Oblique Shocks and Expansion Waves PDF

# Chapter 13 Compressible Flow ## 13.7 Oblique Shocks and Expansion Waves **The process follows the supersonic branch of a Rayleigh line**: * **T** * **P₀₁** * **T₀₁** * **P₁** * **T₁** * **Rayleigh line** * **P₀₂** * **T₀₂...

# Chapter 13 Compressible Flow ## 13.7 Oblique Shocks and Expansion Waves **The process follows the supersonic branch of a Rayleigh line**: * **T** * **P₀₁** * **T₀₁** * **P₁** * **T₁** * **Rayleigh line** * **P₀₂** * **T₀₂** * **P₂** * **T₂** **This problem illustrates the use of the Rayleigh-line equations, Eqs. 13.44, for analyzing frictionless flow of an ideal gas in a duct with heat exchange. The Excel workbook for this Example is convenient for performing the calculations. (Alternatively, the Rayleigh-line Excel add-ins, available on the Web site, are useful for these calculations.)** ## So far we have considered one-dimensional compressible flows. With the understanding we have developed we are ready to introduce some basic concepts of two-dimensional flow: oblique shocks and expansion waves. ### Oblique Shocks **In Section 12.2, we discussed the Mach cone, with Mach angle a, that is generated by an airplane flying at M > 1, as shown (in airplane coordinates) in Fig. 13.26a. The Mach cone is a weak pressure (sound) wave, so weak that, as shown in Fig. 13.26a, it barely disturbs the streamlines—it is the limiting case of an oblique shock. If we zoom in on the airplane we see that at the nose of the airplane we have an oblique shock—a shock wave that is aligned at some angle β<90° to the flow. This oblique shock causes the streamlines to abruptly change direction (usually to follow the surface of the airplane or the airplane’s airfoil). Further away from the airplane we still have an oblique shock, but it becomes progressively weaker (β decreases) and the streamlines experience smaller deflections until, far away from the airplane, the oblique shock becomes a Mach cone (β→ a) and the streamlines are essentially unaffected by the airplane.** **A supersonic airplane does not necessarily generate an oblique shock that is attached to its nose—we may instead have a detached normal shock ahead of the airplane! In fact, as illustrated in Fig. 13.27, as an airplane accelerates to its supersonic cruising speed the flow will progress from subsonic, through supersonic with a detached normal shock, to attached oblique shocks that become increasingly "pressed" against the airplane’s surface.** * **Streamline** * **α** * **(a) Mach cone** * **Oblique shock** * **β** * **Mach cone** * **Weaker oblique shock** * **(b) Oblique shock** * **Fig. 13.26 Mach cone and oblique shock generated by airplane.** * **(a) Subsonic M < 1** * **(b) Low supersonic** * **(c) Increasing M** * **Fig. 13.27 Airplane flow patterns as speed increases.** * **V₁** * **V₂** * **β** * **(β - θ)** * **CV** * **V_2.** * **V_1B** * **V₂** * **y** * **V_₁** * **→ x** * **(a) Oblique shock** * **(b) Oblique shock in shock coordinates** * **Fig. 13.28 Oblique shock control volume.** **We can explain these flow phenomena using concepts we developed in our analysis of normal shocks. Consider the oblique shock shown in Fig. 13.28a. It is at some angle β with respect to the incoming supersonic flow, with velocity V₁, and causes the flow to deflect at some angle θ with velocity V₂ after the shock.** **It is convenient to orient the xy coordinates orthogonal to the oblique shock and decompose V₁ and V₂ into components normal and tangential to the shock, as shown in Fig. 13.28b, with appropriate subscripts. The control volume is assumed to have arbitrary area A before and after the shock, and infinitesimal thickness across the shock (the upper and lower surfaces in Fig. 13.28b). For this infinitesimal control volume, we can write the basic equations: continuity, momentum, and the first and second laws of thermodynamics.** **The continuity equation is** **&frac;∂(ρV )&frac;∂(ρV )&frac;∂(ρV )** **∂t** **+ ∫_CV ρV⋅dA + ∫_CS ρV⋅dA = 0** **∂t** **= 0** **Assumption: (1) Steady flow.