Polymer processing and recycling - Esercizi risolti

EXERCISES

1)​ Shear

A molten layer of PMMA at 190°C is placed between two parallel plates. Suppose that a

shear stress of 100 kPa is applied to the mobile plate and determine the speed of the mobile

plate. The distance between plates is 3 mm. You can use the following graph:

From the graph it is possible to obtain the corresponding apparent viscosity at 190°C:

4 .

η = 4 · 10 ·

The general formula is so we can

τ = η γ

retrieve the shear rate as

−1

τ 100.000 .

γ= = = 2, 5

4

η 4·10 ·

The shear rate is defined as so, in order to

γ =

find the velocity of the plate, we need to integrate γ

over the layer thickness:

= γ

ℎ

∫ = γ ∫ ℎ

0 0

ℎ

= γ ∫ ℎ = γℎ = 2, 5 · 0, 003 = 7, 5

0

2)​ Viscosity and molecular weight

Considering the relationship between molecular weight and viscosity above the critical

molecular weight, what is the % decrease in mean molecular weight to reduce the zero shear

viscosity by 50%.

The relationship between molecular weight and viscosity when we are above the critical

3.4

( )

η

molecular weight is .

2 2

=

η

1 1 0,5η

A decrease in zero shear viscosity is defined by a left hand side that becomes so

1 = 0, 5

η

1

3.4

( )

now we just need to solve .

2

0, 5 =

1

( ) 1

Isolating . This means that the molecular weight must decrease by

2 3,4

= 0, 5 = 0, 81

1

19% to reach a 50% viscosity.

3)​ Viscosity and temperature.

Calculate by how much the temperature should be raised (compared to 100°C) to reduce the

viscosity of PIB by 10 times. The activation energy to obtain viscous flow in PIB is

.

= 11. 4

The relationship between viscosity and temperature is an Arrhenius law .

η () =

0

Since R is we need to convert the activation energy in J/mol:

= 8, 314

.

= 11. 4 · 4186 = 47720, 4

( )

·

η ( )

To compare two viscosities we need to take their ratio: .

0 2 = 2

η ( ) ( )

0 1

· 1

η ( ) 0,1η ( )

Viscosity must decrease by 10 times so .

0 2 0 1

= = 0, 1

η ( ) η ( )

0 1 0 1

The pre-exponential factor does not depend on temperature so

( ) ( )

( )

·

.

= = −

2 2

( ) ( )

2 1

·

1 1

Now we just need to solve the equation with T as the only unknown:

2

( )

0, 1 = −

2 1

0, 1 = −

2 1

0, 1 + =

1 2

( )

1

0, 1 + · =

1 2 −1

( )

⎡⎢ ⎤

= 0, 1 + · ⎥

2 ⎣ ⎦

1 −1

( )

47720 8,314

⎡ ⎤

= 0, 1 + · = 438

⎣ ⎦

8,314·373 47720

2 1

4)​ Capillary rheometer

A polymer melt flows through a cylinder with a 2 mm diameter with a volumetric flow rate Q

-8 3

= 10 m /s. Assuming that under such conditions the polymer behaves as newtonian, with

2

viscosity = 10 Pa s; quantify the pressure drop in the tube for unit of length and the shear

η

stress at the wall of the tube.

From the Hagen Poiseuille equation, the flowrate inside a cylinder (assuming incompressible

fluid, isothermal flow, no slip condition at the walls, newtonian fluid and laminar flow

4

π ∆

regime) is .

= 8η

From this equation we can evaluate the pressure drops in the tube per unit of length:

−8 2 6

·8η 10 ·8·10

∆ .

= = = 2, 5 · 10

4

4

( )

0,002

π π 2 0,002 6

∆

The shear stress at the wall is evaluated as .

2

τ = = · 2, 5 · 10 = 1273

2 2

5)​ Capillary rheometer.

From laboratory tests the following empirical correlation has been assigned to a capillary

viscometer for a polymeric solution under room temperature:

−6 2

-1

where is in .

is in s and

γ = 0. 0257τ + 1. 68 · 10 τ γ τ 2

Calculate the volumetric flow rate in cm³/s for such polymeric solution through a 200 m long

6

tube with D (inner diameter) = 2 cm and 100 bar pressure drop. (Note: 1 bar = 10

i

dine/cm²).

We can evaluate the shear stress at the wall thanks to the equation

0,02 5

100·10

∆ .

2

τ = = · = 250

2 200

2

−3 −3 6

250

2

Since

250 = = 2, 5 · 10 = 2, 5 · 10 · 10 = 2500

5 2

10

the shear stress can be expressed as .

τ = 2500 2

The shear rate can be evaluated from the empirical correlation

−6 2

γ = 0. 0257τ + 1. 68 · 10 τ

−6 2 −1

.

γ = 0. 0257 · 2500 + 1. 68 · 10 · 2500 = 74, 75

Since we have both shear stress and shear rate at the wall we can evaluate the apparent

viscosity τ 250 .

η = = = 3, 34 ·

74,75

γ

From the Hagen Poiseuille equation (assuming isothermal and laminar flow of a newtonian

fluid with no slip condition) we obtain:

4 4 5 3

−5

π ∆ π·0.01 ·100·10

= = = 5, 87 · 10

8η 8·3,34·200 2

6)​ Brinkmann number.

Let’s consider a single screw extruder with D = 60 mm and a channel depth of 4 mm,

i

rotating at a speed 60 rpm. The molten polymer is PC with a viscosity of 100 Pa s and a

thermal conductivity of the melt k = 0.2 W/(m K). The heated sections are at 300°C while the

mean melt temperature is 270°C. Steady-state conditions are met. Determine the

adimensional Brinkman number.

From its definition, the Brinkmann number is computed as

2 2

η η

(

π

)

= = =

∆ ∆

so its value in this application is 2

( )

60

100 π0,06· 60

= = 0, 59

0,2·

(

300−270

)

7)​ Required energy to heat up a polymer.

Quantify the amount of energy necessary to heat 1 kg of PP from ambient temperature up to

the processing temperature of 200°C knowing that , ,

= 165° = 2, 34 °

and . The monomer of polypropylene has a molecular

= 1, 61 ∆ = 8, 7

°

weight of 42,08 g/mol.

The amount of energy needed to increase the temperature of a polymer is defined as the

variation of enthalpy between the starting and the ending point. From 25°C to 200°C we need

to account for:

●​ sensible heat to bring PP to 165°C

●​ latent heat of fusion

●​ sensible heat to bring PP to 200°C

[ ( ) ( ) ]

∆ · = − + ∆ + − ·

1 2

We need to convert the latent heat of fusion in kJ per unit mass:

∆ = 8, 7 /42, 08 = 0, 207 = 207

∆ · = [ 2, 34

( 165 − 25 ) + 207 + 1, 61 ( 200 − 165

) ] · 1 = 590

8)​ Conversion of mechanical energy in extrusion.

Calculate the temperature increase of a molten polymer due to a pressure drop of 68 atm

through the die assuming no h