Interleaved — Phase 3

Physics interleaved practice

printable — key stays hidden on paper

Instructions: Work each problem fully. Use $\gamma = 1.4$ and $R = 287\ \mathrm{J/(kg\cdot K)}$ for air unless stated otherwise. Show method selection reasoning. Marks in brackets. No calculators of shock tables provided — use the closed-form relations where possible.

1. [8] Air flows at $V = 250\ \mathrm{m/s}$ with static temperature $T = 280\ \mathrm{K}$ . Compute (a) the local speed of sound, (b) the Mach number, (c) the stagnation temperature $T_0$ , and (d) classify the regime (subsonic/transonic/supersonic/hypersonic).

2. [6] Starting from the steady-flow energy equation for an adiabatic, no-work open system, derive the stagnation temperature relation $\dfrac{T_0}{T} = 1 + \dfrac{\gamma-1}{2}M^2$ . State every assumption.

3. [7] A normal shock stands in a duct with upstream Mach number $M_1 = 2.0$ . Find $M_2$ , the static pressure ratio $P_2/P_1$ , and the stagnation pressure ratio $P_{02}/P_{01}$ .

4. [8] A converging–diverging nozzle has throat area $A^* = 10\ \mathrm{cm}^2$ . At a section where $A = 15\ \mathrm{cm}^2$ the flow is supersonic. Using the area–Mach relation, find the Mach number there (the supersonic root), and hence $P/P_0$ at that section.

5. [7] A supersonic stream at $M_1 = 3.0$ expands isentropically around a convex corner, turning through a deflection of $\Delta\theta = 15^\circ$ . Using the Prandtl–Meyer function, find the downstream Mach number $M_2$ .

6. [6] Derive the area–velocity relation $\dfrac{dA}{A} = (M^2 - 1)\dfrac{dV}{V}$ from continuity, the isentropic momentum (Euler) relation, and the definition of sound speed. Use it to explain physically why a de Laval nozzle needs a diverging section to accelerate flow beyond $M=1$ .

7. [7] A symmetric thin airfoil (zero camber) operates at angle of attack $\alpha = 4^\circ$ in incompressible flow. (a) Using thin-airfoil theory, compute the lift coefficient $c_\ell$ . (b) A finite wing of aspect ratio $AR = 6$ and elliptic loading has the same $c_\ell$ ; compute its induced drag coefficient $c_{D,i}$ .

8. [7] A converging (only) nozzle is fed from a reservoir at $P_0 = 500\ \mathrm{kPa}$ , $T_0 = 300\ \mathrm{K}$ , and discharges to a back pressure $P_b$ . (a) Determine the back pressure at which the nozzle first becomes choked. (b) If $P_b$ is lowered further, what happens to the exit Mach number and mass flow rate, and why?

9. [6] An oblique shock forms on a wedge in a $M_1 = 2.5$ flow with wave angle $\beta = 40^\circ$ . (a) Find the normal component $M_{n1}$ . (b) Find the deflection angle $\theta$ from the $\theta$ – $\beta$ – $M$ relation. (c) State what happens to $\beta$ if the wedge half-angle exceeds $\theta_{max}$ .

10. [6] A de Laval nozzle is designed for perfectly expanded exit Mach $M_e = 2.4$ . It is operated at a back pressure higher than the design exit pressure but low enough that the flow remains supersonic just inside the exit. (a) Name the operating condition (over- or under-expanded). (b) Describe the wave structure at the exit plane and the sign of $P_e - P_b$ .

Answer keyMark scheme & solutions

1. Tests 3.1.3 (sound speed) + 3.1.4 (Mach) + 3.1.2 (stagnation T). Why: Given $V$ and $T$ directly — you must build $a$ first, then $M$ , then $T_0$ . Recognizing the chain (no area/nozzle info) is the method choice.

(a) $a = \sqrt{\gamma R T} = \sqrt{1.4 \cdot 287 \cdot 280} = \sqrt{112476} = 335.4\ \mathrm{m/s}$ . (b) $M = V/a = 250/335.4 = 0.745$ . (c) $T_0 = T\left(1 + \tfrac{\gamma-1}{2}M^2\right) = 280(1 + 0.2\cdot0.555) = 280(1.111) = 311.1\ \mathrm{K}$ . (d) $M < 1$ → subsonic.

