Physiscs

Chapter 14: Fluid Mechanics

Imron Rosyadi

Chapter 14.1: Fluids, Density, and Pressure

Learning Objectives

By the end of this section, you will be able to:

• State the different phases of matter.
• Describe the characteristics of the phases of matter at the molecular or atomic level.
• Distinguish between compressible and incompressible materials.
• Define density and its related SI units.
• Compare and contrast the densities of various substances.
• Define pressure and its related SI units.
• Explain the relationship between pressure and force.
• Calculate force given pressure and area.

Today, we’re diving into the fundamental concepts of fluid mechanics, starting with the basics of fluids, density, and pressure. We’ll cover the different phases of matter and their characteristics at the molecular level. You’ll learn to differentiate between compressible and incompressible materials, define density and pressure, understand their SI units, and perform basic calculations involving them.

Phases of Matter

Matter exists commonly as:

• Solid: Rigid, specific shape, definite volume.
  • Atoms/molecules are in close proximity with strong forces, forming a lattice structure.
  • Resist shearing forces.
• Liquid: Indefinite shape, definite volume.
  • Molecules bonded but can move past each other.
  • Yield to shearing forces; they flow.
  • Resist compression like solids.
• Gas: Indefinite shape, indefinite volume.
  • Molecules far apart, weak forces.
  • Easily compressible; they flow and expand to fill containers.

Think about ice, water, and steam. These are all the same substance, H2O, but they behave very differently because of their phase. Solids are like a well-organized team, holding their positions. Liquids are like a crowd that can move around but still stays together. Gases are like individuals freely moving in a large open space. The key difference is the strength of intermolecular forces and the freedom of molecular movement.

Molecular Arrangement in Phases

Figure 14.2: (a) Solid (b) Liquid (c) Gas

• Solid (a): Atoms are in close contact, held in fixed positions by strong forces (like springs).
  • Limited movement.
• Liquid (b): Atoms are in close contact but can slide over one another.
  • Fewer bonds, allowing for flow.
• Gas (c): Atoms move about freely and are separated by large distances.
  • Very weak forces; expands to fill container.

This figure visually demonstrates the molecular differences. In solids, notice the tight, almost fixed arrangement. In liquids, they are still close but can slip past each other, which is why liquids flow. In gases, there’s a lot of empty space, making them highly compressible and allowing them to expand freely.

Compressible vs. Incompressible Fluids

• Incompressible Fluid: Requires an extremely large force to change its volume.
  • Liquids (e.g., water) are generally considered incompressible.
  • Molecules are closely packed.
• Compressible Fluid: Volume changes significantly with pressure.
  • Gases (e.g., air) are highly compressible.
  • Molecules are far apart, with weak interactions.

Note

In fluid mechanics, we often treat liquids as incompressible, which simplifies many calculations. Gases, however, are almost always treated as compressible.

The distinction between compressible and incompressible is crucial. For liquids, like water in a hydraulic system, assuming incompressibility is often a very good approximation. For gases, like air in a tire, compression is a fundamental aspect of their behavior.

Density (\(\rho\))

• Definition: Mass per unit volume.

\[ \rho = \frac{m}{V} \]

• SI Unit: Kilograms per cubic meter (\(\text{kg/m}^3\)).
  • Common cgs unit: \(\text{g/cm}^3\) (\(1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3\)).
• Density determines if an object floats or sinks.
  • Example: Wood floats in water because its density is less than water’s. Brass sinks because its density is greater.

Density is a fundamental property. It’s not just about how heavy something is, but how much mass is packed into a given space. This is why a small brass object can weigh the same as a large piece of wood, yet behave differently in water. The brass has much more mass per unit volume.

Density and Floating

Figure 14.3: (a) Same mass, different volumes. (b) Brass sinks, wood floats.

• Even if a block of brass and a block of wood have the same mass (and thus weight):
  • Brass has a smaller volume, meaning higher density.
  • Wood has a larger volume, meaning lower density.
• When placed in water:
  • Brass (\(\rho_{brass} > \rho_{water}\)) sinks.
  • Wood (\(\rho_{wood} < \rho_{water}\)) floats.

This image beautifully illustrates the concept of density’s role in buoyancy. Despite having equal mass, their differing volumes lead to different densities, which dictates their behavior in water.

Densities of Common Substances

Substance	\(\rho (\text{kg/m}^3)\)	Phase
Gold	\(1.93 \times 10^4\)	Solid
Lead	\(1.13 \times 10^4\)	Solid
Mercury	\(1.36 \times 10^4\)	Liquid
Water (\(4.0^\circ\text{C}\))	\(1.000 \times 10^3\)	Liquid
Ice (\(0^\circ\text{C}\))	\(9.17 \times 10^2\)	Solid
Oak	\(7.10 \times 10^2\)	Solid
Air (\(0^\circ\text{C}\))	\(1.29 \times 10^0\)	Gas
Hydrogen (\(0^\circ\text{C}\))	\(9.00 \times 10^{-2}\)	Gas

Tip

Densities of liquids and solids are roughly comparable and much greater than gases. Gases are highly dependent on temperature and pressure. Water has a unique density behavior, reaching maximum density at \(4.0^\circ\text{C}\), explaining why ice floats.

Table 14.1 in the textbook has a more comprehensive list. This condensed version highlights some key points: note the vast difference between solids/liquids and gases. Also, the intriguing property of water’s density peaking at 4 degrees Celsius is why lakes freeze from the top down, insulating life below.

Homogeneous vs. Heterogeneous Substances

• Homogeneous Substance: Density is constant throughout.
  • Example: A solid iron bar.
• Heterogeneous Substance: Density varies within the substance.
  • Example: Swiss cheese (cheese + gas voids).
  • Local Density: \(\rho = \lim_{\Delta V \to 0} \frac{\Delta m}{\Delta V}\)

Figure 14.4: Local density in a heterogeneous mixture.

While we often assume uniform density for simplicity, it’s important to recognize that real-world materials can be heterogeneous. Local density helps describe variations within such materials.

Specific Gravity

• Definition: Ratio of a material’s density to the density of water at \(4.0^\circ\text{C}\) (\(1000 \text{ kg/m}^3\)). \[ \text{Specific gravity} = \frac{\text{Density of material}}{\text{Density of water}} \]
• Dimensionless Quantity: Useful for comparing densities without units.
• Example: Aluminum has a specific gravity of 2.7, meaning it’s 2.7 times denser than water.

