Fluid Mechanics

Fluid Mechanics in physics is the branch of science that studies the behavior of fluids (liquids, gases, and plasmas) and the forces acting on them. It explains how fluids move, how they interact with solid boundaries, and how pressure, density, and velocity affect their motion.

It is divided into two main parts:

1. Fluid Statics (Hydrostatics): The study of fluids at rest, focusing on pressure, buoyancy, and density. Example: Why objects float or sink in water.
2. Fluid Dynamics (Hydrodynamics): The study of fluids in motion, dealing with flow, velocity, viscosity, turbulence, and energy in fluid systems. Example: How water flows through a pipe or how air moves over an airplane wing.

Density

The density of a substance is defined as the mass per unit volume.

ρ=m/v

Relative Density

The relative density of a substance is the ratio of the density of the substance to the density of water at 4∘C. It is also called specific gravity.
The density of water at  4 degrees Celsius is 1 g/cm³.
Thrust: A force acting perpendicularly to a surface is called thrust.

Pascal’s Law of Pressure

Pascal’s Law states that: When a pressure is applied to an enclosed liquid, it is equally transmitted to every portion of the liquid.

​Let p1,p2,p3​ be the pressure of a liquid at three valves of a vessel having cross-sectional areas A1, A2, A3,​ respectively. Then, according to Pascal’s Law:

Upthrust or Buoyancy

The upward force exerted by a fluid on an object that is completely or partially immersed in the fluid is called upthrust (buoyancy).

When a body is completely immersed in water as in figure, pressure at its bottom B is greater than at its top T. So, a net upward force acts on the body due to the pressure difference and upthrust is produced in liquid.
Upthrust of the liquid,

U = Wₐ – Wᵥ
∴ U = Wₐ – Wᵥ

Hence, upthrust is the loss in weight of the object in a liquid.

Archimedes’ Principle

Archimedes’ Principle states that “When a body is fully or partially immersed in a fluid, it experiences an upthrust which is equal to the weight of the fluid displaced by the body.”

Consider a solid block of cross sectional area A and thickness h completely immersed in a fluid of constant density ρ. The horizontal thrust (F) on the block balance each other. Let us now compare the vertical thrusts.

Suppose h₁ and h₂ be the depth of the upper and lower face of the block from the surface of the fluid. Let Pₐₜₘ be the atmospheric pressure.

Total downward thrust on the top face
F₁ = (Pₐₜₘ + h₁ρg)A —- (i)

Total downward thrust on the bottom face,
F₂ = (Pₐₜₘ + h₂ρg)A —- (ii)

Resultant upthrust = F₂ – F₁
= (Pₐₜₘ + h₂ρg)A – (Pₐₜₘ + h₁ρg)A
= (h₂ – h₁)ρgA
= hρgA
∴ F = ρgV (V = A.h)

Where V is the volume of the block, which is equal to the volume of displaced fluid, as As Vρ represents the mass of displaced fluid, (Vρg)(V\rho g)(Vρg) represents the weight of the displaced fluid.

∴ Upthrust = wt. of displaced fluid.

Principle of Flotation:

The law of flotation states, “The weight of a floating body is equal to the weight of liquid displaced.” If W is the weight of the floating body and v is the volume of liquid displaced, then,

W=ρVg

Where ρ is the density of the liquid. If ρ′ is the density of the body and V′ is the total volume of the body, then

W=V′ρ′g

According to this principle,

ρV=ρ′V′

Consider a body of volume V and density ρ′ immersed in a fluid of density ρ. The following two vertical forces act on it:

1. Its true weight W, which acts vertically downward through its center of gravity.
2. Upthrust U, which acts vertically upward through the center of buoyancy.

Conditions of Floatation

i) When W > U
The net force (W − U) acts downward and the body sinks.

W>U    ⇒ Vρg =vσg  ⇒ ρ =σ

ii) When W = U
No net force acts on the body and it floats fully immersed.

$$W=U$$

iii) When W < U
The body floats partially immersed.

W<U    ⇒ Vρg =vσg  ⇒ ρ =σ

Equilibrium of Floating Bodies

Centre of Buoyancy (C.B.): It is the point at which the centre of gravity of the displaced liquid lies.

Meta Centre (M.C.): It is the point of intersection of the vertical line passing through C.B. and the original vertical line.

There are two possible cases for the equilibrium of a floating body:

i) If the M.C. lies above the C.G., the couple tends to rotate the body back to its original position. In this condition, the floating body is in stable equilibrium.
ii) If the M.C. lies below the C.G., the couple tends to rotate the body away from the original position. In this condition, the floating body is in unstable equilibrium.

1.A geologist finds that a moon rock whose mass is 7.2 kg has an apparent mass 5.88 kg when submerged in water. What is the density of the rock?

Solution:
Mass of rock (m) = 7.2 kg
Apparent mass in water (ma) = 5.88 kg
Density of rock (ρ) = ?

We have,

Upthrust = Loss in weight = Weight of rock − Apparent weight
∴ Weight of displaced water = (7.2 − 5.88) × 10
= 1.32 × 10
= 13.2

Mass of displaced water = 1.32
or Density of water × Volume of displaced water = 1.32

1000 × Volume of displaced water = 1.32
∴ Volume of displaced water = 1.32 × 10⁻³ m³

Now,

2.A string supports a solid iron object of mass 200 gm totally immersed in a liquid of specific gravity 0.9. Calculate the tension in the string if the density of iron is 8000 kg/m³.

