A falling-cylinder viscometer consists of a long vertical cylindrical tube (of radius R), capped at both ends, and a sol

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A Falling Cylinder Viscometer Consists Of A Long Vertical Cylindrical Tube Of Radius R Capped At Both Ends And A Sol 1 (465.18 KiB) Viewed 52 times

A falling-cylinder viscometer consists of a long vertical cylindrical tube (of radius R), capped at both ends, and a solid cylindrical slug (of radius KR). The system is sketched in Fig. 1. The slug is equipped with fins so that its axis coincides with that of the tube. One can observe the rate of descent of the slug in the cylindrical container when the latter is filled with fluid. The objective of this exercise is to find an equation that gives the viscosity of the fluid (assumed to be incompressible and Newtonian) in terms of the terminal velocity Vof the slug and of the various geometric quantities shown in Fig. 1. COO Cylindrical slug descends with speed V through the liquid H Liquid fills the cylindrical cavity Figure 1: Sketch of the falling-cylinder viscometer. Assume that the flow is laminar and neglect the entrance and end effects that are present when the fluid enters and leaves the gap between the tube and the slug. Therefore, assume that the only nonzero component of the fluid velocity vector is v, and that in space this depends solely on r, where r denotes the radial coordinate of a cylindrical coordinate system.
a) From the point of view of an observer integral with the laboratory where the viscometer is placed, is this fluid dynamic problem steady-state or not? Justify your answer. [2/50] b) Select an appropriate observer and control volume. Then, derive the linear momentum balance equation governing the function vz(r), proving that this reads: dar dr dlou) = (429) AP L r with P(2) = p(2) + pgz (1.1) Here, AP is the (positive) dynamic pressure change over a generic length L, g is the magnitude of the gravitational field and p is the fluid density. To answer this question, do not employ the general linear momentum balance equations.
c) Using the Newtonian constitutive equation that relates the viscous stress tensor to the velocity gradient, integrate Eq. 1.1 and show that: 4μLV Uz AIR2 In(1/5) In(1/8) (1 - 82) – (1 – k?) +1 (1.2) V In(1/K)] In(1/K) where u denotes the fluid viscosity and g = r/R. State the boundary conditions used, briefly explaining μ the physical grounds on which they are based. [6/50] d) In Eq. 1.2, the quantity AP/L is unknown; thus, for now the velocity profile is also unknown. But with a mass balance equation written on a macroscopic control volume appropriately chosen, it can be proved (the passages are quite lengthy) that this relation holds: 1 APR 4uLV (1.3) (1 – k2) – (1+k2) In(1/6) so that Eq. 1.2 becomes: Uz (1 – 82) – (1+k?) ln(1/8) +1 (1.4) V (1 – K2) - (1+k2) In(1/6) Identify the macroscopic control volume. Then, derive (without solving) the mass balance equation that gives Eq. 1.3. Justify your answer. [6/50] e) Derive an expression for the mean fluid velocity in the gap between the tube and the slug. For k = 0.95, would taking V as velocity scale be correct? [8/50] f) Using Eq. 1.4, calculate the frictional force (in the axial direction) F, exerted by the fluid on the slug. In particular, show that it is: FZ = [-cs (1 - x^)TR?] (P. – P.) 2 where Po - Ph is the change in dynamic pressure between the bottom and top of the slug. Then, make н a force balance on the slug and obtain an expression relating the fluid viscosity to the terminal velocity V of the slug and to the various geometric quantities characterizing the system. [18/50]