Specific gravity is a handy tool. Since water’s density is conveniently 1 g/cm\(^3\), the specific gravity often corresponds directly to the numerical value of density in g/cm\(^3\). It removes the need to worry about different unit systems when comparing materials.

Pressure (p)

• Definition: Normal force F per unit area A over which the force is applied. \[ p = \frac{F}{A} \]
• Pressure at a point: \(p = \frac{dF}{dA}\) for an infinitesimal area.
• SI Unit: Pascal (Pa), where \(1 \text{ Pa} = 1 \text{ N/m}^2\).
• Scalar Quantity: Pressure has no direction, but the force due to pressure is always perpendicular to the surface.

Pressure is a scalar because it describes a distributed force over an area. However, when we talk about the force exerted by pressure, it’s always perpendicular to the surface. Think about a sharp needle vs. a finger: the same force, but the much smaller area of the needle tip results in immense pressure, enough to pierce skin.

Pressure and Force

Figure 14.5: (a) Finger (b) Needle.

• The effect of a force depends heavily on the area over which it’s applied.
• Example:
  • A finger applies force over a large area, resulting in low pressure.
  • A needle applies the same force over a tiny area, resulting in very high pressure.

This is a simple yet powerful illustration of the pressure definition. It’s why snowshoes work (distribute weight over a large area, reducing pressure) and why sharp knives cut well (concentrate force over a small area, increasing pressure).

Pressure in a Fluid of Constant Density

• Pressure in a static fluid increases with depth.
• The pressure at a depth h below the surface of a fluid with constant density \(\rho\) is: \[ p = p_0 + \rho g h \]
  • \(p_0\): atmospheric pressure at the surface.
  • \(\rho\): fluid density.
  • \(g\): acceleration due to gravity.
  • \(h\): depth.

Figure 14.6: Pressure at the bottom of a container.

This equation is fundamental for static fluids. Imagine diving deeper into a swimming pool; you feel the pressure increase. That’s due to the weight of the water column above you, added to the atmospheric pressure pushing down on the surface. The pressure acts on all surfaces at that depth equally.

Example: Dam Force

A dam retaining a reservoir.

Problem: A dam is 500-m wide, and water is 80.0-m deep.

1. What is the average pressure on the dam due to water?
2. Calculate the force against the dam.

Solution:

1. Average depth \(h = 40.0 \text{ m}\).

\(p = \rho g h = (10^3 \text{ kg/m}^3)(9.80 \text{ m/s}^2)(40.0 \text{ m}) = 3.92 \times 10^5 \text{ N/m}^2 = 392 \text{ kPa}\).

2. Area \(A = (80.0 \text{ m})(500 \text{ m}) = 4.00 \times 10^4 \text{ m}^2\).

\(F = p A = (3.92 \times 10^5 \text{ N/m}^2)(4.00 \times 10^4 \text{ m}^2) = 1.57 \times 10^{10} \text{ N}\).

For the dam example, since pressure increases linearly with depth, we use the pressure at the average depth to find the average pressure. Then, applying the definition of pressure, we calculate the total force. This force is massive, highlighting the engineering challenges in dam construction.

Pressure and Container Shape

Figure 14.9: Fluid levels in connected containers of different shapes.

• The pressure in a fluid depends only on the depth from the surface, not on the shape of the container.
• This is why liquids seek the same level in interconnected containers, regardless of their shape.

This phenomenon, often called the “hydrostatic paradox,” demonstrates that the pressure at a given depth is uniform regardless of the volume or shape of the container above it. It’s a direct consequence of pressure being a scalar quantity and depending only on depth.

Atmospheric Pressure Variation with Height

• Unlike liquids, air density changes significantly with height.
• Assuming constant air temperature and ideal gas behavior, atmospheric pressure drops exponentially with height: \[ p(y) = p_0 \exp(-\alpha y) \]
  • \(p_0\): pressure at sea level.
  • \(\alpha = \frac{mg}{k_B T}\) is a constant.
  • \(1/\alpha\) is the pressure scale height (approx. 8800 m for Earth’s atmosphere).

Caution

This exponential drop is an approximation assuming constant temperature and gravity, which isn’t perfectly true over large distances.

This exponential decay of pressure is why mountain climbers need oxygen at high altitudes and why weather balloons burst at very high altitudes. The air becomes progressively thinner as you ascend.

Pressure Direction in Fluids

Figure 14.10: (a) Tire pressure (b) Swimmer.

• Fluid pressure is a scalar, but the forces due to pressure are always exerted perpendicular to any surface.
• Fluids cannot withstand shearing forces.
• This is evident in:
  • Tire walls (air pushes outwards perpendicularly).
  • Swimmers (water pushes inward perpendicularly from all directions, with greater force from below due to depth).

The inability of fluids to resist shear stress is fundamental. It means that any force exerted by a fluid on a surface must be normal to that surface. This is why a balloon takes a spherical shape when inflated and why forces on a submerged object are always perpendicular to its surface.

Chapter 14.2: Measuring Pressure

Learning Objectives

By the end of this section, you will be able to:

• Define gauge pressure and absolute pressure.
• Explain various methods for measuring pressure.
• Understand the working of open-tube barometers.
• Describe in detail how manometers and barometers operate.

Now that we understand pressure, let’s explore how it’s actually measured. We’ll differentiate between two important types of pressure readings: gauge and absolute. Then, we’ll delve into the mechanics of manometers and barometers, which are classic tools for pressure measurement.

Gauge Pressure vs. Absolute Pressure

• Gauge Pressure (\(p_g\)): Pressure relative to atmospheric pressure.
  • Most common pressure gauges (e.g., tire pressure, scuba tanks) read zero at atmospheric pressure.
  • Can be positive (above atmospheric) or negative (below atmospheric, like in a vacuum chamber).
• Absolute Pressure (\(p_{abs}\)): Total pressure, including atmospheric pressure. \[ p_{abs} = p_g + p_{atm} \]
  • Cannot be negative for fluids (fluids push, not pull). Minimum absolute pressure is zero (perfect vacuum).

Important

Always be clear whether a pressure value refers to gauge or absolute pressure, especially in problem-solving. Unless stated otherwise, assume gauge pressure in everyday contexts.

