Fluid Mechanics

Question 1 (10)

(a) Consider a small element, area $A$, thickness $dz$, in a fluid at rest. Vertical force balance: pressure up on bottom, down on top, weight down. $p A - (p+dp)A - \rho g A\,dz = 0$ (2) $\Rightarrow -dp\,A - \rho g A\,dz = 0 \Rightarrow \dfrac{dp}{dz} = -\rho g$ (z upward). (2) Integrating from surface ($p_0$) down a depth $h$ (so $z=-h$): $p = p_0 + \rho g h$. (1)

(b) Cube side $a$, top face at depth $d$. Pressure on top: $p_{top}=p_0+\rho g d$; on bottom: $p_{bot}=p_0+\rho g(d+a)$. (2) Side forces cancel by symmetry. Net upward force: $F_B=(p_{bot}-p_{top})a^2=\rho g a\cdot a^2=\rho g a^3=\rho g V.$ (2) This equals weight of displaced fluid → Archimedes' principle. (1)

Question 2 (12)

(a) Element mass $dm=\rho\,dA\,ds$. Newton's 2nd law along streamline $s$: $dm\,\frac{dv}{dt}= -\frac{\partial p}{\partial s}\,dA\,ds - \rho g\,dA\,ds\,\frac{dz}{ds}$ (2) Weight component along $s$ is $-\rho g\,dA\,ds\,(dz/ds)$ (assumption: gravity only body force). (1) Steady flow: $\dfrac{dv}{dt}=v\dfrac{dv}{ds}$ (assumption: steady, so local $\partial/\partial t=0$). (1) Divide by $\rho\,dA\,ds$: $v\frac{dv}{ds}=-\frac1\rho\frac{dp}{ds}-g\frac{dz}{ds}$ (1) $\frac{d}{ds}\Big(\tfrac12 v^2+\tfrac{p}{\rho}+gz\Big)=0 \quad(\rho=\text{const, incompressible})$ (1) Integrate along streamline (inviscid → no friction term): $p+\tfrac12\rho v^2+\rho g z=\text{const}.$ (2)

(b) (1 each): (4)

• Steady — violated by pulsating/unsteady flow (e.g. water hammer).
• Inviscid — violated near walls / boundary layers (viscous drag).
• Incompressible — violated by high-Mach gas flow.
• Along a streamline (irrotational needed to cross streamlines) — violated across a rotational shear flow / vortex.

Question 3 (10)

(a) Continuity: $A_1v_1=A_2v_2 \Rightarrow v_1=(A_2/A_1)v_2$. (1) Bernoulli (horizontal): $p_1+\tfrac12\rho v_1^2=p_2+\tfrac12\rho v_2^2$. (1) $\Delta p=\tfrac12\rho v_2^2\big(1-(A_2/A_1)^2\big)\Rightarrow v_2=\sqrt{\frac{2\Delta p/\rho}{1-(A_2/A_1)^2}}$ (1) $Q=A_2v_2=A_2\sqrt{\dfrac{2\Delta p/\rho}{1-(A_2/A_1)^2}}$. (1)

(b) Manometer: $\Delta p=(\rho_m-\rho)g\Delta h=(13600-1000)(9.81)(0.20)$ $=12600\times9.81\times0.20=24721.2\ \mathrm{Pa}$. (2)

Areas: $A_2=\tfrac{\pi}{4}(0.05)^2=1.9635\times10^{-3}\ \mathrm{m^2}$. $(A_2/A_1)=(D_2/D_1)^2=(0.5)^2=0.25 \Rightarrow (A_2/A_1)^2=0.0625$. (1) $v_2=\sqrt{\frac{2(24721.2)/1000}{1-0.0625}}=\sqrt{\frac{49.4424}{0.9375}}=\sqrt{52.7385}=7.262\ \mathrm{m/s}$ (2) $Q=1.9635\times10^{-3}\times7.262=0.01426\ \mathrm{m^3/s}\approx1.43\times10^{-2}\ \mathrm{m^3/s}.$ (1)

Question 4 (10)

