In Example 1.2 we replaced a cylindrical problem with a linear approximation. The velocity distribution for this flow, taking the cylindrical character into account V₁ = ∞ (₁ ²²₂ ²) · ( ²² - , ) where R is the radius of the outer cylinder, r is the local radius, k = rinner cylinder / R, and @ is the angular velocity of the inner cylinder. (a) Verify that this distribution shows a zero velocity at the radius of the outer, non-moving cylinder and shows V₂ = kR at the surface of the inner, rotating cylinder. (b) The shear rate in cylindrical coordinates, for a fluid whose velocity depends only on r (equivalent to dV/ dy in rectangular coordinates), is given by 6= shear rate cylindrical coordinates rical) = r 2 " ( 1₁ ²= R²) 6 = w d dr Show that for the above velocity distribution, the shear rate at the surface of the inner cylinder is given by Ve (-;-) (c) Show that the shear rate is 12.26/s, which is 1.15 times the value for the flat approximation in Example 1.2. The
VO is a function of r.
Small r refers to the variable r, as in the radial direction in the cylindrical coordinates. Capital R refers to the radius of the outer cylinder and is a constant. Radius of the inner cylinder is defined as kR, where k (kappa) is also a constant.
In part (a), you are tasked to verify that the profile given in Eqn 1 is correct. You are evaluating the equation at two locations (outer cylinder, inner cylinder).
Recall that angular velocity is w (omega), and w # Ve. VO refers to the local linear velocity in the 0 direction (i.e., tangential velocity).
In part (b), you can ignore the resulting negative sign. You are looking for an expression of the magnitude of the shear rate o (sigma) at the inner cylinder.
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