Compressible Flow & Aerodynamics

Question 1 (10)

(a) Speed of sound (6)

Consider a weak pressure pulse moving at speed $a$ into still gas; work in the wave frame (steady). Gas enters at $a$, $\rho$, $p$ and leaves at $a-dV$, $\rho+d\rho$, $p+dp$.

• Continuity: $\rho a = (\rho+d\rho)(a-dV) \Rightarrow a\,d\rho = \rho\,dV$ (1)
• Momentum: $p - (p+dp) = \rho a[(a-dV)-a] \Rightarrow dp = \rho a\, dV$ (2)
• Combine: eliminate $dV$: $dp = \rho a \cdot \frac{a\,d\rho}{\rho} = a^2 d\rho \Rightarrow a^2 = dp/d\rho$. (1)
• Disturbance is weak & fast ⇒ negligible gradients & no time for heat transfer ⇒ isentropic; for $p=\rho R T$ with $p\rho^{-\gamma}=$const, $dp/d\rho = \gamma p/\rho = \gamma R T$. (1)
• Hence $a=\sqrt{\gamma R T}$. (1)

(b) Stagnation T (4) Adiabatic steady flow energy: $h + V^2/2 = h_0$; with $h=c_pT$: $c_pT_0 = c_pT + V^2/2$. (1) Divide by $c_pT$, use $c_p=\gamma R/(\gamma-1)$, $a^2=\gamma RT$, $M=V/a$: $\frac{T_0}{T}=1+\frac{V^2}{2c_pT}=1+\frac{\gamma-1}{2}M^2.$ (2) At $M=2$: $T_0/T = 1+0.2(4)=1.8 \Rightarrow T_0 = 450\ \mathrm{K}$. (1)

Question 2 (10)

(a) (6)

• Continuity (log-differentiate $\rho A V=$const): $\frac{d\rho}{\rho}+\frac{dA}{A}+\frac{dV}{V}=0$. (2)
• Euler (inviscid, steady): $dp + \rho V\,dV = 0 \Rightarrow \frac{dp}{\rho}=-V\,dV$. (1)
• Isentropic: $dp = a^2 d\rho \Rightarrow \frac{d\rho}{\rho}=\frac{dp}{\rho a^2}=-\frac{V\,dV}{a^2}=-M^2\frac{dV}{V}$. (2)
• Substitute into continuity: $-M^2\frac{dV}{V}+\frac{dA}{A}+\frac{dV}{V}=0 \Rightarrow \frac{dA}{A}=(M^2-1)\frac{dV}{V}$. (1)

(b) (4)

• Subsonic ($M<1$): $(M^2-1)<0$, so $dA$ and $dV$ opposite signs → to accelerate ($dV>0$) need $dA<0$ (converging). (1)
• Supersonic ($M>1$): same sign → to accelerate need $dA>0$ (diverging). (1)
• Therefore continuous subsonic→supersonic acceleration requires converge then diverge = de Laval. (1)
• At throat $dA=0$: either $dV=0$ or $M=1$; for accelerating flow the sonic condition $M=1$ occurs at the throat. (1)

Question 3 (14)

(a) (6) Conservation across shock:

• Mass: $\rho_1 u_1 = \rho_2 u_2$ (1)
• Momentum: $p_1+\rho_1 u_1^2 = p_2+\rho_2 u_2^2$ (1)
• Energy: $h_1+u_1^2/2=h_2+u_2^2/2$ (so $T_0$ constant). (1) Combining with $p=\rho RT$, $a^2=\gamma RT$ and algebra yields (3): $M_2^2 = \frac{1+\frac{\gamma-1}{2}M_1^2}{\gamma M_1^2 - \frac{\gamma-1}{2}}.$

(b) (6) With $M_1=2$, $\gamma=1.4$:

• $M_2^2 = \frac{1+0.2(4)}{1.4(4)-0.2}=\frac{1.8}{5.4}=0.3333 \Rightarrow M_2=0.5774.$ (1.5)
• $p_2/p_1 = 1+\frac{2\gamma}{\gamma+1}(M_1^2-1)=1+\frac{2.8}{2.4}(3)=1+3.5=4.5.$ (1.5)
• $T_2/T_1 = \frac{[1+\frac{\gamma-1}{2}M_1^2][\frac{2\gamma}{\gamma-1}M_1^2-1]}{\frac{(\gamma+1)^2}{2(\gamma-1)}M_1^2}=1.6875.$ (1.5)
• $p_{0_2}/p_{0_1}=0.7209.$ (from $=\left[\frac{(\gamma+1)M_1^2/2}{1+\frac{\gamma-1}{2}M_1^2}\right]^{\frac{\gamma}{\gamma-1}}\left[\frac{2\gamma M_1^2-(\gamma-1)}{\gamma+1}\right]^{-\frac{1}{\gamma-1}}$) (1.5)

(c) (2) Shock is adiabatic ⇒ energy conserved ⇒ $T_0$ unchanged (1). But it is irreversible (entropy rises), so stagnation pressure is lost: $p_{0_2}<p_{0_1}$ (1).