This distinction is critical in engineering and physics. When a scuba tank reads 3000 psi, it’s gauge pressure. The actual pressure inside the tank is 3000 psi + 14.7 psi (atmospheric pressure). For fluids, negative absolute pressure is physically impossible because fluids can’t ‘pull’.

Pressure Measuring Devices

Figure 14.11: (a) Gas cylinder gauge (b) Tire gauge (c) Ionization gauge.

Various devices measure pressure:

• Mechanical gauges: (a) Gas cylinder gauges, (b) tire gauges.
• Strain gauges: Measure material shape change due to pressure.
• Capacitance pressure gauges: Measure capacitance change due to shape.
• Piezoelectric pressure gauges: Generate voltage from pressure difference.
• Ion gauges: Measure pressure by ionizing molecules in vacuum systems (c).

The principle behind most pressure gauges is that some physical property changes predictably with pressure. From simple mechanical deformation to complex electrical effects, engineers leverage various phenomena to get a pressure reading.

Manometers

Figure 14.12: Open-tube manometers.

• U-shaped tube filled with a fluid (often mercury).
• Measures gauge pressure by comparing fluid levels.
• One side open to atmosphere (\(p_{atm}\)).
• Gauge pressure: \(p_g = h \rho g\).
  • 2. If \(p_{abs} > p_{atm}\), \(p_g\) is positive.
  • 3. If \(p_{abs} < p_{atm}\), \(p_g\) is negative.
  • The difference in height, \(h\), indicates the pressure difference.

Manometers are simple but effective. The key is that pressure at the same horizontal level within a continuous fluid must be equal. By measuring the height difference in the fluid column, we directly measure the pressure difference.

Barometers

Figure 14.13: A mercury barometer.

• Invented by Torricelli in 1643.
• Measures atmospheric pressure.
• A glass tube, closed at one end and filled with mercury, inverted into a pool of mercury.
• A near-perfect vacuum exists above the mercury column in the tube.
• Atmospheric pressure supports the column of mercury: \(h \rho g = p_{atm}\).
• Used by meteorologists for weather forecasting (rising mercury = improving weather).

Barometers provide a direct measurement of atmospheric pressure. The height of the mercury column is a direct consequence of the external atmospheric pressure pushing down on the mercury in the dish, balancing the weight of the mercury in the tube. The vacuum above the mercury is crucial for accurate measurement.

Fluid Heights in an Open U-Tube Example

Figure 14.14: Two liquids in a U-tube.

Problem: Two immiscible liquids of densities \(\rho_1\) and \(\rho_2\) (\(\rho_2 < \rho_1\)) are in an open U-tube. Derive a formula for the height difference \(h_2 - h_1\).

Solution: Equate pressures at the interface level in both arms: \(p_0 + \rho_1 g h_1 = p_0 + \rho_2 g h_2\) \(\rho_1 h_1 = \rho_2 h_2\) \(h_1 = \frac{\rho_2}{\rho_1} h_2\) Height difference: \(h_2 - h_1 = h_2 - \frac{\rho_2}{\rho_1} h_2 = (1 - \frac{\rho_2}{\rho_1}) h_2\).

This example uses the principle that points at the same horizontal level within the same continuous fluid have equal pressure. By choosing the interface as our reference level, we can relate the heights of the two liquids and their densities to atmospheric pressure.

Units of Pressure

Unit	Definition
Pascal (Pa)	\(1 \text{ N/m}^2\)
pounds per square inch (psi)	\(1 \text{ psi} = 6.895 \times 10^3 \text{ Pa}\)
Atmosphere (atm)	\(1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} = 760 \text{ mm Hg} = 14.7 \text{ psi}\)
Millibar (mbar)	\(1 \text{ mbar} = 100 \text{ Pa}\) (\(1000 \text{ mbar} = 1 \times 10^5 \text{ Pa}\))
Torr	\(1 \text{ torr} = 1 \text{ mm Hg} = 133.3 \text{ Pa}\)
Bar	\(1 \text{ bar} = 10^5 \text{ Pa}\)

It’s important to be familiar with these different units, as they are used in various fields. Meteorologists often use millibars, while medical professionals might refer to blood pressure in mm Hg. Converting between them is a common task in fluid mechanics.

Chapter 14.3: Pascal’s Principle and Hydraulics

Learning Objectives

By the end of this section, you will be able to:

• State Pascal’s principle.
• Describe applications of Pascal’s principle.
• Derive relationships between forces in a hydraulic system.

We’re moving on to a very practical aspect of fluid mechanics: Pascal’s Principle. This principle is the cornerstone of hydraulics, which powers many machines we use daily, from car brakes to construction equipment. We’ll learn what the principle states, see its applications, and derive key relationships.

Pascal’s Principle

• Statement: When a change in pressure is applied to an enclosed fluid, it is transmitted undiminished to all portions of the fluid and to the walls of its container.
• This principle applies to the change in pressure, not necessarily the absolute pressure being the same everywhere (as pressure still varies with depth).

Figure 14.15: Pressure change in a fluid with a piston.

• If you add a weight \(Mg\) to a piston of area \(A\), the pressure at the top increases by \(\Delta p = \frac{Mg}{A}\).
• According to Pascal’s principle, this \(\Delta p\) is transmitted throughout the entire fluid.
  • \(\Delta p_{top} = \Delta p_{bottom} = \Delta p_{everywhere}\).

Pascal’s principle is a profound insight. It means that if you squeeze a sealed container filled with fluid, that squeeze force is distributed evenly throughout the fluid, regardless of where or how hard you squeeze. This uniform transmission of pressure change is what makes hydraulics so powerful.

Applications of Pascal’s Principle: Hydraulic Systems

Figure 14.16: A typical hydraulic system.

• Hydraulic systems use Pascal’s principle to multiply forces.
• An input force \(F_1\) on a small area \(A_1\) creates a pressure \(p_1 = F_1/A_1\).
• This pressure is transmitted to a larger area \(A_2\), creating a larger output force \(F_2\).
• Since \(p_1 = p_2\) (assuming same height, negligible friction): \[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \]
  • If \(A_2 > A_1\), then \(F_2 > F_1\).

This equation is the core of hydraulic force multiplication. A small force applied over a small area can generate a much larger force over a larger area. This is analogous to a lever, but with the added flexibility of being able to transmit this force through curved pipes and to multiple points simultaneously, using an incompressible fluid like oil.

Hydraulic Jack

Figure 14.17: (a) Hydraulic jack principle (b) Car on a hydraulic jack.