Solution:
Mass of iron (m) = 200 gm = 0.2 kg
Density of iron (S) = 8000 kg/m³
Specific density (Sl) = 0.9

Density of liquid (ρl) = Sl × 1000 = 0.9 × 1000 = 900 kg/m³
Tension in the string (T) = ?
We know,
T=Weight−UpthrustT 
T=mg−Weight of displaced liquid
=0.2×10−(Mass of displaced liquid×9)
=2−(Density of liquid×Volume of displaced liquid×9)

T≈1.74 N
3.An alloy of mass 588 g and volume 100 cc is made of iron of density 8.0 g/cc and aluminum of density 2.7 g/cc. Calculate the proportion by
i) volume, and
ii) mass of the constituents of the alloy.
Solution:

Mass of alloy (m) = 588 g
Volume of alloy (V) = 100 cc

Density of iron (ρi) = 8 g/cc
Density of aluminum (ρal) = 2.7 g/cc

Proportion by volume (Vi:Val)=?
Proportion by mass (mi:mal)=?

We have,

mi+mal=588
or,  m_i​+m_al​=588
or, ρ_i​V_i​+ρ_al​V_al​=588

4. A 25 cm thick block of ice floating on fresh water can support an 80 kg man standing on it, what is the smallest area of the ice block? (sp. gr. of ice = 0.917)

Solution:
Thickness of ice (d) = 25 cm = 0.25 m
mass of the man (m) = 80 kg
Smallest area of the ice block (A) = ?
Sp. gr. of ice = 0.917
Density of ice = 0.917 × 1000 = 917 kg/m³
For floating,
wt. of ice + wt. of man = wt. of displaced water

Surface Tension

Surface Tension is The property of liquid at rest by virtue of which its surface behaves like a stretched membrane and tries to occupy minimum possible surface area is called surface tension.

Mathematically, surface tension is the force per unit length on an imaginary line drawn in the plane of the liquid surface acting at right angles to this line.

Surface Tension (T) = F / length(L)

Surface Energy:
The potential energy per unit area of the surface film is called surface energy. It may also be defined as the amount of work done in increasing the area of a surface film through unity.

Surface energy = work done in increasing surface area/increase in surface area

It is also called free surface energy because the mechanical work done can be released when the surface contracts.

Relation between Surface Tension and Surface Energy

Consider a rectangular frame of wire ABCD in which wire BC is movable. If we dip the frame in a soap solution, a thin film is formed which pulls the wire BC inward left due to surface tension.

If T is surface tension of the film and l is the length of the wire, then the force F on BC due to surface tension is:

F = T × 2l
Suppose the wire is now moved through a distance x from BC to B’C’ against surface tension force F so that surface area of the film increase. In order to increase the film area, work has to be done against F. This surface energy is work.

Work done in increasing surface area,
W = F × distance = T × 2l × x
where 2lx is the increase in surface area.

Excess Pressure inside a Liquid Drop

Consider a drop of liquid of radius R. Let T be the surface tension and P be the excess pressure inside the drop. Suppose due to excess pressure, there will be an increase in the radius of the drop by the quantity dR. In such a case, we write,

W = Force × displacement
= (Excess pressure × area) × dt increase in radius.

Angle of Contact

The angle θ that the tangent to the liquid surface at the point of contact makes with the solid surface inside the liquid is called the angle of contact or capillary angle.

• If adhesive force > cohesive force, the free surface of the liquid is concave in shape. This liquid wets the solid.
•  If cohesive force > adhesive force, the free surface of the liquid is convex in shape. This liquid doesn’t wet the solid.
• If adhesive force = cohesive force, the free surface of the liquid is a plane.

Capillarity

A tube whose bore is comparable to the diameter of a hair is called a capillary tube. The phenomenon of rise or fall of a liquid in a capillary tube in comparison to the surrounding is called capillary action or capillarity.

Consider a capillary tube of radius r open at both ends and dipped into water having a concave meniscus. Let θ be the angle of contact, h be the height of the liquid raised, ρ be the density of the liquid, and T be the surface tension.

The horizontal components of T cancel each other, whereas the vertical components are added, which pulls the liquid upward. The component Tcosθ acts along the whole circumference of the meniscus.

Total upward force = Tcosθ × 2πr
Volume of liquid in the tube above the free surface of liquid,

V = vol. of cylinder of height h and radius r + vol. of cylinder of height and radius r – volume of hemisphere of radius r.

1. Angle of contact of mercury with glass is 135∘ A narrow tube of glass having diameter 2 mm is dipped in a beaker containing mercury. By what height does the mercury go down in the tube relative to the level of mercury outside?

Solution:

Angle of contact (θ)=135∘

2. A capillary tube of 0.4 mm diameter is placed vertically inside a liquid of density 800 kg/m3. Surface tension =5×10−2 N/m and angle of contact =30∘. Calculate the height to which the liquid rises in the capillary tube.

Compiled by: Er. Basant Kumar Yadav