Question 3.5: A ramjet has low-speed stagnation pressure losses given by r......
A ramjet has low-speed stagnation pressure losses given by r_{\text{d}}=0.90,r_{\text{c}}=0.95,r_{\text{n}}=0.96. The ambient and maximum temperatures are 222 K and 1950 K, respectively. The fuel-to-air ratio is assumed to be very small such that f << 1. What is the minimum flight Mach number at which the engine develops positive thrust? Assume the nozzle is unchoked.
The thrust equation is expressed by Equation 3.29 as follows:
\frac{T}{\dot{m}_{\text{a}}} =(1+f)\sqrt{\frac{2\gamma_{\text{e}}R_{\text{e}}T_{04}(m-1)}{(\gamma-1)m} }-M\sqrt{\gamma RT_{\text{a}}}+\frac{P_{\text{e}}A_{\text{e}}}{\dot{m}_{\text{a}}} \left(1-\frac{P_{\text{a}}}{P_{\text{e}}} \right)
where
m=\left(1+\frac{\gamma-1}{2}M^2 \right) \left(r_{\text{d}}r_{\text{c}}r_{\text{n}}\frac{P_{\text{a}}}{P_{\text{c}}} \right) ^{\frac{\gamma -1}{\gamma} }
for an unchoked nozzle P_{\text{a}}=P_{\text{c}} , also assuming small fuel to air ratio; f << 1, then
\frac{T}{\dot{m}_a } =\sqrt{\frac{2\gamma RT_{04}(m-1)}{(\gamma -1)m} } -M\sqrt{\gamma RT_{a}}
Rewrite m as
m=\left(1+\alpha_1M^2\right) \alpha_2
where \alpha_1=\frac{\gamma-1}{2} \text{ and }\alpha_2=(r_{\text{d}}r_{\text{c}}r_{\text{n}})^{\frac{\gamma-1}{\gamma} } also define
\alpha_3=\sqrt{\gamma RT_{\text{a}}}
and
\alpha_4=\frac{2\gamma RT_{04}}{\gamma-1} \\ \therefore \frac{T}{\dot{m}_{\text{a}}} =\sqrt{\frac{\alpha_4[\alpha_2(1+\alpha_1M^2)-1]}{\alpha_2(1+\alpha_1M^2)} }-\alpha_3M
when T/ \dot{m}_{\text{a}}=0
\therefore \frac{\alpha_4[\alpha_2(1+\alpha_1M^2)-1]}{\alpha_2(1+\alpha_1M^2)} =\alpha_3^2 M^2 \\ \therefore \alpha_2 \alpha_4(1+\alpha_1 M^2)-\alpha_4=\alpha_2 \alpha_3^2M^2(1+ \alpha_1M^2) \\ \alpha_1\alpha_2\alpha_3^2M^4+\alpha_2\alpha_3^2M^2-\alpha_1\alpha_2\alpha_4M^4-\alpha_2\alpha_4+\alpha_4=0
Writing the above equation in the general form
AM^4+BM^2+C=0
where A=\alpha_1\alpha_2\alpha_3^2, \quad B=\alpha_2\alpha_3^2-\alpha_1\alpha_2\alpha_4, \quad C=\alpha_4-\alpha_2\alpha_4=(1-\alpha_2)\alpha_4, \text{ then: } M^2=-B\pm \sqrt{B^2-4AC}/2A
Now substitute the numerical values:
\alpha_1=0.2, \quad \quad \alpha_2=0.945, \quad \quad \alpha_3=298.7, \quad \quad \alpha_4=3917 \times 10^3
then A=16.86 \times 10^3,B=-655.98 \times 10^3, \text{ and } C=215 \times 10^3
M^2=\frac{656 \times 10^3 \pm 10^3 \sqrt{430336-14499}}{2 \times 16.86 \times 10^3}
M² = 0.33, which gives M = 0.574 or M² = 38.55, which yields M = 6.2088.
Thrust starts to build up when the flight Mach number M = 0.574 and not zero as depicted in Example 4. A plot for the specific thrust versus the flight Mach number is given in Figure 3.13.
Spike motion
Supersonic ramjet engine is fitted with a spike that moves axially in and out of the engine. This motion is necessary to adjust the oblique shock wave originating from the spike to the intake lips and avoid supersonic spillage and also penetrating shocks into the intake. This is the case for the turboram engine powering the SR-71 aircraft.
The motion of spike is automatically adjusted by the engine control unit (ECU). The following example illustrates this motion.
