2 consider two concentric cylinders of radius 05 m and 1 m...

2. Consider two concentric cylinders of radius 0.5 m and 1 m, respectively, and an equal height 1 m. Between the cylinders is the same oil in Problem 1, and a laminar unidirectional steady Couette flow can be induced by rotating the cylinders. If the inner cylinder is rotating counterclockwise with an angular velocity $\Omega$, what should be the angular velocity of the outer cylinder to make the flow irrotational? (4 points) $\Omega_o = \frac{\Omega}{4}$ 3. (Continued from Problem 2) If the outer cylinder is fixed (no rotation), what is the torque required to rotate the inner cylinder with $\Omega$? (4 points)

Transcript

00:01 Here, ro is the radius of outer cylinder and ri is the radius of inner cylinder.

00:21 Since there is no flow parallel to the axis of cylinder, therefore, the z component of velocity vz is equal to 0 and there is no radial flow, therefore, vr also equal to 0.

00:37 Consider the equation, equation, dou vr divided by dou r plus dou vtheta divided by dou theta plus dou vz divided by dou z is equal to zero.

00:54 Therefore, dou vtheta divided by dou theta is equal to zero or vtheta not equal to a function of theta.

01:05 According to navier -stokes equation in theta direction is, rho multiplied by dou vtheta divided by dou t plus vr dou vtheta divided r plus vtheta divided by r dou vtheta divided by dou theta plus vr vtheta divided by r plus plus vz dou vtheta divided by dou z is equal to minus 1 by r dou rho divided by dou theta theta plus rho g theta plus mu multiplied by dou by dou r of 1 by r dou by dou r of r v theta plus 1 by r square dou square v theta divided by dou theta square plus 2 by r square dou vry dou theta plus dou square v theta divided by dou z square.

02:48 Since vz equal to 0 and vr equal to 0, the equation will become reduced to d by dr of of 1 by r d by dr of r v theta is equal to 0.

03:09 Therefore, d by dr of r v theta is equal to a r and v theta is equal to a r divided by 2 plus b divided by r where a and b are constant at mu is equal to ri, b theta is equal to ohm ri, therefore, ohm ri is equal to a ri divided by 2 plus b divided by ri.

03:55 And at r equal to ro, b theta is equal to 0, therefore, 0 is equal to a ro divided by 2 plus b divided by ro.

04:11 Therefore, a ro divided by 2 is equal to minus b divided by ro or a is equal to minus b divided by 2 ro square and omega ri is equal to minus b divided by 2ro2 plus b divided by ri...

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