Question

Yeast cells are recovered from a fermentation broth by using a tubular centrifuge. Sixty percent of the cells are recovered at a flow rate of 12 l/min with a rotational speed of 4000 rpm. Recovery is inversely proportional to flow rate. a. To increase the recovery of cells to 95% at the same flow rate, what should be the rpm of the centrifuge?

Answer 1

Answer:

$\\ \omega = 5033.22 \ rpm$

Explanation:

Given that:

$R_1 = 0.6\\\\R_2 = 0.95\\\\\omega_1 = 4000 \ rpm\\\\\omega_2 = ???\\\\Q_1 = 12\ l/min\\\\Q_2 = 12\ l/min$

For a continuous centrifuge ; solid recovery R ∝ $\frac{\omega^2}{Q}$

where;

ω = angular velocity of the centrifuge in rpm

Q = flow rate

SO;

$R_1 \alpha \frac{4000^2}{12}$

$R_2 \alpha \frac{\omega^2}{12}$

$\frac{R_1}{R_2} = \frac{4000^2}{12}$  ÷  $\frac{\omega^2 }{12}$

$\frac{R_1}{R_2} = \frac{4000^2}{12} * \frac{12}{\omega^2}$

$\frac{R_1}{R_2} = \frac{4000^2}{\omega^2}\\\\\frac{0.6}{0.95} = \frac{4000^2}{\omega^2}$

${\omega^2} = \frac{4000^2*0.95}{0.6}\\\\\\{\omega} = \sqrt{ \frac{4000^2*0.95}{0.6}}\\\\\\ \omega = 5033.22 \ rpm$

Answer 2

The rotational speed when the recovery of cells is increased to 95% at the same flow rate is; ω₂ = 5033.22 rpm

We are given;

Flow rate; Q₁' = Q'₂ = 12 l/min

Rotational speed 1; ω₁ = 4000 rpm

Recovery 1; R₁ = 60% = 0.6

Recovery 2; R₂ = 95% = 0.95

Now, the formula for recovery of tubular centrifuge is;

R = kω²/Q'

where k is a constant of proportionality

For R₁, we have;

0.6 = k * 4000²/12

k = (0.6 * 12)/4000²

When R₂ is 0.95, we have;

0.95 = ((0.6 * 12)/4000²) * ω₂²/Q₂'

⇒ 0.95 = ((0.6 * 12)/4000²) * ω₂²/12

0.95 * 4000²/0.6 = ω₂²

ω₂² = 25,333,333.33

ω₂ = √25,333,333.33

ω₂ = 5033.22 rpm

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