• A small input force (\(F_1\)) over a small piston area (\(A_1\)) lifts a large load (\(F_2\)) on a large piston area (\(A_2\)).
• The ratio of forces is directly proportional to the ratio of areas: \[ F_2 = F_1 \frac{A_2}{A_1} \]
• Commonly used to lift heavy objects like cars.

Hydraulic jacks are a prime example of force multiplication. A person can lift a car with relatively little effort by repeatedly applying a small force over a small piston area, gradually raising the larger piston. While force is multiplied, work is conserved; the smaller piston moves a greater distance than the larger one.

Example: Calculating Force on Wheel Cylinders (Automobile Brakes)

Figure 14.18: Hydraulic brake system.

Problem: A 500 N force is exerted on a pedal cylinder (master cylinder) with diameter 0.500 cm. Each wheel cylinder has a diameter of 2.50 cm. Calculate the force \(F_2\) at each wheel cylinder.

Solution:

\(F_1 = 500 \text{ N}\)

\(d_1 = 0.500 \text{ cm} \Rightarrow r_1 = 0.250 \text{ cm}\)

\(d_2 = 2.50 \text{ cm} \Rightarrow r_2 = 1.25 \text{ cm}\)

Using \(\frac{F_1}{A_1} = \frac{F_2}{A_2}\):

\(F_2 = F_1 \frac{A_2}{A_1} = F_1 \frac{\pi r_2^2}{\pi r_1^2} = F_1 \frac{r_2^2}{r_1^2}\)

\(F_2 = (500 \text{ N}) \frac{(1.25 \text{ cm})^2}{(0.250 \text{ cm})^2} = (500 \text{ N}) \frac{1.5625}{0.0625} = 1.25 \times 10^4 \text{ N}\).

This calculation shows how a relatively small force on the brake pedal translates into a very large force at each wheel, enabling effective braking. Note that the force is multiplied by a factor equal to the square of the ratio of the radii. This large increase in force is why hydraulic brakes are so effective.

Chapter 14.4: Archimedes’ Principle and Buoyancy

Learning Objectives

By the end of this section, you will be able to:

• Define buoyant force.
• State Archimedes’ principle.
• Describe the relationship between density and Archimedes’ principle.

We’re now shifting gears to understand why some objects float and others sink – the concept of buoyancy. This section will introduce Archimedes’ principle, a cornerstone of fluid statics, and explain how an object’s density relates to whether it floats or sinks.

Buoyant Force (\(F_B\))

• Definition: The upward force on any object in any fluid.
• Arises because pressure increases with depth.
  • Upward force on the bottom of an object is greater than the downward force on the top.
• Result: Net upward force on the object.
• Present whether an object floats, sinks, or is suspended.

Figure 14.20: Net buoyant force on an object.

Think about pushing a beach ball underwater. You feel that upward push – that’s buoyant force. It’s a direct consequence of the pressure difference between the top and bottom of the object. The deeper parts experience higher pressure, leading to a greater upward force.

Archimedes’ Principle

• Statement: The buoyant force on an object equals the weight of the fluid it displaces.

\[ F_B = w_{fl} \]

-   $F_B$: buoyant force.
-   $w_{fl}$: weight of the fluid displaced by the object.

Figure 14.21: Buoyant force equals weight of displaced fluid.

• Named after Archimedes of Syracuse.
• When an object is submerged, it pushes aside, or displaces, a certain volume of fluid. The weight of that displaced fluid is exactly the buoyant force acting on the object.

This is one of the oldest and most elegant principles in physics. It elegantly links the buoyant force to something measurable: the weight of the fluid pushed out of the way. This principle is true whether the object floats, sinks, or is completely submerged.

Density and Archimedes’ Principle

• An object’s average density determines if it floats:
  • If \(\rho_{obj} < \rho_{fl}\): Object floats.
  • If \(\rho_{obj} > \rho_{fl}\): Object sinks.
  • If \(\rho_{obj} = \rho_{fl}\): Object remains suspended (neutrally buoyant).
• Fraction submerged: For a floating object, the fraction of its volume submerged is: \[ \text{fraction submerged} = \frac{V_{sub}}{V_{obj}} = \frac{\rho_{obj}}{\rho_{fl}} \]

Figure 14.22: Unloaded vs. loaded ship.

This is a powerful relationship. It explains why a massive steel ship floats (its average density, including the air inside, is less than water) while a small pebble sinks. The fraction of a floating object that is submerged is a direct ratio of its density to the fluid’s density.

Example: Calculating Average Density

Problem: A 60.0-kg woman floats in fresh water with 97.0% of her volume submerged when her lungs are full of air. What is her average density?

Given:

\(m_{person} = 60.0 \text{ kg}\)

Fraction submerged = 0.970

\(\rho_{water} = 1.000 \times 10^3 \text{ kg/m}^3\)

Solution:

Using the formula for fraction submerged:

\(\text{fraction submerged} = \frac{\rho_{obj}}{\rho_{fl}}\)

\(\rho_{person} = (\text{fraction submerged}) \times \rho_{water}\)

\(\rho_{person} = (0.970) \times (1.000 \times 10^3 \text{ kg/m}^3) = 970 \text{ kg/m}^3\).

Significance: Her average density is indeed less than water, which is why she floats.

This example demonstrates how we can use the fraction of an object submerged to determine its average density. It’s a practical application of Archimedes’ principle and the concept of density.

Measuring Density with Apparent Weight

Figure 14.23: Weighing a coin in air and water.

• An object’s apparent weight when submerged is less than its actual weight in air.
• Apparent weight loss = Weight of fluid displaced = Buoyant force.
• By weighing an object in air (\(W_{air}\)) and then submerged in a fluid (\(W_{app}\)): \(F_B = W_{air} - W_{app}\) Since \(F_B = \rho_{fl} g V_{obj}\) and \(W_{air} = \rho_{obj} g V_{obj}\): \(\rho_{obj} = \rho_{fl} \frac{W_{air}}{W_{air} - W_{app}}\)
• This technique can be used to find either \(\rho_{obj}\) (if \(\rho_{fl}\) is known) or \(\rho_{fl}\) (if \(\rho_{obj}\) is known).

This is a common laboratory technique. By measuring the “lost” weight when an object is submerged, you’re essentially measuring the buoyant force, which in turn tells you about the volume of displaced fluid and thus the object’s (or fluid’s) density.

