Fluids are substances that can flow and do not possess a fixed shape of their own. This category encompasses both liquids and gases. While solids and liquids have a nearly constant volume under standard conditions, gases expand to fill the entire volume of their container. The fundamental distinction lies in their response to shear stress and their relative compressibility.
When an object is submerged in a fluid at rest, the fluid exerts a force perpendicular (normal) to the object's surface at every point. If there were a parallel component of force, the fluid would flow, violating the static equilibrium condition.
The average pressure $P_{av}$ is defined as the normal force acting per unit area:
$$P_{av} = \frac{F}{A}$$In a limiting sense, the pressure $P$ at a point is given by:
$$P = \lim_{\Delta A \to 0} \frac{\Delta F}{\Delta A}$$The density $\rho$ of a fluid of mass $m$ occupying volume $V$ is defined as:
$$\rho = \frac{m}{V}$$The SI unit of density is kg m⁻³ . For liquids, density is nearly constant across wide pressure ranges, whereas gas density varies significantly.
The relative density of a substance is the dimensionless ratio of its density to the density of water at 4 °C (which is 1.0 \times 10³ kg m⁻³ ):
$$\text{Relative Density} = \frac{\rho_{\text{substance}}}{\rho_{\text{water at 4 °C}}}$$Mathematical Problem Statement: Estimate the average pressure sustained by the two femurs (thigh bones) of a human body of mass 40 kg, assuming each femur has a cross-sectional area of 10 cm². Take g = 10 m s⁻².
Step-by-Step Algorithmic Solution Trace:
- Calculate the total cross-sectional area of both femurs: $A = 2 \times 10 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2$.
- Determine the force acting on the femurs due to body weight: $F = m \times g = 40 \times 10 = 400 \text{ N}$.
- Compute the average pressure: $P_{av} = \frac{F}{A} = \frac{400}{20 \times 10^{-4}} = 2 \times 10^5 \text{ N m}^{-2}$.
Final Evaluation Output: $P_{av} = 2 \times 10^5 \text{ N m}^{-2}$
According to Pascal's law , the pressure in a fluid at rest is identical at all points that lie at the same horizontal height. Furthermore, any change in pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel.
Consider a fluid element of area $A$ and height $h$ at rest. The pressure difference between two vertical points separated by height $h$ is given by:
$$P_2 - P_1 = \rho g h$$If point 1 is shifted to the top surface of the liquid open to the atmosphere, then $P_1$ is replaced by atmospheric pressure $P_a$, and the absolute pressure $P$ at depth $h$ becomes:
$$P = P_a + \rho g h$$The excess pressure $P - P_a = \rho g h$ is defined as the gauge pressure .
The pressure at a certain depth depends solely on the height of the liquid column above it, not on the cross-sectional area or shape of the container. This is demonstrated by the hydrostatic paradox , where connected vessels of different shapes hold liquid up to the same height because the pressure at the base of each vessel is identical.
Hydraulic devices leverage Pascal's law to achieve mechanical advantage. In a hydraulic lift, two pistons of different cross-sectional areas $A_1$ and $A_2$ ($A_1 < A_2$) are connected by an enclosed liquid. Applying a small force $F_1$ on the smaller piston produces a pressure $P = \frac{F_1}{A_1}$. This pressure is transmitted undiminished, resulting in an upward force $F_2$ on the larger piston:
$$F_2 = P \times A_2 = F_1 \left(\frac{A_2}{A_1}\right)$$The factor $\frac{A_2}{A_1}$ represents the mechanical advantage of the system, allowing heavy loads to be lifted with minimal effort.
Fluid dynamics is the study of fluids in motion. The nature of fluid flow can be categorized based on velocity profiles and path consistency.
For the steady flow of an incompressible fluid through a pipe of varying cross-section, the mass of fluid entering one end must equal the mass leaving the other end. This leads to the equation of continuity :
$$A v = \text{constant}$$Where $A$ is the cross-sectional area and $v$ is the fluid speed. Consequently, fluid speed increases at narrower constrictions and decreases in wider regions.
Bernoulli's principle is a statement of the conservation of energy applied to steady, non-viscous, incompressible fluid flow.
Along any streamline, the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume remains constant:
$$P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$$This principle \implies that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy.
The speed of efflux $v_1$ of a liquid from a small hole in an open tank at a depth $h$ below the liquid surface is given by Torricelli's law :
$$v_1 = \sqrt{2gh}$$This speed is identical to the speed acquired by a freely falling body dropped from rest through a height $h$.
Viscosity represents the internal friction within a fluid that opposes relative motion between its adjacent layers.
When a fluid layer is sandwiched between a fixed plate and a plate moving with velocity $v$, the fluid experiences a velocity gradient. The coefficient of viscosity $\eta$ is defined as the ratio of shearing stress to strain rate:
$$\eta = \frac{F/A}{v/l} = \frac{F l}{v A}$$According to Stokes' law , the viscous drag force $F$ acting on a small spherical body of radius $a$ moving with velocity $v$ through a fluid of viscosity $\eta$ is:
$$F = 6 \pi \eta a v$$When a sphere falls through a viscous medium, it eventually reaches a constant velocity called the terminal velocity $v_t$, where the downward gravitational force is perfectly balanced by the sum of the upward buoyant force and the viscous drag force:
$$v_t = \frac{2 a^2 (\rho - \sigma) g}{9 \eta}$$Where $\rho$ is the density of the sphere and $\sigma$ is the density of the fluid.
Surface tension is a membrane-like behavior exhibited by liquid surfaces, arising from intermolecular attractive forces.
Molecules inside a liquid are attracted equally in all directions, whereas molecules at the surface experience a net inward cohesive force. Consequently, surface molecules possess extra potential energy. The work required to increase the surface area of a liquid film by an amount $\Delta A$ is stored as surface energy :
$$S = \frac{W}{\Delta A}$$Alternatively, surface tension $S$ is defined as the force per unit length acting in the plane of the interface along a line of unit length:
$$S = \frac{F}{2l}$$The angle of contact $\theta$ is the angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid. It determines whether a liquid will wet a solid surface or form droplets:
Due to surface tension, the pressure inside a curved liquid interface is always greater than the pressure outside. The excess pressure is given by:
When a narrow tube (capillary) is dipped in a liquid, the liquid rises or falls in the tube. The height $h$ of the capillary rise is given by:
$$h = \frac{2 S \cos\theta}{\rho g a}$$Where $a$ is the radius of the capillary tube and $\theta$ is the contact angle. If $\theta$ is acute, the liquid rises ($h > 0$); if $\theta$ is obtuse, the liquid is depressed ($h < 0$).
The following matrix summarizes the key physical quantities associated with the mechanical properties of fluids:
| Physical Quantity | Symbol | Dimensions | SI Unit | Remarks |
|---|---|---|---|---|
| Pressure | $P$ | $[ML^{-1}T^{-2}]$ | Pa (N m⁻²) | Scalar quantity; force normal to area. |
| Density | $\rho$ | $[ML^{-3}]$ | kg m⁻³ | Nearly constant for liquids; varies for gases. |
| Relative Density | - | Dimensionless | None | Ratio of substance density to water density at 4 °C. |
| Coefficient of Viscosity | $\eta$ | $[ML^{-1}T^{-1}]$ | Pl (Pa s) | Measures internal fluid friction. |
| Surface Tension | $S$ | $[MT^{-2}]$ | N m⁻¹ | Equivalent to surface energy per unit area. |
Historical Chronicle: Blaise Pascal & Daniel Bernoulli