The shear stress acting on the element increases by d

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Unformatted text preview: x 0.05/0.12 ∆p= 2.88 Pascals. u = {∆p/4Lµ}(R2 - r2) which is made equal to the average velocity 0.05 m/s 0.05 = (2.88/4 x 1 x 0.018)(0.052 - r2) r = 0.035 m or 35.3 mm. © D.J.DUNN 12 2.6. FLOW BETWEEN FLAT PLATES Consider a small element of fluid moving at velocity u with a length dx and height dy at distance y above a flat surface. The shear stress acting on the element increases by dτ in the y direction and the pressure decreases by dp in the x dp d 2u =µ 2 direction. It was shown earlier that − dx dy It is assumed that dp/dx does not vary with y so it may be regarded as a fixed value in the following work. Fig.2.5 dp du =µ +A Integrating once - y dy dx y 2 dp = µu + Ay + B............(2.6 A) 2 dx A and B are constants of integration. The solution of the equation now depends upon the boundary conditions that will yield A and B. Integrating again - WORKED EXAMPLE No.2.4 Derive the equation linking velocity u and height y at a given point in the x direction when the flow is laminar between two stationary flat parallel plates distance h apart. Go on to derive the volume flow rate and mean velocity. SOLUTION When a fluid touches a surface, it sticks to it and moves with it. The velocity at the flat plates is the same as the plates and in this case is zero. The boundary conditions are hence u = 0 when y = 0 Substituting into equation 2.6A yields that B = 0 u=0 when y=h Substituting into equation 2.6A yields that A = (dp/dx)h/2 Putting this into equation 2.6A yields u = (dp/dx)(1/2µ){y2 - hy} (The student should do the algebra for this). The result is a parabolic distribution similar that given by Poiseuille's equation earlier only this time it is between two flat parallel surfaces. © D.J.DUNN 13 FLOW RATE To find the flow rate we consider flow through a small rectangular slit of width B and height dy at height y. Fig.2.6 dQ = u Bdy =(dp/dx)(1/2µ){y2 - hy} Bdy The flow through the slit is Integrating between y = 0 and y = h to find Q yields The mean velocity is Q = -B(dp/dx)(h3/12µ) um = Q/Area = Q/Bh hence um = -(dp/dx)(h2/12µ) (The student should do the algebra) 2.7 CONCENTRIC CYLINDERS This could be a shaft rotating in a bush filled with oil or a rotational viscometer. Consider a shaft rotating in a cylinder with the gap between filled with a Newtonian liquid. There is no overall flow rate so equation 2.A does not apply. Due to the stickiness of the fluid, the liquid sticks to both surfaces and has a velocity u = ωRi at the inner layer and zero at the outer layer. Fig 2.7 © D.J.DUNN 14 If the gap is small, it may be assumed that the change in the velocity across the gap changes from u to zero linearly with radius r. τ = µ du/dy But since the change is linear du/dy = u/(Ro-Ri) = ω Ri /(Ro-Ri) τ = µ ω Ri /(Ro-Ri) Shear force on cylinder F = shear stress x surface area F = 2πRi hτ = 2πRi2 hµω Ro − Ri Torque = F x R i 2πRi3 hµω Ro − Ri In the case of a rotational viscometer we rearrange so that T ( Ro − R ) µ= 2πRi3 hω In reality, it is unlikely that t...
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This document was uploaded on 02/07/2014.

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