Question

A long circular cylinder of diameter 2a meters is set horizontally in a steady stream (perpendicular to the cylinder axis) of velocity U m/s. The cylinder is caused to rotate at ω rad/s around its axis. Obtain an expression in terms of ω and U for the ratio of the pressure difference between the top and bottom of the cylinder divided by the dynamic pressure of the stream (i.e., the pressure coefficient difference).

Answer 1

Answer:

The ratio of the difference of the pressure at the top and bottom of the cylinder to dynamic pressure is given as

$\dfrac{4a (U\omega- g)}{U^2}$

Explanation:

As the value of the diameter is given as d=2a

The velocity is given as v=U

The rotational velocity is given as ω rad/s

Point A is at the top of the cylinder and point B is at the bottom of the cylinder

Such that the point A is at the highest point on the circumference and point B is at the bottom of the cylinder

Now the velocity at point A is given as

$v_A=U-\dfrac{d}{2}\omega\\v_A=U-\dfrac{2a}{2}\omega\\v_A=U-a\omega$

Now the velocity at point B is given as

$v_B=U+\dfrac{d}{2}\omega\\v_B=U+\dfrac{2a}{2}\omega\\v_B=U+a\omega$

Considering point B as datum and applying the Bernoulli's equation between the point A and B gives

$\dfrac{P_A}{\rho g}+\dfrac{v_A^2}{2 g}+z_A=\dfrac{P_B}{\rho g}+\dfrac{v_B^2}{2 g}+z_B$

Here P_A and P_B are the local pressures at the point A and point B.

v_A and v_B are the velocities at the point A and B

z_A and z_B is the height of point A which is 2a and that of point B is 0

Now rearranging the equation of Bernoulli gives

$\dfrac{P_A-P_B}{\rho g}=\dfrac{v_B^2-v_A^2}{2 g}+z_B-z_A$

Putting the values

$\dfrac{P_A-P_B}{\rho g}=\dfrac{v_B^2-v_A^2}{2 g}+z_B-z_A\\\dfrac{P_A-P_B}{\rho g}=\dfrac{(U+a\omega)^2-(U-a\omega)^2}{2 g}+0-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{(U^2+a^2\omega^2+2Ua\omega)-(U^2+a^2\omega^2-2Ua\omega)}{2g}-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{U^2+a^2\omega^2+2Ua\omega-U^2-a^2\omega^2+2Ua\omega)}{2g}-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{4Ua\omega}{2g}-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{2Ua\omega}{g}-2a\\P_A-P_B=\dfrac{2Ua\omega}{g}*\rho g-2a*\rho g$

$P_A-P_B=2Ua\omega\rho-2a\rho g$

Now the dynamic pressure is given as

$P_D=\dfrac{1}{2}\rho U^2$

$\dfrac{P_A-P_B}{P_D}=\dfrac{2Ua\omega\rho-2a\rho g}{1/2 \rho U^2}\\\dfrac{P_A-P_B}{P_D}=\dfrac{2a\rho (U\omega- g)}{1/2 \rho U^2}\\\dfrac{P_A-P_B}{P_D}=\dfrac{4a\rho (U\omega- g)}{\rho U^2}\\\dfrac{P_A-P_B}{P_D}=\dfrac{4a (U\omega- g)}{U^2}$

So the ratio of the difference of the pressure at the top and bottom of the cylinder to dynamic pressure is given as

$\dfrac{4a (U\omega- g)}{U^2}$