What is the maximum relative centrifugal force applied when starch suspension is sedimented at 2500 pm in a rotor of maximum sample radius equal to 10 cm? Determine the centrifugal effect if the rotor is operated at a rotational speed of 450 s+

The formula for Relative Centrifugal Force (RCF) is:

$$RCF = 1.118 \times 10^{-5} \times r \times (rpm)^2$$

where $r$ is the radius in cm and $rpm$ is the rotational speed in revolutions per minute.

Given:

• Radius $r = 10 \text{ cm}$
• Rotational speed $rpm = 2500 \text{ rpm}$

Substitute the values into the formula:

$$RCF = 1.118 \times 10^{-5} \times 10 \text{ cm} \times (2500 \text{ rpm})^2$$ $$RCF = 1.118 \times 10^{-5} \times 10 \times 6,250,000$$ $$RCF = 1.118 \times 10^{-5} \times 62,500,000$$ $$RCF = 698.75$$

The maximum relative centrifugal force applied is $\boxed{\text{698.75 g}}$.

The rotational speed is given as $450 \text{ s}^{-1}$, which means 450 revolutions per second (rps). We need to convert this to revolutions per minute (rpm).

$$rpm = 450 \text{ rps} \times 60 \text{ s/min} = 27000 \text{ rpm}$$

Assuming the same maximum sample radius $r = 10 \text{ cm}$.

Substitute the new rpm and radius into the RCF formula:

$$RCF = 1.118 \times 10^{-5} \times r \times (rpm)^2$$ $$RCF = 1.118 \times 10^{-5} \times 10 \text{ cm} \times (27000 \text{ rpm})^2$$ $$RCF = 1.118 \times 10^{-5} \times 10 \times 729,000,000$$ $$RCF = 1.118 \times 10^{-5} \times 7,290,000,000$$ $$RCF = 81490.2$$

The centrifugal effect is $\boxed{\text{81490.2 g}}$.