This set of Bioseparation Processes Multiple Choice Questions & Answers (MCQs) focuses on “Preparative Centrifuge”.

1. Which centrifuge can handle large volume, ranging from 1litres to thousand litres?

a) Laboratory centrifuge

b) Preparative centrifuge

c) Both laboratory and preparative centrifuge

d) All types of centrifuge

View Answer

Explanation: Preparative centrifuges can handle large liquid volumes in comparison to laboratory centrifuges and it ranges from 1 litre to many thousands litre. These centrifuges have range of designs and have common feature of tubular rotating chamber, the suspension which needs to be centrifuged are fed in the device from one end and the supernatant as well as precipitate are collected from other end.

2. Which is the simplest type of preparative centrifuge?

a) Tubular bowl centrifuge

b) Disk centrifuge

c) Basket centrifuge

d) Chamber bowl centrifuge

View Answer

Explanation: Tubular bowl centrifuge is the most common type of centrifuge which has rotating speed from 500-2000rpm. These have high centrifugal force and are easy to clean. It allows the flow of fluid to enter at one end and exit at the opposite end and the particles move toward the wall of the cylinder and capture both at wall and exit.

3. What is the equation which governs the radial direction of particles within tubular centrifuge?

a) \([\frac{4}{3} \pi(\frac{d}{2})^3\) (ρ_{S} – ρ_{L})\(]\) [rω^{2}] = -6πμ(\(\frac{d}{2}v\))

b) –\([\frac{4}{3} \pi(\frac{d}{2})^3\) (ρ_{S} – ρ_{L})\(]\) [rω^{2}] = 6πμ(\(\frac{d}{2}v\))

c) \([\frac{4}{3} \pi(\frac{d}{2})^3\) (ρ_{S} – ρ_{L})\(]\) [rω^{2}] = 6πμ(\(\frac{d}{2}v\))

d) \([\frac{4}{3} \pi(\frac{d}{2})^3\) (ρ_{S} – ρ_{L})\(]\) – [rω^{2}] = 6πμ(\(\frac{d}{2}v\))

View Answer

Explanation: The motion of a particle at a point within the tubular centrifuge in a radial direction is governed by a force balance equation \([\frac{4}{3} \pi(\frac{d}{2})^3\) (ρ

_{S}– ρ

_{L})\(]\) [rω

^{2}] = 6πμ(\(\frac{d}{2}v\)) where, d is the diameter of the particle, ρ

_{S}is the density of particle, ρ

_{L}is the density of the liquid, r is the distance of the particle from the axis of rotation, ω is the angular velocity, μ is the viscosity of the liquid and v is the velocity of the moving particles.

4. Describe the LHS and RHS of the equation \([\frac{4}{3} \pi(\frac{d}{2})^3\) (ρ_{S} – ρ_{L})\(]\) [rω^{2}] = 6πμ(\(\frac{d}{2}v\)).

a) Stokes Drag and Buoyancy

b) Constant Velocity and moving velocity

c) Supernatant and pellet

d) Buoyancy and Stokes drag

View Answer

Explanation: The term at left hand side of the equation is Buoyancy and the term on the right hand side of the equation is Stokes drag. In this equation \([\frac{4}{3} \pi(\frac{d}{2})^3\) (ρ

_{S}– ρ

_{L})\(]\) [rω

^{2}] = 6πμ(\(\frac{d}{2}v\)) the particles are assumed to be moving at a constant velocity in the radial direction and it results into v = \(\frac{d^2(\rho_S – \rho_L)r\omega^2}{18\mu}\).

5. What is the final equation obtained for describing the motion of a particle in the radial direction of a centrifuge?

a) ln\(\frac{r_2}{r_1}\) = \(\frac{d^2(\rho_S – \rho_L)\omega^2 \,t}{18\mu}\)

b) \(\frac{r_2}{r_1}\) = \(\frac{d^2(\rho_S – \rho_L)\omega^2 \,t}{18\mu}\)

c) ln\(\frac{r_2}{r_1}\) = –\(\frac{d^2(\rho_S – \rho_L)\omega^2 \,t}{18\mu}\)

d) \(\frac{r_2}{r_1}\) = –\(\frac{d^2(\rho_S – \rho_L)\omega^2 \,t}{18\mu}\)

View Answer

Explanation: The equation ln\(\frac{r_2}{r_1}\) = \(\frac{d^2(\rho_S – \rho_L)\omega^2 \,t}{18\mu}\) is obtained by integrating \(\int_{r_1}^{r_2}\frac{dr}{r} = \int_0^1 \frac{d^2(\rho_S – \rho_L)\omega^2}{18\mu}\) dt from radial location r

_{1}to r

_{2}. The flow of feed in the tubular centrifuge is annular i.e. there is an empty cylindrical shell near the axis of rotation and it is due to the forcing of liquid in the centrifuge towards the wall of the tubular bowl due to centrifugal force.

