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Level 5 — MasteryFluid Mechanics

Fluid Mechanics

⏱ 2 minutes60 marksprintable — key stays hidden on paper

Time limit: 2 hours 30 minutes Total marks: 60 Instructions: Attempt all THREE questions. Full derivations, dimensional consistency, and (where indicated) a short computational scheme are required. Use g=9.81 m/s2g = 9.81\ \mathrm{m/s^2} unless stated. Show all reasoning.

Question 1 — Boundary Layers: Theory, Integral Method, and Computation (22 marks)

A steady, incompressible, laminar flow of a Newtonian fluid (density ρ\rho, dynamic viscosity μ\mu) passes over a semi-infinite flat plate at zero incidence. Free-stream velocity is U∞U_\infty, and xx is measured from the leading edge.

(a) Starting from Prandtl's boundary-layer scaling arguments, show that the boundary-layer thickness scales as δ(x)∼μxρU∞.\delta(x) \sim \sqrt{\frac{\mu x}{\rho U_\infty}}. State clearly which terms of the Navier–Stokes equations you retain and which you discard, and justify each. (5)

(b) Define the displacement thickness δ∗\delta^* and momentum thickness θ\theta as integrals of the velocity profile. Using the von Kármán momentum integral equation with an assumed cubic profile uU∞=32η−12η3,η=yδ,\frac{u}{U_\infty} = \frac{3}{2}\eta - \frac{1}{2}\eta^3,\qquad \eta = \frac{y}{\delta}, derive expressions for δ(x)\delta(x), θ(x)\theta(x), and the local skin-friction coefficient Cf(x)C_f(x) in terms of the local Reynolds number Rex=ρU∞x/μRe_x = \rho U_\infty x/\mu. (9)

(c) The Blasius exact solution gives δ/x=5.0/Rex\delta/x = 5.0/\sqrt{Re_x} and Cf=0.664/RexC_f = 0.664/\sqrt{Re_x}. Compute the percentage errors of your cubic-profile results in (b) relative to Blasius, and comment on why the integral method performs better for CfC_f than for δ\delta. (4)

(d) Write a short pseudocode / Python-style routine that numerically solves the Blasius ODE 2f′′′+ff′′=0,f(0)=f′(0)=0, f′(∞)=12f''' + f f'' = 0,\qquad f(0)=f'(0)=0,\ f'(\infty)=1 by a shooting method (RK4 + secant on the missing initial slope f′′(0)f''(0)). State the target you iterate on and the expected converged value of f′′(0)f''(0). (4)

Question 2 — Bernoulli, Continuity, and a Venturi with Real-Fluid Correction (18 marks)

(a) Derive Bernoulli's equation from F=ma\mathbf{F}=m\mathbf{a} applied to a fluid element along a streamline. List all four assumptions and identify at exactly which step each assumption enters. (6)

(b) A horizontal Venturi meter has inlet area A1=40 cm2A_1 = 40\ \mathrm{cm^2} and throat area A2=10 cm2A_2 = 10\ \mathrm{cm^2}. Water (ρ=1000 kg/m3\rho = 1000\ \mathrm{kg/m^3}) flows through it. A mercury (ρHg=13600 kg/m3\rho_{Hg} = 13600\ \mathrm{kg/m^3}) differential manometer connected across inlet and throat reads a height difference h=25 cmh = 25\ \mathrm{cm}. Using continuity and Bernoulli, derive the ideal volume flow rate QidealQ_{ideal} and evaluate it numerically. (7)

(c) The measured discharge is Qactual=7.9 L/sQ_{actual} = 7.9\ \mathrm{L/s}. Compute the discharge coefficient CdC_d. Explain physically, in terms of viscous boundary-layer effects and the vena contracta, why Cd<1C_d < 1. (5)

Question 3 — Potential Flow, Circulation, and Lift (20 marks)

Consider 2-D incompressible, irrotational flow. Basic building blocks: uniform stream U∞U_\infty, doublet of strength κ\kappa, and a point vortex of circulation Γ\Gamma.