Question 4 (10)

(a) (6) $\theta$–$\beta$–$M$: $\tan\theta = 2\cot\beta\dfrac{M_1^2\sin^2\beta - 1}{M_1^2(\gamma+\cos2\beta)+2}$. (1)

• $M_{n1}=M_1\sin\beta = 3\sin32.24^\circ = 3(0.5336)=1.601.$ (2)
• Normal-shock relation: $M_{n2}^2=\frac{1+0.2 M_{n1}^2}{1.4 M_{n1}^2-0.2}=\frac{1+0.2(2.563)}{1.4(2.563)-0.2}=\frac{1.5126}{3.388}=0.4465\Rightarrow M_{n2}=0.6682.$ (2)
• $M_2 = \frac{M_{n2}}{\sin(\beta-\theta)}=\frac{0.6682}{\sin17.24^\circ}=\frac{0.6682}{0.2964}=2.254.$ (1)

(b) (4) Expansion: $\nu(M_2)=\nu(M_1)+\theta = 49.76+15 = 64.76^\circ$. (1) Interpolate between $\nu(3.6)=60.09$ and $\nu(3.7)=61.15$... target 64.76 is above these; extrapolating the given trend (~1.06°/0.1M): $M_2\approx 3.6+\frac{64.76-60.09}{10.6}\approx 4.04$ (accept $M_2\approx 4.0$). (2) Expansion occurs through infinitely many weak Mach waves, each isentropic, so no entropy generation; the shock compresses abruptly across a finite discontinuity, generating entropy → not isentropic. (1)

Question 5 (10)

(a) (3) Thin-airfoil: $L' = \pi\rho V^2 c\,\alpha$ (symmetric, camber term zero) ⇒ with $L'=\tfrac12\rho V^2 c\,c_\ell$: $c_\ell = 2\pi\alpha$. (2) $\alpha=5^\circ = 0.0873$ rad ⇒ $c_\ell = 2\pi(0.0873)=0.548.$ (1)

(b) (4) Finite wing: trailing tip vortices induce downwash, tilting the local lift vector backward → a drag component (induced drag). (1.5) $c_{D,i}=\dfrac{c_L^2}{\pi e AR}$. (1) $=\dfrac{0.9^2}{\pi(1)(7)}=\dfrac{0.81}{21.99}=0.0368.$ (1.5)

(c) (3) Critical Mach number = free-stream $M$ at which local flow first reaches $M=1$ somewhere on the airfoil. (1) Beyond it, embedded supersonic pockets terminate in shocks → wave drag rises sharply (drag divergence). (1) Whitcomb's area rule smooths the cross-sectional-area distribution of the whole aircraft to delay/reduce transonic wave drag. (1)

Question 6 (6)

import numpy as np
def isentropic(M, gamma=1.4):
    g = gamma
    T = (1 + 0.5*(g-1)*M**2)**-1              # T/T0
    p = T**(g/(g-1))                          # p/p0
    rho = T**(1/(g-1))                        # rho/rho0
    Astar = (1.0/M)*((2/(g+1))*(1 + 0.5*(g-1)*M**2))**((g+1)/(2*(g-1)))
    return {"T/T0": T, "p/p0": p, "rho/rho0": rho, "A/Astar": Astar}

Marks: correct $T/T_0$ (1), $p/p_0$ (1), $\rho/\rho_0$ (1), $A/A^*$ formula (2), valid returnable/self-contained code (1). $A/A^*$ formula: $\dfrac{A}{A^*}=\dfrac1M\left[\dfrac{2}{\gamma+1}\left(1+\tfrac{\gamma-1}{2}M^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$.

[
{"claim":"T0 = 450 K at M=2, T=250 K","code":"T=250; M=2; g=1.4; T0=T*(1+(g-1)/2*M**2); result = abs(T0-450)<1e-6"},
{"claim":"Normal shock M1=2 gives M2=0.5774","code":"g=1.4; M1=2; M2=sqrt((1+(g-1)/2*M1**2)/(g*M1**2-(g-1)/2)); result = abs(float(M2)-0.5774)<1e-3"},
{"claim":"p2/p1=4.5 and T2/T1=1.6875 for M1=2","code":"g=1.4; M1=2; p=1+2*g/(g+1)*(M1**2-1); T=(1+(g-1)/2*M1**2)*(2*g/(g-1)*M1**2-1)/((g+1)**2/(2*(g-1))*M1**2); result = abs(p-4.5)<1e-6 and abs(float(T)-1.6875)<1e-3"},
{"claim":"p02/p01=0.7209 for M1=2 normal shock","code":"g=1.4; M1=2.0; pr=((g+1)*M1**2/2/(1+(g-1)/2*M1**2))**(g/(g-1))*((2*g*M1**2-(g-1))/(g+1))**(-1/(g-1)); result = abs(float(pr)-0.7209)<1e-3"},
{"claim":"Induced drag coefficient = 0.0368 for cL=0.9, AR=7, e=1","code":"cL=0.9; AR=7; e=1; cdi=cL**2/(pi*e*AR); result = abs(float(cdi)-0.0368)<1e-3"},
{"claim":"Oblique shock Mn1=1.601, M2=2.254 for M1=3, beta