Question

(3) Consider the steady laminar flow between the coaxial cylinders shown below. The inner cylinder rotates with angular velocity 2 and the outer cylinder is stationary. The no-slip condition applies at the inner and outer cylinder surfaces and we are considering the cylinders to be very long in the 2-direction hence we may ignore edge effects near the top and bottom surfaces. - R2 Assume that gravity is negligible, v, is zero and that are zero for all variables.

Fluid is Non-Newtonian.

Answer 1

Given -- - Az - 2 The cylinder is coaxial as shown in figure above : - Assume Vzo 20 go for all variables Let us write boundary conditions on the Surface of cylinder. Atr=R, gi0 vo=0 Atr=r wr=0 Vo =WR2 We know that in cylindrical coordinate continuity equation is given by :- ) ta (rv) + Ovo tavz 20 since ave o ovz.co go

Put the above value in equation (i), weget I 2 (& Vr) to to=0 I. Coro)=0 av (VV) 20 (13) on integrating equation (ii) weget- rur = constant at r=R. Riur=constant r=ri voto avr 20 (1) How assume gravity is negligble and solve the momentum equation. o co , Vrzo, vzt0 [for steady state at condition] ovo Op = 0 7 (iv) go =0 zavo =0 ază o we know that Navier stokes equation in cylindrical coordinate is given by : - at ove + (vo ove tvo dove + uz ve + uz ove) = -1 op & go tle [oo (tarve t 3200 Ve +3349] fazvo >2 ovo azz

put the value of equation (iv) in (v), we get u o [ 2 (rvo)] =0 əzvo Lavo -vo o 282 r ar 82 on integration equation (iv, we get OVO = -o croln) gy Multiply by (ar) on both sides & (volodr = -0 % (vol.de on integrating are =-vo tai talovo) - G a crve) - ar ४ Integrating again vo-ar tr Now apply boundary conditions cataRi 0 c2=-CIR? 2 + q R2 = WR2 ..c = WR2-C2 - 2 WR2-cz c = - WR22 R² R₃²-R,2 R22 - R12 rar RI RO proved Vo(r) - WR R₃²-RRL