EGME 333 FALL 2026 Week 4 Lecture 1

Overview

Justin Tran finishes fluid statics with buoyancy, flotation, and stability, then introduces Bernoulli’s equation as conservation of pressure, kinetic, and gravitational potential energy along a streamline. Examples include a life jacket requiring 0.376 ft³ of foam, a heated hot-air balloon, and wind stagnating against a person’s face; he also clarifies Homework 2’s gate geometry and lists Bernoulli’s four assumptions.

Key takeaways

Buoyant force uses the surrounding fluid’s density and only the submerged volume, while object weight uses the object’s density and its total volume.
The life-jacket example requires approximately 0.376 ft³ of foam: a 23.3 lb total load divided by the 62 lb/ft³ difference between water and foam specific weights.
A low center of gravity relative to the center of buoyancy creates a restoring moment; placing heavy equipment low in a boat helps it recover after tilting.
Bernoulli’s equation balances pressure, kinetic, and gravitational energy terms along a streamline: P + ½ρu² + ρgh = constant.
Bernoulli calculations require streamline-based comparison, constant density, steady flow, and negligible viscosity; the inviscid assumption is often the hardest to justify in real flows.
At a stagnation point, flow speed falls to zero and upstream kinetic energy becomes additional pressure, producing an increase of ½ρV² at equal elevation.

Chapters

0:00 Room Chatter Before the EGME 333 Lecture Begins

The recording begins with brief, fragmented greetings and room noise rather than technical instruction.
Justin Tran starts the lecture at approximately 9:13, after the pre-class chatter.

9:13 Lecture Plan, Campus Evacuation Drill, and Homework 2 Deadline

Justin Tran plans to finish buoyancy and flotation, discuss stability, and introduce Bernoulli’s equation.
Thursday’s campuswide evacuation drill at 5:00 PM will shorten the in-person lecture by about 15 minutes.
Homework 2 is due Sunday and is longer than Homework 1; Problem 3 is a former exam problem that can help indicate exam difficulty.

11:19 Homework 2 Problem 2: Effective Area on the L-Shaped Gate

For the horizontal gate plate, water pressure acts on both its top and bottom, so most opposing forces cancel.
The uncancelled horizontal-plate force acts only over the area covering the air pipe; do not use the full plate area.
In Part B, the vertical section—not the horizontal section—is angled at 45°, with water on the gate’s left side.
For the angled section, adjust the hydrostatic force magnitude, area, and force location.

16:41 Buoyant Force Versus Object Weight

For a fully submerged object, buoyant force is the surrounding-fluid density times gravity times object volume: ρfluid gV.
Object weight is calculated using the object’s own density: ρobject gV.
The net vertical force is gV(ρfluid − ρobject), determining whether the object rises, sinks, or is neutrally balanced.

23:38 Partial Submersion Changes Buoyant Volume, Not Total Weight

Buoyancy acts only on the submerged portion, so use submerged volume when an object is only partly in the fluid.
Weight still uses the object’s total volume, whether or not the object is fully submerged.
Air also produces buoyancy, but its effect is usually negligible compared with water for ordinary calculations.

25:29 How Heating Air Makes a Hot-Air Balloon Rise

A burner heats the air inside the balloon, lowering its density relative to the colder surrounding air.
The upward buoyant force depends on cold-air density, while the hot air contributes to the balloon’s weight.
With buoyancy greater than weight, the balloon rises; for a tethered balloon, rope tension balances the net upward force.

31:12 Life-Jacket Design Requirements and Material Properties

The design example uses a minimum required supported load of 22 lb under the stated U.S. Coast Guard requirement.
The foam has a specific weight of 2.0 lb/ft³, while water’s specific weight is 64.0 lb/ft³.
The jacket’s non-foam components add 1.3 lb, so the foam must support a total downward load of 23.3 lb.