Chapter 14.5: Fluid Dynamics

Learning Objectives

By the end of this section, you will be able to:

• Describe the characteristics of flow.
• Calculate flow rate.
• Describe the relationship between flow rate and velocity.
• Explain the consequences of the equation of continuity to the conservation of mass.

So far, we’ve focused on fluids at rest. Now, we’re transitioning to fluid dynamics – the study of fluids in motion. This is where things get more complex and exciting! We’ll start with fundamental concepts like flow types, flow rate, and how velocity changes in moving fluids, all tied together by the crucial principle of mass conservation.

Characteristics of Flow: Streamlines

• Streamline: Represents the path of a small volume of fluid as it flows.
  • The velocity vector is always tangential to the streamline.
• Laminar Flow (Steady Flow): Smooth, parallel streamlines.
  • Layers flow without mixing.
  • Velocity may vary across the flow (e.g., faster in the center of a pipe due to viscosity).
• Turbulent Flow: Irregular, chaotic streamlines that change over time.
  • Fluid mixes, forming eddies and whirlpools.
  • Occurs at high speeds or due to obstructions.

Figure 14.25: (a) Laminar flow (b) Turbulent flow.

Understanding these flow characteristics is essential. Laminar flow is predictable and easier to model, like a calm river. Turbulent flow is complex and chaotic, like rapids. Many real-world systems experience both, and the transition point is important for engineering design.

Flow Rate (\(Q\))

• Volume Flow Rate: Volume of fluid passing a given location through an area per unit time. \[ Q = \frac{V}{t} \]
• Relationship to Velocity: \[ Q = Av \]
  • \(A\): cross-sectional area of flow.
  • \(v\): average speed of the fluid.
• SI Unit: \(\text{m}^3/\text{s}\). (Liters/minute are also common).

Figure 14.26: Volume flow rate.

Flow rate isn’t just about how fast the fluid is moving; it’s also about how much fluid is moving. A small pipe with very fast flow might have the same flow rate as a large pipe with slow flow. This formula is fundamental for analyzing fluid transport.

The Equation of Continuity

• For an incompressible fluid (constant density) in steady flow, the flow rate is constant throughout the pipe.
  • No fluid is created or destroyed.
  • Mass conservation implies volume flow rate conservation. \[ Q_1 = Q_2 \] \[ A_1 v_1 = A_2 v_2 \]
  • If the cross-sectional area decreases, the fluid speed must increase (and vice versa).

Figure 14.27: Fluid speed increases when a tube narrows.

This principle is very intuitive. Think about putting your thumb over the end of a garden hose. The water comes out faster. You’re decreasing the area, so the velocity must increase to maintain the same flow rate. This is a direct consequence of mass conservation for incompressible fluids.

Example: Calculating Fluid Speed through a Nozzle

Problem: A nozzle with diameter 0.500 cm is attached to a hose with radius 0.900 cm. The flow rate is 0.500 L/s. Calculate the speed of water (a) in the hose and (b) in the nozzle.

Given:

\(Q = 0.500 \text{ L/s} = 0.500 \times 10^{-3} \text{ m}^3/\text{s}\)

Hose: \(r_1 = 0.900 \text{ cm} = 0.00900 \text{ m}\)

Nozzle: \(d_2 = 0.500 \text{ cm} \Rightarrow r_2 = 0.250 \text{ cm} = 0.00250 \text{ m}\)

Solution:

1. Speed in the hose (\(v_1\)):

\(A_1 = \pi r_1^2 = \pi (0.00900 \text{ m})^2\)

\(v_1 = \frac{Q}{A_1} = \frac{0.500 \times 10^{-3} \text{ m}^3/\text{s}}{\pi (0.00900 \text{ m})^2} = 1.96 \text{ m/s}\).

2. Speed in the nozzle (\(v_2\)):

Using the equation of continuity \(A_1 v_1 = A_2 v_2\):

\(v_2 = v_1 \frac{A_1}{A_2} = v_1 \frac{\pi r_1^2}{\pi r_2^2} = v_1 \left(\frac{r_1}{r_2}\right)^2\)

\(v_2 = (1.96 \text{ m/s}) \left(\frac{0.900 \text{ cm}}{0.250 \text{ cm}}\right)^2 = (1.96 \text{ m/s}) (3.6)^2 = 25.5 \text{ m/s}\).

This example shows the significant impact of narrowing a pipe. A relatively small reduction in radius leads to a large increase in fluid speed, demonstrating the \(r^2\) dependence in the continuity equation.

Mass Flow Rate and General Continuity Equation

• Mass Flow Rate (\(\frac{dm}{dt}\)): The rate at which mass of fluid moves past a point.

  \[ \frac{dm}{dt} = \rho A v \]

• General Continuity Equation: For any fluid (compressible or incompressible), mass flow rate is conserved:

  \[ \left(\frac{dm}{dt}\right)_1 = \left(\frac{dm}{dt}\right)_2 \]

  \[ \rho_1 A_1 v_1 = \rho_2 A_2 v_2 \]

• If the fluid is incompressible (\(\rho_1 = \rho_2\)), then \(\rho\) cancels, returning \(A_1 v_1 = A_2 v_2\).

While the \(Av\) form of continuity is popular, the \(\rho Av\) form is more general and accounts for situations where fluid density might change, such as with gases or in extreme conditions. It’s a statement of the fundamental principle of mass conservation in fluid flow.

Chapter 14.6: Bernoulli’s Equation

Learning Objectives

By the end of this section, you will be able to:

• Explain the terms in Bernoulli’s equation.
• Explain how Bernoulli’s equation is related to the conservation of energy.
• Describe how to derive Bernoulli’s principle from Bernoulli’s equation.
• Perform calculations using Bernoulli’s principle.
• Describe some applications of Bernoulli’s principle.

Now we combine fluid motion with energy conservation. Bernoulli’s equation is a powerful tool that links pressure, velocity, and height in moving fluids. It’s essentially an energy conservation law for fluids and explains many phenomena, from why shower curtains billow inward to how airplane wings generate lift.

Bernoulli’s Equation

• Based on the conservation of energy for an incompressible, frictionless fluid in laminar flow.
• It states that along a streamline, the sum of pressure, kinetic energy density, and potential energy density is constant: \[ p + \frac{1}{2} \rho v^2 + \rho g y = \text{constant} \]
  • \(p\): absolute pressure.
  • \(\frac{1}{2} \rho v^2\): kinetic energy density (dynamic pressure).
  • \(\rho g y\): potential energy density (hydrostatic pressure).

