Introduction-Fluid-Mechanics-Solution-Chapter-08

# Introduction-Fluid-Mechanics-Solution-Chapter-08 - Problem...

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Q 0.396 m 3 min = Q Re crit π ⋅ν D 4 = or Re crit Q π 4 D 2 D ν = Hence L turb 25 D 40 D = or, for turbulent, L laminar 0.06 Re crit D = Q π 4 D 2 V = Re crit 2300 = Re VD ν = The governing equations are ν 1.46 10 5 m 2 s = From Fig. A.3 D 0.25 m = The given data is Solution Find: Volume flow rate for turbulence; entrance length Given: Data on air flow in duct Problem 8.2
L laminar 0.06 Re crit D = L laminar 34.5m = or, for turbulent, L min 25 D = L min 6.25m = L max 40 D = L max 10m =

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V air 0.0334 m 2 s D = V air 2300 1.45 × 10 5 × m 2 s D = Hence for air V Re crit ν D = For the average velocity Re crit 2300 = Re VD ν = The governing equations are ν w 1.14 10 6 × m 2 s = ρ w 999 kg m 3 = ν air 1.45 10 5 × m 2 s = ρ air 1.23 kg m 3 = From Tables A.8 and A.10 Solution Find: Plots of average velocity and volume and mass flow rates for turbulence for air and water Given: That transition to turbulence occurs at about Re = 2300 Problem 8.3
These results are plotted in the associated Excel workbook m w 2.06 kg ms D × = m w ρ w Q w = m air 0.0322 kg D × = m air ρ air Q air = Finally, the mass flow rates are obtained from volume flow rates For water Q w 0.00206 m 2 s D × = Q w π 4 2300 × 1.14 × 10 6 m 2 s D = Q air 0.0262 m 2 s D × = Q air π 4 2300 × 1.45 × 10 5 m 2 s D = Hence for air QA V = π 4 D 2 V = π 4 D 2 Re crit ν D = π Re crit ⋅ν 4 D = For the volume flow rates V w 0.00262 m 2 s D = V w 2300 1.14 × 10 6 × m 2 s D = For water

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Problem 8.3 (In Excel) 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 The relations needed are From Tables A.8 and A.10 the data required is ρ air = 1.23 kg/m 3 ν air = 1.45E-05 m 2 /s ρ w = 999 kg/m 3 ν w = 1.14E-06 m 2 /s D (m) 0.0001 0.001 0.01 0.05 1.0 2.5 5.0 7.5 10.0 V air (m/s) 333.500 33.350 3.335 0.667 3.34E-02 1.33E-02 6.67E-03 4.45E-03 3.34E-03 V w (m/s) 26.2 2.62 0.262 5.24E-02 2.62E-03 1.05E-03 5.24E-04 3.50E-04 2.62E-04 Q air (m 3 /s) 2.62E-06 2.62E-05 2.62E-04 1.31E-03 2.62E-02 6.55E-02 1.31E-01 1.96E-01 2.62E-01 Q w (m 3 /s) 2.06E-07 2.06E-06 2.06E-05 1.03E-04 2.06E-03 5.15E-03 1.03E-02 1.54E-02 2.06E-02 m air (kg/s) 3.22E-06 3.22E-05 3.22E-04 1.61E-03 3.22E-02 8.05E-02 1.61E-01 2.42E-01 3.22E-01 m w (kg/s) 2.06E-04 2.06E-03 2.06E-02 1.03E-01 2.06E+00 5.14E+00 1.03E+01 1.54E+01 2.06E+01 Re crit 2300 = V ν D = Q π ⋅ν 4 D = m rate ρ Q =
Average Velocity for Turbulence in a Pipe 1.E-04 1.E-02 1.E+00 1.E+02 1.E+04 1.E-04 1.E-03 1.E-02 1.E-01 1.E+00 1.E+01 D (m) V (m/s) Velocity (Air) Velocity (Water) Flow Rate for Turbulence in a Pipe 1.E-07 1.E-05 1.E-03 1.E-01 1.E+01 1.E-04 1.E-03 1.E-02 1.E-01 1.E+00 1.E+01 D (m) Q (m 3 /s) Flow Rate (Air) Flow Rate (Water)

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Mass Flow Rate for Turbulence in a Pipe 1.E-06 1.E-04 1.E-02 1.E+00 1.E+02 1.E-04 1.E-03 1.E-02 1.E-01 1.E+00 1.E+01 D (m) m flow (kg/s) Mass Flow Rate (Air) Mass Flow Rate (Water)
Q 1 0.0786 m 3 min = Q 1 Re crit π ⋅ν D 1 4 = Then the flow rates for turbulence to begin in each section of pipe are Q Re π D 4 = or Re VD ν = Q π 4 π D 2 D ν = Writing the Reynolds number as a function of flow rate Re crit 2300 = The critical Reynolds number is D 3 10 mm = D 2 25 mm = D 1 50 mm =

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Introduction-Fluid-Mechanics-Solution-Chapter-08 - Problem...

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