6. How will you estimate the velocity of a particle along the z-axis?

a) \(\frac{dz}{dt} = -\frac{Q}{\pi(r_t^2 – r_a^2)}\)

b) \(\frac{dz}{dt} = \frac{-Q}{\pi(r_t^2 – r_a^2)}\)

c) \(\frac{dz}{dt} = \frac{Q}{\pi(r_t^2 – r_a^2)}\)

d) \(\frac{dz}{dt} = \frac{Q}{-\pi(r_t^2 – r_a^2)}\)

View Answer

Explanation: The velocity can be obtained by using \(\frac{dz}{dt} = \frac{Q}{\pi(r_t^2 – r_a^2)}\) where, Q is the flow rate of feed. It can be integrated and divided to obtain \(\int_{r_a}^{r_t}\frac{dr}{r} = \int_0^z\frac{\pi d^2(\rho_S – \rho_L)ω^2 (r_t^2 – r_a^2)}{18\mu Q}\) dt and the final equation resulting from the previous ones are Q = [\(\frac{d^2(\rho_S – \rho_L)g}{18\mu}\) ][ \(\frac{\pi z(r_t^2 – r_a^2)}{g\, ln(\frac{r_t}{r_a})}\) ] where, g is the acceleration due to gravity.

7. What is the residence time for the fluid in the bowl?

a) τ = \(\frac{\pi L(R_0^2 – R_1^2)}{Q}\)

b) τ = \(\frac{L(R_0^2 – R_1^2)}{Q}\)

c) τ = \(\frac{\pi(R_0^2 – R_1^2)}{Q}\)

d) τ = \(\frac{\pi L(R_0^2 – R_1^2)}{Q}\)

View Answer

Explanation: The residence time can be calculated by using τ = \(\frac{\pi L(R_0^2 – R_1^2)}{Q}\) as solid build up on the wall of the bowl, the inner radius of the bowl is efficiently decreased by reducing the residence time in the bowl and reducing the sedimentation distance as well as the maximum sedimentation velocity.

8. What does second and first term in the equation Q = [\(\frac{d^2(\rho_S – \rho_L)g}{18\mu}\) ][ \(\frac{\pi z(r_t^2 – r_a^2)}{g\, ln(\frac{r_t}{r_a})}\) ] represents?

a) Sigma factor and sedimentation velocity

b) Sedimentation velocity and sigma factor

c) Sigma factor

d) Sedimentation velocity

View Answer

Explanation: The first term in the equation i.e. [\(\frac{d^2(\rho_S – \rho_L)g}{18\mu}\) ] represents sedimentation velocity, the velocity of a particle settling due to gravity alone and is denoted by v

_{g}. The second term [ \(\frac{\pi z(r_t^2 – r_a^2)}{g\, ln(\frac{r_t}{r_a})}\) ] in the equation represents the sigma factor of the centrifuge. The feed flow rate in tubular centrifuge is expressed in terms of sedimentation velocity and sigma factor. The first part refers to the properties of the material being handled and second refers to those of the equipments.

9. Which is the special type of preparative centrifuge?

a) Tubular bowl centrifuge

b) Disk stack centrifuge

c) Basket centrifuge

d) Chamber bowl centrifuge

View Answer

Explanation: Disc stack centrifuge is a special type of preparative centrifuge, which is a compact design and gives better solid-liquid separation than the standard tubular bowl centrifuge. The feed enters from the top of the device and is distributed at the bottom of the disk bowl centrifuge through a hollow drive shaft. The particles move out and come in contact with the angled disc stack.

10. Estimate the maximum flow rate of the fermentation broth when the fermentation broth is subjected to the pilot scale tubular centrifuge having radius of 0.5μm and the density is 1.10 g/cm^{3}, the speed of the centrifuge is 5000rpm and the diameter of the bowl is 10cm, the length of the bowl is 100cm with the outlet opening of bowl has diameter of 4cm.

a) 5.45 l/min

b) 2.01 l/min

c) 0.5 l/min

d) 0.66 l/min

View Answer

Explanation: The flow rate can be estimated by calculating the sedimentation velocity and the sigma factor. v

_{g}= [\(\frac{2a^2(\rho_S – \rho_L)g}{18\mu}\) ] = \(\frac{2(0.5 × 10^{-6})^2 × (1.10 – 1.00) × 9.81 × 10^6}{9(0.01)}\) = 5.45 × 10

