Bioprocess Engineering Multiple Choice Questions on “Factors Affecting Broth Viscosity”.
1. In the following Vand’s equation what does “Φ”, represents?
μ = μL (1+ 2.5 Φ + 7.25 Φ2)
A. Mass fraction of solids
B. Mass fraction of liquids
C. Volume fraction of solids
D. Volume fraction of liquids
Answer: C
Explanation: The viscosity of a suspension of spheres in Newtonian liquid can be predicted using the Vand equation, where μL is the viscosity of the suspending liquid and Φ is the volume fraction of solids.This equation has been found to hold for yeast and spore suspensions up to 14 vol% solids.
2. The filamentous cells will become flexible under which of the following condition?
A. Increased osmotic pressure and low turgor pressure
B. Decreased osmotic pressure and high turgor pressure
C. Increased osmotic pressure and high turgor pressure
D. Decreased osmotic pressure and low turgor pressure
Answer: A
Explanation: Osmotic pressure of the culture medium affects cell turgor pressure. This in turn affects the hyphal flexibility of filamentous cells; increased osmotic pressure gives a lower turgor pressure making the hyphae more flexible. Improved hyphal flexibility reduces broth viscosity, and can also have a marked effect on yield stress.
3. The temperature is directly proportional to viscosity.
A. True
B. False
Answer: B
Explanation: A fluid’s viscosity strongly depends on its temperature. Along with the shear rate, temperature really is the dominating influence. The higher the temperature is, the lower a substance’s viscosity is. Consequently, decreasing temperature causes an increase in viscosity. The relationship between temperature and viscosity is inversely proportional for all substances. A change in temperature always affects the viscosity – it depends on the substance just how much it is influenced by a temperature change. For some fluids a decrease of 1°C already causes a 10 % increase in viscosity.
4. The pressure is directly proportional to viscosity?
A. True
B. False
Answer: A
Explanation: Viscosity increases with increasing pressure because the amount of free volume in the internal structure decreases due to compression. Consequently, the molecules can move less freely and the internal friction forces increase. The result is an increased flow resistance.
5. Convert 5 cP to Pa.s and estimate the correct answer.
A. 0.5
B. 0.25
C. 0.05
D. 0.005
Answer: D
Explanation: 1 centipoise [cP] = 0.001 pascal second [Pa•s]
so,
5 centipoise = 0.005 pascal second.
6. If the particle size is small in fluid than viscosity will ___________
A. Increase
B. Decrease
C. Remain the same
D. Slightly increase
Answer: B
Explanation: The size of the particles of a substance will greatly affect its viscosity. Small particles can move more easily past each other and can therefore flow faster, meaning they have a lower viscosity. Large particles would mean a higher viscosity.
7. The force of attraction between the particles is directly proportional to viscosity.
A. True
B. False
Answer: A
Explanation: Particles of the same substance have an attractive force on one another. Some substances have a strong attraction while some substances have a weaker attraction. The stronger the attraction of particles, the higher the viscosity.
8. Conversion of 5 Pa-s into Poise will be ____________
A. 0.5
B. 50
C. 500
D. 0.05
Answer: B
Explanation: Ten poise equal one pascal second [Pa s] making the centipoise [cP] and millipascal second [mPa s] identical.
1 Pa-s = 10 P
So,
5 Pa-s = 50 P.
9. When a gas is heated its viscosity will ___________
A. Increase
B. Decrease
C. Remain same
D. Slightly increase
Answer: A
Explanation: A gas’s viscosity increases when heated (more energy causing more movement + therefore more friction because particles are hitting each other). Decreases when cooled.
10. A fluid is flowing between two layers. Calculate the shearing force if the shear velocity is 0.25 m/s and has length 2 m and dynamic viscosity is 2Ns/m2.
A. 0.2 N/m2
B. 0.4 N/m2
C. 0.5 N/m2
D. 0.6 N/m2
Answer: C
Explanation: Given: Shear velocity u = 0.25 m/s,
shear stress y = 0.125/s
dynamic viscosity μ = 2 Ns/m2
The shearing stress is given by,
F = μ A u/y
F = 2 Ns/m2 × 0.125 /s
F = 0.5 N/m2
Therefore, the shearing stress is 0.5 N/m2.
11. Calculate the density of fluid having an absolute viscosity of 0.89 Ns/m2 and kinematic viscosity of 2 m2/s.
A. 0.445 kg/m3
B. 0.440 kg/m3
C. 0.544 kg/m3
D. 0.550 kg/m3
Answer: A
Explanation: Known: Absolute viscosity μ = 0.89 Ns/m2, kinematic viscosity ν = 2 m3/s.
The density is given by,
(p=frac{u}{v})
(p=frac{0.89 Ns^2}{2 m^2})
p = 0.445kg/m3
Therefore, the density of fluid is 0.445kg/m3.