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Chapter 2: Problem 162

A closed, 0.4 -m-diameter cylindrical tank is completely filled with oil $(S G=0.9)$ 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.

Short Answer

Expert verified
The pressure difference just under the vessel cover between a point on the circumference and a point on the axis is 57600 Pa (Pascals).

Step by step solution

01

Convert SG to Fluid Density

The Specific Gravity (SG) is given as 0.9. This number is the ratio of the density of a fluid to the density of a reference fluid. Since the reference fluid is typically water at 4 degrees Celsius, which has a density of 1000 kg/m^3, one can calculate the fluid (oil) density as follows: Density = SG * Density of reference fluid = 0.9 * 1000 kg/m^3 = 900 kg/m^3.
02

Identify given variables

The radius (r) of the cylindrical vessel can be found from the given diameter. As the diameter is 0.4 m, the radius would be r = 0.4 m / 2 = 0.2 m. The angular velocity (omega) is given as 40 rad/s.
03

Calculate pressure difference

From fluid dynamics, the pressure difference can be calculated using the formula \(\Delta P = 0.5 * \rho * \omega ^ 2 * r^2\). Here, \(\Delta P\) is the pressure difference, \(\rho\) is the density of the fluid, \(\omega\) is the angular velocity, and \(r\) is the radius of the cylindrical vessel. When we plug in our numbers, we get \(\Delta P = 0.5 * 900 kg/m^3 * (40 rad/s)^2 * (0.2 m)^2\).
04

Calculation

Using the above formula, \(\Delta P = 0.5 * 900 kg/m^3 * 1600 rad^2/s^2 * 0.04 m^2 = 57600 Pa\). The units come from the SI units: kg/(m·s^2), which is equal to the Pascal (Pa).

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