Question 3.7: Figure 3.17 illustrates a ramjet traveling in level flight a......

At an altitude of 55,000 ft, the ambient conditions are

T_{\text{a}}=-56.5^\circ C =216.5 \text{ K}, \quad P_{\text{a}}=9.122 \text{ KPa}

The sonic speed is a=\sqrt{\gamma RT_{\text{a}}}=295 \text{ m/s}

The flight Mach number is \begin{matrix} M_1&=& \frac{V_{\text{f}}}{a} =1.818 \\ \frac{T_{0\text{a}}}{T_{\text{a}}}&=&1+\frac{\gamma-1}{2}M^2=1.66 \\ \frac{P_{0\text{a}}}{P_{\text{a}}}&=&\left(1+\frac{\gamma+1}{2}M^2 \right)^{\frac{\gamma}{\gamma-1} }=5.9 \end{matrix}

The ambient total conditions are T_{0\text{a}}=360 \text{ K}

P_{0\text{a}}=53.86 \text{ kPa}

Just upstream the engine, the static and total conditions are equal to the ambient conditions or

P_{01}=P_{0\text{a}}, \quad P_1=P_{\text{a}}, \quad T_{1}=T_{\text{a}}, \quad T_{01}=T_{0\text{a}}

The density is \rho_1=\frac{P_1}{RT_1} =\frac{9122}{287 \times 216.5} =0.147 \text{ kg/m}^3

Normal shock wave
Normal shock wave exists between states (1) and (2). From normal shock wave tables or normal shock relations with M_1=1.82,then

\therefore M_2=0.6121,\quad \frac{T_{02}}{T_2} =1.0773,\quad \frac{P_{02}}{P_{01}} =0.803, \quad \frac{T_2}{T_1} =1.5465 \\ P_{02}=43.25 \text{ kPa}, \quad T_2=334.8 \text{ K}, \quad T_{02}=360 \text{ K}\equiv T_{01}

The mass flow rate m^\circ=\rho_1 V_1A_1

m^\circ=0.147 \times 536 \times 0.0929=7.31 \text{ kg/s}

Intake
From state (2) to state (3), the flow is isentropic, thus

P_{03}=P_{02}=43.25 \text{ kPa } \quad \text{ and } \quad T_{03}=T_{02}=360 \text{ K}

From isentropic flow tables or equations in Section 3.5, with M_2=0.6121

\frac{A_2}{A^\ast } =1.16565

since \frac{A_3}{A^\ast } =\frac{A_3}{A^\ast }\frac{A^\ast }{A_2} =\frac{A_3}{A_2}\frac{A_2}{A^\ast } =2 \times 1.16565

\frac{A_3}{A^\ast } =2.3313

From isentropic flow table or equation in Section 3.5, the Mach number at the inlet of combustion chamber is

M_3=0.26

This subsonic flow is appropriate for combustion.

Combustion chamber
From state (3) to state (4), a frictionless, heat addition constant area duct exists; thus the process represents a Rayleigh flow. From Rayleigh flow tables or equations in Section 3.5, with M_3=0.26

\therefore \frac{T_{03}}{T_{0}^\ast} =0.27446, \quad \frac{P_{03}}{P_{0}^\ast} =1.214

Next, with T_{04} = 1280 K

\therefore \frac{T_{04}}{T_{0}^\ast} =\frac{T_{04}}{T_{03}}\frac{T_{03}}{T_{0}^\ast} =\frac{1280}{360} \times 0.27446=0.9759

∴ From Rayleigh flow tables (equations) M_4=0.83 \text{ and }\frac{P_{04}}{P_0^\ast} =1.014

\begin{matrix} \therefore \frac{P_{04}}{P_{03}} &=&\frac{P_{04}}{P_0^\ast} \frac{P_0^\ast}{P_{03}} =\frac{1.014}{1.214} =0.8352 \\ P_{04}&=&36.12 \text{ kPa} \end{matrix}

The pressure drop in the combustion chamber is

\Delta P_{\text{CC}}=P_{03}-P_{04}=43.25-36.12=7.13 \text{ kPa}

This is a rather large pressure drop.

Nozzle
From state (4) to state (5), the flow in nozzle is isentropic flow:

At M_4=0.83, \text{ then }\frac{A_4}{A^\ast} =1.02696

The throat area is A^\ast and calculated as

A_5\equiv A^\ast=\frac{A_4}{(A_4/A^\ast)} =\frac{0.1858}{1.02696} =0.1809 \text{ m}^2

Since \frac{P_6}{P_{06}} =\frac{P_6}{P_{04}} =\frac{P_1}{P_{04}} =\frac{9.122}{36.12} =0.2525

From isentropic tables or the governing equations in Section 3.5, M_6=1.555

Then A_6/A^\ast=1.211 \text{ and }T_6/T_{06}=0.67545.

