Single particle motion
Single particle motion
Assumption ·Dilute two phase flow -Solve continuous phase u and P ignoring dispersed phase. 。Normal approach Force balance Fi to mpdVi/dt Problem with fluid acceleration Start with following equations 0i=0 8xi +}- 825 Oppcvcj 8xi Rectilinear relative motion
Assumption • Dilute two phase flow – Solve continuous phase u and P ignoring dispersed phase. • Normal approach – Force balance Fi to mpdVi/dt – Problem with fluid acceleration • Start with following equations – Rectilinear relative motion
Flows around a sphere ·High Reynolds 卡品a+品 (ue sin)=0 +o{（）+如。品 1 沿 2 0 2十r0r ue ue rug 10p PCa r80 + r r80 +ec{情品(r器）+ Sue 2 Our r2 sin20
Flows around a sphere • High Reynolds
Define stream fuction 10b 18p 4,=p2sin080 u8=一 rsine or For inviscid potential flow 2simn20-Psin20 Wr2 4,=-Wcos0- 2D T3 cos 0 D uo =+W sine-s sin D 6=-W'r cos+cose
• Define stream fuction • For inviscid potential flow
。High Reynolds number -1000->3x105 Laminar layer and wake development Larger than 3x105 Turbulent transition and drag Less than 1000,unsteady and shedding vortices 15 8 1.0 0.5 0 0 102 103 104
• High Reynolds number – 1000->3x105 • Laminar layer and wake development – Larger than 3x105 Turbulent transition and drag – Less than 1000, unsteady and shedding vortices