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

Step-by-step explanation: To determine the pressure of a liquid in a rotating tank,it is used:

p =
`(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)` =
`(p_(fluid).w^(2).r_(B) ^(2) )/(2)` - γfluid.
`z_(B)` - (
`(p_(fluid).w^(2).r_(A) ^(2) )/(2)` - γfluid.
`z_(A)`)

`p_(B) - p_(A)` =
`(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)` =
`(p_(fluid).w^(2))/(2) (r_B^(2) - r_A^(2) )`

`p_(B) - p_(A) = (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