Chapter 14 — Flashcards

Flashcards — Chapter 14: Fluids 🃏

Fluids in motion and at rest — Master pressure, buoyancy, Bernoulli, viscosity, and flow. Test yourself! 💧

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📋 Flashcard Topics: Definitions, Equations & Concepts

§14.1–14.2: Pressure & Depth

Card 1: What is pressure? Pressure is force per unit area: \( P = F/A \). Units: Pascals (Pa) = N/m². In fluids, pressure acts equally in all directions.

Card 2: How does pressure change with depth in a static fluid? Pressure increases linearly with depth: \( P = P_0 + \rho g h \). At 10 m depth in water (ρ = 1000 kg/m³), pressure is roughly 1 atm higher than at the surface.

Card 3: What is the difference between gauge pressure and absolute pressure? Gauge pressure = pressure above atmospheric (P_gauge = P_absolute − P_atm). Absolute pressure includes atmospheric contribution. A tire gauge reads gauge pressure.

Card 4: Why does pressure at the bottom of a dam depend on depth but not the shape of the reservoir? Only depth matters for pressure (P = P₀ + ρgh). The shape determines total force on the dam (F = P_avg × Area), but pressure at any depth is the same regardless of reservoir width or length.

§14.4: Archimedes & Buoyancy

Card 5: State Archimedes' principle. The buoyant force on an object immersed in a fluid equals the weight of the fluid displaced: \( F_b = \rho_{fluid} \cdot g \cdot V_{displaced} \). Acts upward.

Card 6: When does an object float? An object floats when its weight equals the buoyant force. This happens when the average density of the object is less than the fluid density. Condition: \( \rho_{object} < \rho_{fluid} \).

Card 7: A 2 kg block of wood floats in water. Why does it float, and what fraction of its volume is submerged? Wood floats because ρ_wood < ρ_water. At equilibrium, Weight = Buoyant force: \( m_{wood} \cdot g = \rho_{water} \cdot g \cdot V_{submerged} \). If ρ_wood ≈ 600 kg/m³ and ρ_water = 1000 kg/m³, then V_submerged/V_total ≈ 0.6, so 60% is underwater.

Card 8: A ship made of steel (ρ_steel ≈ 7800 kg/m³) floats because it has an air cavity inside. Explain. The ship's average density (steel hull + air inside) is less than water. The large air cavity lowers the overall density below 1000 kg/m³. Buoyant force on the displaced water volume exceeds the weight.

§14.5: Continuity Equation & Conservation of Mass

Card 9: State the continuity equation for an incompressible fluid. \( A_1 v_1 = A_2 v_2 \). The product of cross-sectional area and flow speed is constant. Narrower pipes → faster flow.

Card 10: Water flows through a pipe that narrows from diameter 10 cm to 5 cm. If the speed in the wide section is 2 m/s, what is the speed in the narrow section? Use continuity: \( A_1 v_1 = A_2 v_2 \). Area ratio: \( (10/5)^2 = 4 \). So \( v_2 = v_1 \times 4 = 8 \) m/s. Narrower = faster.

Card 11: Why does water speed up when it exits a nozzle attached to a hose? The nozzle reduces the cross-sectional area. By continuity (A₁v₁ = A₂v₂), when A decreases, v must increase to conserve the volume flow rate.

§14.6: Bernoulli's Principle & Energy Conservation

Card 12: State Bernoulli's equation. \( P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant} \) along a streamline. Sum of pressure, kinetic energy density, and gravitational potential energy density is conserved.

Card 13: Use Bernoulli's equation to explain why pressure is lower where fluid moves faster. In Bernoulli's equation, if v increases, then P must decrease to keep the sum constant. High-speed regions have low pressure. This is the basis of lift on aircraft wings and how spray bottles work.

Card 14: A Venturi tube narrows from 4 cm to 2 cm diameter. If pressure in the wide section is 100 kPa and flow speed is 3 m/s, estimate pressure in the narrow section. (Use ρ = 1000 kg/m³; neglect height difference.) Speed ratio by continuity: \( (4/2)^2 = 4 \), so \( v_2 = 12 \) m/s. Bernoulli (no height change): \( P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \). \( 100000 + 0.5(1000)(9) = P_2 + 0.5(1000)(144) \). \( 100000 + 4500 = P_2 + 72000 \), so \( P_2 ≈ 32.5 \) kPa. Pressure drops in the narrow section.

