Geometrical Explanation

Figure 14.3: The schematic of the turning flow.

The change in the flow direction is assume to be result of the change in the tangential component. Hence, the total Mach number increases. Therefore, the Mach angle increase and result in a change in the direction of the flow. The velocity component in the direction of the Mach line is assumed to be constant to satisfy the assumption that the change is a result of the contour only. Later, this assumption will be examined. The typical simplifications for geometrical functions are used:

$$d\nu \sim \sin (d\nu) ;$$ (14.3)

$$\cos (d\nu) \sim 1$$

These simplifications are the core reasons why the change occurs only in the perpendicular direction ($d\nu « 1$ ). The change of the velocity in the flow direction, $dx$ is

$$dx = (U + dU) \cos\nu -U = dU$$ (14.4)

In the same manner, the velocity perpendicular to the flow, $dy$ , is

$$dy = (U + dU) \sin(d\nu) = U d\nu$$ (14.5)

The $\tan \mu$ is the ratio of $dy/dx$ (see Figure (14.3))

$$\tan \mu = {dx \over dy} = { dU \over U d\nu }$$ (14.6)

The ratio $dU/U$ was shown to be

$${dU \over U } = { dM^2\over 2M^2 \left( 1 + {k -1 \over 2} M^2 \right) }$$ (14.7)

Combining equations (14.6) and (14.7) transforms it into

$$d\nu = - { \sqrt{M^2 - 1} dM^2 \over 2M^2 \left( 1 + {k -1 \over 2} M^2 \right) }$$ (14.8)

After integration of equation (14.8) becomes

$$\nu (M) = - \sqrt{ k+1\over k-1 } \tan^{-1} \sqrt{{k-1\over k+1} \left( M^2 -1\right)} + \tan^{-1} \sqrt{ \left( M^2 -1\right)} + constant$$ (14.9)

The constant can be chosen in a such a way that $\nu= 0$ at $M=1$ .