Fluid Mechanics

Q1 (12)

(a) Efflux velocity: $v = C_d\sqrt{2gy}$. (1) Outflow: $Q_{out} = A_0 v = C_d A_0 \sqrt{2gy}$. (2)

(b) Free-surface area at depth $y$: $r=0.4y$, so $A_s = \pi r^2 = \pi(0.4y)^2 = 0.16\pi y^2$. (1) Continuity (mass conservation): rate of volume loss = outflow: $A_s\frac{dy}{dt} = -Q_{out}\Rightarrow 0.16\pi y^2 \frac{dy}{dt} = -C_d A_0\sqrt{2g}\,\sqrt y.$ (2) $\Rightarrow \frac{dy}{dt} = -\frac{C_d A_0\sqrt{2g}}{0.16\pi}\,y^{-3/2}.$ (1)

(c) Separate: $y^{3/2}dy = -k\,dt$ with $k = \dfrac{C_d A_0\sqrt{2g}}{0.16\pi}$. (1) Integrate $0\to H$: $t = \frac{1}{k}\int_0^H y^{3/2}dy = \frac{2H^{5/2}}{5k}.$ (2) Numbers: $A_0 = 8\times10^{-6}\ \mathrm{m^2}$, $\sqrt{2g}=\sqrt{19.62}=4.429$. $k = \dfrac{0.62\times8\times10^{-6}\times4.429}{0.16\pi} = \dfrac{2.1966\times10^{-5}}{0.50265}=4.370\times10^{-5}$. (1) $H^{5/2}=0.15^{2.5}=0.008714$. $t = \frac{2\times0.008714}{5\times4.370\times10^{-5}} = \frac{0.017428}{2.185\times10^{-4}} \approx 79.8\ \mathrm{s}.$ (1)

Q2 (14)

(a) Uniform stream $\psi_1 = Uy$ plus a doublet $\psi_2 = -Ua^2 y/(x^2+y^2)$. Superposition → flow past a circular cylinder of radius $a$. (3)

(b) In polar-free Cartesian form, $u = \partial\psi/\partial y$, $v = -\partial\psi/\partial x$. $u = U\left(1 - a^2\frac{x^2 - y^2}{(x^2+y^2)^2}\right),\quad v = -U\,a^2\frac{2xy}{(x^2+y^2)^2}.$ (3) Irrotationality: $\omega_z = \partial v/\partial x - \partial u/\partial y$. Since $\psi$ satisfies Laplace's equation ($\nabla^2\psi=0$ for both uniform stream and doublet), $\omega_z = -\nabla^2\psi = 0$. Explicit check confirms $\partial v/\partial x=\partial u/\partial y$. (2)

(c) Stagnation on the $x$-axis ($y=0$): $v=0$ automatically; set $u=0$: $U(1 - a^2/x^2)=0 \Rightarrow x = \pm a = \pm0.5\ \mathrm{m}$, $y=0$. (3)

(d) $\psi=0$ contains the $x$-axis and the circle $x^2+y^2=a^2$, i.e. the cylinder surface (a streamline / solid boundary). $a$ is the cylinder radius; the doublet strength $Ua^2$ is fixed so the body radius is $a=0.5\ \mathrm m$. (3)

Q3 (12)

(a) Reynolds-number similarity: $Re_m = Re_p$. Viscous forces dominate for a deeply submerged body (no free surface → Froude number irrelevant), so matching $Re$ ensures dynamic similarity. (2)

(b) $\dfrac{\rho_m V_m L_m}{\mu_m} = \dfrac{\rho_p V_p L_p}{\mu_p}$, with $L_m = L_p/20$. $V_m = V_p\frac{\rho_p}{\rho_m}\frac{\mu_m}{\mu_p}\frac{L_p}{L_m} = 6\cdot\frac{1025}{998}\cdot\frac{1.00\times10^{-3}}{1.07\times10^{-3}}\cdot 20.$ (2) $= 6\times1.02705\times0.93458\times20 = 115.2\ \mathrm{m/s}.$ (2)

(c) Equal drag coefficient: $\dfrac{F}{\tfrac12\rho V^2 L^2}$ equal. $F_p = F_m\frac{\rho_p V_p^2 L_p^2}{\rho_m V_m^2 L_m^2}.$ (1) Ratios: $\rho_p/\rho_m=1.02705$; $(V_p/V_m)^2=(6/115.2)^2=2.713\times10^{-3}$; $(L_p/L_m)^2=400$. $F_p = 210\times1.02705\times2.713\times10^{-3}\times400 \approx 234\ \mathrm{N}.$ (3)

