CHE218_solutions7

# CHE218_solutions7 - Set 07 6.13 r02 r 2 P Recall the...

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Set 07 6.13 Recall the velocity profile for Poisueille Flow: Δ Δ - - = x P r r r V μ 4 ) ( 2 2 0 (eqn 6.8) and maximum velocity occurs at r = 0: Δ Δ - = Δ Δ - = = x P D x P r V V μ μ 16 4 ) 0 ( 2 0 2 0 max Subsequent work showed average velocity as: Δ Δ - = = x P D V V avg μ 32 2 2 0 max which, upon solving for the pressure differential: 2 0 32 D V x P avg μ = Δ Δ - Also, D xV f D xV avg avg 2 2 2 or Δ = Δ where the proportionality constant is 2 f . Using a horizontal run, the BE dictates that the total friction be: ρ P Δ - = . Equating the two, we have: Δ - = Δ ρ P D xV f avg 2 2 and upon solving for f and using the previously derived pressure differential term, we have: = = Δ Δ - = 16 2 32 2 2 0 2 0 2 0 avg avg avg V D D V V D x P f ρ μ ρ . 6.14* a) The flow is laminar, so doubling the flow rate doubles the pressure drop to 20 psi/ 1000 ft.

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