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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:
• 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.
Effect of temperature on viscosity• For amorphous polymers above
the glass transition, WLF equation can be applied:
wMk0 s
s
s TTC
TTC
2
1lg
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 3.9105 Pas at temperature 122 oC. For processing, the viscosity should be below 2104 Pas, but at least 5000 Pas.
• At what temperature should the processing be done?
1)
• Amorphous PVC follows WLF equation in the temperature range T = Ts 50 °C.
• Solve for the temperature at which the viscosity is at most 2104 Pas :
s
s
s TT
TT
6.101
86.8lg
s
s
s TT
lg86,8
lg6,101
CT
KK
sPa
sPasPa
sPa
T
o139
3.412395
109.3
102lg86.8
109.3
102lg6.101
5
4
5
4
1)
• When the lowest acceptable viscosity is 5000 Pas, the temperature is:
• Processing should be done within the temperature range 139 - 149 oC.
CKK
sPa
sPasPa
sPa
T o1496.422395
109.3
5000lg86.8
109.3
5000lg6.101
5
5
2. Viscosity and Mw
Zero viscosity of a linear polyethylene was determined to be 676000 Pas at 190°C. For polyethylene the constants for comparison of Mw and zero viscosity are k = 3.410-15 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:
wMk0
mol
g
mol
g
Pas
Pas
kM w 630000
10*4.3
6760005.3
150
2b)
• Temperature dependence for the viscosity by Arrhenius:
• When viscosity is reduced by half by altering temperature:
RT
Ek exp
21
2
1
2
1 11exp
exp
exp
1
2
TTR
E
RT
Ek
RT
Ek
2b)• Activation energy for HDPE is 27 kJ/mol, the temperature can be
solved:
=1/0.001946
T2=513.8K = 241°C
• The temperature should be increased by 51 °C in order to reduce the viscosity by half.
2ln27000
314.8
463
112ln
11
212 molJmolKJ
KTR
E
TT
3. Stress-strain (Creep)
• Tensile stress shear stress • Strain
• Shear rate
• Viscosity
• Creep compliance
0l
lt
dt
dy
t
tJ
A
F
A
F
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:
• Creep at the moment t is obtained from the Strain:
• Where
26
2
109.30250.00030.0
81.930
m
N
mms
mkg
A
F
tJt
11 783.0min6
min2exp5.1
GPaGPatJ
3)
• Creep at two minutes:
• PP rod has strained during the two minutes time:
0031.0109.31
10783.0 69 PaPa
t
mmmml 62.00031.0200
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:
where K is a constant
4.30 wKZ
4)
• The usual processing temperature of polystyrene cups is 160 oC and the melt viscosity is then 1.5102 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 1.5102 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:
• Viscosity of the novel polymer grade at 160oC:
sPasPa
ZKKZ
ww
8
4.3
2
4.31,
1,04.30 1002.2
800
105.1
sPasPaKZw 24.384.32,2,0 1069.29501002.2
4)
• The viscosity of this polymer at the glass transition temperature can be obtained using WLF equation:
• The new processing temperature T2 can be solved from WLF equation:
sPa
sPa
TT
TT
KK
KK
TT
TTgg
g
gg
g
12
3734336.51
)373433(44.17
2
6.51
44.17 1040.6
10
1069.2
106.51
44.17lg
g
g
g TT
TT
6.51
44.17lg
CK
sPa
sPa
KKsPa
sPa
T
TT
T
TTTT
TT
TT
o
g
ggg
ggg
g
g
g
g
6.1638.436
44.171040.6
105.1lg
37344.176.513731040.6
105.1lg
44.17lg
44.176.51lg
lg6.5144.17lg
6.51
44.17lg
12
2
12
2
2
2
22
2
2
4)
The processing temperature should be about 4oC higher so that the viscosity would remain the same.