Solved Questions of Basic Concepts and Definitions of Fluids | Fluid Mechanics Solved Assignment 1 | Fluid Mechanics | Solved Questions of Fluid Mechanics | By Akhand Dutta

Solved Questions of Basic Concepts and Definitions of Fluids

Q1: What is fluid? Discuss continuum.

Fluid may be defined as a substance which is capable of flowing. It has no definite shape of its own, but conforms to the shape of the containing vessel. Example- Gas and Liquid are fluids.

Continuum is the continuous distribution of matter with no voids or empty spaces. The concept of continuum in fluid mechanics is the continuous distribution of mass in a body of matter i.e. without spaces. The concept of continuum is general applied looking at the fluid in a macroscopic level and not a microscopic level. This gets helpful in determining the fluid as a point function, like plotting the velocity profile of a fluid, measuring the temperature of a particular point in the fluid etc,. This assumption is justifiable because ordinarily the fluids involved in most of the engineering problems have a large number of molecules and the distances between them are small.

Fluid

Q2: Discuss some physical properties of fluids in brief.

Some of the physical properties of fluids are as follows:

Mass Density or Density-

Density or mass density of a fluid is defined as the ratio of the mass of a fluid to its volume. Thus mass per unit volume of a fluid is called density. It is denoted the symbol ρ (rho). The unit of mass density in SI unit is kg per cubic meter, i.e ., kg/m3.

Specific Weight or Weight Density-

Specific weight or weight density of a fluid is the ratio between the weight of a fluid to its volume. 

Thus weight per unit volume of a fluid is called weight density and it is denoted by the symbol w. 

Specific Volume-

Specific volume of a fluid is defined as the volume of a fluid occupied by a unit mass or volume per unit mass of a fluid is called specific volume. 

Specific Gravity-

Specific gravity is defined as the ratio of the weight density (or density) of a fluid to the weight density (or density) of a standard fluid.

Viscosity-

Viscosity is defined as the property of a fluid which offers resistance to the movement of one layer of fluid over another adjacent layer of the fluid.

Q3: What are the difference between dynamics viscosity and kinematic viscosity?

Dynamic viscosity μ, may be defined as the shear stress required to produce a unit rate of angular deformation. In SI units μ is expressed in N.s/m2, or kg/m.s. It is the constant of proportionality and is also known as the coefficient of viscosity.

On the other hand, The ratio of the dynamic viscosity μ and the mass density is known as Kinematic viscosity. The dimensions of the kinematic viscosity only involves the magnitudes of length and time. The name kinematic viscosity has been given to the ratio (μ/p) because kinematics is defined as the study of motion without regard to the cause of the motion and hence it is concerned with length and time only.

Q4: An oil of viscosity 5 poise is used for lubrication between a shaft and sleeve. The diameter of the shaft is 0.5 m and it rotates at 200 rpm .calculate the power lost in oil for the sleeve length of 100mm, the thickness of the oil film is 1 mm.

Given:

Viscosity, μ= 5 Poise = 0.5 Ns/m^2
Diameter, D= 0.5 m
Rotation, N= 200 rpm
Sleeve Length, L=100 mm= 0.1 m
Thickness, t= 1 mm= 10^-3 m

Tangentia Velocity, u= πDn/60

= (3.14 * 0.5 *200)/60= 5.23 m/s

Using, τ= μdu/dy

du= change in velocity= u-0= 5.23 m/s
dy= change in distance= t= 10^-3 m

Shear Stress, τ= 0.5 * 5.23 * 10^3= 2615 N/m^2

Shear Force= Shear Stress * Area= 2615 * πDL

= 2615 * π * 0.5 * 10^-1
= 410.76 N

Power lost= F*u= 410.76 * 5.23= 2148.27 watt.

Q5: Draw and Explain the Rheological diagram.

Rheological Diagram or Rheogram

Rheological diagram or rheogram is genrerally graphical representation of the rheological characteristics of a material, it can be explained as a graph of shear flow versus shear stress.

Various types of fluids can be depicted using Rheological diagram, the major types are given below with their characteristics:

(a.) Ideal fluid: A fluid which has no viscosity and no shear stress and which is incompressible, is called an ideal fluid.

By Newton's law of viscosity,

τ = μ*(du/dy)

Definition of ideal fluid

μ = 0

τ = 0

(b.) Newtonian fluid: Fluid which obeys Newton's law of Viscosity.


(c.) Non-Newtonian fluid: Fluid which does not obey Newton's law of Viscosity.

τ = A*(du/dy)^n + B


(d.) For Dilatant fluid:

n > 1 & B = 0

Example:- Butter and Quicksand


(e.) For Bringham plastic fluid:

n = 1 & B ≠ 0

Example:- Drilling mud and Toothpaste


(f.) For Pseudoplastic fluid:

n < 1 & B = 0

Example:- Lipsticks, Paints and Blood


(g.) For Thixotropic fluid:

n < 1 & B ≠ 0

Example:- Printer's ink


(h.) For Rheopetic fluid:

n > 1 & B ≠ 0

Example: Gypsum solution in water

Q6: State Newton’s law of viscosity and derive the same.

It states that the shear stress (τ) on a fluid element layer is directly proportional to the rate of shear strain. The constant of proportionality is called the co-efficient viscosity. Newton’s law of Viscosity relates shear stress in a fluid flow to velocity gradient in the direction perpendicular to the flow of fluid.

This is mathematically expressed as Ԏ=μ(du/dy) where Ԏ is the shear stress, μ is coefficient of viscosity and du/dy gives the velocity gradient in the Y direction for fluid flowing in the X direction.