36:15 Calculating the Required Life-Jacket Foam Volume

The free-body diagram includes upward buoyancy and downward forces from the foam, the 22 lb load, and 1.3 lb of other components.
Setting the vertical forces to zero gives (64 − 2)V = 23.3, using water and foam specific weights in lb/ft³.
The required foam volume is 23.3/62 ≈ 0.376 ft³, assuming the jacket is fully submerged.

42:05 Floating-Body Stability and What Happens After a Disturbance

A floating object can be in equilibrium yet still be unstable: a small breeze or rotation may make it flip.
Stability depends on the relative positions of the object’s center of gravity and center of buoyancy.
A restoring moment returns a disturbed object toward its original orientation; an overturning moment drives it farther away.

45:56 Center of Gravity, Center of Buoyancy, and Restoring Moments

The center of gravity is the effective location of the object’s weight; for uniform-density objects it coincides with the centroid.
The center of buoyancy is the centroid of the submerged volume, where the buoyant force acts.
With heavy material low in the object, the weight acts below the buoyant force; a slight rotation creates a restoring moment and stable equilibrium.

52:39 Top-Heavy Floating Objects and Boat Stability

When an object’s center of gravity is above its center of buoyancy, a small disturbance can create a moment that continues the rotation.
The object may capsize until it reaches a more stable orientation, even though its undisturbed position initially had no net moment.
Boat design promotes stability by placing engines and other heavy equipment low in the hull, helping the boat recover in choppy water.

56:17 Bernoulli’s Equation: The Shift from Static to Moving Fluids

Bernoulli’s equation begins the course’s fluid-dynamics unit, where fluids have velocity and kinetic energy.
Applications include airflow over an airfoil, environmental wind patterns, and flow through a converging nozzle.
The equation provides an initial tool for relating fluid velocity and pressure; pressure can also drive fluid from one region to another.
The unit introduces conservation of mass alongside the energy relationship.

1:03:57 Streamlines Trace Fluid-Particle Paths

A streamline represents the path a fluid particle follows as it moves with the flow, such as air passing over an airfoil.
A particle stays on its streamline rather than jumping to a neighboring one.
At each point, a streamline is tangent to the local fluid-velocity direction.
Bernoulli’s equation is applied between locations connected along the same streamline.

1:08:36 Bernoulli’s Equation Links Pressure, Speed, and Height

Between two points on a streamline, P₁ + ½ρu₁² + ρgh₁ = P₂ + ½ρu₂² + ρgh₂.
The terms represent pressure energy, kinetic energy per unit volume, and gravitational potential energy per unit volume.
The equation states that their sum remains constant along the streamline under its required assumptions.

1:14:26 Energy Conversion in Bernoulli Flow and the Hydrostatic Limit

Fluid energy can shift between pressure, kinetic, and gravitational forms while their total remains constant when no external work is added.
A falling solid converts gravitational potential energy into kinetic energy; Bernoulli describes an analogous balance in flowing fluid.
Setting velocity to zero removes the kinetic term and recovers the pressure–height relationship used for hydrostatics.

1:17:17 The Four Conditions for Applying Bernoulli’s Equation

Apply the equation along a streamline, not between arbitrary points that are unconnected by the flow.
Assume incompressible flow, meaning fluid density is constant, and steady flow, meaning conditions do not change with time.
Assume inviscid flow, so viscosity effects are negligible; Tran identifies this as a frequently difficult assumption in practical flows.
Steady does not mean static: a fluid can move with constant conditions over time.

1:21:49 Wind Stagnation Pressure and the Lecture’s Closing Notes

For wind approaching a face, compare an upstream point with the stagnation point at the face; at equal height, the velocity there falls to zero.
Bernoulli converts upstream kinetic energy into pressure, giving stagnation pressure as upstream pressure plus ½ρV².
Tran illustrates the application with a student project that simulated 100 mph wind on a professor’s AI-generated CAD face model.
The lecture ends with a preview of further Bernoulli examples for Thursday, followed by brief post-class administrative discussion.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Justin Tran.