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p06114

# p06114 -

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Unformatted text preview: 908 CHAPTER 6. CONTROL-VOLUME METHOD 6.114 Chapter 6, Problem 114 Problem: Near the entrance to a pipe, for laminar viscous flow the velocity distribution changes from uniform (u1 = U ) to parabolic as shown. At the fully-developed outlet section, the velocity varies according to u2 (r) = Umax [1 − 4(r/D)2 ]. Derive a formula for the net resisting frictional force, Fτ , on the pipe section as a function of U , p1 , p2 , D and fluid density, ρ. ..................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................................................... ......................... ...... ... ........................................................................................................................................................................................................................................................................................ ........................................................................................................................................................................................................................................................................................ . ............................................................................................................................................................................................................. ...... ................. V ou u1 p1 D r ow u2 p2 x V ou ow ..................................................................................................................................................................................................................................................................................... ... . ... . ... . ... . ... . ... . ... . ................................................................................................................................................................................................................................. ........ . ..................................................................................................................................................................................................................................................................................... .......................................................................................................................................................................................................................................................................................................... ... ... ... ... ... ... ... . ....................................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................. ...... ................. ...................................................................................................................................................................................................................................... ........................................................................................................................................................................................................................................................................... . Solution: Use a stationary control volume that lies entirely within the pipe as shown in the figure. Mass Principle: In general, the mass principle tells us that d dt ρ dV + V S ρ ure · n dS = 0 The flow is incompressible and the control volume’s size is constant. Thus, the instantaneous rate of change of the mass in the control volume vanishes. Then, since ure = u, we have ρ u · n dS = 0 S Now, because n is the outer unit normal, we know that u · n is negative at the inlet and positive at the outlet Thus we have π ρ −U D2 + ρ 4 2π R 0 0 In et Umax 1 − r R 2 rdrdθ = 0 Out et where R = D/2 is pipe radius. Then, introducing the dimensionless variable η = r/R and performing the θ integration, we have 2πρ Umax R2 1 0 1 − η 2 η dη = π ρ U D2 4 6.114. CHAPTER 6, PROBLEM 114 909 Performing the indicated integration, there follows: π η2 η4 ρ Umax D2 − 2 2 4 η =1 = η =0 π ρ UD2 4 π π ρ Umax D2 = ρ UD2 8 4 =⇒ Therefore, the maximum velocity at the outlet is Umax = 2U x-Momentum Principle: Including the viscous term, the x-momentum principle is S ρ u(u · n)dS = −i · S p n dS − Fτ S ρ u(u · n)dS So, the friction force is Fτ = −i · S p n dS − Now, the pressure integral is −i · π π π p n dS = − i · p1 D2 (−i) − i · p2 D2 (i) = (p1 − p2 ) D2 4 4 4 S Inlet Outlet The momentum-flux integral is π ρ u(u · n)dS = (ρ U ) −U D2 + 4 S 2π 0 R 0 2 ρ Umax Inlet 22 rdrdθ Outlet π 2 = − ρ U 2 D2 + 2πρ Umax R2 4 π = − ρ U 2 D2 + 2πρ U 2 D2 4 r 1− R 1 0 1 0 2 1 − η 2 η dη η − 2η 3 + η 5 d η where we use the fact that Umax = 2U and R = D/2 in the last line. So, performing the indicated integration, π η2 η4 η6 ρ u(u · n)dS = − ρ U 2 D2 + 2πρ U 2 D2 − + 4 2 2 6 S η =1 η =0 π 1 π = − ρ U 2 D2 + 2πρ U 2 D2 = ρ U 2 D2 4 6 12 910 CHAPTER 6. CONTROL-VOLUME METHOD Thus, the viscous force is Fτ = π π (p1 − p2 ) D2 − ρ U 2 D2 4 12 which can be rewritten as Fτ = π2 1 D p1 − p2 − ρ U 2 4 3 ...
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