The Rheology Handbook. Thomas Mezger

means that the viscosity values depend on the measuring geometry used. This may be the case if tests with a rotational rheometer are performed by use of a too large or a too small measuring gap. By the way, this term is sometimes selected too when testing with capillary viscometers (and melt index testers) if too small or too large capillary diameters are selected (according to standards).

      Note 1: Shear-thinning, time-dependent and independent of time

      Sometimes, the term “shear-thinning” is used to describe time-dependent flow behavior at a constant shear load (see Figure 3.37: no. 2, Figure 3.38: left-hand interval, and Figure 3.41: medium interval). There is a difference between time-dependent shear-thinning behavior (see Chapter 3.4) and shear-thinning behavior which is independent of time (as explained in this section). If no other information is given, the term should be understood in common usage as the latter one.

      Note 2: Very simple evaluation methods

      1) Speed-dependent viscosity ratio (VR) and shear-thinning index

      Some users still perform the following simple test and analysis method which consists of two intervals. In the first part a constantly low rotational speed n1 [min-1] is preset, and in the second part a constantly high speed n2 (usually with n2 = 10 ⋅ n1, for example, at n1 = 3 min-1 and n2 = 30 min-1). Afterwards the “viscosity ratio” (VR) is calculated as follows [3.6]:

      VR = η1(n1)/η2(n2).

      Sometimes this ratio is called the shear-thinning index (ASTM D2196). For ideal-viscous (Newtonian) flow behavior VR =1, for shear-thinning VR > 1, and for shear-thickening VR < 1.

      Example: with η1 = 250 mPas at n1 = 3 min-1, and η2 = 100 mPas at n2 = 30 min-1, results:

      VR = 250/100 = 2.5

      In order to avoid confusion, this ratio should better be called “speed-dependent (or shear rate-

      dependent) viscosity ratio”. Sometimes in out-of-date literature, this speed-dependent ratio is named thixotropy index, TI (ASTM D2556: as thixotropic index). But this term is misleading, since VR quantifies non-Newtonian behavior independent of time, but not thixotropic behavior which is a time-dependent effect. For TI, see also Note 3 in Chapter 3.4.2.2a, and Chapter 3.4.2.2c.

      2) Pseudoplastic index (PPI)

      Some users still use the following simple test and analysis method consisting of two intervals, presetting in the first part a constantly high rotational speed nH [min-1] for a period of t10 = 10 min, and in the second part a constantly low speed nL for another 10 min until reaching time point t20 = 20 min (e. g. when testing ceramic suspensions, with nL = nH/10, for example, at nH = 100 min-1 and nL = 10 min-1). Afterwards the pseudoplastic index (PPI) is calculated as follows [3.7]:

      PPI = [lg ηL(nL, t20) – lg ηH(nH, t10)] / (lg nL – lg nH)

      For ideal-viscous (Newtonian) flow behavior PPI = 0, for shear-thinning (pseudoplastic) PPI < 0, and for shear-thickening (dilatant) PPI > 0.

      Example: with ηH = 0.3 Pas at nH = 100 min-1, and ηL = 1.2 Pas at nL = 10 min-1, then:

      PPI = (lg 1.2 – lg 0.3) / (lg 10 – lg 100) = [0.0792 – (–0.523)] / (1 – 2) = 0.602 / (–1) ≈ –0.6

      Please be aware that η-values are relative viscosity values if the test is performed using a spindle (which is a relative measuring system, see also Chapter 10.6.2). Here, instead of the shear stress often is taken the dial reading DR which is the relative torque value Mrel in %. Then, the viscosity value is calculated simply as η = DR/n (with the rotational speed n in min-1). Usually here, all units are ignored.

      Thus, here: PPI = [lg (DRL/nL) – lg (DRH/nH)] / (lg nL – lg nH)

      Example: with nH = 100 and nL = 10, and with DRH = 50 and DRL = 40, results:

      PPI = [lg (40/10) – lg (50/100)] / (lg 10 – lg 100) = (lg 4 – lg 0.5) / (1 – 2)

      PPI = [0.602 – (–0.301)] / (–1) ≈ –0.9

      Comment: Both determinations, as well VR as well as PPI are not scientific methods.

       3.3.2.1Structures of polymers showing shear-thinning behavior

      The entanglement model

      Example: A chain-like macromolecule of a linear polyethylene (PE) with a molar mass of M = approx. 100,000 g/mol shows a length L of approx. 1 μm = 10-6 m = 1000 nm and a diameter of approx. d = 0.5 nm [3.8]. Macromolecule: Greek makros means large. Therefore, the ratio L/d = 2000:1. Using an illustrative dimensional comparison, this corresponds to a single spaghetti noodle being 1 mm thick – and 2000 mm or 2 m long! So, it is easy to imagine that in a polymer melt or solution these relatively long molecules would entangle loosely with others many times. As a second comparison: A hair with the dia­meter of d = 50 µm showing the length L = 10 cm. For an ultra-high molecular weight PE (UHMW) with M = 3 to 6 mio g/mol, then L/d = 50,000:1 to 100,000:1 approximately. Here, using the illustrative comparison, the piece of spaghetti would be approx. 50 m to 100 m long, and the hair with 2.5 m to 5 m would be permanently out of control!

      At rest, each individual macromolecule can be found in the state of the lowest level of energy consumption: Therefore, without any external load it will show the shape of a three-

      dimensional coil (see Figure 3.8). Each coil shows an approximately spherical shape and each one is entangled many times with neighboring macromolecules.

      During the shear process, the molecules are more or less oriented in shear direction, and their orientation is also influenced by the direction of the shear gradient. When in motion, the molecules disentangle to a certain extent which reduces their flow resistance. For diluted polymer solutions, the chains may even become completely disentangled finally if they are oriented to a high degree. Then, the individual molecules are no longer in the same close contact as before, therefore moving nearly independently of each other (see Figures 3.9 and 3.32: no. 2).

       Figure 3.8: Macromolecules at rest, showing coiled and entangled chains

       Figure 3.9: Macromolecules under high shear load, showing oriented and partially disentangled chains

      Using this concept, the result of the double-tube test can be explained now as follows (see Chapter 2.3, Experiment 2.2 and Figure 2.4): Fluid F1 shows shear-thinning and fluid F2 ideal- viscous flow behavior. At the beginning, there is a certain load on the molecules at the bottom of each tube due to the weight of the column of liquid which is compressing vertically onto them due to the hydrostatic pressure. Regarding the polymer molecules of F1, shortly after the beginning of the Experiment they are moving faster because they are stretched into flow direction now. As a consequence, they are disentangled to a high degree then. Hence, they are able to glide off each other more easily also during the passage through the valve.

      Along with the falling liquid levels, also the shear load or shear stress, respectively, is decreasing continuously. Therefore, the macromolecules are recoiling more and more, and as a consequence, the viscosity of F1 is increasing now. However, F2, the mineral oil, still shows constant viscosity, independent of the continuously changing shear load, since for this ideal-viscous liquid with its very short molecules counts, that there is no significant shear load-dependent change in the flow resistance.

      Figure 3.10 presents the viscosity function of a polymer displaying three intervals on a double logarithmic scale. These three distinct ranges of the viscosity curve only occur for uncrosslinked and unfilled polymers