确实是用力平衡就可以导了 中间参杂了一些小技巧 推导如下 （可能会看的有点辛苦，请见谅） from drag force : F(D)=C(D)*A*ρ*(v^2)/2.........(A) where F(D)=drag force C(D)=drag coeff. A=cross section area ρg=gas density v=vertical settling velocity from free falling equation: F(I)=(ρp-ρg)*g*V............(B) where F(I)=impelling force or frr falling force in the fluid medium ρp=particle density g=acceleration const. V=particle vol. We must postulate that the settling follow the free settling mode, that is there is no force other than drag force and impelling force will participate in the settling process; or we can say that particles will not interact with each other during the settling process. if the settling follow the free settling, than (A) will equal to (B) ∴(ρp-ρg)*g*V=C(D)*A*ρ*(V^2)/2............(C) and V=(π/6)d^3 (d=particle diameter) A=(π/4)d^2 insert A and V in to (C) ∴v={4g*d* [(ρp-ρg)/ρg] / [3*C(D)] }^0.5..........(D) if Re<1 we can assume it as the laminar flow (for external flow field) so, C(D)=24/Re=24μ/ρp*v*d.....Stoke's Law insert this into (D) ∴v=g*(ρp-ρg)*d^2 /18*μ............(E) and u=Q/H*W where u=horizontal velocity=Q/H*W (H*W=cross section area of the settling tank) Q=flow rate H=settling tank height W=settling tank width ∴Q = W*H*u.............(F) and t=H/v(高度/沈降速)=L/u(长度/水平速)=Q/H*W*L(流率/槽体积) where t=overflow time or detention time ∴HWL/Q=H/v v=Q/L*W............(G) combine (G)and(F) get v=H*u/L finally we insert v into (E) after equation operation we can get d=[18μ*u*H / g*L*(ρp-ρg)]^0.5 # Perhaps you can find out related equations in the "particle operation text book" or the last few chapters in the "McCabe & Smith Unit Operation" ※ 引述《o310177o (shain)》之铭言： : 某一沉淀室中，其高度H、长度L，ρp、ρg分别为粒子与气体的密度，μ为黏度， : 要如何推导该室可沉淀的最小粒径为下式 : —————————— : D(min) = ∕ 18μHu : ∕—————————— : ﹨∕ gL(ρp－ρg) : 其中u指该粒子水平移动速度，另假设粒子之速度极慢， : 即 Re< 0.1 (符合Stoke's law) : 有人知道怎麽导吗?? : 谢谢~~ --

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