Question

A closed, 0.4-m-diameter cylindrical tank is completely filled with oil (SG 0.9)and rotates about its vertical longitudinal axis with an angular velocity of 40 rad/s. Determine the difference in pressure just under the vessel cover between a point on the circumference and a point on the axis

Answer 1

Answer: $p_{B} - p_{A}$ = 28800 Pa or 28.8 kPa

Explanation: To determine the pressure of a liquid in a rotating tank,it is used:

p = $\frac{p_{fluid}.w^{2}.r^{2} }{2}$ - γfluid . z + c

where:

$p_{fluid}$ is the liquid's density

w is the angular velocity

r is the radius

γfluid.z is the pressure variation due to centrifugal force.

For this question, the difference between a point on the circumference and a point on the axis will be:

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}.r_{B} ^{2} }{2}$ - γfluid.$z_{B}$ - ($\frac{p_{fluid}.w^{2}.r_{A} ^{2} }{2}$ - γfluid.$z_{A}$)

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} )$ - γfluid($z_{B}$ -$z_{A}$)

Since there is no variation in the z-axis, z = 0 and that the density of oil is 0.9.10³kg/m³:

$p_{B} - p_{A}$ = $\frac{p_{fluid}.w^{2}}{2} (r_B^{2} - r_A^{2} )$

$p_{B} - p_{A} = \frac{0.9.10^3.40^2}{2}(0.2^2 - 0)$

$p_{B} - p_{A}$ = 28800

The difference in pressure between two points, one on the circumference and the other on the axis is $p_{B} - p_{A}$ = 28800 Pa or 28.8 kPa