Question 3.2: A ramjet has a flight speed of Ma = 2.0 at an altitude of 16......

The total conditions for air before entering the engine are

\begin{matrix} \frac{T_{0\text{a}}}{T_{\text{a}}} &=&1+\frac{\gamma-1}{2}M^2_{\text{a}} =1.8 \\ \frac{P_{0\text{a}} }{P_{\text{a}}}&=&\left(1+\frac{\gamma-1}{2}M^2_{\text{a}} \right)^{\frac{\gamma}{\gamma-1} }=7.824 \end{matrix}

Thus, T_{0\text{a}}=390 \text{ K and }P_{0\text{a}}=7.824P_{\text{a}}

The total temperature is constant in the intake section and so

T_{0\text{a}}=T_{01}=T^\prime_{01}=T_{02}=390 \text{ K}

The flight speed is U=M\sqrt{\gamma RT_{\text{a}}}=590 \text{ m/s}.
Now, let us analyze the intake. Since the flight speed is supersonic two shock waves must be generated in the intake to deliver air at a subsonic speed to the combustion chamber as shown in Figure 3.10. The first shock is an oblique shock wave attached to the apex of the spike, which is to be followed by a normal shock wave.

1. Oblique shock wave
The complete governing equations for oblique and normal shock waves are found in Section 3.5; some are rewritten here.
With a free stream Mach number M_{\text{a}} = 2.0 and a deflection angle δ = 12° , the shock wave angle will be either determined from gas dynamics tables or iteratively from the following relation:

\tan \delta=\frac{2 \cot \theta (M_1^2 \sin^2 \theta -1)}{(\gamma+1)M_1^2-2(M_1^2\sin^2 \theta -1)}

Both result in the value θ = 41.56°. As shown in Figure 3.11, the Mach number upstream the oblique shock wave can be decomposed into normal and tangential components. The normal component of the free stream Mach number is

M_{\text{an}}=M_{\text{a}} \sin \theta=2.0 \sin 41.56^\circ=1.326

The normal component of the Mach number downstream the normal shock wave will also be determined either from normal shock wave tables or from the relation

M^2_{1\text{n}} =\frac{(\gamma -1)M^2_{\text{an}}+2 } { 2\gamma M^2_{\text{an}}-(\gamma-1) } \quad \quad \quad (1)

The result is M_{1\text{n}} = 0.773.
The total pressure ratio also may be determined from normal shock wave tables or from the relation

\frac{P_{01}}{P_{0\text{a}}} =\left[\frac{(\gamma+1)M^2_{\text{an}}}{2+(\gamma-1)M^2_{\text{an}}} \right] ^{\frac{\gamma}{\gamma -1} } \left[\left(\frac{2\gamma}{\gamma+1} \right) M^2_{\text{an}}-\left(\frac{\gamma-1}{\gamma+1} \right) \right] ^{-\left(\frac{1}{\gamma-1} \right) } \quad \quad (2)

Thus the pressure ratio is P_{01}/P_{0\text{a}} = 0.9746.
The Mach number for flow leaving the oblique shock wave is

M_1=\frac{M_{1\text{n}}}{\sin (\theta-\delta)} =\frac{0.773}{\sin (41.56-12)} =1.567

2. Normal shock wave
The second (normal) shock wave originates at the engine cowl (nose lips). With M_1 = 1.567, from normal shock wave tables, or Equation 1, M_1^\prime = 0.678, and also the total pressure ratio either from tables or Equation 2 is

\frac{P^\prime_{01}}{P_{01}} =0.90719

The pressure ratio is then

\frac{P^\prime_{01}}{P_{0\text{a}}} =\frac{P^\prime_{01}}{P_{01}} \frac{P_{01}}{P_{0\text{a}}} =0.90719 \times 0.9746=0.884

Since there are no losses in the diffuser, then

P_{02}=P^\prime_{01}

The stagnation pressure ratio in the intake is

r_{\text{d}}=\frac{P_{02}}{P_{0\text{a}}} =0.884

Combustion Chamber
The flow in the combustion chamber is a flow with heat transfer or a Rayleigh flow. Now, if M_2 = 0.25, then from Rayleigh flow tables or the governing equations in Section 3.5

\frac{T_{02}}{T_0^\ast } =0.2568

Then

T^\ast _0=T_{02}\frac{T^\ast _0}{T_{02}} =\frac{390}{0.2568} =1518 \text{ K} \\ \frac{T_{03}}{T^\ast _0}=\frac{2500}{1518}=1.64615 \gt 1.0

which is impossible.
Thus, adding fuel to make T_{03} = 2500 K means that the flow will be choked (M_3 = 1.0) and M_2 < 0.25. Thus to find M_2

