p0616 - ρ w − U π 4 d 2 W ±,² 1 Inlet ρ w β U π 4...

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6.16. CHAPTER 6, PROBLEM 16 547 6.16 Chapter 6, Problem 16 Use a stationary control volume surrounding the main tank as shown in Figure 6.16. Figure 6.16: Cylindrical tank. Because the control volume is stationary, u rel = u , and we can write the mass-conservation principle as d dt 888 V ρ dV ± 1 =0 + 8 s 8 S ρ u · n dS =0 whe reweob se rvetha ttheamoun to fwa te rinthetanki scon s
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Unformatted text preview: ρ w − U π 4 d 2 W ± ,² 1 Inlet + ρ w β U π 4 9 4 d 2 W ± ,² 1 Outlet A + ρ w α U π 4 16 25 d 2 W ± ,² 1 Outlet B = 0 Simplifying and dividing through by πρ Ud 2 / 4 , − 1 + 9 4 β + 16 25 α = 0 which tells us that α = 100 − 225 β 64 Finally, if β = 3 / 5 , then α = 100 − 135 64 = − 35 64 This means the “outlet” velocity is less than zero. Hence, fluid enters at “Outlet” B with a speed of 35 64 U ....
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This note was uploaded on 10/12/2009 for the course AME 309 at USC.

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