Alternative Approach to Governing Equations

Figure 14.4:

The schematic of the coordinate based on the mathematical description.

It must be pointed out that the total velocity isn't at the speed of sound, but only the tangential component. In fact, based on the definition of the Mach angle, the component shown in Figure (14.3) under $U_y$ is equal to the speed of sound, $M=1$ .

After some additional rearrangement, equation (14.15) becomes

$${U_\theta \over r} \left( {\partial U_r \over \partial \theta} - U_\theta \right) = 0$$ (14.10)

If $r$ isn't approaching infinity, $\infty$ and since $U_\theta \neq 0$ leads to

$${\partial U_r \over \partial \theta } = {U_\theta}$$ (14.11)

In the literature, these results are associated with the characteristic line. This analysis can be also applied to the same equation when they are normalized by Mach number. However, the non-dimensionalization can be applied at this stage as well.

The energy equation for any point on a stream line is

$$h(\theta) + {{U_\theta}^2 + {U_r}^2 \over 2} = h_0$$ (14.12)

Enthalpy in perfect gas with a constant specific heat, $k$ , is

$$h(\theta) = C_p T = C_p{ R \over R } T = {1 \over (k-1)} \overbrace{ \overbrace{C_p \over C_v}^k RT}^{c(\theta)^2 } = { c^2 \over k-1}$$ (14.13)

and substituting this equality, equation (14.24), into equation (14.23) results in

$${ c^2 \over k-1} + {{U_\theta}^2 + {U_r}^2 \over 2} = h_0$$ (14.14)

Utilizing equation (14.20) for the speed of sound and substituting equation (14.22) which is the radial velocity transforms equation (14.25) into

$${ {\left(\partial U_r \over \partial \theta \right)}^2 \over k-1} + {\left(\partial U_r \over \partial \theta \right)^2 + {U_r}^2 \over 2} = h_0$$ (14.15)

After some rearrangement, equation (14.26) becomes

$${ k+1 \over k-1} \left(\partial U_r \over \partial \theta \right)^2 + {U_r}^2 = 2 h_0$$ (14.16)

Note that $U_r$ must be positive. The solution of the differential equation (14.27) incorporating the constant becomes

$$U_r = \sqrt{2h_0} \sin \left( \theta \sqrt{k-1 \over k+1} \right)$$ (14.17)

which satisfies equation (14.27) because $\sin^2\theta + \cos^2\theta = 1$ . The arbitrary constant in equation (14.28) is chosen such that $U_r (\theta=0) =0$ . The tangential velocity obtains the form

$$U_\theta = c = {\partial U_r \over \partial \theta} = \sqrt{k-1 \over k+1 } \sqrt{2\;h_0} \;\;\cos \left( \theta \sqrt{k-1 \over k+1} \right)$$ (14.18)

The Mach number in the turning area is

$$M^2 = {{U_\theta}^2 + {U_r}^2 \over c^2} = {{U_\theta}^2 + {U_r}^2 \over {U_\theta}^2 } = 1 + \left( {U_r} \over U_\theta \right) ^2$$ (14.19)

Now utilizing the expression that was obtained for $U_r$ and $U_\theta$ equations (14.29) and (14.28) results for the Mach number is

$$M^2 = 1 + {k+1 \over k-1 } \tan^2 \left( \theta \sqrt{k-1 \over k+1} \right)$$ (14.20)

or the reverse function for $\theta$ is

$$\theta = \sqrt{k+1 \over k-1 } \tan^{-1} \left( \sqrt{k-1 \over k+1 } \left( M^2 -1 \right) \right)$$ (14.21)

What happens when the upstream Mach number is not 1? That is when the initial condition for the turning angle doesn't start with $M=1$ but is already at a different angle. The upstream Mach number is denoted in this segment as $M_{starting}$ . For this upstream Mach number (see Figure (14.2))

$$\tan \nu = \sqrt{{M_{starting}}^2 - 1}$$ (14.22)

The deflection angle $\nu$ , has to match to the definition of the angle that is chosen here ($\theta =0$ when $M=1$ ), so

$$\nu (M) = \theta(M) - \theta(M_{starting})$$ (14.23)

$$%\nonumber = \sqrt{k+1\over k-1} \tan^{-1} \left( \sqrt{k-1\over k+1} \sqrt{ M^2 -1}\right) - \tan^{-1} \sqrt{ M^2 -1}$$ (14.24)

These relationships are plotted in Figure (14.6).