2. Tests 3.1.1 (first law open system) + 3.1.2 (stagnation derivation). Why: Pure derivation — must start from energy conservation, not memorized formula.

Steady-flow energy equation for adiabatic ( $q=0$ ), no shaft work ( $w=0$ ), negligible gravity: $h + \tfrac{1}{2}V^2 = h_0 = \text{const}.$ Stagnation state defined by $V=0$ : $h_0 = h + \tfrac12 V^2$ . For a calorically perfect gas $h = c_p T$ , $c_p = \dfrac{\gamma R}{\gamma-1}$ : $c_p T_0 = c_p T + \tfrac12 V^2 \Rightarrow \frac{T_0}{T} = 1 + \frac{V^2}{2 c_p T}.$ Since $c_p T = \dfrac{\gamma R T}{\gamma-1} = \dfrac{a^2}{\gamma-1}$ and $M^2 = V^2/a^2$ : $\frac{T_0}{T} = 1 + \frac{V^2(\gamma-1)}{2 a^2} = 1 + \frac{\gamma-1}{2}M^2. \;\blacksquare$ Assumptions: adiabatic, no work, steady, calorically perfect gas, negligible potential energy.

3. Tests 3.1.12 (normal shock properties) + 3.1.11 (R-H relations). Why: "Normal shock, given $M_1$ " triggers closed-form R–H relations, not isentropic tables (stagnation P is lost).

$M_2^2 = \dfrac{1 + \frac{\gamma-1}{2}M_1^2}{\gamma M_1^2 - \frac{\gamma-1}{2}} = \dfrac{1 + 0.2\cdot4}{1.4\cdot4 - 0.2} = \dfrac{1.8}{5.4} = 0.3333 \Rightarrow M_2 = 0.577$ .

$\dfrac{P_2}{P_1} = 1 + \dfrac{2\gamma}{\gamma+1}(M_1^2 - 1) = 1 + \dfrac{2.8}{2.4}(3) = 1 + 3.5 = 4.5$ .

Stagnation pressure ratio: $\frac{P_{02}}{P_{01}} = \left[\frac{(\gamma+1)M_1^2/2}{1+\frac{\gamma-1}{2}M_1^2}\right]^{\frac{\gamma}{\gamma-1}}\left[\frac{2\gamma M_1^2 - (\gamma-1)}{\gamma+1}\right]^{-\frac{1}{\gamma-1}}.$ Term A $=\left(\dfrac{2.4\cdot4/2}{1.8}\right)^{3.5}=\left(\dfrac{4.8}{1.8}\right)^{3.5}=(2.6667)^{3.5}=30.9$ . Term B $=\left(\dfrac{11.2-0.4}{2.4}\right)^{-2.5}=(4.5)^{-2.5}=0.0233$ . $P_{02}/P_{01} = 30.9\cdot0.0233 = 0.720$ . (Standard table value ≈ 0.7209.)

4. Tests 3.1.6 (area–Mach) + 3.1.7 (isentropic P/P₀). Why: $A/A^*$ given → invert area–Mach; "supersonic" tells you which of two roots. Then use isentropic (no shock) for $P/P_0$ .

$\dfrac{A}{A^*} = 1.5 = \dfrac{1}{M}\left[\dfrac{2}{\gamma+1}\left(1+\tfrac{\gamma-1}{2}M^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$ . Solving the supersonic branch: $M \approx 1.854$ .

$\dfrac{P}{P_0} = \left(1+\tfrac{\gamma-1}{2}M^2\right)^{-\frac{\gamma}{\gamma-1}} = (1+0.2\cdot3.437)^{-3.5} = (1.6875)^{-3.5} = 0.1602$ .

5. Tests 3.1.16/3.1.17 (Prandtl–Meyer expansion). Why: "Expands around convex corner" = isentropic turning → use $\nu(M)$ , not shock relations.

