Compressible Flow & Aerodynamics

Q1. (6 marks)

• (a) $T_0$ = temperature the flowing gas would attain if brought to rest adiabatically (isentropically); it accounts for both static temperature and kinetic energy: $T_0 = T + V^2/(2c_p)$. (2)
• (b) Mach number $M = V/a$, ratio of local flow speed to local speed of sound. (2)
• (c) Critical Mach number = the free-stream Mach number at which the flow first reaches $M=1$ locally somewhere on the airfoil surface. (2)

Q2. (4 marks)

• $a = \sqrt{\gamma R T} = \sqrt{1.4 \times 287 \times 250} = \sqrt{100450} = 316.9\ \mathrm{m/s}$. (2)
• $M = V/a = 340/316.9 = 1.073$. (1)
• $M>1 \Rightarrow$ transonic/supersonic (close to 1, effectively transonic). (1)

Q3. (4 marks)

• Adiabatic energy eq: $c_p T + \tfrac{1}{2}V^2 = c_p T_0$ (const). (1)
• Divide by $c_p T$: $1 + \dfrac{V^2}{2c_p T} = \dfrac{T_0}{T}$. (1)
• Use $c_p = \dfrac{\gamma R}{\gamma-1}$ and $a^2 = \gamma R T$, so $\dfrac{V^2}{2c_pT} = \dfrac{(\gamma-1)V^2}{2\gamma R T} = \dfrac{(\gamma-1)}{2}M^2$. (1)
• Hence $\dfrac{T_0}{T} = 1 + \dfrac{\gamma-1}{2}M^2$. (1)

Q4. (5 marks)

• $\dfrac{T_0}{T} = 1 + 0.2(2)^2 = 1.8 \Rightarrow T = 500/1.8 = 277.8\ \mathrm{K}$. (2)
• $\dfrac{P_0}{P} = (1.8)^{3.5} = 7.824 \Rightarrow P = 600/7.824 = 76.7\ \mathrm{kPa}$. (3)

Q5. (5 marks)

• State relation correctly. (1)
• Subsonic ($M<1$): $(M^2-1)<0$, so to increase $V$ ($dV>0$) requires $dA<0$ → area must converge. (2)
• Supersonic ($M>1$): $(M^2-1)>0$, so to increase $V$ requires $dA>0$ → area must diverge. Sonic condition ($M=1$) forces $dA=0$ (throat). Therefore accelerate subsonic→sonic in converging section, sonic→supersonic in diverging section → C-D nozzle. (2)

Q6. (4 marks)

• Choked flow: condition where mass flow rate through the nozzle reaches its maximum and becomes independent of downstream (back) pressure. (2)
• Occurs when $M = 1$ at the throat (minimum area). (1)
• Once sonic at throat, further lowering back pressure cannot increase $\dot m$; maximum mass flux is fixed by $P_0, T_0$ and throat area. (1)

Q7. (4 marks)

• $M_2^2 = \dfrac{1 + \frac{\gamma-1}{2}M_1^2}{\gamma M_1^2 - \frac{\gamma-1}{2}} = \dfrac{1 + 0.2(4)}{1.4(4) - 0.2} = \dfrac{1.8}{5.4} = 0.3333 \Rightarrow M_2 = 0.577$. (2)
• $\dfrac{P_2}{P_1} = 1 + \dfrac{2\gamma}{\gamma+1}(M_1^2 - 1) = 1 + \dfrac{2.8}{2.4}(3) = 1 + 3.5 = 4.5$. (2)

Q8. (4 marks)

• Correct sketch with chord (LE–TE straight line), camber line (mid-thickness curve), leading edge, trailing edge labelled. (2)
• Angle of attack $\alpha$ = angle between chord and free-stream velocity. (1)
• $C_L = \dfrac{L}{\tfrac{1}{2}\rho V^2 S}$ (lift / dynamic pressure × reference area). (1)

Q9. (4 marks)

• Induced drag coefficient (elliptic loading): $C_{D,i} = \dfrac{C_L^2}{\pi\,\mathrm{AR}}$ (general: $\dfrac{C_L^2}{\pi e\,\mathrm{AR}}$). (2)
• Increasing aspect ratio $\mathrm{AR}$ decreases induced drag (inversely proportional); higher AR wings have weaker tip vortices/smaller downwash. (2)

[
  {"claim":"Speed of sound at 250 K is ~316.9 m/s","code":"a=sqrt(1.4*287*250); result = abs(float(a)-316.86) < 0.5"},
  {"claim":"M=1.073 for V=340 at T=250K","code":"a=sqrt(1.4*287*250); M=340/a; result = abs(float(M)-1.073) < 0.01"},
  {"claim":"Static T at M=2 with T0=500 is 277.8 K","code":"T0=500; T=T0/(1+0.2*4); result = abs(float(T)-277.78) < 0.1"},
  {"claim":"Static P at M=2 with P0=600 kPa is ~76.7 kPa","code":"P0=600; P=P0/((1.8)**3.5); result = abs(float(P)-76.68) < 0.3"},
  {"claim":"M2=0.577 for M1=2 normal shock","code":"M1=2; M2sq=(1+0.2*M1**2)/(1.4*M1**2-0.2); result = abs(float(sqrt(M2sq))-0.5774) < 0.001"},
  {"claim":"P2/P1=4.5 for M1=2 normal shock","code":"M1=2; r=1+(2.8/2.4)*(M1**2-1); result = abs(float(r)-4.5) < 0.001"}
]