Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

We analyze a straight oblique shock in Fig. 17–38 by decomposing the

velocity vectors upstream and downstream of the shock into normal and tan-

gential components, and considering a small control volume around the

shock. Upstream of the shock, all fluid properties (velocity, density, pres-

sure, etc.) along the lower left face of the control volume are identical to

those along the upper right face. The same is true downstream of the shock.

Therefore, the mass flow rates entering and leaving those two faces cancel

each other out, and conservation of mass reduces to

(17–41)

where Ais the area of the control surface that is parallel to the shock. Since

Ais 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 the

oblique shock) does not change across the shock (i.e.,V1,t
V2,t). This is

easily proven by applying the tangential momentum equation to the control

volume.

When we apply conservation of momentum in the direction normalto the

oblique shock, the only forces are pressure forces, and we get

(17–42)

Finally, since there is no work done by the control volume and no heat

transfer into or out of the control volume, stagnation enthalpy does not

change across an oblique shock, and conservation of energy yields

But since V1,t
V2,t, this equation reduces to

(17–43)

Careful comparison reveals that the equations for conservation of mass,

momentum, and energy (Eqs. 17–41 through 17–43) across an oblique

shock are identical to those across a normal shock, except that they are writ-

ten in terms of the normalvelocity component only. Therefore, the normal

shock relations derived previously apply to oblique shocks as well, but must

be written in terms of Mach numbers Ma1,nand Ma2,nnormal to the oblique

shock. This is most easily visualized by rotating the velocity vectors in

Fig. 17–38 by angle p/2 b, so that the oblique shock appears to be verti-

cal (Fig. 17–39). Trigonometry yields

(17–44)

where Ma1,n
V1,n/c 1 and Ma2,n
V2,n/c 2. From the point of view shown

in Fig. 17–40, we see what looks like a normal shock, but with some super-

posed tangential flow “coming along for the ride.” Thus,

All the equations, shock tables, etc., for normal shocks apply to oblique shocks

as well, provided that we use only the normalcomponents of the Mach number.

In fact, you may think of normal shocks as special oblique shocks in

which shock angle b
p/2, or 90°. We recognize immediately that an

oblique shock can exist only if Ma1,n1, and Ma2,n1. The normal shock

Ma1,n
Ma 1 sin b¬and¬Ma2,n
Ma 2 sin 1 bu 2

h 1 

1

2

V^2 1,n
h 2 

1

2

V^2 2,n

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,t

P 1 AP 2 A
rV2,nAV2,nrV1,nAV1,n S P 1 P 2 
r 2 V^2 2,nr 1 V^2 1,n

r 1 V1,nA
r 2 V2,nA S r 1 V1,n
r 2 V2,n

Chapter 17 | 853

→

V2,n

Oblique

shock

Control

volume

V1,n

V 1

→

V 2

V1,t P^1 P 2

V2,t

u

b

FIGURE 17–38

Velocity vectors through an oblique

shock of shock angle band deflection

angle u.

V1,n

P 1 P 2

P 1 P 2

V 1

V1,t

Ma1,n  1

Ma2,n  1

Oblique

shock

V2,n

V2,t

u

b

bu

→

V 2

→

FIGURE 17–39

The same velocity vectors of Fig.

17–38, but rotated by angle p/2 – b,

so that the oblique shock is vertical.

Normal Mach numbers Ma1,nand

Ma2,nare also defined.

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