$\nu(M) = \sqrt{\dfrac{\gamma+1}{\gamma-1}}\tan^{-1}\!\sqrt{\dfrac{\gamma-1}{\gamma+1}(M^2-1)} - \tan^{-1}\!\sqrt{M^2-1}$ . At $M_1=3$ : $\nu(3) = 49.76^\circ$ . Downstream: $\nu(M_2) = \nu(M_1) + \Delta\theta = 49.76 + 15 = 64.76^\circ$ . Inverting: $M_2 \approx 3.94$ . (Flow accelerates, as expected for expansion.)

6. Tests 3.1.5 (area–velocity derivation). Why: Pure derivation; connects continuity + Euler + sound speed.

Continuity (log-differentiate $\rho A V =$ const): $\dfrac{d\rho}{\rho} + \dfrac{dA}{A} + \dfrac{dV}{V} = 0$ . Euler (inviscid, isentropic): $dP = -\rho V\,dV$ . Sound speed: $a^2 = dP/d\rho \Rightarrow d\rho = dP/a^2 = -\rho V\,dV/a^2$ , so $\dfrac{d\rho}{\rho} = -M^2\dfrac{dV}{V}$ . Substitute: $-M^2\frac{dV}{V} + \frac{dA}{A} + \frac{dV}{V} = 0 \Rightarrow \frac{dA}{A} = (M^2-1)\frac{dV}{V}. \;\blacksquare$ Physics: For $M<1$ , $(M^2-1)<0$ : to accelerate ( $dV>0$ ) area must decrease (converging). For $M>1$ , $(M^2-1)>0$ : to accelerate area must increase (diverging). Hence to pass through $M=1$ (throat) and continue accelerating supersonically, the nozzle must first converge then diverge — the de Laval geometry.

7. Tests 3.1.21 (thin airfoil) + 3.1.22/3.1.23 (induced drag / AR). Why: Symmetric = zero camber so lift is purely $\alpha$ -driven; then finite-wing correction uses elliptic result.

(a) Thin airfoil: $c_\ell = 2\pi\alpha$ (radians). $\alpha = 4^\circ = 0.0698\ \mathrm{rad}$ . $c_\ell = 2\pi(0.0698) = 0.4386$ . (b) Elliptic loading: $c_{D,i} = \dfrac{c_\ell^2}{\pi\,AR} = \dfrac{0.4386^2}{\pi\cdot6} = \dfrac{0.1924}{18.85} = 0.01021$ .

8. Tests 3.1.9 (converging nozzle) + 3.1.8 (choked flow). Why: Converging-only nozzle: exit Mach caps at 1; choke condition uses critical pressure ratio.

(a) Choking begins when $M_e=1$ , i.e. $P_e = P^*$ : $\dfrac{P^*}{P_0} = \left(\dfrac{2}{\gamma+1}\right)^{\frac{\gamma}{\gamma-1}} = (0.8333)^{3.5} = 0.5283$ . $P_b = 0.5283 \cdot 500 = 264.1\ \mathrm{kPa}$ . (b) Lowering $P_b$ below $264$ kPa: exit stays at $M_e=1$ (cannot exceed 1 in converging duct), so mass flow rate stays constant (maximum, choked) and exit pressure remains $P^*$ ; the further expansion occurs outside the nozzle. Exit Mach unchanged.

9. Tests 3.1.13/3.1.14 (oblique shock, θ–β–M). Why: Given $\beta$ and $M_1$ → forward θ–β–M gives deflection; normal component drives shock strength.

(a) $M_{n1} = M_1\sin\beta = 2.5\sin40^\circ = 2.5(0.6428) = 1.607$ . (b) $\tan\theta = 2\cot\beta\dfrac{M_1^2\sin^2\beta - 1}{M_1^2(\gamma+\cos2\beta)+2}$ . $M_1^2\sin^2\beta = 6.25\cdot0.4132 = 2.583$ ; numerator $= 2.583-1 = 1.583$ . $\cos2\beta = \cos80^\circ = 0.1736$ ; denominator $= 6.25(1.4+0.1736)+2 = 6.25(1.5736)+2 = 11.835$ . $\cot40^\circ = 1.1918$ . $\tan\the