\frac{T_{02}}{T^\ast _0}=\frac{T_{02}}{T_{03}}\frac{T_{03}}{T^\ast _0}=\frac{390}{2500} (1)=0.156

From Rayleigh tables or equations, then M_2=0.189 , \text{ and }P_{02}/P^\ast _0=1.2378.
Mach number at state 3 is unity, thus P_{03}=P^\ast_0=P^\ast_0/P_{02}P_{02}

P_{03}=\frac{1}{1.2378} (0.884 P_{0\text{a}})=0.714 \ P_{0\text{a}}=5.586P_{\text{a}}

The stagnation pressure ratio in the combustion chamber is

r_{\text{c}}=\frac{P_{03}}{P_{02}} =\frac{1}{1.2378} =0.808

The fuel-to-air ratio

\begin{matrix} f&=&\frac{C_{\text{P}}(T_{03}-T_{0\text{a}})}{\eta_{\text{b}}Q_{\text{R}}-C_{\text{R}}T_{0\text{a}}} \\ f&=&\frac{1.005(2500-390)}{(0.98)(45,000)-(1.005)(2500)} =0.05099 \end{matrix}

Nozzle
Since M_3 = 1.0 the nozzle diverges immediately. The gases expand in the nozzle and leave at the same ambient pressure. The nozzle efficiency is

\begin{matrix} \eta_{\text{n}} &=&\frac{T_{03}-T_5}{T_{03}-T_{5\text{s}}} \\ T_5 &=&T_{03}\left[1-\eta_{\text{n}}\left\{1-\frac{T_{5\text{s}}}{T_{03}} \right\} \right] \\ T_5&=&T_{03}\left[1-\eta_{\text{n}}\left\{1-\left(\frac{P_{\text{a}}}{P_{03}} \right) ^{\frac{\gamma-1}{\gamma} } \right\} \right] \end{matrix}

With P_{03}/P_{\text{a}}=5.586

T_5=2500\left[1-0.96\left\{1-(5.58)^{-0.2857}\right\} \right] =1567 \text{ K}

To calculate the stagnation pressure ratio in the nozzle, the isentropic total pressure at the nozzle outlet is then

\frac{P_{05}}{P_5} =\left(\frac{T_{05}}{T_5} \right) ^{\frac{\gamma}{\gamma -1} }=\left(\frac{2500}{1567} \right) ^{3.5}=5.129 \\ \therefore P_{05}=5.129P_{\text{a}}

The nozzle total pressure ratio is r_{\text{n}}=P_{05}/P_{03}=0.918.
The engine overall pressure ratio is

\frac{P_{05}}{P_{0\text{a}}} =r_{\text{d}}r_{\text{c}}r_{\text{n}}=0.655

Since T_5/T_{05} = 1567/2500 = 0.6268, then from isentropic flow tables, the exhaust Mach number M_5 = 1.725.
The exhaust speed is then V_5=M_5a_5=1.725\sqrt{(1.4)(287)(1567)}=1368 \text{ m/s}.
The mass flow rate is calculated from the known area at state (1), then with M_1 = 1.567, the static temperature and pressure at state (1) are given from normal shock tables as

\begin{matrix} T_1&=&\frac{T_1}{T_{\text{a}}}T_{\text{a}}=1.207 \times 216.66 =261.5 \text{ K} \\ P_1&=& \frac{P_1}{P_{\text{a}}}P_{\text{a}}=1.8847 \times 10.01 =18.8659 \text{ kPa} \end{matrix}

The air density at state (1) is then \rho_1=P_1/RT_1=0.2513 \text{ kg/m}^3 .
Moreover, the air speed is given by V_1=M_1 \sqrt{\gamma RT_1}=429.8 \text{ m/s} .

\begin{matrix} \dot{m}&=&\rho_1V_1A_1 \\ \dot{m}&=& \rho_1V_1A_1&=&(0.2513) (0.2)(429.8)=21.6 \text{ kg/s} \end{matrix}

The thrust force is given by the relation

\begin{matrix} T &=& \dot{m}_{\text{a}}[(1+f)V_5-V_{\text{a}}]=21.6[1.05099 \times 1368-590] \\ &=& 19897 \text{ N} \\ \text{TSFC}&=&\frac{\dot{m}_{\text{f}}}{T}=\frac{f \dot{m}_{\text{a}}}{T} \\ &=& \frac{(0.05099)(21.6)}{19897}=0.00005535 \text{ kg/N.s} \\ &=& 0.199 \text{ kg/N.h} \end{matrix}