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Fluid mechanics Bernoulli's theorem viscosity

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Introduction to Fluid Mechanics

Fluid mechanics is a branch of physics that studies the behavior of fluids-liquids and gases-both at rest and in motion. Unlike solids, fluids do not have a fixed shape; they conform to the shape of their containers. Understanding fluid mechanics is essential in many fields such as engineering, meteorology, medicine, and even everyday life, like water supply systems and blood flow in our bodies.

Before diving deeper, let's define some key terms:

  • Fluid: A substance that can flow and does not have a fixed shape. Examples include water, air, oil, and mercury.
  • Pressure (P): The force exerted per unit area by the fluid on a surface, measured in pascals (Pa).
  • Density (ρ): The mass per unit volume of a fluid, measured in kilograms per cubic meter (kg/m³).

Fluid mechanics helps us analyze how fluids exert forces, how they move, and how energy is transferred within them. This knowledge is crucial for designing hydraulic machines, aircraft, pipelines, and many other systems.

Pressure in Fluids

Pressure in a fluid arises because fluid molecules constantly collide with surfaces. The deeper you go into a fluid, the greater the pressure due to the weight of the fluid above.

Mathematically, pressure at a depth h in a fluid of density ρ under gravity g is given by:

Pressure at Depth

\[P = P_0 + \rho g h\]

Pressure increases with depth due to fluid weight

P = Pressure at depth (Pa)
\(P_0\) = Atmospheric pressure at surface (Pa)
\(\rho\) = Fluid density (kg/m³)
g = Acceleration due to gravity (9.8 m/s²)
h = Depth below surface (m)

Here, P₀ is the atmospheric pressure acting on the fluid surface. This means even at the surface, there is pressure from the air above.

Consider a water tank open to the atmosphere. At the surface, pressure is atmospheric pressure (~1.01 x 105 Pa). At 2 meters depth, the pressure increases due to the weight of the water above.

Surface (P₀) Depth h P₀ P = P₀ + ρgh

Pascal's Principle

Pascal's principle states that any change in pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid. This principle is the foundation of hydraulic systems like car brakes and hydraulic lifts.

For example, if you apply a force on a small piston in a hydraulic press, the pressure increase is transmitted to a larger piston, resulting in a larger force output, allowing heavy loads to be lifted with less effort.

Bernoulli's Theorem

Bernoulli's theorem is a fundamental principle describing the behavior of a moving fluid. It relates the pressure, velocity, and height of fluid particles along a streamline.

Statement and Assumptions

Bernoulli's theorem states: For an incompressible, non-viscous fluid flowing steadily along a streamline, the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant.

Key assumptions:

  • The fluid is incompressible (density remains constant).
  • The fluid is non-viscous (no internal friction).
  • The flow is steady (velocity at any point does not change with time).
  • The flow is along a streamline.

Derivation

Consider a fluid flowing through a pipe with varying cross-sectional area as shown below:

A₁ (larger area) A₂ (smaller area) A₃ (larger area) P₁, v₁ P₂, v₂ P₃, v₃

Applying conservation of energy between two points 1 and 2 along the streamline:

  • Pressure energy per unit volume: \( P \)
  • Kinetic energy per unit volume: \( \frac{1}{2} \rho v^2 \)
  • Potential energy per unit volume: \( \rho g h \)

Bernoulli's equation is:

Bernoulli's Equation

\[P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant}\]

Sum of pressure, kinetic, and potential energy per unit volume is constant along a streamline

P = Pressure (Pa)
\(\rho\) = Fluid density (kg/m³)
v = Fluid velocity (m/s)
g = Acceleration due to gravity (9.8 m/s²)
h = Height above reference point (m)

This means if the velocity of fluid increases, the pressure decreases if height remains constant, and vice versa.

