Fluid Mechanics

When a container of liquid accelerates horizontally with acceleration a₀:

• The free surface of the liquid forms an angle θ with the horizontal
• This angle is related to acceleration by: tan(θ) = a₀/g
• The surface slopes upward in the direction opposite to acceleration
• The greater the horizontal acceleration, the steeper the angle of the surface

Example: Water in a car's cup holder tilts backward when the car accelerates forward, with the angle increasing proportionally to the acceleration rate.

Pascal's Law: A change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid.

Physical implications:

• Pressure changes are transmitted equally throughout a fluid regardless of distance
• The direction of transmission is omnidirectional (in all directions)
• The magnitude of pressure change remains constant throughout the fluid
• It enables pressure amplification in hydraulic systems

Example application: In a hydraulic lift, a small force F₁ on a small piston (area A₁) creates a pressure change F₁/A₁. This same pressure change acts on a larger piston (area A₂), resulting in a larger force F₂ = (A₂/A₁)·F₁, allowing small forces to lift heavy loads.

Note: While force is multiplied, work is conserved - the smaller piston must move a proportionally greater distance.

For horizontally accelerating liquid:

1. Pressure difference derivation:

   • Consider a horizontal cylinder of fluid with length l
   • Forces: P₁ΔS rightward, P₂ΔS leftward
   • Net force = (P₁-P₂)ΔS must equal mass × acceleration = (ρΔSl)a₀
   • Therefore: (P₁-P₂)ΔS = ρΔSla₀
   • Simplifying: P₁-P₂ = ρla₀
   • Points at same horizontal level have different pressures when accelerating horizontally

2. Free surface inclination:

   • For points A and B on the free surface: P₁ = P₀+h₁ρg and P₂ = P₀+h₂ρg
   • From above: P₁-P₂ = ρla₀
   • Therefore: h₁ρg-h₂ρg = ρla₀
   • Simplifying: (h₁-h₂)/l = a₀/g
   • This equals tan(θ) where θ is the angle of inclination of the free surface
   • Therefore: tan(θ) = a₀/g

The free surface becomes inclined, perpendicular to the apparent gravity (vector sum of g and a₀).

The equation of continuity: A₁v₁ = A₂v₂

Derivation:

1. Consider a fluid flowing through a tube with cross-sections A₁ and A₂
2. In time Δt, fluid at speed v₁ travels distance v₁Δt through A₁
3. Volume entering through A₁ = A₁v₁Δt
4. Volume exiting through A₂ = A₂v₂Δt
5. For incompressible fluid, these volumes must be equal: A₁v₁Δt = A₂v₂Δt
6. Therefore: A₁v₁ = A₂v₂

Physical significance:

• Expresses conservation of mass for fluid flow
• Speed increases where cross-section decreases (like water from a nozzle)
• Product of area and speed (Av) represents volume flow rate (m³/s)
• For incompressible fluids, the volume flow rate is constant throughout the tube
• Demonstrates why fluid speeds up in constricted areas (e.g., blood flow in narrowed arteries)

Example: If a pipe narrows from 10 cm² to 5 cm², the fluid speed doubles through the narrow section.

Bernoulli's equation: P + ρgh + ½ρv² = constant

Derivation:

1. Consider fluid flowing between points A and B
2. Work done on fluid segment:
   • By pressure at A: W₁ = P₁(Δm/ρ)
   • By pressure at B: W₂ = -P₂(Δm/ρ)
   • By weight: W₃ = (Δm)g(h₁-h₂)
3. Change in kinetic energy: ΔKE = ½(Δm)v₂² - ½(Δm)v₁²
4. By work-energy theorem: W₁ + W₂ + W₃ = ΔKE
5. Substituting and simplifying: P₁/ρ + gh₁ + ½v₁² = P₂/ρ + gh₂ + ½v₂²
6. Therefore at any point: P + ρgh + ½ρv² = constant

Limitations:

