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p0614

# p0614 - m o is constant we have d dt w ρ π 4 D 2 h m o W...

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6.14. CHAPTER 6, PROBLEM 14 545 6.14 Chapter 6, Problem 14 Use a stationary control volume surrounding the main tank as shown in Figure 6.14. Figure 6.14: Cylindrical tank. Because the control volume is stationary, u rel = u , and we can write the mass-conservation principle as follows. d dt 888 V ρ dV + 8 s 8 S ρ u · n dS = 0 Now, because n is the outer unit normal, we know that u · n is positive at the outlets, but negative at the inlet. Thus, since the mass of fluid in the inlet and outlet pipe sections,
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Unformatted text preview: m o , is constant, we have d dt w ρ π 4 D 2 h + m o W + ρ X − U π 4 D 2 100 ~ ± ,² 1 Inlet + ρ X 1 4 U π 4 D 2 100 ~ ± ,² 1 Horizontal Outlet + ρ X 1 8 U π 4 D 2 25 ~ ± ,² 1 Vertical Outlet = 0 Simplifying and dividing through by πρ D 2 / 4 , dh dt − 1 100 U + 1 400 U + 1 200 U = 0 which simplifies to dh dt = 1 400 U Therefore, since dh/dt > , the tank is filling....
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