Worked Example 1: Calculating velocity using Bernoulli's theorem

Example 1: Velocity of water exiting a pipe Medium
Water flows from a large tank through a pipe open to the atmosphere. The water surface is 5 m above the pipe outlet. Assuming the velocity of water at the surface is negligible, find the velocity of water exiting the pipe.

Step 1: Identify points: Point 1 at water surface, Point 2 at pipe outlet.

Step 2: Apply Bernoulli's equation between points 1 and 2:

\( P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \)

Since both points are open to atmosphere, \( P_1 = P_2 = P_{\text{atm}} \), and \( v_1 \approx 0 \) (surface velocity negligible).

Let \( h_1 = 5 \, m \), \( h_2 = 0 \, m \) (reference level at outlet).

Equation reduces to:

\( \rho g h_1 = \frac{1}{2} \rho v_2^2 \)

Step 3: Solve for \( v_2 \):

\( v_2 = \sqrt{2 g h_1} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.9 \, m/s \)

Answer: The velocity of water exiting the pipe is approximately 9.9 m/s.

Viscosity

While Bernoulli's theorem assumes no internal friction, real fluids resist motion due to viscosity. Viscosity is a measure of a fluid's resistance to flow or deformation.

Imagine honey flowing slowly compared to water; honey has a higher viscosity.

Definition and Physical Meaning

Viscosity arises because fluid layers slide over each other at different speeds, creating internal friction.

Top layer (fastest) Middle layer Bottom layer (slowest)

Each layer moves at a different velocity, creating a velocity gradient. Viscosity quantifies the force needed to maintain this gradient.

Types of Viscosity

  • Dynamic viscosity (η): Measures the fluid's internal resistance to flow, with units Pascal-second (Pa·s).
  • Kinematic viscosity (ν): Ratio of dynamic viscosity to fluid density, units m²/s. It describes how fast a fluid flows under gravity.

Kinematic Viscosity

\[ u = \frac{\eta}{\rho}\]

Ratio of dynamic viscosity to density

\( u\) = Kinematic viscosity (m²/s)
\(\eta\) = Dynamic viscosity (Pa·s)
\(\rho\) = Density (kg/m³)

Viscosity causes energy loss in fluid flow, often appearing as heat. This is why oil is used as a lubricant-to reduce friction between moving parts.

Worked Example 2: Calculating viscous force using Newton's law of viscosity

Example 2: Force to move a plate over viscous fluid Medium
A flat plate of area 0.5 m² is moved over a stationary plate separated by a 2 mm thick layer of oil with dynamic viscosity \( \eta = 0.1 \, \text{Pa·s} \). If the plate moves at 0.2 m/s, calculate the force needed to maintain this motion.

Step 1: Use Newton's law of viscosity:

\( F = \eta A \frac{v}{d} \)

Where:

  • \( F \) = viscous force (N)
  • \( \eta = 0.1 \, \text{Pa·s} \)
  • \( A = 0.5 \, \text{m}^2 \)
  • \( v = 0.2 \, \text{m/s} \)
  • \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

Step 2: Substitute values:

\( F = 0.1 \times 0.5 \times \frac{0.2}{2 \times 10^{-3}} = 0.05 \times 100 = 5 \, \text{N} \)

Answer: The force required is 5 newtons.

Fluid Flow Types and Continuity Equation

Fluid flow can be broadly classified into two types:

  • Laminar flow: Smooth, orderly flow where fluid particles move in parallel layers.
  • Turbulent flow: Chaotic, irregular flow with eddies and vortices.

The Reynolds number (Re) helps predict flow type:

Reynolds Number

\[Re = \frac{\rho v d}{\eta}\]

Dimensionless number indicating flow regime

Re = Reynolds number
\(\rho\) = Fluid density (kg/m³)
v = Velocity (m/s)
d = Characteristic length or diameter (m)
\(\eta\) = Dynamic viscosity (Pa·s)

Typically, for flow in pipes:

  • \( Re < 2000 \) indicates laminar flow
  • \( Re > 4000 \) indicates turbulent flow
  • Between 2000 and 4000 is transitional flow

The continuity equation expresses conservation of mass for incompressible fluids:

Continuity Equation

\[A_1 v_1 = A_2 v_2\]

Cross-sectional area times velocity is constant along a streamline

A = Cross-sectional area (m²)
v = Fluid velocity (m/s)

This means if the pipe narrows, velocity increases to maintain constant flow rate.