Figure 14.30: Geometry for deriving Bernoulli’s equation.

Bernoulli’s equation is a statement of mechanical energy conservation applied to a fluid. The “constant” means that if you pick any two points along a streamline, the sum of these three terms will be the same. This equation has immense practical applications.

Analyzing Bernoulli’s Equation

• Energy Transformation: Pressure, kinetic energy density, and potential energy density can interconvert, but their sum remains constant.
• Pressure Drop with Speed: As fluid speed (v) increases, the pressure (p) tends to decrease (assuming constant height), to keep the sum constant.
  • Examples: Shower curtain bulging inward, cars pulling towards trucks on a highway.

Figure 14.29: Pressure drop between passing vehicles.

This is a key takeaway: faster fluid flow generally means lower pressure. The energy that goes into increasing kinetic energy must come from somewhere, and often it comes from a reduction in pressure. This is a counter-intuitive idea for many, but it’s crucial for understanding fluid dynamics.

Special Cases of Bernoulli’s Equation

1. Static Fluids (\(v_1 = v_2 = 0\)):

   \[ p_1 + \rho g y_1 = p_2 + \rho g y_2 \]

   If \(y_2 = 0\), then \(p_2 = p_1 + \rho g y_1\). This confirms pressure increases with depth, as seen in fluid statics.

2. Bernoulli’s Principle (Constant Depth, \(y_1 = y_2\)):

   \[ p_1 + \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2} \rho v_2^2 \]

   This clearly shows that if \(v_2 > v_1\), then \(p_2 < p_1\). Pressure drops as speed increases.

It’s reassuring that Bernoulli’s equation, when simplified for static fluids, reduces to our familiar hydrostatic pressure equation. Bernoulli’s principle is a frequently used simplification for horizontal flow, emphasizing the inverse relationship between speed and pressure.

Example: Calculating Pressure in a Hose

Problem: Water speed in a hose increases from 1.96 m/s to 25.5 m/s in a nozzle. Absolute pressure in the nozzle is \(1.01 \times 10^5 \text{ N/m}^2\) (atmospheric). Assuming level, frictionless flow, calculate the pressure in the hose.

Given:

\(v_1 = 1.96 \text{ m/s}\) (hose)

\(v_2 = 25.5 \text{ m/s}\) (nozzle)

\(p_2 = 1.01 \times 10^5 \text{ Pa}\) (nozzle pressure)

\(\rho_{water} = 1000 \text{ kg/m}^3\)

Level flow implies \(y_1 = y_2\).

Solution:

Using Bernoulli’s principle: \(p_1 + \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2} \rho v_2^2\)

Solving for \(p_1\):

\(p_1 = p_2 + \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2} \rho (v_2^2 - v_1^2)\)

\(p_1 = 1.01 \times 10^5 \text{ Pa} + \frac{1}{2} (1000 \text{ kg/m}^3) [(25.5 \text{ m/s})^2 - (1.96 \text{ m/s})^2]\)

\(p_1 = 1.01 \times 10^5 \text{ Pa} + 500 \text{ kg/m}^3 [650.25 \text{ m}^2/\text{s}^2 - 3.84 \text{ m}^2/\text{s}^2]\)

\(p_1 = 1.01 \times 10^5 \text{ Pa} + 500 \text{ kg/m}^3 (646.41 \text{ m}^2/\text{s}^2)\)

\(p_1 = 1.01 \times 10^5 \text{ Pa} + 3.23 \times 10^5 \text{ Pa} = 4.24 \times 10^5 \text{ Pa}\).

The pressure in the hose is significantly higher than in the nozzle, which makes sense. The higher pressure in the hose is what pushes the water to accelerate as it enters the narrower nozzle, converting pressure energy into kinetic energy.

Applications of Bernoulli’s Principle

1. Entrainment: Using high-velocity fluid to create low pressure, which pulls other fluids into the stream.
   • Bunsen burners entrain air for combustion.
   • Atomizers entrain perfume drops.
   • Aspirators use water to create suction.
   • Chimneys entrain air to draw smoke up.

Figure 14.31: Entrainment devices.

Entrainment is a clever application of Bernoulli’s principle that you encounter frequently without realizing it. It’s about using the energy in one fluid stream to move another, often leveraging a pressure difference.

Applications: Velocity Measurement

• Devices like the Pitot tube use Bernoulli’s principle to measure fluid velocity.
• By creating a “dead spot” (zero velocity) at one opening and measuring pressure at a point of flow, the pressure difference can be related to fluid speed. \[ p_1 = p_2 + \frac{1}{2} \rho v_2^2 \implies \frac{1}{2} \rho v_2^2 = p_1 - p_2 \] \[ v_2 = \sqrt{\frac{2(p_1 - p_2)}{\rho}} \]
• Commonly used as air-speed indicators in aircraft.

Figure 14.32: Pitot tube for velocity measurement.

Pitot tubes are ubiquitous in aviation and industrial flow measurement. By measuring the difference between static and total pressure, they can accurately determine the fluid velocity. It’s a direct application of converting kinetic energy into pressure.

Chapter 14.7: Viscosity and Turbulence

Learning Objectives

By the end of this section, you will be able to:

• Explain what viscosity is.
• Calculate flow and resistance with Poiseuille’s law.
• Explain how pressure drops due to resistance.
• Calculate the Reynolds number for an object moving through a fluid.
• Use the Reynolds number for a system to determine whether it is laminar or turbulent.
• Describe the conditions under which an object has a terminal speed.

Until now, we’ve often idealized fluids as “frictionless.” In reality, fluids have internal friction, known as viscosity, and their flow can become chaotic, or turbulent. This section will delve into these real-world complexities, introducing viscosity, Poiseuille’s law, and the Reynolds number to characterize fluid flow.

Viscosity and Laminar Flow

• Viscosity (\(\eta\)): A measure of internal friction within a fluid and between fluid and surroundings.
  • Juice has low viscosity; maple syrup has high viscosity.
• Laminar Flow: Fluid layers slide smoothly past each other without mixing.
• Turbulence: Chaotic fluid flow with eddies and swirls, layers mix.
  • Caused by obstructions or high fluid speeds.

