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Problem 8.122 [Difficulty: 3] Given: Flow of oil in a pipe Find: Percentage change in loss if diameter is reduced Solution: Basic equations hl f L D ⋅ V 2 2 ⋅= f 64 Re = Laminar 1 f 2.0− log e D 3.7 2.51 Re f⋅+ ⎛⎜⎜⎝ ⎞ ⎠⋅= Turbulent Available data ν 7.5 10 4−⋅ ft 2 s ⋅= L 100 ft⋅= D 1 in⋅= Q 45 gpm⋅= Q 0.100 ft 3 s = Here V Q A = 4 Q⋅ π D2⋅ = V 4 π 0.1× ft 3 s ⋅ 12 1 1 ft ⋅⎛⎜⎝ ⎞ ⎠ 2 ×= V 18.3 ft s ⋅= Then Re V D⋅ ν = Re 18.3 ft s ⋅ 1 12 × ft⋅ s 7.5 10 4−× ft2⋅ ×= Re 2033= The flow is LAMINAR hl f L D ⋅ V 2 2 ⋅= hl 64 Re L D ⋅ V 2 2 ⋅= hl 64 2033 100 1 12 × 18.3 ft s ⎛⎜⎝ ⎞ ⎠ 2 2 ×= hl 6326 ft2 s2 ⋅= When the diameter is reduced to D 0.75 in⋅= V Q A = 4 Q⋅ π D2⋅ = V 4 π 0.1× ft 3 s ⋅ 12 0.75 1 ft ⋅⎛⎜⎝ ⎞ ⎠ 2 ×= V 32.6 ft s ⋅= Re V D⋅ ν = Re 32.6 ft s ⋅ 0.75 12 × ft⋅ s 7.5 10 4−× ft2⋅ ×= Re 2717= The flow is TURBULENT For drawn tubing, from Table 8.1 e 0.000005 ft⋅= Given 1 f 2.0− log e D 3.7 2.51 Re f⋅+ ⎛⎜⎜⎝ ⎞ ⎠⋅= f 0.0449= hl f L D ⋅ V 2 2 ⋅= hl .0449 100 0.75 12 × 32.6 ft s ⎛⎜⎝ ⎞ ⎠ 2 2 ×= hl 3.82 104× ft2 s2 ⋅= The increase in loss is 3.82 10 4× 6326− 6326 504 %⋅= This is a HUGH increase! The main increase is because the diameter reduction causes the velocity to increase; the loss goes as V2, and 1/D, so it increases very rapidly
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