Bioprocess Engineering Questions and Answers – Fluids in Motion

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

1. Which of the following fluid flow can form “eddies”?
a) Streamline flow
b) Laminar flow
c) Turbulent flow
d) Non-turbulent flow
View Answer

Answer: c
Explanation: The slower the flow the more closely the streamlines represent actual motion. Slow fluid flow is therefore called streamline or laminar flow. In fast motion, fluid particles frequently cross and recross the streamlines. This motion is called turbulent flow and is characterized by formation of eddies.

2. What do you mean by “u” in the following equation of Reynolds number and for what type of conduit is it applicable?
\(R_e = \frac{Duρ}{\mu}\)

a) Fluid viscosity, Stirred vessel
b) Average linear velocity, Pipe
c) Average linear velocity, Stirred vessel
d) Fluid viscosity, Pipe
View Answer

Answer: b
Explanation: A parameter used to characterise fluid flow is the Reynolds number. For full flow in pipes with circular cross-section, Reynolds number Re is defined as:
\(R_e = \frac{Duρ}{\mu}\)

where D is pipe diameter, u is average linear velocity of the fluid, ρ is fluid density, and μ is fluid viscosity. For stirred vessels there is another definition of Reynolds number:
\(R_{ei} = \frac{N_i D_i^2 \rho}{\mu}\)

where Rei is the impeller Reynolds number, Ni is stirrer speed, Di is impeller diameter, ρ is fluid density and μ is fluid viscosity.

3. In smooth pipes, Laminar flow is encountered at what value of Reynolds number?
a) More than 2100
b) 2100-4000
c) More than 4000
d) Less than 2100
View Answer

Answer: d
Explanation: One of the most significant outcomes of Reynolds’ experiments is that there is a critical Reynolds number which marks the upper boundary for laminar flow in pipes. In smooth pipes, laminar flow is encountered at Reynolds numbers less than 2100. Under normal conditions, flow is turbulent at Re above about 4000. Between 2100 and 4000 is the transition region where flow may be either laminar or turbulent depending on conditions at the entrance of the pipe and other variables.

4. What do you mean by incompressible flow?
a) Temperature is constant
b) Density is constant
c) Pressure is constant
d) Velocity is constant
View Answer

Answer: b
Explanation: In fluid mechanics or more generally continuum mechanics, incompressible flow (isochoric flow) refers to a flow in which the material density is constant within a fluid parcel an infinitesimal volume that moves with the flow velocity.

5. Is water incompressible?
a) True
b) False
View Answer

Answer: b
Explanation: An important concept in fluid mechanics is that liquids, like water and oil, cannot be compressed much when you push down on them in an enclosed container. So, liquids are considered to be incompressible fluids.

6. A garden hose of diameter 2 cm is used to fill a 20 liter bucket. If it takes 1 minute to fill the bucket, what is the speed at which water enters the hose?
a) 106.1 cm/s
b) 106 cm/s
c) 106.5 cm/s
d) 106.8 cm/s
View Answer

Answer: a
Explanation: The cross-sectional area of the hose will be given by:
A1 = π r2 = π (2 cm/2)2 = π cm2
To find the velocity, v1, we use
Flow rate = A1 v1 = 20.0 L/min = \(\frac{20.0 × 10^3 cm^3}{60.0 s}\)
v1 = \(\frac{20.0 × \frac{10^3 cm^3}{60.0s}}{\pi cm^2}\)
= 106.1 cm/s

7. Refer to Q6 and estimate a practical joker pinches the open end of the hose down to a diameter of 5 mm, and sprays his neighbor with it. What is the speed at which water comes out of the hose?
a) 1690 cm/s
b) 1660 cm/s
c) 1698 cm/s
d) 1668 cm/s
View Answer

Answer: c
Explanation: The flow rate (A1v1) of the water approaching the constriction must be equal to the flow rate leaving the hose (A2v2). This gives:
v2 = \(\frac{A_1 v_1}{A_2}\)
= \(\frac{(\pi)(106.1)}{(\pi)(0.5/2)^2}\)
= 1698 cm/s.