Card 15: How does Bernoulli's principle explain why a curve ball curves in baseball? The spinning ball drags air on one side faster than the other (Magnus effect). High-speed air has low pressure; low-speed air has high pressure. Pressure imbalance pushes the ball toward the low-pressure region.

§14.7: Viscosity & Drag

Card 16: Define viscosity. Viscosity η is a fluid's resistance to flow. Higher viscosity = thicker, more sluggish fluid (honey vs. water). Units: Pa·s (pascal-seconds). Temperature affects viscosity: liquids thin when heated, gases thicken.

Card 17: What is Stokes drag force, and when does it apply? \( F_d = 6\pi\eta r v \). Applies to small objects moving slowly through viscous fluids (low Reynolds number). Drag is proportional to velocity, not velocity squared. Example: falling raindrops reach terminal velocity in ~1 second.

Card 18: A small sphere (radius 1 mm) falls through oil. The drag force is 6π×η×r×v. If the sphere reaches terminal velocity (Stokes regime), what is the balance of forces? At terminal velocity, drag balances weight: \( mg = 6\pi\eta r v_{terminal} \). Acceleration becomes zero. Small objects reach terminal velocity quickly; large objects do not (they remain accelerating until much higher speeds).

Card 19: Why is viscosity higher in some fluids than others? Viscosity arises from intermolecular forces. Molecules in thicker fluids (honey, oil) have strong interactions and move past each other slowly. Thin fluids (water, alcohol) have weaker interactions. Temperature increases molecular motion, reducing viscosity of liquids but increasing viscosity of gases.

§14.3 & General: Measuring Pressure

Card 20: What is a barometer, and why does mercury rise to ~760 mm in a tube? A barometer measures atmospheric pressure. Mercury (ρ = 13,600 kg/m³) in a sealed tube above a reservoir: P_atm = ρ_Hg × g × h. At sea level, P_atm ≈ 101,325 Pa, so h ≈ 760 mm. The mercury column is held up by atmospheric pressure pushing on the reservoir.

Card 21: A manometer is a U-tube filled with water. If one end is open to air and the other connects to a low-pressure system, the water level rises on the open side. Why? Atmospheric pressure pushes on the open water surface (P_atm > P_low). The pressure difference ΔP = P_atm − P_low lifts the water column to height h where ρgh = ΔP. Used to measure gauge pressure.

Problem-Solving Scenarios

Card 22: A swimming pool is 2 m deep filled with fresh water (ρ = 1000 kg/m³). Calculate the force on the pool floor if its area is 50 m². At depth h = 2 m, pressure is \( P = P_0 + \rho g h = 101,325 + 1000(10)(2) = 121,325 \) Pa. Force on floor = \( P \times A = 121,325 \times 50 = 6,066,250 \) N ≈ 6.1 MN. (Huge force, which is why pools need strong construction.)

Card 23: An aircraft wing is designed so that air flows faster over the upper surface. Using Bernoulli's principle, explain how this generates lift. Fast air over the top → low pressure (P_top). Slower air under the wing → high pressure (P_bottom). Pressure difference (P_bottom − P_top) acts upward. This pressure difference integrated over wing area = lift. This is how wings generate lift at any speed.

Card 24: Why do submarines need thicker hulls at greater depths, and what is the limit? Pressure increases as \( P = P_0 + \rho g h \). At 1 km depth, P ≈ 100 atm. Hull must withstand enormous force. At ~1000 m, even titanium hulls (strongest practical materials) approach their yield strength. The deepest submarine ever (Trieste, 1960) reached ~11 km depth, but only for extreme-depth exploration, not regular operation.

Key Concepts Summary

What's Next?

👉 §14.7 — Practice problems (worked examples) 👉 §14.8 — Exercises (computational & conceptual) 👉 §14.9 — Real-world applications (dams, pumps, aircraft, plumbing)

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