Fluid Mechanics: Pressure, Buoyancy, and Dynamic Flow Notes

Total Pressure and Force on Pool Walls

• Total Pressure at the Bottom Surface vs. Side Wall:     * The total pressure on the side of a pool at its very bottom is identical to the pressure on the bottom floor of the pool, provided they are at the same depth.     * Numerical Value: At the bottom depth, the pressure is recorded as $113\,kPa$.

• Total Pressure at the Top (Surface) of the Pool:     * At the surface, there is no depth ($h = 0$), meaning there is no fluid pressure ($P_{fluid} = 0$).     * Value: The total pressure consists only of atmospheric pressure, which is $1.01 \times 10^5\,Pa$.

• Estimating Total Force on a Side Wall Using Average Pressure:     * Force on a surface is defined as pressure times area ($F = P \times A$).     * Because the fluid pressure changes linearly with depth (from approximately $12\,kPa$ at the bottom to $0\,kPa$ at the surface), the average pressure ($P_{avg}$) should be used to calculate the total force on the wall.     * Average Calculation: If the pressure range is from $0$ to $12\,kPa$, the average pressure is $6\,kPa$.     * Surface Area of a Cylinder (Side): The equation provided is $S.A. = 2\pi r h$.     * Conceptual Metaphor: If one were to cut a cylinder vertically and unravel it, it would result in a rectangle. The length of the rectangle is the circumference of the circle ($2\pi r$), and the width is the height ($h$). Thus, $A = 2\pi r h$.     * Example Parameters:         * Radius ($r$): $2.7\,m$.         * Height ($h$): $1.2\,m$.         * Average Pressure used in calculation: $6,000\,Pa$ (though the speaker mentions $12,000\,Pa \times 2\pi \times 4 \times h$ in an intermediate, potentially incorrect thought process, they emphasize taking the halfway point: $6\,kPa$).

Introduction to Buoyancy and Archimedes' Principle

• Definition of Buoyancy Force ($F_B$):     * It is the net upward force exerted by a fluid on an object immersed in it. It results specifically from the difference in fluid pressure at different depths.

• Source of the Buoyant Force:     * Pressure increases with depth due to the weight of the water column above. Therefore, the upward pressure at the bottom of an object is greater than the downward pressure at the top of the object.     * Forces acting on the sides of a symmetrical object (like a cylinder) are equal and opposite, canceling each other out. However, the force on the bottom ($F_{bottom}$) is greater than the force on the top ($F_{top}$).

• Derivation of the Buoyant Force Formula:     * $\text{Net Force} = F_{bottom} - F_{top}$.     * Using $F = P \times A$ and $P = \rho_{fluid} g h + P_{atm}$.     * The atmospheric pressure term ($P_{atm}$) cancels out when subtracting top from bottom: $(P_{atm} + \rho g h_2) A - (P_{atm} + \rho g h_1) A$.     * $F_B = \rho_{fluid} g A (h_2 - h_1)$.     * Since $A \times (h_2 - h_1)$ is the volume of the object ($V_{obj}$) for a submerged object:     * The Buoyant Force Formula: $F_B = \rho_{f} g V_{obj}$.

• Archimedes' Principle:     * The buoyant force is equal to the weight of the fluid displaced by the object.     * Equation: $F_B = W_{displaced\,fluid} = m_{f} g$.     * Since mass equals density times volume ($m = \rho V$), this becomes: $F_B = \rho_{fluid} V_{displaced} g$.

• Historical Context: The Golden Crown:     * King Hiero received a crown meant to be pure gold but suspected it was debased with silver. He asked Archimedes to verify this without destroying the crown.     * Archimedes realized that by weighing the object in air and then in water, the difference in apparent weight (the buoyant force) could be used to find the displacement and thus the density of the object.     * Upon his discovery, he reportedly shouted "Eureka!" ("I have found it") while running through the streets.

Specific Gravity and Floating

• Floating Conditions:     * An object floats if its density is less than the density of the fluid (\rho_{obj} < \rho_{fluid}).     * Human Example: Humans can float due to air in their lungs. To sink or stay at the bottom of a pool, one must breathe out (exhale), which reduces the volume of the body and increases net density.

• Specific Gravity ($SG$):     * The ratio of the density of an object to the density of water ($\rho_{H_{2}O} = 1000\,kg/m^3$).     * Formula: $SG = \frac{\rho_{obj}}{\rho_{H_{2}O}}$.

• Hot Air Balloons:     * Operate on the principle of density. Heating air increases its volume while mass remains constant ($V \uparrow$, $m \rightarrow \text{constant}$, therefore $\rho \downarrow$).     * The balloon floats in air because the hot air inside is less dense than the cooler atmospheric air outside.

