A ramjet engine operates at a flight Mach number M_(0)=0.8 at 9000 m altitude where p_(0)=30.14 kPa and T_(0)=229K For an effective exhaust velocity V_(e)=760m/s and inlet cross-sectional area A_(1)=0.187m^2 calculate; (a) the air mass flow rate m_(0) (b) the fuel mass flow rate my for a fuel-to-air ratio of ffa=0.03 (c) the exhaust gas mass flow rate me. (d) the gross thrust, (e) the net thrust, (f) the thrust power, (8) the propulsive efficiency n_(p) (h) the thermal efficiency 7_(th) (i) the specific fuel consumption q (i) the overall efficiency n_(o)

(a) The air mass flow rate $m_{0}$ can be calculated using the formula:<br /><br />$m_{0} = \dot{m}_{a} = \rho_{0} \cdot A_{1} \cdot V_{0}$<br /><br />where $\rho_{0}$ is the air density at the inlet, $A_{1}$ is the inlet cross-sectional area, and $V_{0}$ is the inlet velocity.<br /><br />To find $\rho_{0}$, we can use the ideal gas law:<br /><br />$\rho_{0} = \frac{p_{0}}{R_{a} \cdot T_{0}}$<br /><br />where $p_{0}$ is the inlet pressure, $R_{a}$ is the specific gas constant for air, and $T_{0}$ is the inlet temperature.<br /><br />Substituting the given values, we get:<br /><br />$\rho_{0} = \frac{30.14 \times 10^{3}}{0.287 \times 229} = 0.476 \, \text{kg/m}^3$<br /><br />Now, we can calculate $m_{0}$:<br /><br />$m_{0} = 0.476 \times 0.187 \times 343 = 34.6 \, \text{kg/s}$<br /><br />(b) The fuel mass flow rate $m_{f}$ can be calculated using the formula:<br /><br />$m_{f} = ffa \cdot m_{0}$<br /><br />Substituting the given fuel-to-air ratio $ffa = 0.03$ and the calculated air mass flow rate $m_{0}$, we get:<br /><br />$m_{f} = 0.03 \times 34.6 = 1.038 \, \text{kg/s}$<br /><br />(c) The exhaust gas mass flow rate $m_{e}$ is the sum of the air and fuel mass flow rates:<br /><br />$m_{e} = m_{0} + m_{f} = 34.6 + 1.038 = 35.638 \, \text{kg/s}$<br /><br />(d) The gross thrust $F_{g}$ can be calculated using the formula:<br /><br />$F_{g} = \dot{m}_{e} \cdot (V_{e} - V_{0})$<br /><br />Substituting the given exhaust velocity $V_{e} = 760 \, \text{m/s}$ and the calculated inlet velocity $V_{0} = 343 \, \text{m/s}$, we get:<br /><br />$F_{g} = 35.638 \times (760 - 343) = 19,019.8 \, \text{N}$<br /><br />(e) The net thrust $F_{n}$ can be calculated using the formula:<br /><br />$F_{n} = F_{g} - \dot{m}_{e} \cdot V_{0}$<br /><br />Substituting the calculated values, we get:<br /><br />$F_{n} = 19,019.8 - 35.638 \times 343 = 13,019.8 \, \text{N}$<br /><br />(f) The thrust power $P_{t}$ can be calculated using the formula:<br /><br />$P_{t} = F_{n} \cdot V_{0}$<br /><br />Substituting the calculated values, we get:<br /><br />$P_{t} = 13,019.8 \times 343 = 4,456,000 \, \text{W}$<br /><br />(g) The propulsive efficiency $\eta_{p}$ can be calculated using the formula:<br /><br />$\eta_{p} = \frac{2 \cdot V_{e}}{V_{0} + V_{e}}$<br /><br />Substituting the given exhaust velocity $V_{e} = 760 \, \text{m/s}$ and the calculated inlet velocity $V_{0} = 343 \, \text{m/s}$, we get:<br /><br />$\eta_{p} = \frac{2 \times 760}{343 + 760} = 0.69$<br /><br />(h) The thermal efficiency $\eta_{th}$ can be calculated using the formula:<br /><br />$\eta_{th} = \frac{V_{e} - V_{0}}{V_{e}}$<br /><br />Substituting the given exhaust velocity $V_{e} = 760 \, \text{m/s}$ and the calculated inlet velocity $V_{0} = 343 \, \text{m/s}$, we get:<br /><br />$\eta_{th} = \frac{760 - 343}{760} = 0.55$<br /><br />(i) The specific fuel consumption $q$ can be calculated using the formula:<br /><br />$q = \frac{m_{f}}{F_{n}}$<br /><br />Substituting the calculated fuel mass flow rate $m_{f} = 1.038 \, \text{kg/s}$ and the net thrust $F_{n} = 13,019.8 \, \text{N}$, we get:<br /><br />$q = \