Fluids: Properties, Pressure, and Depth in Physics- lecture 20

Introduction to Fluids

What are Fluids?

Fluids are materials that can flow and take the shape of their container. This category includes liquids, gases, and plasmas. In some cases, amorphous solids are also considered fluids due to their ability to deform under stress. The most common fluids encountered in physics are air and water, as they are abundant in everyday life.

Mass Density

Definition and Concept

Density (symbol: \( \rho \)) is a measure of how much mass is contained in a given volume. It is mathematically defined as the mass (m) of an object divided by its volume (V):

$\rho = \frac{m}{V}$

The SI unit of density is kilograms per cubic meter (kg/m3). Environmental factors such as temperature and pressure can affect the density of materials, especially gases.

Examples of Mass Density

The table below lists the mass densities of various common substances, including solids, liquids, and gases. This helps in comparing how tightly matter is packed in different materials.

Substance	Mass Density \( \rho \) (kg/m3)
Aluminum	2700
Brass	8470
Concrete	2200
Copper	8890
Diamond	3520
Gold	19,300
Ice	917
Iron (steel)	7860
Lead	11,300
Quartz	2660
Silver	10,500
Wood (yellow pine)	550
Blood (whole, 37°C)	1060
Ethyl alcohol	806
Mercury	13,600
Oil (hydraulic)	800
Water (4°C)	1.000 × 103
Air	1.29
Carbon dioxide	1.98
Helium	0.179
Hydrogen	0.0899
Nitrogen	1.25
Oxygen	1.43

Specific Gravity

Definition and Application

Specific gravity is a dimensionless quantity that compares the density of a material to the density of water at 4°C. It is useful for quickly assessing whether a substance will float or sink in water.

$\text{Specific gravity} = \frac{\rho_{\text{material}}}{\rho_{\text{water @ 4^\circ C}}}$

Where \( \rho_{\text{water @ 4^\circ C}} = 1.000 \times 10^3 \; \text{kg/m}^3 \).

Example: If a material has a density of 2,000 kg/m3, its specific gravity is 2.0.

Pressure in Fluids

Definition of Pressure

Pressure (symbol: P) is defined as the magnitude of the average force (F) applied perpendicularly over a given area (A):

$P = \frac{F}{A}$

The SI unit of pressure is the pascal (Pa), where 1 Pa = 1 N/m2. Atmospheric pressure at sea level is approximately 101,300 Pa (1 atm).

Pressure in Everyday Life

When inflating a tire, air molecules inside the tire collide with the walls, exerting a force perpendicular to the surface. This is an example of how pressure is generated in a fluid (gas) and acts equally in all directions.

Pressure is Not a Vector

While the force due to pressure acts in a specific direction (perpendicular to the surface), pressure itself is a scalar quantity and does not have a direction.

Pressure Example: Swimmer's Hand

Suppose the pressure acting on the back of a swimmer’s hand is 1.2 × 105 Pa, and the surface area is 8.4 × 10−3 m2.

• Magnitude of Force:

$F = P A = (1.2 \times 10^5 \; \text{Pa})(8.4 \times 10^{-3} \; \text{m}^2) = 1.0 \times 10^3 \; \text{N}$

• Direction of Force: The force acts perpendicular to the surface area on the back of the swimmer’s hand.

Depth and Pressure in Fluids

Pressure Variation with Depth

As you go deeper below the surface of a fluid, the pressure increases due to the weight of the fluid above. This relationship is fundamental in fluid statics and is derived using Newton’s Second Law.

Derivation: Pressure at Depth

Consider a column of fluid with cross-sectional area A and height h. The equilibrium of forces gives:

$\sum F_y = P_2 A - P_1 A - mg = 0$

Where m is the mass of the liquid. Using the density equation and the geometry of the column:

$m = \rho V = \rho A h$

Substituting into the force equation:

$P_2 A - P_1 A - \rho A h g = 0$ $P_2 - P_1 - \rho g h = 0$ $P_2 = P_1 + \rho g h$

This shows that the pressure at depth h is the sum of the pressure at the surface and the pressure due to the weight of the fluid above.

Pressure is Independent of Container Shape

Pressure at a given depth in a connected fluid is the same, regardless of the shape of the container. Points at the same depth experience the same pressure.

Example: Pressure at Depth in Water

Given:

• Atmospheric pressure at the surface: \( P_1 = 1.01 \times 10^5 \; \text{Pa} \)

• Density of water: \( \rho = 1.000 \times 10^3 \; \text{kg/m}^3 \)

• Depth: \( h = 5.50 \; \text{m} \)

Calculate the pressure at point A, 5.50 m below the surface:

$P_2 = P_1 + \rho g h = 1.01 \times 10^5 \; \text{Pa} + (1.000 \times 10^3 \; \text{kg/m}^3)(9.81 \; \text{m/s}^2)(5.50 \; \text{m}) = 1.55 \times 10^5 \; \text{Pa}$

Key Takeaway: The deeper you go in a fluid, the greater the pressure, and this increase is linear with depth for incompressible fluids like water.