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Content  Turbulent Validation  Thermal Validation  Simple Model Test  Plans for Next Period.

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Presentation on theme: "Content  Turbulent Validation  Thermal Validation  Simple Model Test  Plans for Next Period."— Presentation transcript:

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2 Content  Turbulent Validation  Thermal Validation  Simple Model Test  Plans for Next Period

3 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

4 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 =

5 Comparing with Experiment Data  Picture

6 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

7 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

8 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 =

9 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

10 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

11 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.

12 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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