428 Substitution of 436 into Eq 435b and slight rearrangement of the result

# 428 substitution of 436 into eq 435b and slight

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the same pressure gradient treatment used in the plane Poiseuille case, Eq. (4.28).) Substitution of (4.36) into Eq. (4.35b), and slight rearrangement of the result, yields ∂r parenleftbigg r ∂u ∂r parenrightbigg = p μL r , (4.37) a second-order differential equation needing two boundary conditions in order to determine inte- gration constants and produce a complete solution. The first of these corresponds to the no-slip condition for a viscous flow, applied at the pipe wall r = R ; that is, u ( R ) = 0 . (4.38) The second condition to be employed in the present case is less intuitive, but it is one that arises often when working with differential equations expressed in polar coordinate systems. Namely, we require that the velocity remain bounded at r = 0, the center of the pipe. This is equivalent to imposing the formal mathematical condition ∂u ∂r vextendsingle vextendsingle vextendsingle vextendsingle r =0 = 0 , (4.39) which is a natural choice that will enforce the expected symmetry of the velocity profile depicted in Fig. 4.13, We now integrate Eq. (4.37) to obtain r ∂u ∂r = p 2 μL r 2 + C 1 , or ∂u ∂r = p 2 μL r + C 1 r . (4.40) Next we observe that at r = 0 this formula for the velocity derivative will become unbounded unless C 1 = 0. Thus, on physical grounds we should set C 1 = 0. But formally, we can apply the second boundary condition given in Eq. (4.39) to show that, indeed, if this condition is satisfied C 1 = 0 will hold, thus eliminating the unbounded term. We also remark that the equation preceding (4.40) leads to this result without any assumptions beyond boundedness of ∂u/partialr —just simple evaluation at r = 0. (Note that if the term C 1 /r were retained, a second integration, as will be performed next, would result in u ln r which is unbounded as r 0.) We can now integrate (4.40) with C 1 = 0 to obtain u ( r ) = p 4 μL r 2 + C 2 , (4.41) and application of the no-slip condition imposed by Eq. (4.38) gives 0 = pR 2 4 μL + C 2 , or C 2 = pR 2 4 μL .
132 CHAPTER 4. APPLICATIONS OF THE NAVIER–STOKES EQUATIONS Finally, substitution of this back into Eq. (4.41) yields u ( r ) = p 4 μL ( r 2 R 2 ) = pR 2 4 μL bracketleftbigg 1 parenleftBig r R parenrightBig 2 bracketrightbigg , (4.42) the Hagen–Poiseuille velocity profile for fully-developed flow in a pipe of circular cross section. Some Physical Consequences of the Hagen–Poiseuille Formula We observe from the definition given earlier for ∆ p (= p 1 p 2 ), that we expect ∆ p > 0 to hold; and we see that for this case the flow is indeed in the direction indicated in Fig. 4.13. Furthermore, it is clear from Eq. (4.42) that the variation of u with r is quadratic, and this is quite similar to the velocity profile found for plane Poiseuille flow. Two quantities of interest can be derived directly from the velocity profile obtained above. The first of these is the maximum velocity (actually speed in this case) which from the preceding figures we expect will occur on the centerline of the pipe, i.e. , at r = 0. In fact, we know from calculus that the maximum of a (twice-differentiable) function occurs at the location where the first derivative is zero and, simultaneously, the second derivative is negative. From Eq. (4.42) we see that ∂u ∂r = 1 2 pR 2 μL · r R 2 , and setting this to zero implies that r = 0, as expected, for the location of maximum velocity.

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