• Valid only for: • Steady flow (velocity at any point constant with time) • Incompressible fluids (constant density) • Nonviscous fluids (no energy loss due to friction) • Irrotational flow (no vortices)
• Invalid for: • Turbulent flow • Flows with significant viscous effects • Compressible fluids at high velocities • Flows with heat transfer or energy dissipation

A tube of flow is:

• A bundle of streamlines that form a tubular region in the fluid
• Created by drawing streamlines from all points on the periphery of a cross-sectional area
• A region where fluid entering one end must exit the other end (no fluid crosses the tube walls)
• Bordered by streamlines that act as "walls" (though no physical barrier exists)

The constant property throughout a tube of flow is the product of:

• Area of cross-section (A) × fluid velocity (v)

This is expressed by the equation of continuity: A₁v₁ = A₂v₂

Where:

• A₁, v₁ are the area and velocity at one cross-section
• A₂, v₂ are the area and velocity at another cross-section

This principle reflects conservation of mass in fluid dynamics, meaning the mass flow rate remains constant throughout the tube.

Bernoulli's equation states:

$P + \rho gh + \frac{1}{2}\rho v^2 = \text{constant}$

Where:

• $P$ = pressure
• $\rho$ = fluid density
• $g$ = acceleration due to gravity
• $h$ = height above reference level
• $v$ = fluid velocity

This equation expresses conservation of energy in fluid flow, showing that in steady, irrotational flow of an incompressible, nonviscous fluid, the sum of pressure energy, gravitational potential energy, and kinetic energy per unit volume remains constant along a streamline.

Example: When water flows through a pipe that narrows, the velocity increases while the pressure decreases, maintaining the constant value in the equation.

In a Venturi tube:

• Pressure difference: $(P_1 - P_2) = \frac{1}{2}\rho(v_2^2 - v_1^2)$
• From continuity equation: $A_1v_1 = A_2v_2$
• The height difference in manometer tubes: $P_1 - P_2 = \rho gh$

By measuring the height difference h in the manometer tubes and knowing cross-sectional areas $A_1$ and $A_2$, one can determine:

1. The pressure difference
2. The fluid speeds $v_1$ and $v_2$
3. The flow rate $Q = A_1v_1 = A_2v_2$

This makes the Venturi tube an effective device for measuring flow rates in pipes without disrupting the flow.

A spinning cricket ball changes its plane of motion due to:

1. When a ball spins about the vertical axis while moving horizontally:

   • On one side (A), the air flow speed increases due to spin drag
   • On the opposite side (B), the air flow speed decreases due to opposing drag

2. According to Bernoulli's principle:

   • Lower pressure develops on side A (higher air speed)
   • Higher pressure develops on side B (lower air speed)

3. This pressure difference creates a net force (F) perpendicular to both the direction of motion and the spin axis

4. This force (known as the Magnus effect) causes the ball to deviate from its original plane of motion, enabling "swing" in cricket bowling.

The effect is more pronounced at higher speeds and spin rates.

When we cannot assume $A_2 \ll A_1$:

1. From Bernoulli's equation and continuity: $v_2^2 = \frac{2gh}{1-(\frac{A_2}{A_1})^2}$

2. Physical significance:

   • As $A_2$ approaches $A_1$, the denominator approaches zero, making $v_2$ approach infinity
   • This indicates that the simple model breaks down when hole size approaches tank size
   • In reality, viscous effects and turbulence prevent infinite velocities
   • The equation shows that the actual efflux speed is always greater than $\sqrt{2gh}$ when the hole size is significant

3. Practical implications:

   • For accurate flow calculations, we must consider the ratio of areas
   • The correction factor $\frac{1}{1-(\frac{A_2}{A_1})^2}$ accounts for the kinetic energy of fluid in the tank
   • Neglecting this correction in engineering applications can lead to underestimating flow rates
   • For $A_2/A_1 = 0.1$, the correction increases the speed by approximately 0.5%
   • For $A_2/A_1 = 0.5$, the correction increases the speed by approximately 33%