Level 4 — ApplicationCompressible Flow & Aerodynamics

Compressible Flow & Aerodynamics

printable — key stays hidden on paper

Level 4 — Application (novel problems, no hints) Time: 60 minutes | Total: 60 marks

Use $\gamma = 1.4$ , $R = 287\ \text{J/kg·K}$ for air unless otherwise stated.

Question 1 — De Laval Nozzle Design (14 marks)

Air is stored in a large reservoir at $P_0 = 8\ \text{bar}$ , $T_0 = 600\ \text{K}$ and is expanded through a converging–diverging nozzle. The design exit Mach number is $M_e = 2.4$ .

(a) Determine the exit static pressure $P_e$ and exit static temperature $T_e$ for isentropic flow. (4)

(b) Compute the ratio of exit area to throat area $A_e/A^*$ . (4)

(c) The throat area is $A^* = 10\ \text{cm}^2$ . Given the flow chokes, find the mass flow rate through the nozzle. (4)

(d) State the exit velocity $V_e$ . (2)

Question 2 — Normal Shock in a Diffuser (12 marks)

A supersonic aircraft flies at $M_1 = 2.0$ at an altitude where $P_1 = 26\ \text{kPa}$ and $T_1 = 223\ \text{K}$ . A normal shock stands ahead of an inlet.

(a) Find the Mach number $M_2$ , static pressure $P_2$ , and static temperature $T_2$ immediately behind the shock. (6)

(b) Compute the stagnation pressure ratio $P_{0_2}/P_{0_1}$ across the shock and comment on what it represents physically. (4)

Question 3 — Oblique Shock / Expansion on a Wedge (14 marks)

A symmetric diamond (double-wedge) airfoil with half-angle $5^\circ$ flies at $M_\infty = 3.0$ at zero angle of attack.

(a) On the front (compression) surface the flow is deflected by $\theta = 5^\circ$ . Using the $\theta$ – $\beta$ – $M$ relation, determine the weak-shock wave angle $\beta$ (you may solve iteratively; state your answer to the nearest degree). (6)

(b) Behind the leading-edge shock, compute the downstream Mach number component and hence $M_2$ . (4)

(c) On the rear surface the flow expands through a Prandtl–Meyer expansion of $10^\circ$ total turn (from the shocked flow). If $\nu(M_2) = 26.0^\circ$ , find the Mach number $M_3$ downstream of the expansion given that $\nu(M_3) = 36.0^\circ$ corresponds to $M_3 \approx 2.38$ . (4)

Question 4 — Finite Wing & Induced Drag (12 marks)

A wing with an elliptical lift distribution has aspect ratio $AR = 7$ , span efficiency $e = 1.0$ , and operates at a lift coefficient $C_L = 0.9$ .

(a) Compute the induced drag coefficient $C_{D,i}$ . (3)

(b) The wing is redesigned with $AR = 10$ at the same $C_L$ . Find the new $C_{D,i}$ and the percentage reduction in induced drag. (5)

(c) The airfoil section has a 2D lift-curve slope $a_0 = 2\pi\ \text{per rad}$ . Estimate the 3D lift-curve slope $a$ for the $AR = 7$ wing using $a = \dfrac{a_0}{1 + a_0/(\pi e\, AR)}$ . (4)

Question 5 — Thin Airfoil & Critical Mach (8 marks)

(a) A thin symmetric airfoil (no camber) flies at $\alpha = 4^\circ$ , $V_\infty = 60\ \text{m/s}$ , $\rho = 1.225\ \text{kg/m}^3$ , chord $c = 1.5\ \text{m}$ . Using thin-airfoil theory, compute the lift coefficient $C_L$ and the lift per unit span $L'$ . (5)

(b) Define the critical Mach number and briefly explain how increasing airfoil thickness affects it. (3)

Answer keyMark scheme & solutions

Question 1 (14 marks)

(a) Isentropic relations at $M_e = 2.4$ : $\frac{T_0}{T_e} = 1 + \frac{\gamma-1}{2}M_e^2 = 1 + 0.2(5.76) = 2.152$ $T_e = 600/2.152 = 278.8\ \text{K}$ (2) $\frac{P_0}{P_e} = (2.152)^{3.5} = 14.62$ $P_e = 8/14.62 = 0.547\ \text{bar} = 54.7\ \text{kPa}$ (2)

(b) Area–Mach relation: $\frac{A}{A^*} = \frac{1}{M}\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$ $= \frac{1}{2.4}\left[\frac{1}{1.2}(2.152)\right]^{3}=\frac{1}{2.4}(1.7933)^3=\frac{5.767}{2.4}=2.403$ (4)

(c) Choked mass flow: $\dot m = \frac{P_0 A^*}{\sqrt{T_0}}\sqrt{\frac{\gamma}{R}}\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}$ Coefficient $=\sqrt{1.4/287}\,(0.8333)^{3}=0.06984\times0.5787=0.04042$ $\dot m = \dfrac{800000\times10\times10^{-4}}{\sqrt{600}}\times0.04042 = \dfrac{800}{24.49}\times0.04042 = 1.320\ \text{kg/s}$ (4)

(d) $a_e=\sqrt{\gamma R T_e}=\sqrt{1.4\times287\times278.8}=334.7\ \text{m/s}$ ; $V_e=M_e a_e=2.4\times334.7=803\ \text{m/s}$ (2)

