Question 3.6: A ramjet engine is fitted with a variable geometry intake wh......
A ramjet engine is fitted with a variable geometry intake where a center body moves forward or backward to adjust the oblique shock wave to the intake lips (Figure 3.14). The half angle of the spike is 10°. The flight Mach number M_1 in the supersonic regime varies from M_{\text{minimum}} to 3.5. The inlet radius of the intake is 250 mm.
(a) Find the minimum Mach number M_{\text{min}} \gt 1.0 that corresponds to an attached shock wave to the apex.
(b) Find the distance (S) wave to the apex of the center body.
(c) Does the center body move forward or rearward when the flight Mach number increases?
(d) What would be the value of (S) if the Mach number M_1=2.5? Calculate the distance moved by the spike.
(e) Calculate the total pressure ratio P_{03}/P_{01} in the case of M_1=3.0.
It is known that for a certain deflection angle, there is a minimum Mach number that provides an attached shock wave. If the Mach number is reduced a detached shock wave exists.
(a) From shock wave graphs (Figure 3.15), for a semi-vertex angle of the spike of 10°, then the minimum Mach number that corresponds to an attached shock wave is M_{\text{min}}=1.42 and the corresponding wave angle [\sigma]_{\text{min}}=68 ^\circ.
(b) The distance S corresponding to this minimum Mach number is
\begin{matrix} \tan 68 &=&\frac{R}{S} \\ S &=&\frac{R}{\tan 68}&=&\frac{250}{2.4751}=101.0 \text{ mm} \end{matrix}
(c) Figure 3.15, illustrating the shock wave angle versus the deflection angle (σ versus δ), defines two types of shock waves, namely, strong and weak shock waves. The strong shock wave is close to the normal one and has a large angle, where the flow downstream of the shock is subsonic. The weak shock wave occurs at small angles. The flow downstream of the weak shock wave is supersonic. The angle of the weak shock wave decreases as the Mach number increases. Thus, the spike (or the center body) moves forward to keep the shock wave attached to the intake lips.
(d) At M_1=2.5\Rightarrow \sigma=33 ^\circ
S=\frac{R}{\tan 33} =\frac{250}{0.6494} =384.07 \text{ mm}
Thus, the spike moves forward by a distance (Δ) as shown in Figure 3.16.
Δ = 384.07 − 101.0 = 283.07 mm
(e) Following the same procedure of Example 2, then for flight Mach number M_1 = 3.0, the shock angle is σ = 27°.
The corresponding total pressure ratio is P_{02}/P_{01}=0.97 = 0.97.
Now, to evaluate the pressure ratio between states (3) and (1), the Mach number at state (2) is to be first determined. For the oblique shock wave then with M_1 = 3.0 and δ = 10°, M_2 = 2.5.
A normal shock wave is developed downstream the oblique shock wave.
The corresponding pressure ratio (from tables or equations in Section 3.5) is P_{03}/P_{02}=0.499.
The overall pressure ratio is
\frac{P_{03}}{P_{01}} =\frac{P_{03}}{P_{02}} \frac{P_{02}}{P_{01}} =0.484
