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All questions of Fluid Mechanics for Mechanical Engineering Exam

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A real fluid is any fluid which
  • a)
    has surface tension and is compressible
  • b)
    has zero shear stress
  • c)
    has constant viscosity and density
  • d)
    has viscosity
Correct answer is option 'D'. Can you explain this answer?

Saranya Saha answered
Real Fluids in Mechanical Engineering

Real Fluids are fluids that have viscosity, which is the property of a fluid that causes it to resist flow. Real fluids are common in everyday life, and they are found in many industrial applications.

Viscosity

Viscosity is the measure of a fluid's resistance to deformation or flow. It is a property of the fluid, and it determines how easily the fluid can flow. Real fluids have viscosity, which means that they resist flow.

Examples of Real Fluids

Some examples of real fluids include:

- Water
- Oil
- Gasoline
- Honey
- Blood

Properties of Real Fluids

Real fluids have properties that are different from ideal fluids. Some of the properties of real fluids include:

- Viscosity
- Density
- Compressibility

Applications of Real Fluids

Real fluids are used in many applications in mechanical engineering. Some examples include:

- Lubrication of machine parts
- Cooling of engines
- Hydraulic systems
- Pneumatic systems

Conclusion

Real fluids are an important part of mechanical engineering. They have properties that make them different from ideal fluids, and they are used in many applications. Understanding the properties of real fluids is essential for designing and operating mechanical systems.

When a shear stress is applied to a substance, it is found to resist it by static deformation, the  substance is 
  • a)
    liquid
  • b)
    solid
  • c)
    gas
  • d)
    fluid
Correct answer is option 'B'. Can you explain this answer?

Neha Joshi answered
  • A solid can resist shear stress by a static deformation; a fluid cannot
  • Any shear stress applied to a fluid, no matter how small, will result in motion of that fluid
  • The fluid moves and deforms continuously as long as the shear stress is applied
  • So, a fluid at rest must be in a state of zero shear stress

A perfect fluid is
  • a)
    a real fluid is
  • b)
    the one which obeys perfect gas laws
  • c)
    compressive and gaseous
  • d)
    incompressible and frictionless
Correct answer is option 'D'. Can you explain this answer?

Cstoppers Instructors answered
  • A perfect fluid is incompressible and frictionless and it is explained with the help of Euler's equation.
Euler's equation of motion:
  • The Euler's equation for a steady flow of an ideal fluid along a streamline is a relation between the velocity, pressure, and density of a moving fluid.
  • It is based on Newton's Second Law of Motion which states that if the external force is zero, linear momentum is conserved.
  • The integration of the equation gives Bernoulli's equation in the form of energy per unit weight of the following fluid.
It is based on the following assumptions:
  • The fluid is non-viscous (i,e., the frictional losses are zero)
  • The fluid is homogeneous and incompressible (i.e., the mass density of the fluid is constant)
  • The flow is continuous, steady, and along the streamline.
  • The velocity of the flow is uniform over the section.
  • No energy or force (except gravity and pressure forces) is involved in the flow.

The concept of continuum in fluid flow assumes that the characteristics length of the flow is 
  • a)
    smaller than the mean free path of the molecules.
  • b)
    larger than the mean free path of the molecule
  • c)
    larger than the dimensions of the suspended particles
  • d)
    larger than the wavelength of sound in the medium.
Correct answer is option 'B'. Can you explain this answer?

