  lg
s
T 
 Ts

8,86  lg
s
 101.6  lg
5
3
.
9

10
Pa  s  395 K  412.3K
T 
2  10 4 Pa  s
8.86  lg
3.9  105 Pa  s
 T  139o C
1)
• When the lowest acceptable viscosity is 5000 Pas, the
temperature is:
5000 Pa  s
 101.6  lg
5
3
.
9

10
Pa  s  395K  422.6 K  149 o C
T
5000 Pa  s
8.86  lg
3.9  10 5 Pa  s
• Processing should be done within the
temperature range 139 - 149 oC.
2. Viscosity and Mw
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.
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?
2a)
• The weight average molecular weight for the polymer can be
calculated from equation:

0  kM w
Mw 

0
k

3.5
676000 Pas g
g
 630000
15
3.4 *10 Pas mol
mol
2b)
• Temperature dependence for the viscosity by Arrhenius:
 E 

RT


  k  exp 
• When viscosity is reduced by half by altering temperature:
 E 

k  exp 
 E  1 1 
RT1 
1 2

 
 exp     
2 1
R  T1 T2  
 E 


k  exp 
 RT 2 
2b)
• Activation energy for HDPE is 27 kJ/mol, the temperature can be
solved:
8.314 J
1
1 E
1
1
molK ln 2
  ln 2 


T2 T1 R
T2 463K 27000 J
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.
3. Stress-strain (Creep)
 
• Tensile stress
• Strain
 t  
• Shear rate
• Viscosity
F
A
shear stress  
l
l0
 
dy
dt



• Creep compliance
J t  
 t 

F
A
3. Creep
• 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
3)
• Stress imposed on the cross section of the polymer rod is:
m
2
F
6 N
s
 
 3.9  10
A 0.0030m  0.0250m
m2
30kg  9.81
• Creep  at the moment t is obtained from the Strain:
 t   J t  
• Where
 2 min 
1
1
J t   1.5  exp 
GPa  0.783GPa
 6 min 
3)
• Creep at two minutes:
 t   0.783  10
9
1
 3.9  106 Pa  0.0031
Pa
• PP rod has strained during the two minutes time:
l  200mm  0.0031  0.62mm
4. Viscosity and chain length
• 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:
 0  KZ w3.4
where K is a constant
4)
• The usual processing temperature of polystyrene cups is 160
oC 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.
• 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 100oC.
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
8


2
.
02

10
Pa  s
3.4
800
• 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
4)
• The viscosity of this polymer at the glass transition
temperature can be obtained using WLF equation:
  17.44  T  Tg 
lg

 g 
g
51.6  T  Tg 

10
2.69 102 Pa  s
12


6
.
40

10
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:
  17.44  T  Tg 
lg


g
51.6  T  Tg 
4)
lg
  17.44T2  Tg 

 g 51.6  T2  Tg 
 lg

T2  Tg   17.44  T2  Tg   51.6  lg 
g
g
lg
 T2 

Tg  51.6  17.44Tg
g

lg
 17.44
g
The processing temperature
should be about 4oC higher
so that the viscosity would
remain the same.
1.5  10 2 Pa  s
373K  51.6  17.44  373K
lg
12
 T2  6.40  10 Pa  s 2
 436.8 K  163.6 o C
1.5  10 Pa  s
lg
 17.44
12
6.40  10 Pa  s