Question 2 (12 marks)

(a) For $M_1=2.0$ : $M_2^2=\frac{1+\frac{\gamma-1}{2}M_1^2}{\gamma M_1^2-\frac{\gamma-1}{2}}=\frac{1+0.8}{5.6-0.2}=\frac{1.8}{5.4}=0.3333\Rightarrow M_2=0.5774$ (2) $\frac{P_2}{P_1}=1+\frac{2\gamma}{\gamma+1}(M_1^2-1)=1+\frac{2.8}{2.4}(3)=1+3.5=4.5\Rightarrow P_2=117\ \text{kPa}$ (2) $\frac{T_2}{T_1}=\frac{[1+\frac{\gamma-1}{2}M_1^2][\frac{2\gamma}{\gamma-1}M_1^2-1]}{\frac{(\gamma+1)^2}{2(\gamma-1)}M_1^2}=1.6875\Rightarrow T_2=376.3\ \text{K}$ (2)

(b) $\dfrac{P_{0_2}}{P_{0_1}}=0.7209$ (standard M=2 shock table value). Physically it is a measure of the total-pressure (available energy) loss caused by the irreversible shock. (4)

(c) The shock is adiabatic ( $T_0$ conserved) but irreversible: strong gradients dissipate mechanical energy internally, so entropy rises ( $s_2>s_1$ ) while total enthalpy is unchanged. (2)

Question 3 (14 marks)

(a) $\theta$ – $\beta$ – $M$ : $\tan\theta = 2\cot\beta\dfrac{M_1^2\sin^2\beta-1}{M_1^2(\gamma+\cos2\beta)+2}$ . Iterating with $M_1=3$ , $\theta=5^\circ$ gives weak solution $\beta \approx 23.1^\circ$ ( $\approx 23^\circ$ ). (6)

(b) Normal component $M_{n1}=M_1\sin\beta=3\sin23.1^\circ=1.177$ . $M_{n2}^2=\dfrac{1+0.2(1.385)}{1.4(1.385)-0.2}=\dfrac{1.277}{1.739}=0.7343\Rightarrow M_{n2}=0.857$ . $M_2=\dfrac{M_{n2}}{\sin(\beta-\theta)}=\dfrac{0.857}{\sin18.1^\circ}=\dfrac{0.857}{0.3107}=2.76$ (4)

(c) Prandtl–Meyer: $\nu(M_3)=\nu(M_2)+\Delta\theta=26.0^\circ+10.0^\circ=36.0^\circ$ . Given $\nu=36^\circ \Rightarrow M_3\approx2.38$ . The flow accelerates isentropically through the expansion. (4)

Question 4 (12 marks)

(a) $C_{D,i}=\dfrac{C_L^2}{\pi e\,AR}=\dfrac{0.81}{\pi(1)(7)}=\dfrac{0.81}{21.99}=0.0368$ (3)

(b) $C_{D,i}'=\dfrac{0.81}{\pi(10)}=\dfrac{0.81}{31.42}=0.02578$ . Reduction $=\dfrac{0.0368-0.02578}{0.0368}=0.30=30\%$ (5)

(c) $a=\dfrac{2\pi}{1+2\pi/(\pi\cdot1\cdot7)}=\dfrac{2\pi}{1+2/7}=\dfrac{6.283}{1.2857}=4.887\ \text{per rad}$ (4)

Question 5 (8 marks)

(a) Symmetric ⇒ $C_L=2\pi\alpha$ , $\alpha=4^\circ=0.06981\ \text{rad}$ . $C_L=2\pi(0.06981)=0.4386$ (2) $q=\tfrac12\rho V^2=0.5(1.225)(60^2)=2205\ \text{Pa}$ . $L'=q\,c\,C_L=2205\times1.5\times0.4386=1450\ \text{N/m}$ (3)

(b) Critical Mach number: the free-stream Mach number at which the local flow first reaches $M=1$ somewhere on the airfoil. Increasing thickness raises local surface velocities (larger suction peak), so sonic conditions are reached at lower free-stream Mach — thicker airfoils have lower $M_{cr}$ . (3)

[
{"claim":"Q1a Pe = 8/2.152^3.5 bar ≈ 0.547","code":"Pe=8/(2.152**3.5); result = abs(Pe-0.547)<0.01"},
{"claim":"Q1b A/A* at M=2.4 ≈ 2.403","code":"M=2.4; g=1.4; AAs=(1/M)*((2/(g+1))*(1+(g-1)/2*M**2))**((g+1)/(2*(g-1))); result = abs(AAs-2.403)<0.02"},
{"claim":"Q2a P2/P1 at M1=2 = 4.5","code":"g=1.4;M1=2;r=1+2*g/(g+1)*(M1**2-1); result = abs(r-4.5)<1e-6"},
{"claim":"Q4b induced drag reduction ~30%","code":"a=0.81/(pi*7);b=0.81/(pi*10);red=(a-b)/a; result = abs(red-0.30)<0.01"},
{"claim":"Q5a CL = 2*pi*4deg ≈ 0.4386","code":"CL=2*pi*(4*pi/180); result = abs(CL-0.4386)<0.001"}
]