Figure 14.34: (a) Laminar flow (b) Turbulent flow.

Viscosity is essentially “fluid friction.” It’s why some liquids flow easily and others are “thick.” Laminar flow is ordered and predictable, while turbulent flow is messy and complex. Recognizing these flow types is crucial for understanding fluid behavior in real-world applications.

Measuring Viscosity

• Viscosity (\(\eta\)) is defined by applying a force \(F\) to move a top plate over a stationary bottom plate, with fluid in between. \[ F = \eta \frac{Av}{L} \]
  • \(A\): area of the plates.
  • \(v\): velocity of the top plate.
  • \(L\): distance between plates.
• Solving for viscosity: \[ \eta = \frac{FL}{Av} \]
• SI Unit: \(\text{Pa} \cdot \text{s}\) (Pascal-second).

Figure 14.36: Setup for measuring viscosity.

This setup provides a precise way to quantify viscosity. The force required to maintain a constant velocity directly depends on the fluid’s internal friction. Different fluids, even at the same temperature, can have vastly different viscosities.

Poiseuille’s Law for Laminar Flow in Tubes

• Flow rate (\(Q\)) in a horizontal tube is driven by a pressure difference (\(\Delta p = p_2 - p_1\)) and opposed by resistance (\(R\)). \[ Q = \frac{\Delta p}{R} \]
• For laminar flow in a uniform tube of radius \(r\) and length \(l\) with fluid viscosity \(\eta\), the resistance (\(R\)) is: \[ R = \frac{8 \eta l}{\pi r^4} \]
• Combining these gives Poiseuille’s Law for flow rate: \[ Q = \frac{(p_2 - p_1) \pi r^4}{8 \eta l} \]

Figure 14.38: Laminar flow in a pipe.

Poiseuille’s law is a powerful equation for understanding fluid flow in pipes, like blood flow in arteries. Notice the \(r^4\) term – this means that even small changes in the tube’s radius have a huge impact on flow rate. This is critical in medicine for understanding blood pressure and circulation.

Example: Air Conditioning System Flow Rate

Problem: An AC system delivers air at gauge pressure 0.054 Pa (\(20^\circ\text{C}\)). Conduit: \(D = 18.00 \text{ cm}\), \(L = 20 \text{ m}\). Room: \(12 \text{ m} \times 6 \text{ m} \times 3 \text{ m}\).

1. Volume flow rate assuming laminar flow?
2. Time to replace air in the room?
3. New flow rate if conduit \(D = 9.00 \text{ cm}\)?

Given:

\(\Delta p = 0.054 \text{ Pa}\)

\(\eta_{air} = 0.0181 \times 10^{-3} \text{ Pa} \cdot \text{s}\) (from Table 14.4)

Solution (a):

\(r = D/2 = 9.00 \text{ cm} = 0.09 \text{ m}\)

\(Q = \frac{\Delta p \pi r^4}{8 \eta l} = \frac{(0.054 \text{ Pa}) \pi (0.09 \text{ m})^4}{8 (0.0181 \times 10^{-3} \text{ Pa} \cdot \text{s}) (20 \text{ m})} = 3.84 \times 10^{-3} \text{ m}^3/\text{s}\).

Solution (b):

\(V_{room} = (12 \text{ m})(6 \text{ m})(3 \text{ m}) = 216 \text{ m}^3\)

\(\Delta t = \frac{V_{room}}{Q} = \frac{216 \text{ m}^3}{3.84 \times 10^{-3} \text{ m}^3/\text{s}} = 5.63 \times 10^4 \text{ s} \approx 15.6 \text{ hours}\).

Solution (c):

New radius \(r' = 4.50 \text{ cm} = 0.045 \text{ m}\) (half the original).

Since \(Q \propto r^4\), the new flow rate will be \(\left(\frac{r'}{r}\right)^4 Q = \left(\frac{0.045}{0.09}\right)^4 Q = \left(\frac{1}{2}\right)^4 Q = \frac{1}{16} Q\).

\(Q' = \frac{1}{16} (3.84 \times 10^{-3} \text{ m}^3/\text{s}) = 2.40 \times 10^{-4} \text{ m}^3/\text{s}\).

This example clearly shows the dramatic effect of tube radius on flow rate due to the \(r^4\) dependence in Poiseuille’s law. Halving the diameter reduces the flow to one-sixteenth, significantly impacting the air turnover time in the room. This highlights why pipe sizing is critical in engineering.

Flow and Resistance as Causes of Pressure Drops

• Pressure drops occur due to fluid flow encountering resistance. \[ p_2 - p_1 = R Q \]
• This equation is valid for both laminar and turbulent flows.
• Consequences:
  • During heavy use (high \(Q\)), the pressure drop (\(p_2 - p_1\)) is large, leading to lower pressure (\(p_1\)) for users.
  • Resistance (\(R\)) is much greater in narrow sections, causing larger pressure drops there.
  • Turbulence greatly increases \(R\), leading to significant pressure reduction downstream.

Figure 14.39: Pressure drop in a water main.

This equation quantifies why water pressure drops when many people use water simultaneously, or why a blocked artery reduces blood flow and creates pressure issues. Resistance is the key factor in these pressure losses.

Measuring Turbulence: The Reynolds Number

• Reynolds Number (\(N_R\)): A dimensionless quantity that indicates whether flow is laminar or turbulent. \[ N_R = \frac{2 \rho v r}{\eta} \quad (\text{for flow in a tube}) \]
  • \(\rho\): fluid density.
  • \(v\): fluid speed.
  • \(r\): tube radius.
  • \(\eta\): fluid viscosity.
• Turbulence Thresholds:
  • \(N_R < 2000\): Laminar flow.
  • \(N_R > 3000\): Turbulent flow.
  • \(2000 < N_R < 3000\): Unstable flow (can be laminar but easily becomes turbulent).

Note

Turbulence leads to increased energy dissipation (lost as heat and eddies), making flow less efficient.

The Reynolds number is an incredibly useful parameter in fluid dynamics. It allows engineers to predict and design systems to avoid unwanted turbulence, which can waste energy and cause wear. The transition region is particularly tricky, showcasing chaotic behavior where small disturbances can have large effects.

Example: Turbulent Flow or Laminar Flow?

Problem: For the AC system (conduit \(D = 18.00 \text{ cm}\)) from Example 14.8, we found \(Q = 3.84 \times 10^{-3} \text{ m}^3/\text{s}\).

