The Rheology Handbook. Thomas Mezger

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СКАЧАТЬ rel="nofollow" href="#ulink_9ef43d72-20e2-5d09-bab6-6d90ce680098">[2.7].

      Examples

      2a) Pipeline transport of automotive coatings [2.8] [2.9]

      For a closed circular pipeline with the diameter DN 26 (approx. R = 13 mm = 1.3 ⋅ 10-2 m), and the volume flow rate V ̇ = 1.5 to 12 L/min = 2.51 ⋅ 10-5 to 2.00 ⋅ 10-4 m3/s; results: γ ̇ w = 14.6 to 116 s-1 = approx. 15 to 120 s-1. For a pipeline branch with DN 8 (approx. R = 4 mm = 4 ⋅ 10-3 m), and V ̇ = 0.03 to 1 L/min = 5.06 ⋅ 10-7 to 1.67 ⋅ 10-5 m3/s; results: γ ̇ w = 10.1 to 332 s-1 = approx. 10 to 350 s-1

      2b) Drinking water supply, transport in pipelines [2.10]

      For a pipeline with the diameter DN 1300 (approx. R = 650 mm = 0.65 m), and a volume flow rate of max. V ̇ = 3300 L/s = 3.30 m3/s; and for a second pipeline with DN 1600 (approx.

      R = 800 mm = 0.80 m) with max. V ̇ = 4700 L/s = 4.70 m3/s; results: max. γ ̇ w = 15.3 and 11.7 s-1, respectively.

      2c) Filling bottles using a filling machine (e. g. drinks in food industry):

      Filling volume per bottle: V = 1 L = 0.001 m3; filling time per bottle: t = 5 s, then:

       V ̇ = V/t = 2 ⋅ 10-4 m3/s; diameter of the circular geometry of the injection nozzle: d = 2R = 10 mm; result: γ ̇ w = 2037 s-1 = approx. 2000 s-1

      2d) Squeezing an ointment out of a tube (e. g. pharmaceuticals):

      Pressed out volume: V = 1cm3 = 10-6 m3; time to squeeze out: t = 1 s; then: V ̇ = V/t = 10 -6 m3/s; diameter of the tube nozzle: d = 2R = 6 mm; result: γ ̇ w = 47.2 s-1 = approx. 50 s-1

      2e) Filling ointment into tubes using a filling machine (e. g. medicine):

      Filling volume per tube: V = 100 ml = 10-4 m3; filling time per tube (at 80 work-cycles per minute, where 50 % is filling time): t = (60 s/2)/80 = 0.375 s; then: V ̇ = V/t = 2.67 ⋅ 10-4 m3/s, using an injection nozzle with an annular geometry and a cross-sectional area of A = 24 ⋅ 10-6 m2, which for a rough estimation, corresponds to a circular area showing R = 2.76 ⋅ 10-3 m (since A = π ⋅ R2); result: γ ̇ w = 16,200 s-1

      2f) Transport process of a stucco gypsum suspension during production of architectural plates [2.11]

      Size of the plates to be produced, made of stucco gypsum: Thickness h = 1.2 cm = 0.012 m und width b = 1.20 m; production speed v = L/t = 60 m/min = 1 m/s, with the length L of the plates; thus: necessary volume flow rate V ̇ 1 = V/t = (h · b · L)/t = 0.0144 m3/s; for a mixer with three outlet pipes, each of them with a diameter of d = 2R = 75 mm; thus, for each single pipe counts: V ̇ = (14.4 · 10-3 m3/s) / 3 = 4.80 · 10-3 m3/s, resulting in: γ ̇ w = 116 s-1

      3) Sedimentation of particles in suspensions

      Assumptions: fluid in a state-at-rest; the particles are almost suspended and therefore they are sinking very, very slowly in a steady-state process (laminar flow, at a Reynolds number Re ≤ 1; more about Re numbers: see Chapter 10.2.2.4b); spherical particles; the values of the weight force FG [N] and the flow resistance force FR [N] of a particle are approximately equal in size.

      According to Stokes’ law (Georges G. Stokes, 1819 to 1903 [2.12]):

      Equation 2.6

      FG = Δm ⋅ g = FR = 3 ⋅ π ⋅ dp ⋅ η ⋅ v

      with the mass difference Δm [kg] between a particle and the surrounding fluid, the gravitation constant g = 9.81 m/s2, the mean particle diameter dp [m], the shear viscosity of the dispersion fluid η [Pas], and the particles’ settling velocity v [m/s].

      The following applies: Δm = Vp ⋅ Δρ, with the volume Vp [m3] of a particle, and the density difference Δρ [kg/m3] = (ρp - ρfl) between the particles and the dispersion fluid; particle density ρp [kg/m3] and fluid density ρfl [kg/m3].

      The following applies for spheres: Vp = (π ⋅ dp 3) / 6; and therefore, for the settling velocity

      Equation 2.7

      Assumption for the shear rate: γ ̇ = v/h

      with the thickness h of the boundary layer on a particle surface, which is sheared when in motion against the surrounding liquid (the shear rate occurs on both sides of the particle). This equation is valid only if there are neither interactions between the particles, nor between the particles and the surrounding dispersion fluid.

      Assuming simply, that h = 0.1 ⋅ d, then: γ ̇ = (10 ⋅ v)/d

      Examples

      3a) Sedimentation of sand particles in water

      With dp = 10 µm = 10-5 m, η = 1mPas = 10-3 Pas, and ρp = 2.5 g/cm3 = 2500 kg/m3 (e. g. quartz silica sand), and ρfl = 1 g/cm3 = 1000 kg/m3 (pure water); results: v = approx. 8.2 ⋅ 10-5 m/s

      Such a particle is sinking a maximum path of approx. 30 cm in 1 h (or approx. 7 m per day).

      With h = 1 µm results: γ ̇ = v/h = approx. 80 s-1

      3b) Sedimentation of sand particles in water containing a thickener

      With dp = 1 µm = 10-6 m, and η = 100 mPas (e. g. water containing a thickener, measured at

       γ ̇ = 0.01 s-1), and with the same values for ρp and ρfl as above in Example (3a), results: v = approx. 8.2 ⋅ 10-9 m/s (or v = 0.7 mm per day). With h = 0.1 µm results: γ ̇ = 0.08 s-1 approximately.

      Note 1: Calculation of a too high settling velocity if interactions are ignored

      Stokes’ sedimentation formula only considers a single particle sinking, undisturbed on a straight path. Therefore, relatively high shear rate values are calculated. These values do not mirror the real behavior of most dispersions, since usually interactions are occurring. The layer thickness h is hardly determinable. We know from colloid science: It depends on the strength of the ionic charge on the particle surface, and on the ionic concentration of the dispersion fluid (interaction potential, zeta-potential) [2.28] [2.29]. Due to ionic adsorption, a diffuse double layer of ions can be found on the particle surface. For this reason, in reality the result is usually a considerably lower settling velocity. Therefore, and since the shear rate within the sheared layer is not constant: СКАЧАТЬ