Neha Joshi answered
  • The concept of the continuum is a kind of idealization of the continuous description of matter where the properties of the matter are considered as continuous functions of space variables.
  • In this approach, the matter is continuously distributed with no void present continuum concept be applied if the mean free path is much less than the system dimensions.
  • Although any matter is composed of several molecules, the concept of continuum assumes a continuous distribution of mass within the matter or system with no empty space, instead of the actual conglomeration of separate molecules.
  • Describing a fluid flow quantitatively makes it necessary to assume that flow variables (pressure, velocity etc.) and fluid properties vary continuously from one point to another. 
  • One of the factors considered important in determining the validity of the continuum model is molecular density. It is the distance between the molecules which is characterized by a mean free path ( λ ). It is calculated by finding statistical average distance the molecules travel between two successive collisions.
  • If the mean free path is very small as compared with some characteristic length in the flow domain (i.e., the molecular density is very high) then the gas can be treated as a continuous medium.
  • If the mean free path is large in comparison to some characteristic length, the gas cannot be considered continuous and it should be analysed by the molecular theory.
  • A dimensionless parameter is known as Knudsen number, Kn = λ/L, where λ is the mean free path and L is the characteristic length. It describes the degree of departure from the continuum.
  • Usually when Kn > 0.01, the concept of the continuum does not hold good and when Kn < 0.01, the concept of the continuum does hold good.
  • In the continuum approach, fluid properties such as density, viscosity, thermal conductivity, temperature, etc. can be expressed as continuous functions of space and time.

In the phenomenon of cavitation, the characteristic fluid property involved is
  • a)
    Surface tension
  • b)
    Viscosity
  • c)
    Bulk modulus of elasticity
  • d)
    Vapour pressure
Correct answer is option 'D'. Can you explain this answer?

Avantika Sen answered
In any flow system, if the pressure at any point in the liquid approaches the vapour pressure, vaporization of liquid starts, resulting in the pockets of dissolved gasses and vapours. The bubbles of vapour thus formed are carried by the flowing liquid into a region of high pressure where they collapse, giving rise to high impact pressure. This phenomenon is known as cavitation.

The potential lines in case of a doublet are
  • a)
    circles tangent to the x-axis
  • b)
    circles tangent to the y-axis
  • c)
    concentric circles with center on the x-axis
  • d)
    concentric circles with center on the y-axis
Correct answer is option 'B'. Can you explain this answer?

Sagarika Mukherjee answered
Understanding Doublets in Fluid Mechanics
In fluid mechanics, a doublet is a combination of a source and a sink of equal strength very close to each other. The potential lines or streamlines associated with a doublet reveal important characteristics of fluid flow.
Potential Lines of a Doublet
When analyzing the potential lines of a doublet, we consider how these lines behave in relation to the axes in a Cartesian coordinate system.
Option Analysis
- a) Circles Tangent to the X-axis
This option suggests circular paths that touch the x-axis. However, doublets do not create such potential lines.
- b) Circles Tangent to the Y-axis
This is the correct answer. The potential lines resulting from a doublet resemble circles that are tangent to the y-axis. As the doublet is oriented along the x-axis, the flow lines exhibit this circular characteristic.
- c) Concentric Circles with Center on the X-axis
While concentric circles indicate uniform flow, they do not accurately represent the potential lines of a doublet.
- d) Concentric Circles with Center on the Y-axis
Similar to option c, these circles would not represent the nature of the flow induced by a doublet.
Conclusion
In summary, the potential lines of a doublet are best represented by circles that are tangent to the y-axis. This behavior illustrates the unique flow characteristics created by the interaction of a source and sink, confirming option 'b' as the correct choice. Understanding these flow patterns is essential in fluid mechanics and applications like aerodynamics and hydrodynamics.

When a body, floating in a liquid is given a small angular displacement, it starts oscillating about a point known as
  • a)
    center of pressure
  • b)
    center of gravity
  • c)
    center of buoyancy
  • d)
    meta center
Correct answer is option 'D'. Can you explain this answer?

Ameya Sen answered

Where M is metacentre and θ is small angle of hill through which a body is tilted.
Hence metacenter is a point about which a body oscillate when tilted or given a small angular displacement.

A sprue in a sand mould has a top diameter of 20 mm and height of 200 mm. The velocity of the molten metal at entry of the sprue is 0.5 m/s. Assume acceleration due to gravity as 9.8 m/s2 and neglect all losses. If the mould is well ventilated the velocity (upto 3 decimal points accuracy) of the molten metal at the bottom of the sprue is ____m/s.
Correct answer is '2.042'. Can you explain this answer?