** **Then** **(-P₁V₁A) + (P₂V₂A) = 0 ** **(The tangential velocity components V_1, and V_2, flow through an infinitesimal area so do not contribute to continuity.) Hence,** **P₁V₁ = P₂V₂** **Next we consider the momentum equation for motion normal and tangential to the shock. We get an interesting result if we look first at the tangential, y component.** **∂** **(pVdV , + V, PV dÃ)** **∂t** **+∫_CV ∫_CS = 0** **Assumption: (2) Negligible body forces.** **Then** **0 = V_1,(-P₁V_1,A) + V_2,(P₂V_2,A)** **or, using Eq. 13.45a** **V_1, = V_2, = V** **Hence we have proved that the oblique shock has no effect on the velocity component parallel to the shock ( a result that is perhaps not surprising). The momentum equation for the normal, x direction is** **∂** **F_Sx + F_SX + F_/Bx = ** **∂t** **∫_CV V pdV + ∫_CS VpdA** **For our control volume we obtain** **P₁A-P₂A = V_1,(-P₁V_1,A) + V_2,(P₂V_2,A)** **P₁ + P₁V₁² = P₂ + P₂V₂²** **or, again using Eq. 13.45a,** **The first law of thermodynamics is** **∂** **W** **∂t** **∫_CV ep dV + ∫_CS (e + pv) pv dà = 0** **where** **e=u+&fracV²2+&fracV²2** **e = u +&fracV²2+&fracV²2** **Assumptions: (4) Adiabatic flow.** **(5) No work terms.** **(6) Negligible gravitational effect.** **For our control volume we obtain** **0 = (µ + P₁v₁ + &fracV²_1n2)(P₁V_1,A) + (µ + P₂v₂ + &fracV²_2n2)(P₂V_2,A)** **(Remember that *u* here represents the specific volume.) This can be simplified by using *h* = *u* + *pu*, and continuity (Eq. 13.45a),** **h₁ + &fracV²_1n2 = h₂ + &fracV²_2n2** **But each velocity can be replaced by its Pythagorean sum, so** **h₁ + &fracV₁² + &fracV₁²2 = h₂ + &fracV₂² + &fracV₂²2** **We have already learned that the tangential velocity is constant, V_1, = V_2, = V, so the first law simplifies to** **h₁ + &fracV₁²2 = h₂ + &fracV₂²2** **Finally, the second law of thermodynamics is** **∂** **∂t**∫_CV spdV + ∫_CS spv dà ≥ ∫_CS &frac;∂(µ )dA ** **∂t** **The shock is irreversible, so Eq. 4.58 for our control volume is** **S₁(-P₁V₁A) + S₂(P₂V₂A) > 0** **and, again using continuity,** **S₂ > S₁** **The continuity and momentum equations and the first and second laws of thermodynamics for an oblique shock are given by Eqs. 13.45a through 13.45d respectively. Examination of these equations shows that they are identical to the corresponding equations for a normal shock we derived in Section 13.3. Equations 13.11a through 13.11d, except V₁ and V₂, are replaced with normal velocity components V_1, and V_2, respectively! Hence, we can use all of the concepts and equations of Section 13.3 for normal shocks, as long as we replace the velocities with their normal components only. The normal velocity components are given by** **V₁ = V₁ sin β** **V₂ = V₂ sin(β-θ)** **The corresponding Mach numbers are** **M₁ = &fracV₁C₁ = M₁ sin β** **M₂ = &fracV₂C₂ = M₂ sin(β-θ)** **The oblique shock equations for an ideal gas with constant specific heats are obtained directly from Eqs. 13.20:** **M₂² = &fracM₁² + &frac2k-1 +1&frac2k-1 M₁² -1** **P₀₂P₀₁ =(1+&frack+12M₁²&frack-12)^k/(k-1)(1+&frack+12M₂²&frack-12)^-k/(k-1) **T₂T₁ =(1+&frack+12M₁²&frack-12)^-1(1+&frack+12M₂²&frack-12)** **P₂P₁ = &frack+12M₁²&frack-12+1&frack+12M₂²&frack-12+1** **P₂P₁ = &fracV₁V₂ = &frack+12M₁²&frack-12+1&frack+12M₂²&frack-12+1** **Equations 13.48, along with Eqs. 13.46 and 13.47, can be used to analyze oblique shock problems. Appendix E.5 lists flow functions for M₂ and property ratios P₀₂/P₀₁ , T₂/T₁, P₂/P₁, and P₂/P₁ (V₁/V₂) in terms of M₁ for oblique-shock flow of an ideal gas. A table of values of these property ratios is presented for air (k= 1.4) for a limited range of Mach numbers. The associated Excel workbook, *Oblique-Shock Relations*, can be used to print a larger table of values for air and other ideal gases. In essence, as demonstrated in Example 13.11, an oblique shock problem can be analyzed as an equivalent normal shock problem. The reader is urged to download the normal shock Excel add-ins from the Web site; they apply to these equations as well as Eqs. 13.20 for a normal shock.** ## Example 13.11 COMPARISON OF NORMAL AND OBLIQUE SHOCKS **Air at -2°C and 100 kPa is traveling at a speed of 1650 m/s. Find the pressure, temperature, and speed after the air experiences a normal shock. Compare with the pressure, temperature, and speed (and find the deflection angle θ) if the air instead experiences an oblique shock at angle β = 30°.