# Bioprocess Engineering Questions and Answers – Classification of Fluids

This set of Bioprocess Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Classification of Fluids”.

1. “The density of liquids is practically dependent of pressure”?
a) True
b) False

Explanation: The density of liquids is practically independent of pressure; liquids are incompressible fluids. If the density of a fluid changes with pressure, the fluid is compressible. Gases are generally classed as compressible fluids.

2. What is the property of an ideal or perfect fluid?
a) Compressible and zero viscosity
b) Compressible and zero density
c) Incompressible and zero viscosity
d) Incompressible and zero density

Explanation: Viscosity is the property of fluids responsible for internal friction during flow. An ideal or perfect fluid is a hypothetical liquid or gas which is incompressible and has zero viscosity. The term inviscid applies to fluids with zero viscosity.

3. Fluids which undergo strain rates proportional to the applied shear stress are termed as?
a) Inviscid fluid
b) Newtonian fluid
c) Non- Newtonian fluid
d) Viscous fluid

Explanation: Newtonian fluids undergo strain rates proportional to the applied shear stress. It is defined to be a fluid whose shear stress is linearly proportional to the velocity gradient in the direction perpendicular to the plane of shear. This definition means regardless of the forces acting on a fluid, it continues to flow.

4. Which of the following is not an example of a Newtonian fluid?
a) Water
b) Glycerine
c) Non- Drip Paints
d) Alcohol

Explanation: Stirring a non-Newtonian fluid can cause the viscosity to decrease, so the fluid appears “thinner” (this is seen in non-drip paints). There are many types of non-Newtonian fluids, as they are defined to be something that fails to obey a particular property – for example, most fluids with long molecular chains can react in a non-Newtonian manner.

5. Which of the following is not an example of a Non- Newtonian fluids?
a) Gels
b) Water
c) Suspensions
d) Pudding

Explanation: Water is a Newtonian fluid, because it continues to display fluid properties no matter how much it is stirred or mixed. A slightly less rigorous definition is that the drag of a small object being moved slowly through the fluid is proportional to the force applied to the object. (Compare friction). Important fluids, like water as well as most gases, behave – to good approximation – as a Newtonian fluid under normal conditions on Earth.
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6. What do you mean by the term “Rheology”?
a) Study of materials with both solid and fluid characteristics
b) Study of materials with only solid characteristics
c) Study of materials with only fluid characteristics
d) Study of material with both fluid and gas characteristics

Explanation: It is the study of the flow of matter, primarily in a liquid state, but also as “soft solids” or solids under conditions in which they respond with plastic flow rather than deforming elastically in response to an applied force. It is a branch of physics which deals with the deformation and flow of materials, both solids and liquids.

7. What is the unit of viscosity of fluids in C.G.S?
a) ml/s
b) l/s
c) Poise
d) Newton

Explanation: The unit of viscosity (.in C.G.S) is poise.
One poise = 0.1 N.s / m2,
µwater = 10-3 Ns / m2
µair = 1.81 x 10-5 Ns / m2.

8. If 5 m3 of certain oil weighs 45 kN calculate the specific weight of the oil.
a) 10 kN/m3
b) 9 kN/m3
c) 5 kN/m3
d) 2 kN/m3

Explanation: Given data: Volume = 5 m3
Weight = 45 kN

9. Refer to Q8 and Calculate specific gravity of the oil. (Specific weight of water = 9.807 kN/m3)
a) 0.991
b) 0.918
c) 0.917
d) 0.992

Explanation: Given data: Volume = 5 m3
Weight = 45 kN
Specific weight of water = 9.807 kN/m3

10. A liquid has a mass density of 1550 kg/m3. Calculate its specific weight.
a) 1.50×102 N/m3
b) 1.52×104 N/m3
c) 1.54×104 N/m3
d) 1.50×104 N/m3

Explanation: Given data: Mass density = 1550 kg/m3
Specific weight = mass density × Acceleration due to gravity
= 1550 kg/m3 × 9.8 m/s2
= 1.52×104 N/m3.

11. Refer to Q10 and calculate specific gravity.
a) 1.55
b) 1.56
c) 1.50
d) 1.54

Explanation: Given data: Mass density = 1550 kg/m3

12. Refer to Q 10 and Q11, and calculate specific volume.
a) 6.51 × 10-5 m3/N
b) 6.78 × 10-5 m3/N
c) 6.45 × 10-5 m3/N
d) 6.57 × 10-5 m3/N

Explanation: Specific volume = 1/(Specific weight)
= 1/(1.52×104 N/m3)
= 6.57 × 10-5 m3/N.

13. If the equation of a velocity profile over a plate is v = 5y2 + y (where v is the velocity in m/s) determine the shear stress at y = 0. Given the viscosity of the liquid is 8.35 poise.
a) 0.830
b) 0.832
c) 0.835
d) 0.834

Explanation: Given Data: Velocity profile v = 5y2 + y;
μ = 8.35 poise

Substituting y=0 on the above equation, we get shear stress at respective depths.
τ = 0.835

14. Refer to Q13 and determine the shear stress at y =7.5cm.
a) 1.46 N/m3
b) 1.45 N/m3
c) 1.40 N/m3
d) 1.43 N/m3

Explanation: Given Data: Velocity profile v = 5y2 + y;
μ = 8.35 poise

Substituting y=7.5 on the above equation, we get shear stress at respective depths.
τ = 1.46 N/m3.

15. “Mach is the Dimensionless quantity”.
a) True
b) False

Explanation: In fluid dynamics, the Mach number (M or Ma) is a dimensionless quantity representing the ratio of flow velocity past a boundary to the local speed of sound.

where:
M is the Mach number,
u is the local flow velocity with respect to the boundaries (either internal, such as an object immersed in the flow, or external, like a channel), and
c is the speed of sound in the medium.

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