Baishali Chopra answered
Understanding the Problem
To calculate the velocity of the molten metal at the bottom of the sprue, we can use the principle of conservation of energy, specifically Bernoulli's equation. The equation relates the velocity and height of the fluid in motion.
Given Data
- Top Diameter of Sprue: 20 mm
- Height of Sprue: 200 mm
- Initial Velocity (V1): 0.5 m/s
- Acceleration due to Gravity (g): 9.8 m/s²
- Density and viscosity are neglected due to the assumption of ideal conditions.
Key Concepts
- Potential Energy: At the top of the sprue, the molten metal possesses gravitational potential energy due to its height.
- Kinetic Energy: As it descends, potential energy converts into kinetic energy, increasing the velocity of the molten metal.
Applying Bernoulli's Equation
1. Initial State (Top of the Sprue):
- Velocity (V1) = 0.5 m/s
- Height (h1) = 0.2 m (200 mm)
2. Final State (Bottom of the Sprue):
- Velocity (V2) = ?
- Height (h2) = 0 m
Using Bernoulli’s equation:
V1²/2 + g*h1 = V2²/2 + g*h2
Simplifying, we find:
0.5²/2 + 9.8*0.2 = V2²/2
3. Calculating V2:
- Solve for V2, applying the values.
- V2 = √(0.5² + 2*9.8*0.2)
This gives us V2 ≈ 2.042 m/s.
Conclusion
Thus, the velocity of the molten metal at the bottom of the sprue, considering ideal conditions and neglecting losses, is approximately 2.042 m/s.

Dimensional analysis is useful in
Dimensional analysis deals with the process whereby all the important factors involved in a physical phenomenon are systematically arranged into dimensional group.
  • a)
    checking the correctness of a physical equation
  • b)
    determining the number of variables involved in a particular phenomenon
  • c)
    determining the dimensionless groups from the given variables
  • d)
    the exact formulation of a physical phenomenon
Correct answer is option 'C'. Can you explain this answer?

Kirti Bose answered
Dimensional analysis is a powerful tool used in the field of physics and engineering to analyze and understand the relationships between different physical variables involved in a phenomenon. It involves systematically arranging the important factors into dimensional groups, which helps in simplifying and solving complex equations. Dimensional analysis is particularly useful in determining dimensionless groups from given variables. Let's explore why option 'C' is the correct answer.

1. Checking the correctness of a physical equation:
Dimensional analysis can be used to check the correctness of a physical equation by ensuring that the dimensions of the variables on both sides of the equation are consistent. Each physical quantity has a specific dimension (e.g., length, mass, time), and if the dimensions on both sides of the equation do not match, then it indicates an error or inconsistency in the equation.

2. Determining the number of variables involved in a particular phenomenon:
Dimensional analysis can also help in determining the number of variables involved in a particular phenomenon. By analyzing the dimensions of the physical quantities, we can identify the independent variables that affect the phenomenon. This understanding is crucial in simplifying the problem and focusing on the key variables that need to be considered.

3. Determining the dimensionless groups from the given variables:
One of the primary applications of dimensional analysis is to determine dimensionless groups from given variables. Dimensionless groups are ratios of physical quantities with the same dimensions, and they have special significance in many physical phenomena. These dimensionless groups often represent fundamental relationships and can provide insights into the behavior of the system under study. By identifying and analyzing these dimensionless groups, engineers and scientists can derive meaningful conclusions and make predictions about the phenomenon.

4. The exact formulation of a physical phenomenon:
While dimensional analysis helps in simplifying and understanding the relationships between variables, it does not provide the exact formulation of a physical phenomenon. The exact formulation typically involves mathematical modeling, experimentation, and theoretical analysis. However, dimensional analysis can serve as a valuable initial step in formulating the problem and identifying the key variables and their relationships.