** **Given: Air flow with:** * **P₁ = 100 kPa** * **T₁ = -2°C** * **V₁ = 1650 m/s** **Find: Downstream pressure, temperature, and speed if it experiences (a) a normal shock and (b) an oblique shock at angle β = 30°. Also find the deflection angle θ.** *** **Solution:** **(a) Normal shock** **First compute the speed of sound.** **C₁ = √kRT₁ = √1.4 × 287&fracNmkgK × 271K × &frackgms2N = 330 m/s** **Then the upstream Mach number is** **M₁ = &fracV₁C₁ = &frac1650ms330ms = 5.0** **From the normal-shock flow functions, Eqs. 13.20, at M₁ = 5.0** **M₁ M₂ P₂/P₁ T₂/T₁ V₂/V₁** **5.0 0.4152 29.00 5.800 0.2000** **From these data** **T₂ = 5.800T₁ = (5.800)271K = 1572K = 1299°C** **P₂ = 29.00p₁ = (29.00)100kPa = 2.9MPa** **V₂ = 0.200V₁ = (0.200)1650m/s = 330m/s** **(b) Oblique shock** **First compute the normal and tangential components of velocity** **V_1, = V₁ sin β = 1650 m/s × sin 30° = 825 m/s** **V_1, = V₁ cos β = 1650 m/s × cos 30° = 1429 m/s** **Then the upstream normal Mach number is** **M₁ = &fracV₁C₁ = &frac825ms330ms = 2.5** **From the oblique-shock flow functions, Eqs. 13.48, at M_1,= 2.5** **M₁ M₂ P₂/P₁ T₂/T₁ V₂/V₁** **2.5 0.5130 7.125 2.138 0.300** **From these data** **T₂ = 2.138T₁ = (2.138)271K = 579K = 306°C** **P₂ = 7.125p₁ = (7.125)100kPa = 712.5kPa** **V₂ = 0.300V₁ = (0.300)825 m/s = 247.5 m/s** **V_2, = V_1, = 1429 m/s** **The downstream velocity is given by the Pythagorean sum of the velocity components.** **Note that** **V₂ = √(V₂² + V₂²) = √(247.5² + 1429²) m/s = 1450 m/s** **C₂ = √kRT₂ = √1.4 × 287&fracNmkgK × 579K × &frackgms2N = 482 m/s** **so that the downstream Mach number is** **M₂ = &fracV₂C₂ = &frac1450ms482ms = 3.01** **Although the downstream normal Mach number must be subsonic, the actual downstream Mach number could be subsonic or supersonic (as in this case). The deflection angle can be obtained from Eq. 13.46b** **V₂ = V₂ sin(β – θ)** **or** **θ = β-sin^-1 (&fracV₂V₂)= 30° - sin^-1 (&frac247.51450)= 30° - 9.8° = 20.2°** **This Example illustrates:** * **That an oblique shock involves deflection of the flow through angle θ.** * **Use of normal shock functions for the solution of oblique shock problems.** * **The important result that for a given supersonic flow an oblique shock will always be weaker than a normal shock, because M₁ < M₁.** * **That while M₂ < 1 always, M₂ can be subsonic or supersonic (as in this case).** **The Excel workbook for oblique shocks is convenient for performing these calculations. (Alternatively, the normal shock relations Excel add-ins, available on the Web site, are useful for these calculations.)** **We can gain further insight into oblique shock behavior by combining some of our earlier equations to relate the deflection angle θ, the Mach number M₁, and the shock angle β. From the oblique shock geometry of Fig. 13.28b,** **&fracV₁V₂ = &fracV_1,tanβV_2,tan(β-θ)= &fractanβtan(β-θ)** **We can also relate the two normal velocities from Eq. 13.48e,** **&fracV₁V₂ = &frack+12M₁²&frack-12+1&frack+12M₁²&frack-12+1** **Equating the two expressions for the normal velocity ratio, we have** **&fracV₁V₂ = &fractanβtan(β-θ) = &frack+12M₁²&frack-12+1&frack+12M₁²&frack-12+1** **and** **tan(β-θ) = (1+&frack+12M₁²&frack-12)^-1tanβ** **Finally, if we use M_1, = M₁ sin β in this expression and further simplify, we obtain (after using a trigonometric identity and more algebra)** **tan θ = &frac2 cot(M₁²sin²β-1)M₁²(k+cos2β)+2** **Equation 13.49 relates the deflection angle θ to the incoming Mach number M₁ and the oblique shock angle β. For a given Mach number, we can compute θ as a function of β, as shown in Fig. 13.29 for air (k =1.4). Appendix E.5 presents a table of values of deflection angle θ as a function of Mach number M₁ and oblique shock angle β for air (k = 1.4) for a limited range of Mach numbers. The associated Excel workbook, *Oblique-Shock Relations*, can be used to print a larger table of values for air and other ideal gases. The reader is urged to download the oblique shock Excel add-in from the Web site; it can be used for solving Eq. 13.49 for the deflection angle 0, oblique shock β, or M₁. We should note that we used M₁ and shock angle β to compute θ, but in reality the