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The Rheology Handbook. Thomas MezgerЧитать онлайн книгу.

The Rheology Handbook - Thomas Mezger


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see Chapter 8.7.1.

      3.1.2.1.2b) Viscosity/temperature shift factor aT, and Arrhenius plot

      The following holds for the viscosity/temperature shift factor aT in general:

      Equation 3.8

      aT = η(T) / η(Tref)

      The dimensionless factor aT is the ratio of the two viscosity values at the temperature T and at the reference temperature Tref. For polymers, this relation is only valid for the values of the zero-shear viscosity η0. The following holds for the temperature shift factor according to Arrhenius (with T in [K]):

      Equation 3.9

eq-003-012

      The semi-logarithmic, so-called Arrhenius plot presents the temperature shift factor aT on the y-axis on a logarithmic scale versus the reciprocal temperature 1/T on the x-axis (with the unit: 1/K).

      In order to estimate viscosity values at temperatures at which no measuring values are available, proceed as follows:

      1 Select Tref (e. g. a temperature, at which an η-value is available).

      2 Calculate the shift factor aT for another available η(T)-value (using Equation 3.8).

      3 Calculate the flow activation energy value EA (using Equation 3.9).

      4 Calculate the shift factor aT for the desired η(T)-value (using Equation 3.9).

      5 Result: Calculate the desired η(T)-value (using Equation 3.8).

Table 3.5: Temperature-dependent viscosity values, see the example of Chapter 3.5.4b
T [°C]50607080
T [K]323333343353
η [Pas]2.001.551.261.00

      3.1.2.1.3Example: Calculation of flow activation energy and viscosity/temperature shift factor of a mineral oil, and determination of viscosity values at further temperatures

      From a mineral oil is known: η1 = 2.00 Pas (at T1 = +50 °C = 323 K), and η2 = 1.00 Pas

      (at T2 = +80 °C = 353 K). Desired is the viscosity value η3 at T3 = +60 °C (= 333 K).

      1 Here is selected: Tref = T1 = 50 °C = 323 K

      2 aT is calculated for T2 (as aT2): aT2 = η2 (T2) / η1 (Tref) = 1.00 Pas / 2.00 Pas = 0.5

      3 EA is calculated with Tref = 323 K and T2 = 353 K:

      ln (aT) = ([EA / RG] ⋅ [(T2)-1 – (Tref)-1]), therefore: EA = ([RG ⋅ ln (aT)] / [(T2)-1 – (Tref)-1])

      EA = 8.314 ⋅ 10–3 (kJ/mol ⋅ K) ⋅ ln(0.5) / (2.83 ⋅ 10–3 – 3.10 ⋅ 10–3) 1/K = 21.3 kJ/mol

      1 aT is calculated for T3 = 333 K (as aT3): aT3 = exp ([EA / RG] ⋅ [(T3)-1 – (Tref)-1]aT3 = exp ([21.3/8.314 ⋅ 10–3] ⋅ [3.00 ⋅ 10–3 – 3.10 ⋅ 10–3]) = exp (-0.256) = 1/e0.256 = 0.774

      2 Result: η3 (T3) = aT3 ⋅ η1(Tref) = 0.774 ⋅ 2.00 Pas = 1.55 Pas

      The already mentioned and a further calculated temperature-dependent viscosity value are presented in Table 3.5.

      BrilleEnd of the Cleverly section

      Conversion of pressure units:

      Equation 3.10

      1 bar = 0.1 MPa = 105 Pa = 105 N/m2

      (and 1000 psi = approx. 6.89 MPa, however, psi is not an SI-unit; see also Chapter 15.3i)

      In most cases, the viscosity values of fluids are increasing with increasing pressure (fluids are gases and liquids). However, liquids are influenced very little by the pressure applied since liquids, in contrast to gases, are almost non-compressible at low and medium pressures. In order to explain the reason for this, you can imagine a volume element filled with the following materials: For gases the degree of space filling is only around 0.1 %, however, for liquids it is around 95 %, and crystalline solids are even completely filling the space. The following comparison illustrates this point: For most liquids, the fairly considerable change in pressure of Δp = 0.1 to 30 MPa (= 1 to 300 bar) causes about the same small change in viscosity like the small change in temperature of ΔT = 1 K.

      Even under the enormous pressure difference of Δp = 0.1 to 200 MPa (= 1 to 2000 bar), the viscosity increase of low-molecular liquids is usually showing just a factor of 3 to 7 only. However, for highly viscous mineral oils this factor can be up to 20,000 and for synthetic oils even up to 8 million. Very high peak pressures may occur as well in natural processes as well as for technical application.

      When performing deep drilling processes, per 100 m depth pressure may increase by 1MPa (10 bar) and temperature by 3 K (3 °C). Examples: Drilling fluids or muds in deep oil wells for exploration of crude oil and natural gas, or for thermal energy. In 5000 m depth should be calculated with p = 50 bis 80 MPa (500 bis 800 bar) and T = 150 °C. In HPHT gas fields (high pressure/high temperature; such as the field “Elgin“ in the Northern Sea), there are up to 110 MPa and about 200 °C. Injection pumps of modern combustion motors may reach values of punctual top pressures up to p = 210 MPa. Lubricants in cogwheels, in ball bearings and sliding bearings or in gears may be confronted with more than 1 GPa [3.9] [3.75] [3.76].

      Fluids, in contrast to solids, are able to flow since their atoms and molecules can change places due to the free space in between them [3.27] [3.36]. For most liquids, the viscosity values are increasing with increasing pressure since the amount of free volume within the internal structure is decreasing due to compression, and therefore, the molecules are more and more limited in their mobility. This increases the internal frictional forces and, as a consequence, also the flow resistance. The dependence of η(p) is stronger if the molecules show a higher degree of branching.

      Water does not behave like most other liquids. For T < +32 °C, at a pressure of up to p = 20 MPa (= 200 bar), above all the structure of the three-dimensional network of hydrogen bridges will be destroyed. In this temperature range, the mentioned network is relatively strong compared to the structures of other low molecular liquids. Therefore here, the viscosity values are decreasing with increasing pressure. For T > +32 °C however, water begins to behave like most other liquids, and then, viscosity increases with increasing pressure.

      For further information on pressure-dependent rheological behavior: see [3.10] [3.69].

      BrilleFor “Mr. and Ms. Cleverly”

      3.2.1.1.1Viscosity/pressure shift factor ap

      The following holds for the viscosity/pressure shift factor ap in general:

      Equation 3.11

      ap = η(p) / η(pref)

      The dimensionless factor ap is the ratio of the two viscosity values at the pressure p and at the reference pressure pref. For polymers, this relation is only valid for the values of the zero-shear viscosity η0. The viscosity/pressure coefficient αp is defined as


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