Example Problems and Calculations

Problem 1: Object on the Lake Bottom

• Given:     * Volume of object ($V_{obj}$): $3.0 \times 10^5\,cm^3$.     * Mass of object ($m$): $70\,kg$.

• Unit Conversion:     * $1\,m^3 = 1,000,000\,cm^3$ ($10^6\,cm^3$).     * $V_{obj} = \frac{3.0 \times 10^5}{10^6} = 0.3\,m^3$ (Speaker corrects this to $0.03\,m^3$ in a later step: $30,000\,cm^3 / 1,000,000 = 0.03\,m^3$).

• Forces Involved:     * Gravity ($mg$) pointing down.     * Normal force ($F_N$) pointing up.     * Buoyant force ($F_B$) pointing up.

• Equilibrium Equation:     * $F_B + F_N - mg = 0$.     * $F_N = mg - F_B = mg - \rho_{f} g V_{obj}$.

• Result calculation:     * Weight ($mg$): $70 \times 9.8 = 686\,N$.     * Buoyant Force ($F_B$): $1000 \times 9.8 \times 0.03 = 294\,N$.     * Normal Force ($F_N$): $686 - 294 = 392\,N$.

Problem 2: The Floating Log

• Given:     * Specific Gravity ($SG$): $0.6$.     * Total Volume of log ($V_{total}$): $2\,m^3$.

• Objective: Find the volume submerged ($V_{sub}$).

• Analysis:     * The density of the log is $0.6 \times 1000 = 600\,kg/m^3$.     * For a floating object, $F_B = mg$.     * $\rho_{f} g V_{sub} = \rho_{obj} g V_{total}$.     * The ratio of volume submerged to total volume is the same as the ratio of the densities: $\frac{V_{sub}}{V_{total}} = \frac{\rho_{obj}}{\rho_{f}}$.     * $V_{sub} = 0.6 \times 2 = 1.2\,m^3$.

Fluid Dynamics and Bernoulli’s Equation

• Mass Flow Rate:     * Defined as the mass of fluid passing a point per unit time ($\frac{\Delta m}{\Delta t}$).     * Assuming the pipe is always full (no air bubbles), the mass flow rate must be constant throughout the pipe.

• Equation of Continuity:     * $\rho_1 A_1 v_1 = \rho_2 A_2 v_2$.     * For an incompressible fluid (where $\rho_1 = \rho_2$): $A_1 v_1 = A_2 v_2$.     * This implies that in a thinner section of pipe (smaller area), the fluid velocity ($v$) must increase.

• Bernoulli’s Principle:     * Where the velocity of a fluid is high, the pressure it exerts against the walls of the container is low. Conversely, where velocity is low, pressure is high.

• Bernoulli's Equation:     * Derived from the Work-Energy Theorem ($W_{net} = \Delta KE$).     * Equation: $P_1 + \frac{1}{2} \rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g y_2$.     * Terms represent pressure energy, kinetic energy density, and gravitational potential energy density.

• Application: Water Tank/Spigot Problem:     * Calculating the velocity ($v_2$) of water leaving a spigot at the bottom of a tank of height $h$.     * Assume the surface of the tank is open to the atmosphere ($P_1 = P_{atm}$) and the spigot is open to the atmosphere ($P_2 = P_{atm}$). These cancel out.     * Assume the velocity at the top of the tank is practically zero ($v_1 \approx 0$).     * The equation simplifies to gravitational potential energy converting to kinetic energy: $\rho g h = \frac{1}{2} \rho v_2^2$.     * Velocity calculation: $v_2 = \sqrt{2gh}$.

Questions & Discussion

• Q: Why are the forces on the top and bottom of an object in a fluid not equal?

• A: Because pressure depends on depth ($P = \rho g h$). The bottom is deeper than the top, so the pressure and resulting force are greater at the bottom, creating the net buoyant force.

• Aside on Physics and Tests: The speaker mentions that physics is a difficult class and comments on various students' potential performance. One student mentions taking a photo of a previous class's work (Santos class). Another student is mentioned regarding the "Amy test."

• Q: Why does pressure decrease as velocity increases?

• A: Pressure measures the impact of molecules against the walls of the container. If more of the molecule's velocity component is directed forward (longitudinally through the pipe), the component hitting the walls decreases.

• Discussion on Pipe Fullness: The speaker clarifies that in standard hydrodynamic problems, we assume pipes are constantly full with no air, meaning an amount of water exiting must be replaced immediately by water entering.