(a) Coaxial cylinder radius $r$, length $L$. Pressure force drives, viscous shear resists. $\Delta p\,(\pi r^2)=\tau(2\pi r L),\quad \tau=-\mu\frac{du}{dr}$ (2) $\frac{du}{dr}=-\frac{\Delta p}{2\mu L}r$ (2) Integrate: $u=-\dfrac{\Delta p}{4\mu L}r^2+C$; no-slip $u(R)=0\Rightarrow C=\dfrac{\Delta p}{4\mu L}R^2$. $u(r)=\frac{\Delta p}{4\mu L}(R^2-r^2).$ (2)

(b) $Q=\int_0^R u(r)\,2\pi r\,dr=\frac{2\pi\Delta p}{4\mu L}\int_0^R(R^2-r^2)r\,dr$ (2) $=\frac{\pi\Delta p}{2\mu L}\Big[\frac{R^2r^2}{2}-\frac{r^4}{4}\Big]_0^R=\frac{\pi\Delta p}{2\mu L}\cdot\frac{R^4}{4}=\frac{\pi R^4\Delta p}{8\mu L}.$ (2)

Question 5 (10)

(a) Variables: $F,\ V,\ D,\ \rho,\ \mu$ → $n=5$. Fundamental dims $M,L,T$ → $k=3$. $\pi$ theorem: $n-k=2$ dimensionless groups. (2) Choosing repeating variables $\rho,V,D$: $\pi_1=\frac{F}{\rho V^2 D^2}\ (\text{drag coefficient}),\qquad \pi_2=\frac{\rho V D}{\mu}\ (\text{Reynolds number}).$ (3) Thus $C_D=f(Re)$. (1)

(b) Same fluid ($\rho,\mu$ same). Reynolds similarity: $Re_m=Re_p$: $V_m D_m=V_p D_p \Rightarrow V_m=V_p\frac{D_p}{D_m}=30\times5=150\ \mathrm{m/s}.$ (4)

Question 6 (8)

(a) Prandtl: at high $Re$, viscous effects are confined to a thin boundary layer near the surface; outside it flow is effectively inviscid. (2) Separation: within the boundary layer momentum is already reduced by friction. An adverse pressure gradient ($dp/dx>0$) decelerates fluid further; near the wall the low-momentum fluid is brought to rest and reverses, forming the separation point where $\partial u/\partial y|_{wall}=0$. Without adverse gradient the near-wall fluid keeps enough momentum to stay attached. (2)

(b) Kutta–Joukowski: lift per unit span $L=\rho V_\infty\Gamma$, with $\Gamma$ the circulation. (2) A spinning cylinder drags fluid around it (viscosity), creating circulation $\Gamma$. Superposed on the free stream, velocity is higher on one side and lower on the other; by Bernoulli the higher-speed side has lower pressure → net transverse force (Magnus effect). (2)

[
  {"claim":"Venturi Delta p = 24721.2 Pa","code":"dp=(13600-1000)*9.81*0.20; result = abs(dp-24721.2)<1e-1"},
  {"claim":"Venturi throat velocity approx 7.262 m/s","code":"dp=(13600-1000)*9.81*0.20; ratio=(0.5**2)**2; v2=sqrt(2*dp/1000/(1-ratio)); result = abs(v2-7.262)<1e-2"},
  {"claim":"Venturi Q approx 0.01426 m^3/s","code":"import sympy as sp; dp=(13600-1000)*9.81*0.20; ratio=(0.5**2)**2; v2=sqrt(2*dp/1000/(1-ratio)); A2=sp.pi/4*0.05**2; Q=float(A2*v2); result = abs(Q-0.01426)<1e-4"},
  {"claim":"Poiseuille integral gives pi R^4 dp/(8 mu L)","code":"import sympy as sp; r,R,dp,mu,L=sp.symbols('r R dp mu L',positive=True); u=dp/(4*mu*L)*(R**2-r**2); Q=sp.integrate(u*2*sp.pi*r,(r,0,R)); result = sp.simplify(Q-sp.pi*R**4*dp/(8*mu*L))==0"},
  {"claim":"Model test speed 150 m/s for same-fluid Re similarity","code":"Vm=30*5; result = Vm==150"}
]