Quizzes · Fluid Mechanics
Question bank organized by difficulty. Every concept page also carries its own quiz — these are integrative questions spanning multiple topics.
Easy
E1 (conceptual). Why does pressure in a lake depend on depth but not on the lake's width?
Answer
Hydrostatic equilibrium is a local balance: \(dP/dy = \rho g\) integrates over depth only. Each column of fluid supports its own weight — neighboring columns push sideways equally and cancel.
E2 (computational). A hydraulic lift has pistons of area 0.01 m² and 1 m². What force on the small piston supports a 1000 kg car on the large one?
Answer
Equal pressure (Pascal): \(F_1 = F_2 A_1/A_2 = 9800 \times 0.01 = 98\) N — about 10 kg-force.
E3 (MCQ). Water flows through a pipe that doubles in diameter. The speed:
- (a) doubles (b) halves (c) quarters (d) quadruples
Answer
(c). Continuity: \(A \propto d^2\) quadruples, so \(v\) drops 4×.
E4 (conceptual). A ship made of steel floats, but a steel ball sinks. Reconcile.
Answer
Floating depends on average density of the displaced-volume envelope (buoyancy). The hull encloses mostly air, so the ship displaces its weight in water long before submerging.
E5 (MCQ). The no-slip condition says that at a stationary solid wall:
- (a) pressure vanishes (b) fluid velocity is zero (c) shear stress is zero (d) vorticity is zero
Answer
(b) — see viscosity. Consequently shear stress and vorticity are typically largest at walls.
Medium
M1 (computational). A wind of 15 m/s blows over a flat warehouse roof of 500 m² (\(\rho_\text{air} = 1.2\ \text{kg/m}^3\)). Estimate the net upward force if the interior is at rest.
Answer
Bernoulli: \(\Delta p = \frac12\rho v^2 = 135\) Pa; \(F = 135 \times 500 = 67.5\) kN — about 6.9 tonnes of lift.
M2 (computational). Oil (\(\mu = 0.1\) Pa·s, \(\rho = 900\ \text{kg/m}^3\)) flows at \(Q = 10^{-4}\ \text{m}^3/\text{s}\) through a 2 cm-diameter, 10 m pipe. Find the pressure drop and verify the flow is laminar.
Answer
Hagen–Poiseuille: \(\Delta P = \frac{8\mu L Q}{\pi R^4} = \frac{8\times0.1\times10\times10^{-4}}{\pi\times(0.01)^4} \approx 25.5\) kPa. Mean speed \(v = Q/A = 0.318\) m/s; \(Re = \rho v d/\mu = 900\times0.318\times0.02/0.1 \approx 57 \ll 2300\). Laminar.
M3 (conceptual). Explain why a spinning bucket of water makes a parabolic surface, not a conical or spherical one.
Answer
Rigid-body rotation: the isobar condition gives \(dz/dr = r\omega^2/g\) — slope growing linearly with \(r\) integrates to \(z \propto r^2\). Centripetal demand grows linearly outward; only a parabola's tilt keeps pace.
M4 (computational). Use the Buckingham Pi theorem to show the drag on a submarine depends on only two dimensionless groups, and name them.
Answer
\(F = \phi(\rho, v, L, \mu)\): \(n = 5\), \(k = 3\) ⇒ 2 groups: \(C_D = F/(\rho v^2 L^2)\) and \(Re = \rho v L/\mu\). Hence \(C_D = f(Re)\).
M5 (MCQ). Across a stationary normal shock, which quantity decreases?
- (a) pressure (b) temperature (c) density (d) flow speed
Answer
(d) — see Rankine–Hugoniot; entropy demands compression and deceleration.
Hard
H1 (computational). A 2 mm-wavelength ripple and a 2 m-wavelength wave both travel on deep water. Which is faster and why? (\(\gamma = 0.074\) N/m)
Answer
Gravity–capillary dispersion (interface relation): \(c^2 = g/k + \gamma k/\rho\). For \(\lambda = 2\) m: \(c \approx 1.77\) m/s (gravity-dominated). For \(\lambda = 2\) mm: \(k = 3142\), \(c^2 = 0.0031 + 0.233\), \(c \approx 0.49\) m/s. The long wave is faster; capillary waves are slow but their speed rises again at even shorter wavelengths (minimum ≈ 0.23 m/s at \(\lambda \approx 1.7\) cm).
H2 (computational). Kolmogorov: a kitchen mixer dissipates 50 W in 0.5 kg of water (\(\nu = 10^{-6}\ \text{m}^2/\text{s}\)). Estimate the smallest eddy size.
Answer
\(\varepsilon = 50/0.5 = 100\ \text{W/kg} = 100\ \text{m}^2/\text{s}^3\); \(\eta = (\nu^3/\varepsilon)^{1/4} = (10^{-18}/100)^{1/4} = 10^{-5}\) m — ten microns. See energy cascade.
H3 (conceptual). The linear Kelvin–Helmholtz growth rate increases without bound as \(k \to \infty\) for an ideal vortex sheet. Give two physical effects that regularize this, and the scaling of each.
Answer
(1) Surface tension: adds \(-\gamma k^3/(\rho_1+\rho_2)\) under the radicand — stabilizes \(k\) above a cutoff. (2) Finite shear-layer thickness \(\delta\) (or viscosity \(\sim \mu k^2\)): modes with \(k\delta \gtrsim 1\) see a smooth profile, not a jump, capping growth near \(k \sim 1/\delta\).
H4 (computational). A supernova releases \(10^{44}\) J into interstellar gas of density \(2\times10^{-21}\ \text{kg/m}^3\). Using Sedov–Taylor, estimate the remnant radius after 1000 years.
Answer
\(t = 3.15\times10^{10}\) s; \(R \sim (Et^2/\rho)^{1/5} = \left(\frac{10^{44}\times9.9\times10^{20}}{2\times10^{-21}}\right)^{1/5} = (5\times10^{85})^{1/5} \approx 2\times10^{17}\) m ≈ 6 pc — the right order for young remnants like Tycho's.
H5 (conceptual). Why does the RANS approach never close, no matter how many moment equations you derive? What does every practical model therefore do?
Answer
The Navier–Stokes nonlinearity is quadratic: the equation for the \(n\)-th moment always contains the \((n{+}1)\)-th (\(\overline{u'u'}\) needs \(\overline{u'u'u'}\), etc.) — an infinite hierarchy. Practical models truncate it by modeling some moment in terms of lower ones (e.g. Boussinesq eddy viscosity for \(\overline{u_i'u_j'}\)), importing empirical constants.