Worked Example 3: Determining pressure difference in a pipe using Bernoulli's theorem

Example 3: Pressure difference between two points in a pipe Hard
Water flows through a horizontal pipe that narrows from diameter 0.1 m to 0.05 m. If the pressure at the wider section is 2 x 105 Pa and the velocity there is 2 m/s, find the pressure at the narrower section. (Density of water = 1000 kg/m³)

Step 1: Calculate areas:

\( A_1 = \pi \times (0.1/2)^2 = \pi \times 0.05^2 = 7.85 \times 10^{-3} \, m^2 \)

\( A_2 = \pi \times (0.05/2)^2 = \pi \times 0.025^2 = 1.96 \times 10^{-3} \, m^2 \)

Step 2: Use continuity equation to find \( v_2 \):

\( A_1 v_1 = A_2 v_2 \Rightarrow v_2 = \frac{A_1 v_1}{A_2} = \frac{7.85 \times 10^{-3} \times 2}{1.96 \times 10^{-3}} = 8 \, m/s \)

Step 3: Apply Bernoulli's equation (horizontal pipe, so \( h_1 = h_2 \)):

\( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \)

Rearranged to find \( P_2 \):

\( P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2) \)

Substitute values:

\( P_2 = 2 \times 10^5 + \frac{1}{2} \times 1000 \times (2^2 - 8^2) \)

\( P_2 = 2 \times 10^5 + 500 \times (4 - 64) = 2 \times 10^5 - 30000 = 170000 \, Pa \)

Answer: Pressure at the narrower section is 1.7 x 105 Pa.

Worked Example 4: Estimating flow rate using continuity equation

Example 4: Flow rate in pipes of varying cross-section Easy
Water flows through a pipe with cross-sectional area 0.02 m² at velocity 3 m/s. Find the velocity when the pipe narrows to 0.01 m² and the flow rate in liters per second.

Step 1: Use continuity equation:

\( A_1 v_1 = A_2 v_2 \Rightarrow v_2 = \frac{A_1 v_1}{A_2} = \frac{0.02 \times 3}{0.01} = 6 \, m/s \)

Step 2: Calculate flow rate \( Q \):

\( Q = A_1 v_1 = 0.02 \times 3 = 0.06 \, m^3/s \)

Convert to liters per second:

\( 0.06 \, m^3/s = 60 \, \text{liters/s} \)

Answer: Velocity in narrow pipe is 6 m/s, flow rate is 60 liters per second.

Worked Example 5: Effect of viscosity on flow time

Example 5: Time taken for viscous fluid to flow through a tube Hard
A viscous fluid flows through a horizontal tube of length 1 m and radius 0.005 m under a pressure difference of 1000 Pa. The dynamic viscosity of the fluid is 0.2 Pa·s. Estimate the time taken for 0.001 m³ of fluid to flow through the tube. (Use Poiseuille's law)

Step 1: Poiseuille's law gives volumetric flow rate:

\( Q = \frac{\pi r^4 \Delta P}{8 \eta L} \)

Where:

  • \( r = 0.005 \, m \)
  • \( \Delta P = 1000 \, Pa \)
  • \( \eta = 0.2 \, Pa \cdot s \)
  • \( L = 1 \, m \)

Step 2: Calculate \( Q \):

\( Q = \frac{\pi \times (0.005)^4 \times 1000}{8 \times 0.2 \times 1} = \frac{3.1416 \times 6.25 \times 10^{-10} \times 1000}{1.6} \approx \frac{1.9635 \times 10^{-6}}{1.6} = 1.227 \times 10^{-6} \, m^3/s \)

Step 3: Calculate time \( t \) to flow 0.001 m³:

\( t = \frac{\text{Volume}}{Q} = \frac{0.001}{1.227 \times 10^{-6}} \approx 814 \, s \approx 13.6 \, \text{minutes} \)

Answer: It takes approximately 13.6 minutes for the fluid to flow through the tube.

