p0618 - 694 CHAPTER 6. CONTROL-VOLUME METHOD 6.18 Chapter...

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Unformatted text preview: 694 CHAPTER 6. CONTROL-VOLUME METHOD 6.18 Chapter 6, Problem 18 Problem: The figure illustrates a jet pump. At Section 1, a high-speed jet of fluid is injected into a uniform flow of velocity U1 in a duct of area A. The fluid mixes and, at Section 2, returns to nominally uniform flow with velocity U2 = 5 U1 . If the jet area is 2 Aj = 1 A, what is the jet velocity, Uj ? Assume the flow is steady and incompressible 5 with density ρ. .. .......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... .. ... ... ... ... ... ... ... ... ... ... .... .. ... ... ... ... .. .. ....................................................................................................................................................................................................................................................................................................................................................................... . .. .. .. .. ... .. .. .. .. .. .. .. . . . . .............................................................................................................................................................................................................................................................................................................................................................................................................................................. . . . . .. . . . . . . . . . . . . . ... ..................................................................................................................................................................................................................................................................................................................................................................................... Aj u1 Uj U1 U2 A ............................................................................................ ............................................................................................ ............................................................. ............................................................................................ ............................................................................................ .................................................................................................... .................................................................. ...... ...... ...... ... ............................................................................................ . ......................................................................................................................................................................................................................................................................................... ......................................................................................................................................................................................................................................................................................... ......................................................................................................................................................................................................................................................................................... ........... ...... .................................... ................................................................................................................................................................................................ ........................................................................................................................................................................................................................................................................................... j 0 j 1 j 2 Solution: Our primary interest is in relating conditions at the exit plane (Plane 2) and the plane at which the jet injects fluid (Plane 1). However, we don’t know the velocity, u1 . Rather, we are given U1 on Plane 0 upstream of Plane 1. Thus, we use two control volumes in this problem as shown in the figure. We can use the mass principle on the control volume bounded by Planes 0 and 1 to determine u1 as a function of U1 and the appropriate areas. This accounts for the partial blockage created by the physical presence of the pump, which reduces the area from A to (A − Aj ). Then, we can use the mass principle on the control volume bounded by Planes 1 and 2 to complete the solution. Since the flow is steady and incompressible, for a stationary control volume (so that ure = u) coincident with the wind-tunnel walls between Sections 0 and 1, the mass principle simplifies to S ρ u · n dS = 0 Since there is no flow through the tunnel walls, we have 4 4 ρ u · n dS = ρ (−U1 A) + ρ u1 A = ρA −U1 + u1 = 0 5 5 S Sect on 0 Sect on 1 Therefore, the velocity u1 is 5 u1 = U1 4 6.18. CHAPTER 6, PROBLEM 18 695 Now, for a control volume between Sections 1 and 2, we have 4 1 ρ u · n dS = ρ −u1 A + ρ −Uj A + ρ (U2 A) 5 5 S Section 1 Jet Section 2 4 1 5 1 5 = ρA − u1 − Uj + U1 = ρA −U1 − Uj + U1 = 0 5 5 2 5 2 Therefore, the jet velocity is Uj = 15 U1 2 ...
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