(a) Define the stream function ψ\psi and velocity potential ϕ\phi. Show that for incompressible and irrotational plane flow both satisfy Laplace's equation, and prove that lines of constant ψ\psi and constant ϕ\phi are orthogonal. (5)

(b) Superpose a uniform stream, a doublet, and a vortex to model flow past a rotating cylinder of radius aa. Write ψ(r,θ)\psi(r,\theta), impose the cylinder-surface condition to relate κ\kappa to U∞U_\infty and aa, and find the stagnation-point locations as a function of Γ\Gamma. State the condition for the stagnation points to leave the cylinder surface. (8)

(c) State and prove the Kutta–Joukowski theorem L′=ρU∞ΓL' = \rho U_\infty \Gamma (lift per unit span) for the cylinder, by integrating pressure (from Bernoulli) around the surface. (5)

(d) State Kelvin's circulation theorem and use it to explain the origin of the starting vortex shed by an aerofoil that suddenly begins to move. (2)

End of paper.

Answer keyMark scheme & solutions

Question 1

(a) Prandtl scaling — 5 marks

  • Inside the boundary layer, inertia ∼ρU∞2/x\sim \rho U_\infty^2/x (convective term ρu ∂u/∂x\rho u\,\partial u/\partial x) balances the dominant viscous term μ ∂2u/∂y2∼μU∞/δ2\mu\,\partial^2 u/\partial y^2 \sim \mu U_\infty/\delta^2. (2)
  • Justification: streamwise diffusion μ ∂2u/∂x2\mu\,\partial^2u/\partial x^2 is negligible vs transverse μ ∂2u/∂y2\mu\,\partial^2u/\partial y^2 because δ≪x\delta \ll x; the yy-momentum equation reduces to ∂p/∂y≈0\partial p/\partial y \approx 0 (pressure impressed from outer flow). (1)
  • Balance: ρU∞2/x∼μU∞/δ2⇒δ2∼μx/(ρU∞)\rho U_\infty^2/x \sim \mu U_\infty/\delta^2 \Rightarrow \delta^2 \sim \mu x/(\rho U_\infty), hence δ∼μx/(ρU∞)=x/Rex\delta \sim \sqrt{\mu x/(\rho U_\infty)} = x/\sqrt{Re_x}. (2)

(b) Von Kármán integral with cubic profile — 9 marks

Definitions (1): δ∗=∫0δ ⁣(1−uU∞)dy,θ=∫0δuU∞(1−uU∞)dy.\delta^* = \int_0^\delta\!\Big(1-\tfrac{u}{U_\infty}\Big)dy,\qquad \theta=\int_0^\delta \tfrac{u}{U_\infty}\Big(1-\tfrac{u}{U_\infty}\Big)dy.

With u/U∞=32η−12η3u/U_\infty = \tfrac32\eta-\tfrac12\eta^3:

  • θ/δ=∫01(32η−12η3)(1−32η+12η3) dη=39280\theta/\delta = \int_0^1 (\tfrac32\eta-\tfrac12\eta^3)(1-\tfrac32\eta+\tfrac12\eta^3)\,d\eta = \dfrac{39}{280}. (2)
  • Wall shear: τw=μ ∂u∂y∣0=μU∞32δ\tau_w = \mu\,\dfrac{\partial u}{\partial y}\Big|_0 = \mu U_\infty \dfrac{3}{2\delta}. (1)

Momentum integral (zero pressure gradient): τw=ρU∞2dθdx\tau_w = \rho U_\infty^2 \dfrac{d\theta}{dx}. (1)

