Answer to Question #155931 in General Chemistry for simsim

Question

A pilot-scale disc-track centrifuge is tested for recovery of bacteria. The centrifuge contains 25 discs with inner and outer diameters 2cm and 10cm, respectively. The half-cone angle is 35 degree. When operated at a speed of 3000 rpm with a feed rate of 3.5 litres/min, 70% of the cells are recovered. If a bigger centrifuge is to be used for industrial treatment of 80 litres/min, what operating speed is required to achieve the same sedimentation performance if the larger centrifuge contains 55 discs with outer diameter 15cm, inner diameter 4.7cm, and half-cone angle 45 degrees?

Answer 1

 For small scale centrifuge:

D₁ = 0.02m , R₁ = 0.01m

D₀ = 0.10m , R₀ = 0.05m

n₁ = 25discs

ω₁ = 3000rpm

Q₁ = 3.5 L/min

θ₁ = 35⁰

Large scale centrifuge:

D_0,2 = 0.15m , R_0,2 = 0.075m

D_1,2 = 0.047m , R_1,2 = 0.0235m

n₂ = 55discs

ω₂ = ?

Q₂ = 80 L/min

θ₂ = 45⁰

Q = $v$_g  Σ 

Since the dimensions of particles does not change in both cases , therefore $v$_g remains constant.

Q ∝Σ  $\implies$ $\frac{Q_1}{ Σ_1}$ = $\frac{Q_2}{ Σ_2}$

 Σ₁ = $\frac{2\pi nω}{3g}$ (R$_0^3$ - R$_1^3$ ) cotθ

Σ ₁ = $\frac{2\pi(25)(2\pi\frac{rad}{rev}3000\frac{rev}{min}\frac{1min}{60s})^2}{3(9.81m/s²)}$ (0.05³ --0.01³ )m³ × $\frac{1}{tan35⁰}$

Σ₁ = 106.6m²

Σ₂ = $\frac{Q_2}{Q_1}$ Σ₁

= $\frac{80}{35}$× 106.6

Σ₂ = 2436.5m²

Next we find ω₂

Σ₂ = $\frac{2\pi n_2(\omega_2)^2}{3g}$ ( R$_{0,3} ^3$ - R$_{1,2}^3$ ) cot45⁰

$\omega$₂ = $\sqrt{\frac{3gΣ_2}{2\pi n_2(R_{0,1}^3-R_{1,2}^3)cot45⁰}}$

$\omega$₂ = $\sqrt{\frac{3(9.81m/s²)2436.5m²}{2 \pi(55)(0.075³-0.0235³)m³cot45⁰}}$

$\omega$₂ = 658 $\frac{rad}{s}$ × $\frac{rev}{2\pi rad}$ × $\frac{60s}{1min}$

$\omega$₂ = 6286.7 rpm