The Rheology Handbook. Thomas MezgerЧитать онлайн книгу.
the shear gap. Of course, this applies only if the shear conditions are met as mentioned in the beginning of Chapter 2.2. However, this does not apply to the velocity v which decreases from the maximum value vmax on the upper, movable plate to the minimum value vmin = 0 on the lower, immovable plate. Therefore, when testing pure liquids, sometimes as a synonym for shear rate the term velocity gradient is used (e. g. in ASTM D4092).
b) Calculation of shear rates occurring in technical processes
The shear rate values which are given below are calculated using the mentioned formulas and should only be seen as rough estimations. The main aim of these calculations is to get merely an idea of the dimension of the relevant shear rate range.
1) Coating processes: painting, brushing, rolling or blade-coating
γ ̇ = v/h, with the coating velocity v [m/s] and the wet layer thickness h [m]
Examples
1a) Painting with a brush:
With v = 0.1 m/s and h = 100 µm = 0.1 mm = 10-4 m; result: γ ̇ = 1000 s-1
1b) Buttering bread:
With v = 0.1 m/s and h = 1 mm = 10-3 m; result: γ ̇ = 100 s-1
1c) Applying emulsion paint with a roller
With v = 0.2 m/s (or 5 s per m), and h = 100 µm = 0.1 mm = 10-4 m; result: γ ̇ = 2000 s-1
1d) Blade-coating of adhesive dispersions (e. g. for pressure-sensitive adhesives PSA):
With the application rate AR (i. e. mass per coating area) m/A [g/m2]; for the coating volume V [m3] applies, with the mass m [kg] and the density ρ [1 g/cm3 = 1000 kg/m3]: V = m/ρ
Calculation: h = V/A = (m/ρ)/A = (m/A)/ρ = AR/ρ; with AR = 1 g/m2 = 10-3 kg/m2 holds:
h =10-6 m = 1 µm; and then: γ ̇ = v/h. See Table 2.2 for shear rates occurring in various kinds of blade-coating processes [2.4] [2.5].
| Table 2.2: Shear rates of various kinds of blade-coating processes for adhesive emulsions | |||||
| Coating process | Application rateAR [g/m2] | Coating velocityv [m/min] | Coating velocityv [m/s] | Layer thicknessh [µm] | Approx. shear rate range γ ̇ [s-1] |
| metering blade | 2 to 50 | up to 250 | up to 4.2 | 2 to 50 | 80,000 to 2 mio. |
| roller blade | 15 to 100 | up to 100 | up to 1.7 | 15 to 100 | 10,000 to 100,000 |
| lip-type blade | 20 to 100 | 20 to 50 | 0.33 to 0.83 | 20 to 100 | 3000 to 50,000 |
| present maximum | 2 to 100 | 700 | 12 | 2 to 100 | 120,000 to 6 mio. |
| future plans | up to 1500 | up to 25 | 250,000 to 12.5 mio. | ||
2) Flow in pipelines, tubes and capillaries
Assumptions: horizontal pipe, steady-state and laminar flow conditions (for information on laminar and turbulent flow see Chapter 3.3.3), ideal-viscous flow, incompressible liquid. According to the Hagen/Poiseuille relation , the following holds for the maximum shear stress τw and the maximum shear rate γ ̇ w in a pipeline (index w for “at the wall”):
Equation 2.4
τw = (R ⋅ Δp) / (2 ⋅ L)
Equation 2.5
γ ̇ w = (4 ⋅ V ̇ ) / (π ⋅ R3)
With the pipe radius R [m]; the pressure difference Δp [Pa] between inlet and outlet of the pipe or along the length L [m] of the measuring section, respectively (Δp must be compensated by the pump pressure); and the volume flow rate V ̇ [m3/s]. This relation was named in honor to Gotthilf H. L. Hagen (1797 to 1848) [2.6] and Jean L. M. Poiseuille (1799 to 1869) [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