Formula Bank

Pressure at depth
\[ P = P_0 + \rho g h \]
where: \( P \) = pressure at depth (Pa), \( P_0 \) = atmospheric pressure (Pa), \( \rho \) = fluid density (kg/m³), \( g \) = 9.8 m/s², \( h \) = depth (m)
Bernoulli's Equation
\[ P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \]
where: \( P \) = pressure (Pa), \( \rho \) = fluid density (kg/m³), \( v \) = velocity (m/s), \( g \) = 9.8 m/s², \( h \) = height (m)
Continuity Equation
\[ A_1 v_1 = A_2 v_2 \]
where: \( A \) = cross-sectional area (m²), \( v \) = velocity (m/s)
Newton's Law of Viscosity
\[ F = \eta A \frac{v}{d} \]
where: \( F \) = viscous force (N), \( \eta \) = dynamic viscosity (Pa·s), \( A \) = area (m²), \( v \) = velocity (m/s), \( d \) = thickness of fluid layer (m)
Kinematic Viscosity
\[ u = \frac{\eta}{\rho} \]
where: \( u \) = kinematic viscosity (m²/s), \( \eta \) = dynamic viscosity (Pa·s), \( \rho \) = density (kg/m³)
Poiseuille's Law (Volumetric Flow Rate)
\[ Q = \frac{\pi r^4 \Delta P}{8 \eta L} \]
where: \( Q \) = flow rate (m³/s), \( r \) = tube radius (m), \( \Delta P \) = pressure difference (Pa), \( \eta \) = dynamic viscosity (Pa·s), \( L \) = length of tube (m)

Tips & Tricks

Tip: Always check assumptions before applying Bernoulli's theorem.

When to use: To avoid incorrect application in viscous or turbulent flows.

Tip: Use the continuity equation first to find unknown velocities before applying Bernoulli's equation.

When to use: When pipe cross-section changes affect fluid velocity.

Tip: Convert all units to SI (meters, pascals, seconds) before calculations.

When to use: To avoid unit conversion errors during problem solving.

Tip: Visualize fluid layers and velocity gradients to understand viscosity effects.

When to use: When solving problems involving viscous flow or internal friction.

Tip: Use approximate values \( g = 9.8 \, m/s^2 \) and atmospheric pressure \( 1.01 \times 10^5 \, Pa \) for quick calculations.

When to use: During time-limited entrance exams for faster solving.

Common Mistakes to Avoid

❌ Applying Bernoulli's theorem to viscous or turbulent flows.
✓ Use Bernoulli's theorem only for ideal fluids (non-viscous, incompressible, steady flow).
Why: Viscosity and turbulence cause energy losses not accounted for in Bernoulli's equation.
❌ Ignoring height differences when applying Bernoulli's equation.
✓ Include gravitational potential energy term \( \rho g h \) when height varies.
Why: Height differences affect pressure and velocity in fluid flow.
❌ Confusing dynamic and kinematic viscosity.
✓ Remember dynamic viscosity \( \eta \) measures internal friction, kinematic viscosity \( u \) includes density.
Why: They have different units and physical meanings.
❌ Using inconsistent units, especially mixing cm and m or Pa and atm.
✓ Always convert to SI units (meter, Pascal) before calculations.
Why: Unit inconsistency leads to wrong answers.
❌ Forgetting to apply the continuity equation before Bernoulli's theorem.
✓ Calculate velocities using continuity equation first when cross-section changes.
Why: Velocity changes affect pressure and energy terms in Bernoulli's equation.
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