APPLICATIONS OF THE NAVIERSTOKES EQUATIO which can be solved for the desired

# Applications of the navierstokes equatio which can be

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CHAPTER 4. APPLICATIONS OF THE NAVIER–STOKES EQUATIONS which can be solved for the desired speed, U : U = radicalBigg 2( p 1 p 2 ) ρ . We remark that it is clear that we have had to use two different streamlines to obtain this result, a fact that is often ignored for this problem. But the results are correct provided we are willing to assume that the flow is irrotational. While this may not be completely accurate it provides a reasonable approximation in the present case. (The reader may wish to consider this in detail for both the inviscid approximation used here and the actual, physical viscous cases.) Moreover, it is also common practice to make automatic corrections to pitot tube readouts of speed to at least partially account for the approximate nature of the analysis. Study of Flow in a Syphon In this subsection we consider a somewhat more elaborate example of the use of Bernoulli’s equation, in fact, one that also requires application of the control-volume continuity equation, or actually just a simple formula for volume flow rate derived from this. In particular, we will analyze flow in a syphon, a device that can be used to transfer liquids between two containers without using a pump. EXAMPLE 4.6 Figure 4.7 provides a schematic of a syphon being used to extract liquid from a tank with only the force of gravity. The heights y 2 and y 3 and areas A 1 , A 2 and A 3 are assumed known with A 2 = A 3 , and the pressure acting on the liquid surface in the tank, as well as that at y y x g y 1 3 2 2 3 Figure 4.7: Schematic of flow in a syphon. the end of the hose draining into the bucket, is taken to be atmospheric; that is, p 1 = p 3 = p atm .
4.2. BERNOULLI’S EQUATION 115 It is desired to determine speed of the liquid entering the bucket and the pressure in the hose at location 2. We will assume steady-state behavior, implying that the volume flow rate through the hose is small compared with the total volume of the tank; i.e. , the height of liquid in the tank does not change very rapidly. It is easily seen from the continuity equation that U 3 = U 2 since the fluid is a liquid, and thus incompressible (with assumed known density) and the corresponding areas are equal. Furthermore, again from the continuity equation, it must be the case that U 2 A 2 = U 1 A 1 . We next observe that the figure implies A 1 A 2 , and it follows that U 1 U 2 . Thus, we will be able to neglect U 1 when used together with U 2 . We now apply Bernoulli’s equation. It should be recognized that to find the flow speed at location 2 via Bernoulli’s equation it will be necessary to know the pressure at this location, and this must be found from the pressure at location 3 (known to be atmospheric) again via Bernoulli’s equation and the fact, already noted, that the flow speeds are the same at these locations. We begin by writing the equation between locations 1 and 2. By rearranging Eq. (4.13) we obtain the form p 1 γ + U 2 1 2 g + y 1 = p 2 γ + U 2 2 2 g + y 2 , but we observe from the figure that y 1 = y 2 , and we assume U 1 can be neglected. Since p 1 = p atm , this results in U 2 2 2 g = p atm p 2 γ .

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