fluidmech

Fluidmech

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Unformatted text preview: ILLE'S EQUATION for LAMINAR FLOW Poiseuille did the original derivation shown below which relates pressure loss in a pipe to the velocity and viscosity for LAMINAR FLOW. His equation is the basis for measurement of viscosity hence his name has been used for the unit of viscosity. Consider a pipe with laminar flow in it. Consider a stream tube of length ∆L at radius r and thickness dr. Fig.2.3 du du =− dy dr The shear stress on the outside of the stream tube is τ. The force (Fs) acting from right to left is due to the shear stress and is found by multiplying τ by the surface area. y is the distance from the pipe wall. y = R − r dy = −dr Fs = τ x 2πr ∆L For a Newtonian fluid , τ = µ du du . Substituting for τ we get the following. = −µ dy dr du dr The pressure difference between the left end and the right end of the section is ∆p. The force due to this (Fp) is ∆p x circular area of radius r. Fs = - 2πr∆Lµ Fp = ∆p x πr2 Equating forces we have - 2πrµ∆L du = − du = ∆pπr 2 dr ∆p rdr 2 µ∆L In order to obtain the velocity of the streamline at any radius r we must integrate between the limits u = 0 when r = R and u = u when r = r. u ∫ du = 0 ∆p rdr 2µ∆L ∫ R r ∆p r 2 − R2 4µ∆L ∆p u= R2 − r 2 4µL ( u=− ( ) ) © D.J.DUNN 10 This is the equation of a Parabola so if the equation is plotted to show the boundary layer, it is seen to extend from zero at the edge to a maximum at the middle. Fig.2.4 For maximum velocity put r = 0 and we get ∆pR 2 u1 = 4 µ∆L The average height of a parabola is half the maximum value so the average velocity is um = ∆pR 2 8µ∆L Often we wish to calculate the pressure drop in terms of diameter D. Substitute R=D/2 and rearrange. ∆p = 32 µ∆Lu m D2 The volume flow rate is average velocity x cross sectional area. Q= πR 2 ∆pR 2 πR 4 ∆p πD 4 ∆p = = 8µ∆L 8µ∆L 128µ∆L This is often changed to give the pressure drop as a friction head. The friction head for a length L is found from hf =∆p/ρg hf = 32 µLu m ρgD 2 This is Poiseuille's equation that applies only to laminar flow. © D.J.DUNN 11 WORKED EXAMPLE 2.2 A capillary tube is 30 mm long and 1 mm bore. The head required to produce a flow rate of 8 mm3/s is 30 mm. The fluid density is 800 kg/m3. Calculate the dynamic and kinematic viscosity of the oil. SOLUTION Rearranging Poiseuille's equation we get h ρgD 2 µ= f 32Lu m πd 2 π x 12 = = 0.785 mm 2 4 4 Q 8 um = = = 10.18 mm/s A 0.785 0.03 x 800 x 9.81 x 0.0012 µ= = 0.0241 N s/m or 24.1 cP 32 x 0.03 x 0.01018 µ 0.0241 ν= = = 30.11 x 10 -6 m 2 / s or 30.11 cSt ρ 800 A= WORKED EXAMPLE No.2.3 Oil flows in a pipe 100 mm bore with a Reynolds number of 250. The dynamic viscosity is 0.018 Ns/m2. The density is 900 kg/m3. Determine the pressure drop per metre length, the average velocity and the radius at which it occurs. SOLUTION Re=ρum D/µ. Hence um = Re µ/ ρD um = (250 x 0.018)/(900 x 0.1) = 0.05 m/s ∆p = 32µL um /D2 ∆p = 32 x 0.018 x 1...
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This document was uploaded on 02/07/2014.

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