^{-6}cm/s. The completely recovery can be attained by calculating the sigma factor Σ = [ \(\frac{\pi L(R_0^2 – R_1^2)\omega^2}{g\, ln(\frac{R_0}{R_1})}\) ] = \(\frac{\pi (100)(25-4)(5000 \frac{rev}{min} × \frac{2\pi \,rad}{rev})^2}{9.81 × ln(\frac{5}{2}) × 100 × (60s/min)^2}\) = 2.01 × 10

^{6}cm

^{2}and the flow rate Q = (5.45 × \(\frac{10^{-6}cm}{s})\) × 2.01 × 10

^{6}cm

^{2}× \(\frac{l}{10^3 cm^2}\) × 60 s/min = 0.66 \(\frac{l}{min}\).

11. What is the equation of motion in the direction of x?

a) \(\frac{dx}{dt}\) = v_{0} + v_{ω} sin θ

b) \(\frac{dx}{dt}\) = -v_{0} – v_{ω} sin θ

c) \(\frac{dx}{dt}\) = v_{0} – v_{ω} sin θ

d) –\(\frac{dx}{dt}\) = v_{0} – v_{ω} sin θ

View Answer

Explanation: The equation of motion \(\frac{dx}{dt}\) = v

_{0}– v

_{ω}sin θ in the direction of x is the simplified assumption that v

_{0}is much greater than v

_{ω}sinθ and the average value of v

_{0}is denoted by < v

_{0}> and is given by the flow rate Q divided by the cross-sectional area A perpendicular to the flow for n disks.

12. How to estimate the cross sectional area of the disc stack centrifuge?

a) A = n(2πRl)

b) A = (2πRl)

c) A = n(2πR)

d) A = n(2πl)

View Answer

Explanation: The cross sectional area A which is perpendicular to flow for n discs is A = n(2πRl) where, R is the radial distance from the centre of rotation which varies and l is the space between the discs which are fixed.

13. What is the equation of motion in y direction of centrifugal particle motion?

a) \(\frac{dy}{dt}\) = v_{ω} sin θ = [ \(\frac{2a^2(\rho – \rho_0)\omega^2\, R}{9\mu}\) ] cos θ

b) \(\frac{dy}{dt}\) = v_{ω} cos θ = -[ \(\frac{2a^2(\rho – \rho_0)\omega^2\, R}{9\mu}\) ] cos θ

c) \(\frac{dy}{dt}\) = v_{ω} cos θ = [ \(\frac{2a^2(\rho – \rho_0)\omega^2\, R}{9\mu}\) ] cos θ

d) \(\frac{dy}{dt}\) = v_{ω} – cos θ = -[ \(\frac{2a^2(\rho – \rho_0)\omega^2\, R}{9\mu}\) ] – cos θ

View Answer

Explanation: The equation of motion in y direction of centrifugal particle motion is \(\frac{dy}{dt}\) = v

_{ω}cos θ = [ \(\frac{2a^2(\rho – \rho_0)\omega^2\, R}{9\mu}\) ] cos θ where, centrifugal velocity is obtained from [ \(\frac{2a^2(\rho – \rho_0)\omega^2\, R}{9\mu}\) ] and the slope of the trajectory of particles moving between the pair of discs is determined by Q = [ \(\frac{2a^2(\rho – \rho_0)g}{9\mu}\) ][ \(\frac{2n\pi \omega^2(R_0^3 – R_1^3)cotθ}{3g}\) ] = {v

_{g}}[Σ].

14. How can you estimate the time between discharges in the semi-continuous mode of operation?

a) t_{d} = \(\frac{V_{s \varphi_e}}{Q\varphi_f}\)

b) t_{d} = –\(\frac{V_{s \varphi_e}}{Q\varphi_f}\)

c) t_{d} = \(\frac{-V_{s \varphi_e}}{Q\varphi_f}\)

d) t_{d} = \(\frac{V_{s \varphi_e}}{Q – \varphi_f}\)

View Answer

Explanation: The time required between discharges in the semi-continuous mode of operation of the disc centrifuge with volumetric feed flow rate can be obtained by t

_{d}= \(\frac{V_{s \varphi_e}}{Q\varphi_f}\) where, V

_{s}is the solids holdup volume of the bowl and φ

_{e}as well as φ

_{f}are the volume fractions of solids in the existing sediments and the feed respectively.

**Sanfoundry Global Education & Learning Series – Bioseparation Processes**.

To practice all areas of Bioseparation Processes, __ here is complete set of 1000+ Multiple Choice Questions and Answers__.

**If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]**