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Published byHarry Collins Modified over 8 years ago
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Content Turbulent Validation Thermal Validation Simple Model Test Plans for Next Period
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Turbulent ValidationTurbulent Validation Comparison of normalized mean angular momentum profiles between present simulation (Re=8000) and the experiment of Smith & Townsend (1982). u θ Azimuthal Velocity R 1 Radius of Inner Cylinder R 2 Radius of Outer Cylinder U 0 Tangential Velocity of Inner Cylinder r Distance from Centre Axis
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Boundary ConditionsBoundary Conditions R 1 = 0.1525 m R 2 = 0.2285 m Ω = 22.295 rad/s (Re=17295) Height = 1.80 m End walls are free surfaces k- epsilon and k- omega were chosen to compare Measure points are located along the mid-height of the gap Mesh Density Axial = Circle = Radial =
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Comparing with Experiment Data Picture
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Possible Reasons for Difference Flow time interval is not enough ΔT epsilon =27.68s ΔT omega =20.48s Sampling frequency f experiment =10kHz f simulation =200Hz Mesh density Tip: 文章名称 used k-epsilon as the turbulent model
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Thermal ValidationThermal Validation K eq = -h*r*ln(R 1 /R 2 )/k Re = Ω* (R 1 -R 2 )*R 1 /ν h Convective Heat Transfer Coefficient R 1 Radius of Inner Cylinder R 2 Radius of Outer Cylinder K Thermal Conductivity ν Kinematic Viscosity r Distance from Centre Axis Fluid is air
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Boundary ConditionsBoundary Conditions K eq = -h*r*ln(R 1 /R 2 )/k Re = Ω* (R 1 -R 2 )*R 1 /ν R 1 = 1.252 cm R 2 = 2.216 cm Height = 50.64 Gr= 1000 ΔT= 7.582 K Ti = 293K To= 300.582K End walls are fixed and insulated Re=[40 120 280] Ω=[5.008 15.023 35.054] rad/s Since for η=0.565 Re c = 70, All the three cases are in laminar mode. Mesh Density Axial = Circle = Radial =
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Comparing with Experiment Data Re 2 h(w/m 2 k)k eq Experiment DataResidue 16005.6391.5681.0809.8e-04 144009.7212.7041.5002.0e-03 7840016.0224.4562.1201.8e-03
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Possible Reasons for Difference Boundary condition set-up ideal gas, pressure based, real apparatus error (axial temperature gradient, end walls effect) Wrong understanding of the experiment
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Simple Model TestSimple Model Test R 1 = 96.85 mm R 2 = 97.5 mm Height = 140 mm Q=4 L/min V in = 0.000168 m/s T in = 308K T out = 551K Ω=29.311 rad/s End walls are fixed and insulated Measure points are located in the vertical lines close to the inner cylinder. Since for η=0.975 Re c = 260.978, In this case Re=1837.075 So, it is in laminar mode.
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Plans for Next PeriodPlans for Next Period Keep running both of the turbulent cases Finish the thermal validation Couette flow validation More validation of the thermal part (optional) Finish simple model test Check geometry related paper
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