Thus A_6=1.211 \times 0.1809=0.21906 \text{ m}^2.

Since T_{06}=T_{04}=1280 \text{ K, then }T_6=864.6 \text{ K}.

The jet speed is now calculated as V_6=M_6\sqrt{\gamma RT_6}=916.5 \text{ m/s}.

Thrust Force = m^\circ=[(1+f)V_6-V_{\text{f}}]+P_1(A_6-A_1).

Since f=\frac{1}{40} =0.025 , then the thrust force is

T=(7.31)[(1.025)(916.5)-536]+9.122 \times 10^3(0.219-0.0929)=4099 \text{ N}

If the nozzle is convergent (from state (4) to state (6))
Check for choking

\begin{matrix} \frac{P_{04}}{P_{\text{a}}} &=&\frac{36.12}{9.122} =3.959 \\ \frac{P_{04}}{P_{\text{c}}}&=&\left(\frac{\gamma+1}{2} \right)^{\frac{\gamma}{\gamma-1} }=1.8929 \end{matrix}

Since \frac{p_{04}}{p_a} \gt \frac{p_{04}}{p_c} , then the nozzle, is choked

From isentropic flow tables or equation, with M_4=0.83, \text{ then }\frac{A_4}{A^\ast} =1.02708

A_6=A^\ast=\frac{0.1858}{1.02708} =0.1809 \text{ m}^2

The thrust force equation is now expressed as T=m^\circ [(1+f)V_6-V_1]+P_6A_6-P_1A_1

\begin{matrix} T&=&7.31[(0.125)(654)-536]+19.08 \times 10^3 \times 0.1809-9.122 \times 10^3 \times 0.0929 \\ &=& 3587 \text{ N} \end{matrix}

Propulsive efficiency
The general expression for the propulsive efficiency is Equation 2.7, which is rewritten here

\eta_{\text{p}}=\frac{u\left\{\overset{\bullet}{m}_{\text{a}}[(1+f)u_{\text{e}}-u]+A_{\text{e}}(P_{\text{e}}-P_{\text{a}})\right\} }{u\left\{\overset{\bullet}{m}_{\text{a}}[(1+f)u_{\text{e}}-u]+A_{\text{e}}(P_{\text{e}}-P_{\text{a}})\right\}+(1/2)\overset{\bullet}{m}_{\text{a}}(1+f)(u_{\text{e}}-u)^2 } \quad \quad \quad (2.7) \\ \eta_{\text{p}} =\frac{TV_{\text{f}}}{TV_{\text{f}}+\frac{1}{2}\dot{m}_{\text{a}}(V_{\text{e}}-V_{\text{f}})^2 }

The exhaust mass flow rate \dot{m}_{\text{e}}=\dot{m}_{\text{a}}(1+f)=7.31 \times 1.025=7.493 \text{ kg/s}

Case (1)    Convergent–divergent nozzle (V_{\text{e}}=916.5 \text{ m/s})

\begin{matrix} \eta_{\text{p}}&=& \frac{4099 \times 536}{4090 \times 536+\frac{1}{2}7.493(916.5-536)^2} \\ &=&80.19\% \end{matrix}

Case (2) Convergent nozzle (V_{\text{e}}=654 \text{ m/s})

\begin{matrix} \eta_{\text{p}}&=&\frac{3587 \times 536}{3587 \times 536+\frac{1}{2}\times 7.493(654-536)^2 } \\ &=&97.35 \% \end{matrix}

The TSFC is expressed as TSFC = \frac{m^\circ_{\text{f}}} {T}=\frac{f}{T/m^\circ_{\text{a}}}

Case (1)    With T = 4090N, then \begin{matrix} T/m^\circ_{\text{a}}&=&559.5 \text{ N.s/kg} \\ \text{TSFC} &=& \frac{0.025}{559.5}\frac{\text{kg}}{\text{N.s}}=0.16\frac{\text{kg}}{\text{N.h}} \end{matrix}

Case (2)   Now t = 3587 N

\text{TSFC}=\frac{0.025 \times 3600 }{(3587/7.31)} =0.1834 \frac{\text{kg}}{\text{N.h}}

Comments
1. Convergent nozzle results in less thrust and consequently higher TSFC compared with a convergent–divergent nozzle
2. The propulsive efficiency for the case of convergent nozzle is higher than that of a convergent–divergent nozzle as the exhaust speed is sonic in the former case and supersonic in the latter.

Figure 3.18 illustrates the cycle on the T–S diagram for both convergent and convergent–divergent nozzles.