p0674 - 6.74. CHAPTER 6, PROBLEM 74 809 6.74 Chapter 6,...

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Unformatted text preview: 6.74. CHAPTER 6, PROBLEM 74 809 6.74 Chapter 6, Problem 74 Problem: Sand of density ρ is being loaded onto a barge as shown. The velocity of the sand is U , the cross-sectional area of the pipe is A, and the mass-flow rate through the pipe is m = ρ U A. ˙ (a) If the mass of the barge is Mo , what is the total mass of sand and barge a time t after the pipe starts emitting sand? (b) Ignoring the instantaneous rate of change of momentum in your control volume, determine the tension, T , in the mooring line as a function of m, U and angle ˙ α. ˙ (c) Find the buoyancy force on the barge as a function of Mo , m, t, U , α, and gravitational acceleration, g . ...... .. ....................................................................................... ................................................................................................................................. . ... . .. . . ... .. . ..................... . ........................ .......... ........ .. ..... ... ....................................................................................................................... ... . ..... .. ... . . . .. . .................. .................... .............................................................. . ................. ................................... ...................... ................. .... ........ α z U g = −g k ρ x .................................................................. ................................................................. .. ............................................................................................................................... .................................................................. .. . . . . . . . . ........................................................................... . .................................................................................... . ... .... .... .... .... .... .... .... ...... . . . . . . . . .. ............................................................................................. .................................................................. .............................................................. .................................................................. . .................................................................. . ................................................................. ................................................................... . .................................................................. . .................................................................. .................................................................. .................................................................................... .. ...... ...... ...... ...... ...... ...... ...... ...... ..... .................................................................. .................................................................. .................................................................. .................................................................. . ............................................................ ............................................................................................................................. . ...................................................................... ........................................................................................................................................................................................................................................................................................................................ ..................................................................................................................................................................................................................................................................................................................................................................................................... ................................................................. .... . T • Solution: Use a stationary control volume whose bounding surface encloses the barge and sand as shown in the figure (a) Since the flow is unsteady, the mass principle for this control volume (for which ure = u) is d ρ dV + ρ u · n dS = 0 dt V S Because there is a net flow of sand into the control volume at a rate m, the closed-surface ˙ integral is ρ u · n dS = ρ(−UA) = −m ˙ S where the minus sign corresponds to a negative flux out of the control volume. Denoting the total mass of the barge and sand by M , by definition M= ρ dV V 810 CHAPTER 6. CONTROL-VOLUME METHOD Therefore, the mass principle simplifies to dM =m ˙ dt So, integrating over time and using the fact that M (0) = Mo , the instantaneous mass of the loaded barge is M (t) = Mo + mt ˙ (b) We are given that the instantaneous rate of change of momentum is negligible. Thus, for this control volume, the x component of the momentum principle is S ρu (u · n) dS = −i · S (p − pa ) n dS − T where T is the tension in the mooring line required to hold barge and sand in place. First, the net flux of x momentum out of the control volume is S ρu (u · n) dS = (ρ U cos α)(−UA) = −mU cos α ˙ Now, the net pressure force on the barge is the buoyancy force, which is directed vertically upward. Hence, the horizontal component vanishes, i.e., −i · S (p − pa ) n dS = 0 So, the x-momentum equation simplifies to T = mU cos α ˙ (c) We again use the fact that the instantaneous rate of change of momentum is negligible. So, the z component of the momentum principle is S ρw (u · n) dS = −k · S (p − pa ) n dS − Mg where M g is the weight of the barge and sand. First, the net flux of z momentum out of the control volume is S ρw (u · n) dS = (−ρ U sin α)(−U A) = mU sin α ˙ As noted above, the net pressure force on the barge is the buoyancy force, which is directed vertically upward. Hence, −k · S (p − pa ) n dS = Fbuoy where Fbuoy is the buoyancy force on the barge. So, the z -momentum equation simplifies to mU sin α = Fbuoy − M g ˙ Therefore, the buoyancy force on the barge is Fbuoy = Mo g + m(gt + U sin α) ˙ ...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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