In conclusion, dimensional analysis is a powerful tool that can be used in various ways, including checking the correctness of a physical equation, determining the number of variables involved, and most importantly, determining dimensionless groups from given variables. This analysis helps in simplifying complex equations, understanding the behavior of the system, and making predictions about the phenomenon under study.

If B = Center of buoyancy, G = Center of gravity and M= Metacenter, of afloating body, the body will be in stable equilibrium if
  • a)
    MG = 0
  • b)
    M is below G
  • c)
    BG = 0
  • d)
    M is above G
Correct answer is option 'D'. Can you explain this answer?

Akshita Desai answered
For a floating body
Metacentre above centre of gravity→ Stable Equilibrium
Metacentre coincides centre of gravity→ Neutral Equilibrium
Metacentre below centre of gravity→ Unstable Equilibrium
For a submerged body
Centre of buoyancy above centre of gravity→ Stable Equilibrium
Centre of buoyancy coincides centre of gravity→ Neutral Equilibrium
Centre of buoyancy below centre of gravity→ Unstable Equilibrium

Consider the following statements:
A tube, with the section diverging in the direction of flow, can be used as
1. Supersonic nozzle
2. Subsonic nozzle
3. Supersonic diffuser
4. Subsonic diffuser
Select the correct answer using the codes given below:
  • a)
    1 alone
  • b)
    2 alone
  • c)
    1 and 4
  • d)
    2 and 3
Correct answer is option 'C'. Can you explain this answer?

Anshul Sharma answered
Supersonic and Subsonic Flow in Nozzles and Diffusers

Introduction:
In fluid dynamics, a nozzle is a device that accelerates the flow of a fluid (usually a gas) by increasing its velocity. A diffuser, on the other hand, is a device that slows down the flow of a fluid by increasing its pressure. Both nozzles and diffusers can operate in either subsonic or supersonic flow conditions. Let us analyze the given statements and determine which options are correct.

Analysis:
1. Supersonic Nozzle: A supersonic nozzle is designed to accelerate the flow of a fluid to supersonic speeds. It is characterized by a converging-diverging shape. As the fluid passes through the converging section, its velocity increases, and it reaches its maximum velocity at the throat of the nozzle. After passing through the throat, the flow expands in the diverging section, maintaining supersonic speeds. Therefore, a tube with a section diverging in the direction of flow can be used as a supersonic nozzle. Hence, Statement 1 is correct.

2. Subsonic Nozzle: A subsonic nozzle is designed to accelerate the flow of a fluid to subsonic speeds. It is characterized by a converging shape. The fluid passes through the converging section, and its velocity increases. However, it does not reach supersonic speeds. Therefore, a tube with a section diverging in the direction of flow cannot be used as a subsonic nozzle. Hence, Statement 2 is incorrect.

3. Supersonic Diffuser: A supersonic diffuser is designed to slow down the flow of a supersonic fluid by increasing its pressure. It is characterized by a converging-diverging shape, similar to a supersonic nozzle but in the opposite direction. As the supersonic flow enters the diffuser, it expands in the diverging section, decreasing its velocity and increasing its pressure. Therefore, a tube with a section diverging in the direction of flow can be used as a supersonic diffuser. Hence, Statement 3 is incorrect.

4. Subsonic Diffuser: A subsonic diffuser is designed to slow down the flow of a subsonic fluid by increasing its pressure. It is characterized by a diverging shape. As the fluid enters the diffuser, it expands in the diverging section, decreasing its velocity and increasing its pressure. Therefore, a tube with a section diverging in the direction of flow can be used as a subsonic diffuser. Hence, Statement 4 is correct.

Conclusion:
From the analysis, we can conclude that a tube with a section diverging in the direction of flow can be used as a supersonic nozzle and a subsonic diffuser. Therefore, the correct answer is option 'C' (1 and 4).

