Thermodynamics

1 V 1,n A r 2 V 2,n A S r 1 V 1,n r 2 V 2,n (17–41)Chapter 17 | 853We analyze a straight oblique shock in Fig. 17–38 by decomposing thevelocity vectors upstream and downstream of the shock into normal and tangentialObliqueshockcomponents, and considering a small control volume around thePV 1 1,t P 2shock. Upstream of the shock, all fluid properties (velocity, density, pressure,etc.) along the lower left face of the control volume are identical toV2V 1,n→→Vthose along the upper right face. The same is true downstream of the shock.1uTherefore, the mass flow rates entering and leaving those two faces cancel ControlV 2,tvolumeeach other out, and conservation of mass reduces toV 2,nbwhere A is the area of the control surface that is parallel to the shock. SinceA is the same on either side of the shock, it has dropped out of Eq. 17–41.As you might expect, the tangential component of velocity (parallel to theoblique shock) does not change across the shock (i.e., V 1,t V 2,t ). This iseasily proven by applying the tangential momentum equation to the controlvolume.When we apply conservation of momentum in the direction normal to theoblique shock, the only forces are pressure forces, and we getP 1 A P 2 A rV 2,n AV 2,n rV 1,n AV 1,n S P 1 P 2 r 2 V 2 2,n r 1 V 2 1,n (17–42)Finally, since there is no work done by the control volume and no heattransfer into or out of the control volume, stagnation enthalpy does notchange across an oblique shock, and conservation of energy yieldsFIGURE 17–38Velocity vectors through an obliqueshock of shock angle b and deflectionangle u.h 01 h 02 h 0 S h 1 1 2 V 2 1,n 1 2 V 2 1,t h 2 1 2 V 2 2,n 1 2 V 2 2,tBut since V 1,t V 2,t , this equation reduces toh 1 1 2 V 2 1,n h 2 1 2 V 2 2,n(17–43)Careful comparison reveals that the equations for conservation of mass,momentum, and energy (Eqs. 17–41 through 17–43) across an obliqueshock are identical to those across a normal shock, except that they are writtenin terms of the normal velocity component only. Therefore, the normalshock relations derived previously apply to oblique shocks as well, but mustbe written in terms of Mach numbers Ma 1,n and Ma 2,n normal to the obliqueshock. This is most easily visualized by rotating the velocity vectors inFig. 17–38 by angle p/2 b, so that the oblique shock appears to be vertical(Fig. 17–39). Trigonometry yieldsMa 1,n Ma 1 sin bandMa 2,n Ma 2 sin 1b u2(17–44)where Ma 1,n V 1,n /c 1 and Ma 2,n V 2,n /c 2 . From the point of view shownin Fig. 17–40, we see what looks like a normal shock, but with some superposedtangential flow “coming along for the ride.” Thus,All the equations, shock tables, etc., for normal shocks apply to oblique shocksas well, provided that we use only the normal components of the Mach number.In fact, you may think of normal shocks as special oblique shocks inwhich shock angle b p/2, or 90°. We recognize immediately that anoblique shock can exist only if Ma 1,n 1, and Ma 2,n 1. The normal shockP 1 P 2Ma 2,n 1Ma 1,n 1b uuObliqueshock→V 2V 2,tV 1,nVV 2,n1,tb→V 1P 1 P 2FIGURE 17–39The same velocity vectors of Fig.17–38, but rotated by angle p/2 – b,so that the oblique shock is vertical.Normal Mach numbers Ma 1,n andMa 2,n are also defined.