Question

The lighter liquid layer and the interphase layer in a basket centrifuge, rotating at a speed of 1000 rpm, are 0.1025 m and 0.105 m away from the center, respectively. Considering the densities of lighter and heavier liquids as 920 kg/m\(^3\) and 1015 kg/m\(^3\), the differential pressure in horizontal direction required to maintain the interphase layer in kPa is \underline{\hspace{3cm}} (rounded off to 3 decimal places, \(\pi = 3.14\)).

Hint:

Centrifugal pressure difference depends on density difference, angular velocity squared, and radial distance difference.

Answer 1

Step 1: Convert rpm to angular velocity. \[ \omega = \frac{2\pi N}{60} = \frac{2 \cdot 3.14 \cdot 1000}{60} = 104.67 \, rad/s \]

Step 2: Pressure difference formula in centrifuge. \[ \Delta P = \frac{\Delta \rho \, \omega^2}{2} (r_2^2 - r_1^2) \] Here, \(\Delta \rho = 1015 - 920 = 95 \, kg/m^3\).

Step 3: Substitute values. \[ \Delta P = \frac{95 \times (104.67)^2}{2} \cdot (0.105^2 - 0.1025^2) \] \[ = 95 \cdot 5478.6 \cdot (0.011025 - 0.010506)/2 \] \[ = 95 \cdot 5478.6 \cdot 0.000519/2 \] \[ = 5494.6 \, Pa = 5.495 \, kPa \] \[ \boxed{5.495 \, kPa} \]