ch08 - Problem 8.1 Given: Air entering duct Find: [1] Flow...

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Problem 8.1 [1] Given: Air entering duct Find: Flow rate for turbulence; Entrance length Solution: The governing equations are Re VD ν = Re crit 2300 = Q π 4 D 2 V = The given data is D6 i n = From Table A.9 ν 1.62 10 4 × ft 2 s = L laminar 0.06 Re crit D = or, for turbulent, L turb = 25D to 40D Hence Re crit Q π 4 D 2 D ν = or Q Re crit π ν D 4 = Q 2300 π 4 × 1.62 × 10 4 × ft 2 s 1 2 × ft = Q 0.146 ft 3 s = For laminar flow L laminar 0.06 Re crit D = L laminar 0.06 2300 × 6 × in = L laminar 69.0 ft = For turbulent flow L min 25 D = L min 12.5 ft = L max 40 D = L max 20 ft =
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Problem 8.2 [2]
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Problem 8.3 [3] Given: Air entering pipe system Find: Flow rate for turbulence in each section; Which become fully developed Solution: From Table A.9 ν 1.62 10 4 × ft 2 s = The given data is L5 f t = D 1 1in = D 2 1 2 in = D 3 1 4 in = The critical Reynolds number is Re crit 2300 = Writing the Reynolds number as a function of flow rate Re VD ν = Q π 4 D 2 D ν = or Q Re π ν D 4 = Then the flow rates for turbulence to begin in each section of pipe are Q 1 Re crit π ν D 1 4 = Q 1 2300 π 4 × 1.62 × 10 4 × ft 2 s 1 12 × ft = Q 1 0.0244 ft 3 s = Q 2 Re crit π ν D 2 4 = Q 2 0.0122 ft 3 s = Q 3 Re crit π ν D 3 4 = Q 3 0.00610 ft 3 s = Hence, smallest pipe becomes turbulent first, then second, then the largest. For the smallest pipe transitioning to turbulence ( Q 3 ) For pipe 3 Re 3 2300 = L laminar 0.06 Re 3 D 3 = L laminar 2.87ft = L laminar < L: Not fully developed or, for turbulent, L min 25 D 3 = L min 0.521ft = L max 40 D 3 = L max 0.833ft = L max/min < L: Not fully developed For pipes 1 and 2 L laminar 0.06 4Q 3 πν D 1 D 1 = L laminar 2.87ft = L laminar < L: Not fully developed L laminar 0.06 3 D 2 D 2 = L laminar 2.87ft = L laminar < L: Not fully developed
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For the middle pipe transitioning to turbulence ( Q 2 ) For pipe 2 Re 2 2300 = L laminar 0.06 Re 2 D 2 = L laminar 5.75ft = L laminar > L: Fully developed or, for turbulent, L min 25 D 2 = L min 1.04ft = L max 40 D 2 = L max 1.67ft = L max/min < L: Not fully developed For pipes 1 and 3 L 1 0.06 4Q 2 πν D 1 D 1 = L 1 5.75ft = L 3min 25 D 3 = L 3min 0.521ft = L 3max 40 D 3 = L 3max 0.833ft = L max/min < L: Not fully developed For the large pipe transitioning to turbulence ( Q 1 ) For pipe 1 Re 1 2300 = L laminar 0.06 Re 1 D 1 = L laminar 11.5ft = L laminar > L: Fully developed or, for turbulent, L min 25 D 1 = L min 2.08ft = L max 40 D 1 = L max 3.33ft = L max/min < L: Not fully developed For pipes 2 and 3 L 2min 25 D 2 = L 2min 1.04ft = L 2max 40 D 2 = L 2max 1.67ft = L max/min < L: Not fully developed L 3min 25 D 3 = L 3min 0.521ft = L 3max 40 D 3 = L 3max 0.833ft = L max/min < L: Not fully developed
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Problem 8.4 [2] Given: That transition to turbulence occurs at about Re = 2300 Find: Plots of average velocity and volume and mass flow rates for turbulence for air and water Solution: From Tables A.8 and A.10 ρ air 1.23 kg m 3 = ν air 1.45 10 5 × m 2 s = ρ w 999 kg m 3 = ν w 1.14
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ch08 - Problem 8.1 Given: Air entering duct Find: [1] Flow...

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