16.100 F2004 Homework 7 Solutions

16.100 Homework #7 Solutions Po ↑ Pc Pd 22° E C a) Approach: resolve forces caused by pressure in y-direction to get L', then convert to Cℓ [C_{l} = \frac{L'}{\frac{1}{2}\rho_{m}M_{m}^{2}c}] To get L', need pressures P₀, P₁, P₂, Pd. Face (a): Expansion wave, since flow is turned away from itself (see Anderson). Need to get the pressure above face (a), so we can get the Mach no. using the Prandtl-Meyer function, then get Pa [\gamma_{a}=\gamma_{w}+\theta] [=v(Ma)+2^{*}] turning angle [\nu(M_{a})=\nu(3)=49.76^{\circ}] using Appendix C (Andersen) [=51.76^{\circ}] which Mach no. has this value? Thus, M₆ = 0.31 Assume freestream pressure is p₀₀ [P_{a} = P_{∞}\frac{1}{(\frac{\gamma+1}{2})^{(\gamma-1)/\gamma}} = P_{∞}\frac{1}{(1+\frac{2}{7})^{5/7}}] (or just use Appendix A, isentropic flow properties) [=36.73\cdot\frac{1}{42.45}] so P₄ = 0.8512 p₀ Face (b): Another expansion wave, θ = 20° (Same method as before) [M₆ = 51.76° + 20° = 71.76°] [P₄ = P₀(\frac{1}{289.4})] (using Appendix C, no stag pressure loss in either by expansion wave) [P₄ = 36.73 \cdot \frac{1}{289.4}] [P₄ = 0.1269 p₀] Face (c): Oblique shock wave on lower leading edge, so we expect pressure P₆ to be higher than p₀. To get p₆, need the normal Mach number, since it controls the "jump" in flow properties. "Normal" means perpendicular to the shock (use β, not θ). What is β? See section 9.2 for important O-B-M relation. Using figure 9.7, knowing θ, Ma, can get β = 40°. M_ax = M_∞ sin β = 3 sin 40° = 1.93 Can use normal shock table to get p₆ now, or use [\frac{p_{c}}{p_{∞}} = 1 + \frac{2γ}{γ+1}(M_{\infty}n^{2}-1)] so [p_{c} = 4.179 p₀] We also need the Mach no. over face (c) in order to figure out p₄ after the expansion wave. We know M_∞, the normal Mach no. before the shock. Using normal shock relations, we can find M_cn and then use trigonometry to get M_c. To get M_cn: [M_{cn} = \sqrt{\frac{1+\frac{\gamma-1}{2}M_{\infty}^{2}}{\gamma M_{\infty}^{2}-\frac{\gamma-1}{2}}} = 0.59] (normal to shock) So M_c = sin(β-θ) = 0.59/sin(40°-22°) = 1.91 (total) Face (d): Expansion wave from c → d, same method as before, θ = 20° [M₄ = M_c + θ = M(M_c) + 20° = 23.78° + 20° = 43.70°] To get p₄: Careful! P_out and p_in aren't isentropic. We can get around this: [\frac{p_{d}}{p_{e}} = (\frac{p_{d}}{p_{i}})(\frac{p_{i}}{p_{e}}) = (\frac{1}{23.28})(6.8)] using M_d, M_e from Appendix A so p₄ = 1.221 p₀ Now that we have all pressures, we can find L', the lift per unit span: [L' = \int_{s} p \hat{n} \cdot ds = (\frac{p_{e}}{\frac{1}{2}p_{\infty}M_{\infty}^{2}})c] [L' = \frac{5.5097c}{\frac{1.4}{2}(3)^{2}} = 0.5097c] where ĵ is perpendicular to freestream → ĵ = ĵ [L' = -\int p_{a}\hat{n}{a}\cdot\hat{j}ds-\int p{w}\hat{n}{w}\cdot\hat{j}ds-\int p{d}\hat{n}{d}\cdot\hat{j}ds-\int p{l}\hat{n}_{l}\cdot\hat{j}ds] pressures and normal vectors are constant over each face, so [L = -p_{a}(\hat{a}{e}\cdot j)l-p{w}(\hat{a}{i}\cdot j)l-p{d}(\hat{n}{d}\cdot j)l-p{l}(\hat{a}_{l}\cdot\hat{j})l] [= -p_{s}l~cos~2°-p_{w}l~cos~22°-p_{d}l~cos~15°-p_{l}l~cos~1°] since A-J's are just cosine of angle between ṅ and ĵ [C_{l} = \frac{L}{\frac{1}{2}\rho_{\infty}M_{\infty}^{2}c} = \frac{l}{\frac{1}{2}M_{\infty}^{2}c}(\begin{matrix}-\frac{p_{a}}{p_{\infty}}\omega s~2^{}-\frac{p_{w}}{p_{\infty}}\omega s~22^{}-\frac{p_{d}}{p_{\infty}}\omega s~15^{*}-\frac{p_{l}}{p_{\infty}}c\omega s~j%^{n}\end{matrix})] know all these values [C_{l} = 0.3317] expansion wave before face (a) streamlines are tangent to surfaces expansion shock