8. Water flows through a rubber hose 2 cm in diameter at a velocity of 4 m/s. What must be the diameter of the nozzle in order that the water emerge at 16 m/s?
a) 0.884 cm
b) 0.891 cm
c) 0.881 cm
d) 0.894 cm
View Answer

Answer: d
Explanation: The area is proportional to the square of diameter, so:
v1d12 = v2d22
d22 = v1d12 / v2 = \(\frac{(4 \frac{m}{s})(2cm)^2}{(20 cm)^2}\)
d2 = 0.894 cm

9. Water flows through a rubber hose 2 cm in diameter at a velocity of 4 m/s. What is the rate of flow rate of flow in m3/min?
a) 0.0754 m3/min
b) 0.0750 m3/min
c) 0.0764 m3/min
d) 0.0760 m3/min
View Answer

Answer: a
Explanation: R = v1A1 = v2A2
R = v1A1; A1 = \(\frac{\pi d_2^1}{4}\)

R1 = v1 \(\frac{\pi d_2^1}{4} = \frac{(4 \frac{m}{s})\pi(0.02 m)^2}{4}\)

R1 = 0.00126 m3/s
R1 = \(0.00126 \frac{m^3}{min} (\frac{1 min}{60 s})\)

R1 = 0.0754 m3/min.

10. Water ideally flows through a pipe of radius 6 centimeters at a rate of 5 meters per second. The pipe then narrows to a radius of 2 centimeters. What is the new velocity of the water?
a) 50 m/s
b) 55 m/s
c) 60 m/s
d) 65 m/s
View Answer

Answer: b
Explanation: First convert everything to basic metric units.
(6cm)((1m)/(100cm)) = .06m
(2cm)((1m)/(100cm)) = .02m

Now find the cross-sectional area in both sections of the pipe.
A = π(r)2
π(.06)2 ≈ 0.011m2
π(.02)2 ≈ 0.001m2

Next, use the equation of continuity with the areas you have obtained and the given velocity.
A1v1 = A2v2
(0.011m2)(5m/s) = (.001m2)v2
0.055m3/s = (.001m2)v2
55m/s = v2.

11. Venturi relation is one of the applications of ___________
a) Light equation
b) Bernoulli’s equation
c) Speed equation
d) Equation of continuity
View Answer

Answer: b
Explanation: The flow speed of a fluid can be measured using a device such as a Venturi meter or an orifice plate, which can be placed into a pipeline to reduce the diameter of the flow. For a horizontal device, the continuity equation shows that for an incompressible fluid, the reduction in diameter will cause an increase in the fluid flow speed. Subsequently, Bernoulli’s principle then shows that there must be a decrease in the pressure in the reduced diameter region. This phenomenon is known as the Venturi effect.

12. Simplified equation of continuity is represented as _____________
a) A1V2 = A2V2
b) A1V1 = A1V2
c) A2V1 = A1V1
d) A1V1 = A2V2
View Answer

Answer: d
Explanation: Continuity uses the conservation of matter to describe the relationship between the velocities of a fluid in different sections of a system. The simple observation that the volume flow rate, Av, must be the same throughout a system provides a relationship between the velocity of the fluid through a pipe and the cross-sectional area. Continuity works in tandem with Bernoulli’s principle in the design and construction of systems of irrigation, plumbing, etc.

13. Relative density of mercury is _____________
a) 13.6
b) 1
c) 1000
d) 9.8
View Answer

Answer: a
Explanation: Relative density of a substance is the ratio of its density to that of So relative density of mercury is 13.6 means that density of mercury is 13.6 times the density of water.

14. The unit of pressure one bar is ___________
a) 1 Pascal
b) 1 kilo Pascal
c) 100 kPascal
d) 1000 kPascal
View Answer

Answer: c
Explanation: A bar (b) is a metric measurement unit of pressure. One bar is equivalent to ten newtons (N) per square centimeter (cm2).

1 bar = 100,000 Pascal = 100,000 N/m2 = 100,000 N / (100*100cm2) = 10 N/cm2.

15. Reynolds number signifies the ratio of ____________
a) inertia forces to gravity forces
b) buoyant forces to inertia forces
c) inertial forces to viscous forces
d) gravity forces top viscous forces
View Answer

Answer: c
Explanation: The Reynolds number is the ratio of inertial forces to viscous forces within a fluid which is subjected to relative internal movement due to different fluid velocities, in which is known as a boundary layer in the case of a bounding surface such as the interior of a pipe.

Sanfoundry Global Education & Learning Series – Bioprocess Engineering.

To practice all areas of Bioprocess Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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