Content  Mesh Independence Study  Taylor-Couette Validation  Wavy Taylor Validation  Turbulent Validation  Thermal Validation  Simple Model Test  Plans for Next Period

Mesh Independence StudyMesh Independence Study

Taylor-Couette ValidationTaylor-Couette Validation Wavelength T/Tc Full Length

Wavy Taylor ValidationWavy Taylor Validation η (a/b)a(cm)b(cm)h(cm)R(11Rc) Ω (rad/s) Upper Boundcircumradialaxial Free Fundamental angular frequency ω = s= ω /(m Ω )=0.334

Wavy Taylor ValidationWavy Taylor Validation η (a/b)a(cm)b(cm)h(cm)R(11Rc) Ω (rad/s) Upper Boundcircumradialaxial Free Two fundamental frequencies ω = s= ω /(m Ω )=0.362

Wavy Taylor ValidationWavy Taylor Validation η (a/b)a(cm)b(cm)h(cm)R(11Rc) Ω (rad/s) Upper Boundcircumradialaxial Free Fundamental angular frequency ω = s= ω /(m Ω )=0.458

Comparing with Experiment Data η (a/b)Computed S1Measured S ± ± ±0.001 The difference is located in the reasonable region of uncertainty Need to be calculated longer.

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

Boundary ConditionsBoundary Conditions R 1 = m R 2 = m Ω = 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 = 400 Circle = 100 Radial = 60

Comparing with Experiment Data

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

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

Boundary ConditionsBoundary Conditions K eq = -h*r*ln(R 1 /R 2 )/k Re = Ω* (R 1 -R 2 )*R 1 /ν R 1 = cm R 2 = cm Height = Gr= 1000 ΔT= K Ti = 293K To= K End walls are fixed and insulated Re=[ ] Ω=[ ] rad/s Since for η=0.565 Re c = 70, All the three cases are in laminar mode. Mesh Density Axial = 1000 Circle = 100 Radial = 60

Comparing with Experiment Data Re 2 h(w/m 2 k)k eq Experiment DataResidue e e e-03

Comparing with Experiment Data

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

Simple Model TestSimple Model Test R 1 = mm R 2 = 97.5 mm Height = 140 mm Q=4 L/min V in = m/s T in = 308K T out = 551K Ω= 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 = , In this case Re= So, it is in laminar mode.

Important TipsImportant Tips  Combined fl ows in annular space not only on the operating point (axial Reynolds and Taylor numbers), but also e and strongly e on geometry and, to a lesser degree, on parietal thermal conditions.

Plans for Next PeriodPlans for Next Period  Keep running both of the turbulent cases  Finish the thermal validation  Couette flow validation  Repeat Taylor-couette validation with full length  Wavy validation should be finished with running 0.95 case long enough  More validation of the thermal part (optional)  Keep turbulent case running  Finish simple model test  Check geometry related paper