Question 3.1: A pulsejet engine is employed in powering a vehicle flying a......
A pulsejet engine is employed in powering a vehicle flying at a Mach number of 2 at an altitude of 40,000 ft. The engine has an inlet area 0.084 m² . The pressure ratio at combustion chamber is P_{03}=P_{02}=9 , fuel heating value is 43,000 kJ/kg, and combustion efficiency is 0.96.
Assuming ideal diffuser (P_{02}=P_{0\text{a}}) , it is required to calculate
1. The air mass flow rate
2. The maximum temperature
3. The fuel-to-air ratio
4. The exhaust velocity
5. The thrust force
6. The thrust-specific fuel consumption (TSFC)
At altitude 40,000 ft, the ambient temperature and pressure are 216.65 K and 18.75 kPa, respectively.
The flight speed is
\begin{matrix} U&=&M\sqrt{\gamma RT_\text{a}} =2\sqrt{1.4 \times 287 \times 216.65} \\ &=& 590 \text{ m/s} \end{matrix}
The air mass f low rate is
\dot{m}_{\text{a}}=\rho_{\text{a}}UA_{\text{i}}=\frac{P_{\text{a}}}{RT_{\text{a}}} UA_{\text{i}}=14.99 \text{ kg/s}
Diffuser: The total temperature at the intake is equal to the total upstream temperature, or
\begin{matrix} T_{02} &=& T_{0\text{a}}=T_{\text{a}}\left(1+\frac{\gamma_{\text{c}}-1}{2}M^2 \right) \\ &=& 216.65\left(1+\frac{1.4-1}{2} \times 2^2 \right) \\&=&390 \text{ K} \end{matrix}
The total pressure at the diffuser outlet also equals the free stream total pressure, as the flow through the diffuser is assumed ideal.
\begin{matrix} P_{02}&=&P_{0\text{a}}=P{\text{a}}\left(\frac{T_{0\text{a}}}{T_{\text{a}}} \right)^{\frac{\gamma}{\gamma-1} } \\ P_{02}&=&146.7 \text{ kPa} \end{matrix}
Combustion Chamber: A constant volume process is assumed; then
\begin{matrix} T_{03}&=&T_{02}\left(\frac{P_{03}}{P_{02}} \right) \\ &=&9T_{02}=3510 \text{ K} \end{matrix}
The fuel-to-air ratio is calculated from Equation 3.6 (f=\frac{C_{p_{\text{h}}}T_{03}-C_{p_{\text{c}}}T_{02}}{\eta_{\text{b}}Q_{\text{R}}-C_{p_{\text{h}}}T_{03}} ) as
\begin{matrix} f&=&\frac{1.148 \times 3510-1.005 \times 390}{0.96 \times 43,000-1.148 \times 3510} \\ &=& 0.0976 \end{matrix}
Tail pipe: The exhaust speed is calculated from Equation 3.8 (U_{\text{e}}=\sqrt{2C_{p_{\text{h}}}T_{03}\left[1-\left(\frac{P_{\text{a}}}{P_{03}} \right)^{\frac{\gamma_{\text{h}}-1}{\gamma_{\text{h}}} } \right] }) as
\begin{matrix} U_{\text{e}}&=&\sqrt{2 \times 1148 \times 3510\left[1-\left(\frac{18.75_{\text{a}}}{1320.3} \right)^{0.25} \right] } \\ &=& 2297 \text{ m/s} \end{matrix}
The specific thrust is
\frac{T}{\dot{m}_{\text{a}}} =(1+f)U_{\text{e}}-U \\ \frac{T}{\dot{m}_{\text{a}}} =1931.4\frac{\text{N.s}}{\text{kg}}
The thrust force is then T=\dot{m}_{\text{a}}(T/\dot{m}_{\text{a}})=28955 \text{ N}
The TSFC is calculated from Equation 3.10 (\text{TSFC}=\frac{\dot{m}_{\text{f}}}{T} =\frac{f}{(T/\dot{m}_{\text{a}})} )
\begin{matrix} \text{TSFC}&=&\frac{0.0976}{1931.4} =5.06 \times 10^{-5}\frac{\text{kg}}{\text{N.s}} \\ \text{TSFC}&=& 0.182 \frac{\text{kg}}{\text{N.s}} \end{matrix}