(d) $V_m \approx 115\ \mathrm{m/s}$ is impractically high in water — cavitation and enormous power/structural loads make true $Re$ matching hard (hence real tests relax it or use scaling corrections). (2)

Q4 (12)

(a) Poiseuille profile: $u(r) = \frac{\Delta p}{4\mu L}(R^2 - r^2).$ (2) Centreline max ($r=0$): $u_{max} = \frac{\Delta p\,R^2}{4\mu L} = \frac{800\times(0.005)^2}{4\times10^{-3}\times2} = \frac{800\times2.5\times10^{-5}}{8\times10^{-3}} = 2.5\ \mathrm{m/s}.$ (2)

(b) $Q = \dfrac{\pi\Delta p R^4}{8\mu L} = \dfrac{\pi\times800\times(0.005)^4}{8\times10^{-3}\times2}$. $(0.005)^4=6.25\times10^{-10}$; numerator $=\pi\times800\times6.25\times10^{-10}=1.5708\times10^{-6}$; denom $=1.6\times10^{-2}$. $Q = 9.82\times10^{-5}\ \mathrm{m^3/s}.$ (2) $\bar V = Q/(\pi R^2) = 9.82\times10^{-5}/(\pi\times2.5\times10^{-5}) = 1.25\ \mathrm{m/s}$ (= $u_{max}/2$, as expected). (2)

(c) $Re = \dfrac{\rho \bar V (2R)}{\mu} = \dfrac{1000\times1.25\times0.01}{10^{-3}} = 12500$. This exceeds $\sim2300$, so the assumption of laminar flow is in fact inconsistent — the flow would be turbulent. (Full marks for correctly computing $Re\approx1.25\times10^4$ and noting the inconsistency.) (2)

(d) Force balance on fluid cylinder: $\Delta p\,\pi R^2 = \tau_w\,2\pi R L$ $\tau_w = \frac{\Delta p\,R}{2L} = \frac{800\times0.005}{2\times2} = 1.0\ \mathrm{Pa}.$ (2)

Q5 (10)

(a) $Re_L = \dfrac{U_\infty L}{\nu} = \dfrac{4\times1.2}{1.5\times10^{-5}} = 3.2\times10^{5}$. (2) $\delta_L = \dfrac{5.0\,L}{\sqrt{Re_L}} = \dfrac{5.0\times1.2}{\sqrt{3.2\times10^5}} = \dfrac{6.0}{565.7} = 0.01061\ \mathrm{m} \approx 10.6\ \mathrm{mm}.$ (2)

(b) $C_f = 1.328/\sqrt{3.2\times10^5} = 1.328/565.7 = 2.347\times10^{-3}$ (one side). (1) Drag per side $= C_f\cdot\tfrac12\rho U_\infty^2\cdot(bL)$. $\tfrac12\rho U_\infty^2 = \tfrac12\times1.2\times16 = 9.6\ \mathrm{Pa}$; area $bL=0.5\times1.2=0.6\ \mathrm{m^2}$. One side $= 2.347\times10^{-3}\times9.6\times0.6 = 0.01352\ \mathrm{N}$. (2) Both sides: $F = 2\times0.01352 = 0.0270\ \mathrm{N}$. (1)

(c) $\delta^*$ measures the deficit-equivalent displacement of the outer streamlines caused by momentum loss near the wall; it weights the velocity deficit $(1-u/U)$, which is small over most of $\delta$, so $\delta^* \approx \delta/3$ (Blasius) — always less than the full thickness $\delta$. (2)

[
 {"claim":"Q1 drain time ≈ 79.8 s","code":"Cd=0.62; A0=8e-6; g=9.81; H=Rational(15,100); k=Cd*A0*sqrt(2*g)/(0.16*pi); t=2*H**Rational(5,2)/(5*k); result = abs(float(t)-79.8)<1.0"},
 {"claim":"Q3 model velocity ≈ 115.2 m/s","code":"Vp=6; Vm=Vp*(1025/998)*(1.00e-3/1.07e-3)*20; result = abs(Vm-115.2)<0.5"},
 {"claim":"Q4 Q ≈ 9.82e-5 m^3/s and umax=2.5","code":"dp=800; R=0.005; mu=1e-3; L=2; Q=pi*dp*R**4/(8*mu*L); umax=dp*R**2/(4*mu*L); result = abs(float(Q)-9.82e-5)<1e-6 and abs(float(umax)-2.5)<1e-6"},
 {"claim":"Q5 both-side drag ≈ 0.0270 N","code":"ReL=3.2e5; Cf=1.328/sqrt(ReL); q=0.5*1.2*16; area=0.6; F=2*Cf*q*area; result = abs(float(F)-0.0270)<0.001"}
]