causality is the reverse: it is the deflection θ caused by an object such as the surface of an airplane wing that causes an oblique shock at angle β. We can draw some interesting conclusions from Fig. 13.29:** * **For a given Mach number and deflection angle, there are generally two possible oblique shock angles—we could generate a weak shock (smaller β value, and hence smaller normal Mach number, M₁,) or a strong shock (larger β value, and hence, a larger normal Mach number). In most cases the weak shock appears (exceptions include situations where the downstream pressure is forced to take on a large value as caused by, for example, an obstruction).** * **For a given Mach number, there is a maximum deflection angle. For example, for air (k= 1.4), if M₁ = 3, the maximum deflection angle is θ_max ≈ 34°. Any attempt to deflect the flow at an angle θ>θ_max would cause a detached normal shock to form instead of an oblique shock.** * **For zero deflection (θ→0), the weak shock becomes a Mach wave and β→α = sin^-1(1/M₁).** **Figure 13.29 can be used to explain the phenomena shown in Fig. 13.27. If an airplane (or airplane wing), causing deflection θ, accelerates from subsonic through supersonic speed, we can plot the airplane’s progress on Fig. 13.29 as a horizontal line from right to left, through lines of increasing Mach number. For example, for θ = 10°, we obtain the following results: As M₁ increases from subsonic through about 1.4 there is no oblique shock solution—we instead either have no shock (subsonic flow) or a detached normal shock; at some Mach number the normal shock first attaches and becomes an oblique shock (Problem 13.187 shows that for θ = 10°, the normal shock first attaches and becomes oblique at M₁ ≈ 1.42, with β ~ 67°); as M₁ increases from 1.6 through 1.8, 2.0, 2.5, etc., toward infinity, from Fig. 13.29, β ≈ 51°, 44°, 39°, 32°, toward 12%, respectively—the oblique shock angle progressively decreases, as we saw in Fig. 13.27.** **A problem involving oblique shocks is solved in Example 13.12.** ## Example 13.12 OBLIQUE SHOCKS ON AN AIRFOIL **An airplane travels at a speed of 600 m/s in air at 4°C and 100 kPa. The airplane’s airfoil has a sharp leading edge with included angle δ = 6°, and an angle of attack α = 1°. Find the pressures on the upper and lower surfaces of the airfoil immediately after the leading edge.** **Given: Air flow over sharp leading edge with:** * **P₁ = 100 kPa** * **T₁ = 4°C** * **V₁ = 600 m/s** * **δ = 6°** * **α = 1°** **Find: Pressure on upper and lower surfaces.** **Solution:** * **For an angle of attack of 1° of an airfoil with leading edge angle 6°, the deflection angles are θ_u = 2° and θ_l = 4° as shown.** * **(a) Upper surface** **First compute the speed of sound** **C₁ =√kRT₁ = √1.4 × 287 &fracNmkgK× 277K× &frackgms2N = 334 m/s** **Then the upstream Mach number is** **M₁ = &fracV₁C₁ = &frac600ms334ms = 1.80** **For M₁ = 1.80 and θ_u = 2°, we obtain β_u from** **tanθ_u = &frac2cot(M₁²sin²β_u-1)M₁²(k+cos2β_u)+2** **This can be solved for β_u using manual iteration or interpolation, or by using, for example, Excel’s Goal Seek function.** **Then we can find M_1,(upper)** **β_u = 35.5°** **M₁(upper) = M₁ sin β_u = 1.80 × sin 35.5° = 1.045** **The normal Mach number for the upper oblique shock is close to one—the shock is quite weak.** <start_of_image> Diagrams/Tables/Images: * The document contains many diagram and table elements. * Due to the limit on the length of the text response, I cannot fully describe them. * The key concept is that of an **oblique shock**, which is a shock wave that is aligned *at some angle β<90°* to the flow. * There are several figures depicting the flow around an airplane, including the Mach cone and how it changes as an airplane accelerates to supersonic cruise speed. * The figures are designed to help the reader visualize the concepts of **oblique shocks** and **expansion waves**. * Several tables are used to provide information on the characteristics of oblique shocks, such as the relationships between Mach number, shock angle, and deflection angle. * They illustrate the principles of oblique shock flow and the various equations used to calculate the relevant quantities.