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

Centripetal Acceleration = 88826.44 m/s²

RCF = 9054.7

Step-by-step explanation:

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

`Centripetal\ Acceleration = (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 = ((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 = (Centripetal\ Acceleration)/(g)\\RCF = (88826.44\ m/s^2)/(9.81\ m/s^2)\\`

RCF = 9054.7