\Rightarrow \delta\,d\delta = \frac{140}{13}\frac{\mu}{\rho U_\infty}dx.$$ Integrate ($\delta(0)=0$): **(2)** $$\delta^2 = \frac{280}{13}\frac{\mu x}{\rho U_\infty}\ \Rightarrow\ \boxed{\frac{\delta}{x}=\sqrt{\frac{280}{13}}\frac{1}{\sqrt{Re_x}}=\frac{4.64}{\sqrt{Re_x}}}.$$ Then $\theta = \tfrac{39}{280}\delta \Rightarrow \theta/x = 0.646/\sqrt{Re_x}$. Skin friction **(2)**: $$C_f=\frac{\tau_w}{\tfrac12\rho U_\infty^2}=\frac{3\mu}{\rho U_\infty\delta}=\frac{3}{4.64}\frac{1}{\sqrt{Re_x}}=\boxed{\frac{0.646}{\sqrt{Re_x}}}.$$ ### (c) Errors vs Blasius — 4 marks - $\delta$: $4.64$ vs $5.0 \Rightarrow$ error $= (5.0-4.64)/5.0 = 7.2\%$ (under-predicts). **(1.5)** - $C_f$: $0.646$ vs $0.664 \Rightarrow$ error $= (0.664-0.646)/0.664 = 2.7\%$. **(1.5)** - Comment: $\delta$ is sensitive to the outer-edge tail where any assumed profile is least accurate (definition of $\delta$ is somewhat arbitrary), whereas $C_f$ depends on the near-wall slope and momentum thickness which the cubic captures well; integral quantities average out profile error. **(1)** ### (d) Shooting method — 4 marks ``` Set s = f''(0) guess. Integrate 2f'''+ff''=0 as system: y1=f, y2=f', y3=f'' y1'=y2, y2'=y3, y3'=-0.5*y1*y3 Use RK4 from eta=0 to eta_max(~10), IC (0,0,s). Target: F(s) = y2(eta_max) - 1 = 0 (i.e. f'(inf)=1). Secant iterate on s until |F(s)|<tol. ``` - Correct system + BC target stated. **(3)** - Converged value $f''(0) \approx 0.3321$. **(1)** --- ## Question 2 ### (a) Bernoulli derivation — 6 marks Element of length $ds$, cross-section $dA$ along streamline; forces: pressure $[p-(p+dp)]dA = -dp\,dA$, gravity component $-\rho g\,dA\,ds\,(dz/ds)$. **(2)** Newton's law $= \rho\,dA\,ds\, a_s$, with steady acceleration $a_s = v\,dv/ds$ **(← "steady" assumption enters here)**. **(1)** $$-dp\,dA - \rho g\,dA\,dz = \rho\,dA\,ds\, v\frac{dv}{ds}$$ Divide by $dA$, use $\rho=$const **(← incompressible)** and neglect friction **(← inviscid)**; integrate along a streamline **(← along-streamline)**: **(2)** $$p+\tfrac12\rho v^2+\rho g z = \text{const}.$$ All four assumptions (steady, inviscid, incompressible, along a streamline) correctly located. **(1)** ### (b) Venturi ideal flow — 7 marks Continuity: $A_1v_1=A_2v_2 \Rightarrow v_1=(A_2/A_1)v_2$. **(1)** Bernoulli (horizontal): $p_1-p_2 = \tfrac12\rho(v_2^2-v_1^2)$. **(1)** Manometer: $p_1-p_2=(\rho_{Hg}-\rho)g h$. **(1)** $$v_2=\sqrt{\frac{2(\rho_{Hg}-\rho)gh}{\rho\,[1-(A_2/A_1)^2]}}.