It may be seen from similar triangles in Fig. 1.1 that the ratio V/Y can be replaced by the velocity gradient (dv/dy), which is the rate of angular deformation of the fluid. If a constant of proportionality
μ (Greek ‘mu’) be introduced, the shear stress Ԏ (Greek ‘tau’) equal to (F/A) between any two thin 

sheets of fluid may be expressed as

Ԏ =F/A = μV/Y

= μdv/dy

This expression is called Newton’s equation of viscosity.

Q7: The space between two flat parallel plates is filled with an oil. Each side of the plate is 60 cm. The thickness of the oil plate is 12.5 mm. The upper plate which moves at 2.55 m/s  requires a force of 98.1 N to maintain the speed. Determine: (a) Dynamic viscosity of oil in poise. And (b) Kinematic viscosity of oil in Stokes if the specific gravity of the oil is 0.95.

Given:

Each side of plate= 60  cm= 0.6 m
Thickness of oil film= 12.5 mm= 12.5 * 10^-3 m
Upper plate velocity= 2.5 m/s
Force, τ= 98.1 N

Using, τ= μdu/dy

du= 2.5 - 0= 2.5 m/s
dy= 12.5 * 10^-3 m

μ= τ/(du/dy)

= (98.1 * 12.5)/(2.5 * 1000)

=0.4905 or, 4.905 poise

Specific gravity of oil= 0.95

Mass density of water= 0.95 * 1000= 950


Kinematic Viscosity= 4.905/950

= 5.163 * 10^-3

or, Kinematic Viscosity= 51.63 stoke.

Q8: What is the pressure within a 1 mm diameter spherical droplet of water relative to the droplet of water relative to the atmospheric pressure outside? Assume surface tension for pure water to be 0.073 N/m.

We know that the pressure difference (Δp) = 2σ/R

So, Δp = [(2*7.3*10^-2) / (0.5*10^-3)

Δp = 292 N/m2.

Q9: Calculate the capillary rise in a glass tube of 2.5 mm diameter when immersed vertically in  (a) Water and (b) Mercury. Take surface tension σ = 0.0725 N/m for water and σ = 0.52 N/m  for mercury in contact with air. The specific gravity for mercury is 13.6 and angle of contact is 130°.

Given:

The diameter of tube (d) = 2.5 mm = 2.5 * 10^-3 m

Surface tension, σ for water = 0.0725 N/m

σ for mercury = 0.52 N/m

The specific gravity of mercury = 13.6

Therefore, density = 13.6 * 1000 kg/m3


(a.) Capillarity rise for water (θ =0)


We know that h = (4*σ / ρ*g*h)

So, h = [(4* 0.00725) / (1000*9.81*2.5*10^-3)]

h = 0.0118 m = 1.18 cm.


(b.) For mercury


The angle of a constant between mercury and glass tube  = 130°

We know that h = (4*σ*cosθ / ρ*g*d)

So, h = [(4*0.52*cos 130) / (13.6*1000*9.81*2.5*10^-3)]

h = -0.004 m = -0.4 cm.

Here, the negative sign indicates capillary depression.

Q10: Determine the bulk modulus of elasticity of liquid, if the pressure of liquid is increased from 70 N/cm2 to 130 N/cm2. The Volume of the liquid decrease by 0.15 percent.

Given:

Initial pressure = 70 N/cm2

Final pressure = 130 N/cm2

dp = Increase in pressure = 130 - 70 = 60 N/cm2

Decrease in Volume = 0.15%

(-dV/V) = + (0.15/100)

So, the bulk modulus of elasticity of liquid (K) =  [(dp) / (-dV/V)]

K = (60*100/0.15) = 4 * 10^4 N/cm2.

Q11: Define Newtonian and Non- Newtonian fluids.

Newtonian Fluids- A Newtonian fluid is the one which obeys Newton’s law of viscosity. In a Newtonian fluid, there is a linear relationship between the magnitude of shear stress and the resulting rate of deformation i.e., the constant of proportionality μ  does not change with the rate of deformation. Hence, A Newtonian fluid is a fluid in which the viscous stresses arising from its flow, at every point, are linearly correlated to the local strain rate—the rate of change of its deformation over time.

Non-Newtonian Fluids- A non-Newtonian fluid is a fluid that does not follow Newton's law of viscosity, i.e., constant viscosity independent of stress. In a non-Newtonian fluid, there is a non-linear relation between the magnitude of the applied shear stress and the rate of angular deformation. In the case of a plastic substance which is a non-Newtonian fluid, an initial yield stress is to be exceeded to cause a continuous deformation.

Q12: Explain the following:

i) Compressibility.
ii) Vapour pressure and Cavitation.
iii) Real fluids and Ideal fluid.

i) Compressibility- Compressibility of a fluid is quantitatively expressed as the inverse of the bulk modulus of elasticity K of the fluid, which is defined as:

K = Stress/Strain

= –dp/(dV/V)


ii) Vapour Pressure- Vapor pressure (or vapour pressure in British English; see spelling differences) or equilibrium vapour pressure is defined as the pressure exerted by a vapour in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.


Cavitation- Cavitation is a phenomenon in which rapid changes of pressure in a liquid lead to the formation of small vapour-filled cavities in places where the pressure is relatively low. Cavitation can also be defined as the phenomenon of formation of vapour bubbles of a flowing liquid in a region where the pressure of the liquid falls below its vapour pressure


iii) Ideal Fluids- The type of fluids which don't have viscosity and are incompressible are known as ideal fluid, this type of fluid do not offer shear resistance.


Real Fluids- The type of fluids which have viscosity are known as real fluids. Fluids of this kind always offer shear resistance.










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