The drag coefficient for laminar flow varies with reynolds number (Re) as
  • a)
    Re1/2
  • b)
    Re
  • c)
    Re-1
  • d)
    Re-1/2
Correct answer is option 'D'. Can you explain this answer?

Anshul Basu answered
Explanation:

The drag coefficient is a dimensionless quantity that characterizes the resistance of an object to movement through a fluid. The drag coefficient for laminar flow can be determined by the Reynold's number (Re), which is the ratio of inertial forces to viscous forces in a fluid flow.

The drag coefficient for laminar flow varies with Reynolds number (Re) as Re-1/2. This means that as the Reynolds number increases, the drag coefficient decreases.

The Reynolds number is given by the formula:

Re = (ρVD)/μ

where,

ρ = density of the fluid
V = velocity of the object
D = characteristic length of the object
μ = viscosity of the fluid

The drag coefficient for laminar flow can be determined using the following formula:

Cd = (2F)/(ρV2A)

where,

F = drag force
A = reference area

The drag coefficient for laminar flow is dependent on the shape of the object and the Reynolds number.

Conclusion:

In conclusion, the drag coefficient for laminar flow varies with Reynolds number (Re) as Re-1/2. This means that as the Reynolds number increases, the drag coefficient decreases. The drag coefficient for laminar flow is dependent on the shape of the object and the Reynolds number.

An odd shaped body weighing 7.5 kg and occupying 0.01 m3 volume wiil be completely submerged in a fluid having specific gravity of
  • a)
    1
  • b)
    1.2
  • c)
    0.8
  • d)
    0.75
Correct answer is option 'D'. Can you explain this answer?

Aniket Saini answered
Submergence of Odd Shaped Body in Fluid

Given:

Weight of the body, W = 7.5 kg

Volume of the body, V = 0.01 m3

Specific gravity of the fluid, S.G. = ρ/ρw

where, ρ = Density of the fluid

ρw = Density of water

a) If S.G. = 1

Density of water, ρw = 1000 kg/m3

Weight of the displaced water, WD = V × ρw = 0.01 × 1000 = 10 kg

As the weight of the body (7.5 kg) is less than the weight of the displaced water (10 kg), the body will float on the surface of the fluid. Therefore, the answer is incorrect.

b) If S.G. = 1.2

Density of the fluid, ρ = S.G. × ρw = 1.2 × 1000 = 1200 kg/m3

Weight of the displaced water, WD = V × ρw = 0.01 × 1000 = 10 kg

Buoyancy force acting on the body, FB = WD × g = 10 × 9.81 = 98.1 N

Weight of the body, W = 7.5 × 9.81 = 73.58 N

As the buoyancy force (98.1 N) is greater than the weight of the body (73.58 N), the body will float on the surface of the fluid. Therefore, the answer is incorrect.

c) If S.G. = 0.8

Density of the fluid, ρ = S.G. × ρw = 0.8 × 1000 = 800 kg/m3

Weight of the displaced water, WD = V × ρw = 0.01 × 1000 = 10 kg

Buoyancy force acting on the body, FB = WD × g = 10 × 9.81 = 98.1 N

Weight of the body, W = 7.5 × 9.81 = 73.58 N

As the buoyancy force (98.1 N) is greater than the weight of the body (73.58 N), the body will float on the surface of the fluid. Therefore, the answer is incorrect.

d) If S.G. = 0.75

Density of the fluid, ρ = S.G. × ρw = 0.75 × 1000 = 750 kg/m3

Weight of the displaced water, WD = V × ρw = 0.01 × 1000 = 10 kg

Buoyancy force acting on the body, FB = WD × g = 10 × 9.81 = 98.1 N

Weight of the body, W = 7.5 × 9.81 = 73.58 N

As the buoyancy force (98.1 N) is greater than the weight of the body (73.58 N), the body will sink and be completely submerged in the fluid. Therefore, the correct answer is option 'D'.

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