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.

Pressure is defined as the limit of force per unit area as the area approaches zero: $P = \lim_{\Delta S \rightarrow 0} \frac{F}{\Delta S}$

Key properties:

• It's a scalar quantity (magnitude only)
• For homogeneous and nonviscous fluids, pressure at a point is independent of the orientation of the area
• Measured in pascals (Pa) or N/m²
• Differs from force by being intensive (independent of size) while force is extensive
• Pressure acts perpendicular to any surface in contact with the fluid

Example: 1 Pa equals the pressure exerted by a force of 1 N spread over 1 m²

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.

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

For a small volume element of fluid at rest:

1. Forces acting on the element:

   • Upward force ($F_1$) from fluid below
   • Downward force ($F_2$) from fluid above
   • Weight of the element ($W = \rho g \Delta S \, dz$)

2. Force balance for equilibrium: $F_1 - F_2 - W = 0$ $P_1\Delta S - P_2\Delta S - \rho g \Delta S \, dz = 0$

3. Simplifying: $P_1 - P_2 = \rho g \, dz$

This equilibrium of forces explains why:

• Pressure increases with depth
• The pressure gradient is proportional to fluid density
• In the absence of external forces, fluid pressure acts equally in all directions at a given point

A manometer measures pressure using a U-shaped tube filled with liquid:

• The pressure of gas in the vessel (P) equals: P = P₀ + hρg Where:

  • P₀ = atmospheric pressure
  • h = height difference between liquid columns
  • ρ = density of the manometer liquid
  • g = acceleration due to gravity

• The excess pressure (P - P₀) is called gauge pressure

• The liquid rises higher in the open arm when vessel pressure exceeds atmospheric pressure

• For accurate measurements, the liquid should have high density (often mercury) and low vapor pressure

Example: If the height difference is 10 cm using mercury (ρ = 13600 kg/m³), the gauge pressure would be: (0.1 m)(13600 kg/m³)(9.8 m/s²) = 13,328 Pa

Streamline flow (or steady flow) is characterized by:

• Particles at a specific point all move with the same velocity (speed and direction)
• Each particle follows the same path as previous particles that passed through that point
• Streamlines never intersect each other
• Flow patterns remain stable and predictable over time
• Particles passing through point A will all follow the same path to point B
• Typically occurs at lower flow speeds

Unlike turbulent flow, which features:

• Chaotic, irregular motion
• Erratic velocity changes at the same point
• Unpredictable particle paths
• Formation of eddies and vortices

The work-energy theorem applies to fluid flow as follows:

1. For fluid moving from point 1 to point 2:

   • Work is done by pressure forces at both ends: $W_1 = P_1A_1v_1\Delta t$ and $W_2 = -P_2A_2v_2\Delta t$
   • Work is done by gravity: $W_g = \Delta m g(h_1 - h_2)$

2. This work causes a change in kinetic energy:

   • $\Delta KE = \frac{1}{2}\Delta m(v_2^2 - v_1^2)$

3. By conservation of mass (continuity equation):

   • $\Delta m = \rho A_1v_1\Delta t = \rho A_2v_2\Delta t$

4. Applying work-energy theorem ($W_{total} = \Delta KE$):

   • $P_1\frac{\Delta m}{\rho} - P_2\frac{\Delta m}{\rho} + \Delta m g(h_1 - h_2) = \frac{1}{2}\Delta m(v_2^2 - v_1^2)$

5. Simplifying yields Bernoulli's equation:

   • $\frac{P_1}{\rho} + gh_1 + \frac{1}{2}v_1^2 = \frac{P_2}{\rho} + gh_2 + \frac{1}{2}v_2^2$

When fluid moves from a wider to a narrower section:

1. Velocity changes:

   • Velocity increases as the cross-sectional area decreases
   • This follows from the continuity equation: $A_1v_1 = A_2v_2$
   • If $A_2 < A_1$, then $v_2 > v_1$ by the same proportion

2. Physical consequences:

   • Pressure decreases in the narrower section (Bernoulli's principle)
   • Kinetic energy increases as potential energy and pressure energy decrease
   • Flow rate (volume per time) remains constant
   • Momentum transfer increases in the narrower section
   • Fluid exerts greater force per unit area on the walls of the narrower section

Example: In a river, water speeds up when flowing through a narrow gorge, creating rapids and lower pressure in that section compared to wider, calmer sections.

In a Venturi tube:

• Pressure decreases at the constriction (P₂ < P₁)
• This follows directly from Bernoulli's principle
• The mathematical relationship is: P₁ + ½ρv₁² = P₂ + ½ρv₂²
• Since v₂ > v₁ (velocity increases at constriction), P₂ must decrease
• The pressure difference can be expressed as: P₁ - P₂ = ½ρ(v₂² - v₁²)

This demonstrates energy conservation: as kinetic energy increases at the constriction, pressure energy must decrease to maintain constant total energy along a streamline.

Real-world application: Carburetors use this principle to draw fuel into an air stream.