Question

g 4.86 Separators are used to separate liquids of diff erent densities, such as cream from skim milk, by rotating the mixture at high speeds. In a cream separator, the skim milk goes to the outside while the cream migrates toward the middle. A factor of merit for the centrifuge is the centrifugal acceleration force (RCF), which is the radial acceleration divided by the acceleration due to gravity. A cream separator can operate at 9000 rpm (rev/min). If the bowl of the separator is 20 cm in diameter, what is the centripetal acceleration if the liquid rotates as a solid body, and what is the RCF

Answer 1

Answer:

Centripetal Acceleration = 88826.44 m/s²

RCF = 9054.7

Explanation:

First, we will find the value of the centripetal acceleration by using the following formula:

$Centripetal\ Acceleration = \frac{v^2}{r}$

where,

v = linear speed of liquid or separator = rω

ω = angular speed of liquid or separator = (9000 rpm)(2π rad/rev)(1 min/60 s) = 942.48 rad/s

r = radius of seperator = diameter/2 = 20 cm/2 = 10 cm = 0.1 m

Therefore,

$Centripetal\ Acceleration = \frac{(r\omega)^2}{r}\\Centripetal\ Acceleration = r\omega^2\\Centripetal\ Acceleration = (0.1\ m)(942.48\ rad/s)^2$

Centripetal Acceleration = 88826.44 m/s²

Now, for the RCF:

$RCF = \frac{Centripetal\ Acceleration}{g}\\RCF = \frac{88826.44\ m/s^2}{9.81\ m/s^2}$

RCF = 9054.7