fluidmech

1 2 1 2 y 1 k 2 djdunn 30 the flow of a plastic

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Unformatted text preview: e, chocolate and Mayonnaise). Graph F shows the relationship for a plastic fluid that exhibits shear thickening characteristics. Graph G shows the relationship for a Casson fluid. This is a plastic fluid that exhibits shearthinning characteristics. This model was developed for fluids containing rod like solids and is often applied to molten chocolate and blood. Fig.4.1 © D.J.DUNN 29 MATHEMATICAL MODELS The graphs that relate shear stress τ and rate of shear strain γ are based on models or equations. Most are mathematical equations created to represent empirical data. Hirschel and Bulkeley developed the power law for non-Newtonian equations. This is as follows. & τ = τ y + Kγ n K is called the consistency coefficient and n is a power. In the case of a Newtonian fluid n = 1 and τy = 0 and K = µ (the dynamic viscosity) & τ = µγ For a Bingham plastic, n = 1 and K is also called the plastic viscosity µp. The relationship reduces to & τ = τy + µp γ For a dilatent fluid, τy = 0 and n>1 For a pseudo-plastic, τy = 0 and n<1 The model for both is & τ = Kγ n & The Herchel-Bulkeley model is as follows. τ = τ y + Kγ n This may be developed as follows. τ = τ y + Kγ& n τ − τ y = Kγ& n sometimes written as τ − τ y = µ p γ& n where µ p is called the plastic viscosity. dividing by γ& γ& n τ τy − =K = Kγ& n −1 γ& γ& γ& τ τy = + Kγ& n −1 The ratio is called the apparent viscosity µ app γ& γ& τ τy µ app = = + Kγ& n −1 γ& γ& τy For a Bingham plastic n = 1 so µ app = +K γ& For a Fluid with no yield shear value τ y = 0 so µ app = Kγ& n −1 The Casson fluid model is quite different in form from the others and is as follows. 1 2 1 2 y 1 & τ = τ + Kγ 2 © D.J.DUNN 30 THE FLOW OF A PLASTIC FLUID Note that fluids with a shear yield stress will flow in a pipe as a plug. Within a certain radius, the shear stress will be insufficient to produce shearing so inside that radius the fluid flows as a solid plug. Fig. 4.2 shows a typical situation for a Bingham Plastic. Fig.4.2 MINIMUM PRESSURE The shear stress acting on the surface of the plug is the yield value. Let the plug be diameter d. The pressure force acting on the plug is ∆p x πd2/4 The shear force acting on the surface of the plug is τy x π d L Equating we find ∆p x πd2/4 = τy x π d L d = τy x 4 L/∆p or ∆p = τy x 4 L/d The minimum pressure required to produce flow must occur when d is largest and equal to the bore of the pipe. ∆p (minimum) = τy x 4 L/D The diameter of the plug at any greater pressure must be given by d = τy x 4 L/∆p For a Bingham Plastic, the boundary layer between the plug and the wall must be laminar and the velocity must be related to radius by the formula derived earlier. u= ∆p ∆p R2 − r 2 = D2 − d 2 4 µL 16 µL ( ) ( ) FLOW RATE The flow rate should be calculated in two stages. The plug moves at a constant velocity so the flow rate for the plug is simply Qp = u x cross sectional area = u x πd2/4 The flow within the boundary layer is found in the usual way as follows. Consider an elementary ring radius r and width dr. dQ = u x 2πr dr = Q= ∆p R 2 − r 2 x 2πr dr 4 µL ( ) ∆pπ r 2 3 ∫ rR − r dr 2 µL R ( ) ∆pπ ⎡ r 2 R 2 r 4 ⎤ ∆pπ ⎡⎛ R 4 R 4 ⎞ ⎛ r 2 R 2 r 4 ⎞⎤ ⎟−⎜ − ⎥= − − ⎟⎥ Q= ⎢⎜ ⎢ 2 µL ⎣ 2 4 ⎦ r 2µL ⎣⎜ 2 4⎟ ⎜ 2 4 ⎟⎦ ⎝ ⎠⎝ ⎠ R ∆pπ ⎡⎛ R 4 Q= ⎢⎜ 2 µL ⎣⎜ 4 ⎝ ⎞ r 2R2 r 4 ⎤ ⎟− +⎥ ⎟ 2 4⎦ ⎠ The mean velocity as always is defined as um = Q/Cross sectional area. © D.J.DUNN 31 WORKED EXAMPLE 4.1 & The Herchel-Bulkeley model for a non-Newtonian fluid is as follows. τ = τ y + Kγ n . Derive an equation for the minimum pressure required drop per metre length in a straight horizontal pipe that will produce flow. Given that the pressure drop per metre length in the pipe is 60 Pa/m and the yield shear stress is 0.2 Pa, calculate the radius of the slug sliding through the middle. SOLUTION Fig. 3.3 The pressure difference p acting on the cross sectional area must produce sufficient force to overcome the shear stress τ acting on the surface area of the cylindrical slug. For the slug to move, the shear stress must be at least equal to the yield value τy. Balancing the forces gives the following. p x πr2 = τy x 2πrL p/L = 2τy /r 60 = 2 x 0.2/r r = 0.4/60 = 0.0066 m or 6.6 mm © D.J.DUNN 32 WORKED EXAMPLE 4.2 A Bingham plastic flows in a pipe and it is observed that the central plug is 30 mm diameter when the pressure drop is 100 Pa/m. Calculate the yield shear stress. Given that at a larger radius the rate of shear strain is 20 s-1 and the consistency coefficient is 0.6 Pa s, calculate the shear stress. SOLUTION For a Bingham plastic, the same theory as in the last example applies. p/L = 2τy /r 100 = 2 τy/0.015 τy = 100 x 0.015/2 = 0.75 Pa A mathematical model for a Bingham plastic is & τ = τ y + Kγ = 0.75 + 0.6 x 20 = 12.75 Pa © D.J.DUNN 33 ASSIGNMENT 4 1. Research has shown that tomato ketchup has the following viscous properties at 25oC. Consistency coefficient K = 18.7 Pa sn Power n = 0.27 Shear yield stress = 32 Pa Calculate the apparent viscosity when the rate of shear is 1, 10, 100 and 1000 s-1 and conclude on the effect of the shear rate on the apparent viscosity. Answers γ=1 µapp = 50.7 γ = 10 µapp = 6.682 γ = 100 µapp = 0.968 γ = 1000 µapp = 0.153 2. A Bingham plastic fluid has a viscosity of 0.05 N s/m2 and yield stress of 0.6 N/m2. It flows in a tube 15 mm bore diameter and 3 m long. (i) Evaluate the minimum pressure drop required to produce flow. (480 N/m2 ) The actual pressure drop is twice the minimum value. Sketch the velocity profile and calculate the following. (ii) The radius of the solid core. (3.75 mm) (iii) The velocity of the core. (67.5 mm/s) (iv) The volumetric flow rate. (7.46 cm3/s) n 3. ⎛ du ⎞ A non-Newtonian fluid is modelled by the equation τ = K ⎜ ⎟ where n = 0.8 and ⎝ dr ⎠ K = 0.05 N s0.8/m2. It flows through a tube 6 mm bore diameter under the influence of a pressure drop of 6400 N/m2 per metre length. Obtain an expression for the velocity profile and evaluate the following. (i) The centre line velocity. (0.953 m/s) (ii) The mean velocity. (0.5 m/s) © D.J.DUNN 34...
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