1. Was the assumption of laminar flow valid?
2. At what velocity would the flow become turbulent?

Given:

\(Q = 3.84 \times 10^{-3} \text{ m}^3/\text{s}\)

\(r = 0.09 \text{ m}\)

\(\rho_{air} = 1.23 \text{ kg/m}^3\) (from Table 14.1 at \(0^\circ\text{C}\), but usually taken as \(1.29 \text{ kg/m}^3\) for air, use \(1.23 \text{ kg/m}^3\) as given in prompt)

\(\eta_{air} = 0.0181 \times 10^{-3} \text{ Pa} \cdot \text{s}\)

Solution (a):

First, find velocity \(v = \frac{Q}{\pi r^2} = \frac{3.84 \times 10^{-3} \text{ m}^3/\text{s}}{\pi (0.09 \text{ m})^2} = 0.15 \text{ m/s}\).

Now, calculate Reynolds number:

\(N_R = \frac{2 \rho v r}{\eta} = \frac{2 (1.23 \text{ kg/m}^3) (0.15 \text{ m/s}) (0.09 \text{ m})}{0.0181 \times 10^{-3} \text{ Pa} \cdot \text{s}} = 1835\).

Since \(N_R = 1835 < 2000\), the flow is laminar. The assumption was valid.

Solution (b):

Flow becomes turbulent when \(N_R \approx 2000\).

\(v = \frac{N_R \eta}{2 \rho r} = \frac{2000 (0.0181 \times 10^{-3} \text{ Pa} \cdot \text{s})}{2 (1.23 \text{ kg/m}^3) (0.09 \text{ m})} = 0.16 \text{ m/s}\).

This example shows how to use the Reynolds number to verify assumptions about flow behavior. It’s important for designers to keep flow laminar if efficiency is desired, as exceeding the critical Reynolds number will lead to turbulent flow and wasted energy.

Key Takeaways

• Phases of Matter: Solids are rigid, liquids flow and maintain volume, gases expand and are compressible.
• Density (\(\rho = m/V\)): Mass per unit volume; determines floating/sinking. Pressure always pushes perpendicular to surfaces.
• Pressure (\(p = F/A\)): Force per unit area. \(p = p_0 + \rho g h\) for static fluids of constant density.
• Pascal’s Principle: Pressure changes in enclosed fluids transmit undiminished; basis for hydraulics (\(F_1/A_1 = F_2/A_2\)).
• Archimedes’ Principle: Buoyant force equals weight of fluid displaced (\(F_B = w_{fl}\)).
• Flow Rate (\(Q = Av\)): Volume of fluid per unit time.
• Equation of Continuity (\(A_1 v_1 = A_2 v_2\)): For incompressible fluids, speed increases as area decreases.
• Bernoulli’s Equation (\(p + \frac{1}{2} \rho v^2 + \rho g y = \text{constant}\)): Conservation of energy for ideal fluids, linking pressure, velocity, and height.
• Viscosity (\(\eta\)): Fluid friction; higher viscosity means slower flow.
• Poiseuille’s Law (\(Q = \frac{\Delta p \pi r^4}{8 \eta l}\)): Describes laminar flow rate in tubes; flow is highly sensitive to radius (\(r^4\)).
• Reynolds Number (\(N_R = \frac{2 \rho v r}{\eta}\)): Predicts laminar (\(<2000\)) vs. turbulent (\(>3000\)) flow.

This slide summarizes the core concepts of this chapter. It’s a quick reference for the most important definitions, principles, and equations. Make sure you understand each of these points.

Key Equations

Equation	Description
\(\rho = \frac{m}{V}\)	Definition of density
\(p = \frac{F}{A}\)	Definition of pressure
\(p = p_0 + \rho g h\)	Pressure at a depth h in a fluid of constant density
\(p_{abs} = p_g + p_{atm}\)	Relationship between absolute and gauge pressure
\(\frac{F_1}{A_1} = \frac{F_2}{A_2}\)	Pascal’s principle in a hydraulic system
\(F_B = w_{fl}\)	Archimedes’ principle (buoyant force)
\(\text{fraction submerged} = \frac{\rho_{obj}}{\rho_{fl}}\)	Fraction of a floating object submerged
\(Q = Av\)	Volume flow rate
\(A_1 v_1 = A_2 v_2\)	Equation of continuity (for incompressible fluids)
\(p + \frac{1}{2} \rho v^2 + \rho g y = \text{constant}\)	Bernoulli’s equation
\(Q = \frac{(p_2 - p_1) \pi r^4}{8 \eta l}\)	Poiseuille’s law for laminar flow
\(N_R = \frac{2 \rho v r}{\eta}\)	Reynolds number (for flow in a tube)

This table brings together all the major equations we’ve covered. It’s a handy cheat sheet for solving problems and remembering the mathematical relationships between fluid properties.

Key Terms

Term	Definition
Absolute Pressure	The sum of gauge pressure and atmospheric pressure.
Archimedes’ Principle	The buoyant force on an object equals the weight of the fluid it displaces.
Bernoulli’s Equation	States that the sum of pressure, kinetic energy density, and potential energy density is constant along a streamline for an ideal fluid.
Buoyant Force	The upward force on an object in a fluid.
Density (\(\rho\))	Mass per unit volume.
Fluid Dynamics	The study of fluids in motion.
Flow Rate (\(Q\))	Volume of fluid passing a point per unit time.
Gauge Pressure	Pressure relative to atmospheric pressure.
Incompressible Fluid	A fluid whose volume changes negligibly with pressure.
Laminar Flow	Smooth, orderly fluid flow in layers without mixing.
Manometer	A device for measuring pressure using a column of fluid.
Pascal’s Principle	A change in pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid.
Poiseuille’s Law	Describes the laminar flow rate of a viscous fluid through a tube.
Pressure (\(p\))	Force per unit area.
Reynolds Number (\(N_R\))	A dimensionless quantity that predicts the onset of turbulence.
Specific Gravity	Ratio of a substance’s density to water’s density at \(4.0^\circ\text{C}\).
Streamline	The path followed by a small element of fluid.
Turbulence	Chaotic, irregular fluid flow with eddies and swirls.
Viscosity (\(\eta\))	A measure of a fluid’s internal friction or resistance to flow.

Familiarize yourself with these terms. A strong vocabulary is essential for understanding the concepts and communicating effectively in physics.