$$ **(1)** Numbers: $(\rho_{Hg}-\rho)gh = 12600\times9.81\times0.25 = 30901.5\ \mathrm{Pa}$. $(A_2/A_1)^2=(10/40)^2=0.0625$; $1-0.0625=0.9375$. $v_2=\sqrt{2\times30901.5/(1000\times0.9375)}=\sqrt{65.92}=8.119\ \mathrm{m/s}$. **(2)** $Q_{ideal}=A_2 v_2 = 10\times10^{-4}\times8.119 = 8.12\times10^{-3}\ \mathrm{m^3/s}=8.12\ \mathrm{L/s}$. **(1)** ### (c) Discharge coefficient — 5 marks $C_d = Q_{actual}/Q_{ideal} = 7.9/8.12 = 0.973$. **(2)** Physical explanation **(3)**: real flow has (i) viscous losses in boundary layers reducing effective velocity/energy, (ii) a **vena contracta** — the effective throat area is smaller than geometric $A_2$, and (iii) non-uniform velocity profile. Hence measured $Q$ is below ideal, giving $C_d<1$ (typically 0.95–0.98 for Venturis, higher than orifice plates because smooth contraction minimises separation). --- ## Question 3 ### (a) Laplace + orthogonality — 5 marks $u=\partial\psi/\partial y$, $v=-\partial\psi/\partial x$ (satisfies continuity identically). $u=\partial\phi/\partial x$, $v=\partial\phi/\partial y$. **(1)** Irrotational $\Rightarrow \partial v/\partial x-\partial u/\partial y=0 \Rightarrow \nabla^2\psi=0$. Incompressible $\Rightarrow \nabla^2\phi=0$. **(2)** Orthogonality: $\nabla\phi=(u,v)$, $\nabla\psi=(-v,u)$; dot product $=-uv+vu=0$, so equipotentials ⟂ streamlines. **(2)** ### (b) Rotating cylinder — 8 marks $$\psi = U_\infty r\sin\theta\Big(1-\frac{a^2}{r^2}\Big) - \frac{\Gamma}{2\pi}\ln\frac{r}{a}.$$ Here doublet strength $\kappa=U_\infty a^2$ makes $r=a$ a streamline. **(2 for form, 1 for κ relation)** Velocities: $v_r=\tfrac1r\partial\psi/\partial\theta$, $v_\theta=-\partial\psi/\partial r$. On $r=a$: $v_r=0$, and $$v_\theta|_{r=a} = -2U_\infty\sin\theta - \frac{\Gamma}{2\pi a}.$$ **(2)** Stagnation ($v_\theta=0$): $\sin\theta_s = -\dfrac{\Gamma}{4\pi a U_\infty}$. **(2)** Two surface stagnation points exist while $|\Gamma|\le 4\pi a U_\infty$. When $|\Gamma|>4\pi a U_\infty$ the stagnation point **leaves the surface** and moves into the fluid. **(1)** ### (c) Kutta–Joukowski — 5 marks Surface speed $v_\theta=-2U_\infty\sin\theta-\Gamma/2\pi a$. Bernoulli: $p=p_\infty+\tfrac12\rho U_\infty^2-\tfrac12\rho v_\theta^2$. **(1)** Lift $=-\oint p\,\sin\theta\, a\,d\theta$ (component). Only the cross term survives integration: **(2)** $$L' = -\int_0^{2\pi}\!\Big[-\tfrac12\rho\big(2U_\infty\sin\theta\big)\big(\tfrac{\Gamma}{2\pi a}\big)2\Big]\sin\theta\, a\,d\theta$$ Constant and $\sin^2\theta\to$const terms integrate to zero except the $\sin^2\theta$ cross term giving $$\boxed{L' = \rho U_\infty \Gamma}.$$ Drag $=0$ (d'Alembert). **(2)** ### (d) Kelvin's theorem — 2 marks $\dfrac{D\Gamma}{Dt}=0$ for a material circuit in inviscid, barotropic flow with conservative body forces. Initially $\Gamma=0$ around a large circuit enclosing the aerofoil; when the aerofoil starts and develops bound circulation $\Gamma_b$ (to satisfy Kutta condition), an equal-and-opposite **starting vortex** $-\Gamma_b$ is shed downstream so total remains zero. **(2)** --- ```verify [ {"claim":"Cubic profile momentum thickness ratio theta/delta = 39/280","code": "import sympy as sp\neta=sp.symbols('eta')\nu=sp.