p0630

# p0630 - 8 s 8 S ( u u cv ) n dS = w 5 3 W k W k 4 p D 2 d 2...

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6.30. CHAPTER 6, PROBLEM 30 565 6.30 Chapter 6, Problem 30 To solve this problem, we select a deforming control volume as shown in Figure 6.30. Figure 6.30: Cylindrical plunger advancing into a cone. The mass-conservation principle tells us that d dt 888 V ρ dV + 8 s 8 S ρ u rel · n dS =0 In a time t , the control volume decreases in size by an amount V = π 4 D 2 W t = dV dt = π 4 WD 2 Therefore, the unsteady term is d dt 888 V ρ dV = ρ dV dt = π 4 ρ WD 2 Turning to the net mass-flux integral, the absolute velocity between the plunger and the cone is u = 2 3 W k , the control volume velocity is u cv = W k and the unit normal is n = k . Thus, letting D denote the diameter of the cone when the plunger velocity matches the velocity of the fluid forced out of the cone,

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Unformatted text preview: 8 s 8 S ( u u cv ) n dS = w 5 3 W k W k 4 p D 2 d 2 Q = 5 12 W p D 2 d 2 Q Thus, the mass-conservation equation simplifies to 4 WD 2 + 5 12 W p D 2 d 2 Q = D 2 = 5 2 d 2 566 CHAPTER 6. CONTROL-VOLUME METHOD Therefore, the plunger velocity matches the velocity of the fluid forced out of the cone when D = 5 2 d Finally, the cone shape is such that D = 0 at z = 0 and D = 2 d at z = 80 d . Thus, in general D = 1 40 z Hence, we have 1 40 z = 5 2 d = z = 40 5 2 d = 63 . 2 d...
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## This note was uploaded on 10/12/2009 for the course AME 309 at USC.

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p0630 - 8 s 8 S ( u u cv ) n dS = w 5 3 W k W k 4 p D 2 d 2...

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