Question 3.4: For an ideal ramjet engine, prove that the thrust is express......
For an ideal ramjet engine, prove that the thrust is expressed by the relation
T=mu(\sqrt{\tau_{\text{b}}}-1), \quad \tau_{\text{b}}=\frac{T_{04}}{T_{02}}
For a flow at stratosphere region of the standard atmosphere, where T_{\text{a}} = constant, and if the maximum temperature T_{04} = constant (full-throttle operation), prove that the thrust attains a maximum value when the flight Mach number M_0 is given by:
M_0=\sqrt{\frac{2}{\gamma-1} \left(\tau_{\lambda}^{\frac{1}{3} }-1\right) }, \quad \tau_{\lambda}=\frac{T_{04}}{T_{\text{a}}}
Plot the relation
\left(\frac{T}{\dot{m}_{\text{a}}a_0} \right) \text{ versus } M_0 \quad \text{ for } \quad \tau_{\lambda}=9.0
where is the sonic speed at the flight altitude.
The specific thrust for an ideal ramjet engine with negligible fuel-to-air ratio f and unchoked nozzle is given by the relation:
\frac{T}{\dot{m}} =(u_{\text{e}}-u)=u\left(\frac{u_{\text{e}}}{u}-1 \right)
From Equation 3.18b (\therefore u_{\text{e}}=u\sqrt{\frac{T_{04}}{T_{0\text{a}}} }\equiv u\sqrt{\frac{T_{04}}{T_{02}} }), the ratio between the exhaust and flight speeds is equal to the ratio between the maximum temperature and the total free stream temperature
\frac{u_{\text{e}}}{u} =\sqrt{\frac{T_{04}}{T_{0\text{a}}} }
Define this temperature ratio as
\begin{matrix} \tau_{\text{b}}&=&\frac{T_{04}}{T_{02}} \\ \therefore\frac{u_{\text{e}}}{u}&=& \sqrt{\tau_{\text{b}}} \\ \frac{T}{\dot{m}_{\text{u}}} &=& \left(\frac{u_{\text{e}}}{u} -1\right) =(\sqrt{\tau_{\text{b}}}-1) && (\text{a}) \end{matrix}
Since \tau_{\lambda}=\frac{T_{04}}{T_{\text{a}}} \text{ and } \tau_{\text{b}}=\frac{T_{04}}{T_{02}}
\begin{matrix} \therefore \tau_{\lambda} &=& \frac{T_{04}}{T_{02}}\frac{T_{02}}{T_{\text{a}}}=\tau_{\text{b}}\left(1+\frac{\gamma -1}{2}M^2_0 \right) \\ \tau_{\text{b}} &=& \frac{\tau_{\lambda}}{\left(1+\frac{\gamma-1}{2}M^2_0 \right) } && \text{(b)} \end{matrix}
Since the flight speed (u) is equal to M_0a_0 , then from Equations a and b,
\begin{matrix} \therefore \frac{T}{\dot{m}} &=& M_0a_0\left\lgroup\sqrt{\frac{\tau_{\lambda}}{\left(1+\frac{\gamma-1}{2}M^2_0 \right) } }-1 \right\rgroup \\ \frac{\partial (T/\dot{m})}{\partial M_0}&=& a_0\left\lgroup\sqrt{\frac{\tau_{\lambda}}{\left(1+\frac{\gamma-1}{2}M^2_0 \right) } }-1\right\rgroup + M_0a_0 \left\lgroup(-0.5)\left(\frac{\gamma-1}{2} \right)(2M_0)\frac{\sqrt{\tau_{\lambda}}}{\left(1+\frac{\gamma-1}{2}M^2_0 \right)^{3/2} } \right\rgroup \end{matrix}The specific thrust is a maximum when
\frac{\partial (T/\dot{m})}{\partial M_0} =0 \\ \sqrt{\tau_{\lambda}} \left(1+ \cancel{\frac{\gamma-1}{2} }M_0^2\right) -\left(1+\frac{\gamma-1}{2} M_0^2\right) ^{3/2}-\left(\cancel{\frac{\gamma-1}{2} }M_0^2\right) \sqrt{\tau_{\lambda}}=0
\begin{matrix} \sqrt{\tau_{\lambda}} &=& \left(1+\frac{\gamma -1}{2} M_0^2 \right)^{3/2} \\ \tau_{\lambda}^{1/3} &=& \left(1+\frac{\gamma-1}{2} M_0^2\right) \\ M_o^2 &=&\frac{2}{\gamma-1}\left(\tau_{\lambda}^{1/3}-1\right) \\ (M_0)_{\text{max.thrust }} &=&\sqrt{\frac{2}{\gamma-1}\left(\tau_{\lambda}^{1/3}-1\right) } \end{matrix}
Now to plot the relation (T/ \dot{m}a_0) against M_0 , since
\frac{T}{\dot{m}a_0} =M_0(\sqrt{\tau_{\text{b}}}-1)
From the above relation, the thrust, or T/ \dot{m}a_0=0, when (1) M_0=0, and (2) \tau_{\text{b}}=1
Since \tau_{\lambda}=\tau_{\text{b}}\left(1+\frac{\gamma-1}{2}M^2_o \right) , then at \tau_{\text{b}}=1,
\tau_{\lambda}=\left(1+\frac{\gamma -1}{2} M_0^2 \right)
and
M_0=\sqrt{\frac{2}{\gamma-1}(\tau_{\lambda}-1) }=\sqrt{\frac{2}{0.4}(9-1) }=6.32
Thus the thrust is also equal to zero when the flight Mach number M_0 = 6.32. Since the thrust is equal to zero at M_0 = 0 and M_0 = 6.32, it will attain a maximum value at an intermediate value calculated from the relation
M_0= \sqrt{\frac{2}{\gamma-1}(\tau_{\lambda}^{1/3}-1) }=\sqrt{\frac{2}{0.4}(9^{1/3}-1) }=2.322
Thus the thrust is maximum at M_0 = 2.322
A plot for the relation (T/\dot{m}_{\text{a}}a_0) versus M_0 for \tau_{\lambda} = 9.0 is illustrated in Figure 3.12.
Comment
The above example illustrates the range of operations for an ideal ramjet engine. The engine can develop a thrust force when the flight Mach number is some value close to zero. However, this is normally not the case due to the different sources of irreversibilities in the engine. Example 5 clarifies that a real ramjet engine can develop thrust force only at higher values of Mach number that is close to or greater than 0.5 depending on the total pressure ratios within its three modules.
