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 f/a=0.03 (c) the exhaust gas mass flow rate me. (d) the gross thrust. (e) the net thrust, (f) the thrust power, (g) the propulsive efficiency n_(p) (h) the thermal efficiency n_(th) (i) the specific fuel consumption q (i) the overall efficiency n_(0)

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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 f/a=0.03 (c) the exhaust gas mass flow rate me. (d) the gross thrust. (e) the net thrust, (f) the thrust power, (g) the propulsive efficiency n_(p) (h) the thermal efficiency n_(th) (i) the specific fuel consumption q (i) the overall efficiency n_(0)

Answer 1

To solve this problem, we will use the given data and apply the principles of thermodynamics and fluid mechanics. Let's calculate each part step by step. --- ### **Given Data:** - Flight Mach number \( M_0 = 0.8 \) - Altitude: \( h = 9000 \, \text{m} \) - Ambient pressure: \( p_0 = 30.14 \, \text{kPa} \) - Ambient temperature: \( T_0 = 229 \, \text{K} \) - Effective exhaust velocity: \( V_e = 760 \, \text{m/s} \) - Inlet cross-sectional area: \( A_1 = 0.187 \, \text{m}^2 \) - Fuel-to-air ratio: \( f/a = 0.03 \) **Constants:** - Specific gas constant for air: \( R = 287 \, \text{J/kg·K} \) - Ratio of specific heats for air: \( \gamma = 1.4 \) --- ### **(a) Air mass flow rate (\( m_0 \)):** The air mass flow rate is calculated using the continuity equation: \[ m_0 = \rho_0 V_0 A_1 \] where: \[ \rho_0 = \frac{p_0}{R T_0}, \quad V_0 = M_0 a_0, \quad a_0 = \sqrt{\gamma R T_0} \] #### Step 1: Calculate speed of sound (\( a_0 \)): \[ a_0 = \sqrt{\gamma R T_0} = \sqrt{1.4 \cdot 287 \cdot 229} = \sqrt{91958.6} \approx 303.24 \, \text{m/s} \] #### Step 2: Calculate flight velocity (\( V_0 \)): \[ V_0 = M_0 a_0 = 0.8 \cdot 303.24 = 242.59 \, \text{m/s} \] #### Step 3: Calculate air density (\( \rho_0 \)): \[ \rho_0 = \frac{p_0}{R T_0} = \frac{30.14 \times 10^3}{287 \cdot 229} = \frac{30140}{65623} \approx 0.4594 \, \text{kg/m}^3 \] #### Step 4: Calculate air mass flow rate (\( m_0 \)): \[ m_0 = \rho_0 V_0 A_1 = 0.4594 \cdot 242.59 \cdot 0.187 \approx 20.85 \, \text{kg/s} \] So, the air mass flow rate is: \[ \boxed{m_0 = 20.85 \, \text{kg/s}} \] --- ### **(b) Fuel mass flow rate (\( m_f \)):** The fuel mass flow rate is related to the air mass flow rate by the fuel-to-air ratio: \[ f/a = \frac{m_f}{m_0} \] Rearranging for \( m_f \): \[ m_f = (f/a) \cdot m_0 = 0.03 \cdot 20.85 \approx 0.6255 \, \text{kg/s} \] So, the fuel mass flow rate is: \[ \boxed{m_f = 0.6255 \, \text{kg/s}} \] --- ### **(c) Exhaust gas mass flow rate (\( m_e \)):** The exhaust gas mass flow rate is the sum of the air mass flow rate and the fuel mass flow rate: \[ m_e = m_0 + m_f = 20.85 + 0.6255 \approx 21.4755 \, \text{kg/s} \] So, the exhaust gas mass flow rate is: \[ \boxed{m_e = 21.48 \, \text{kg/s}} \] --- ### **(d) Gross thrust (\( F_g \)):** Gross thrust is given by: \[ F_g = m_e V_e \] Substitute the values: \[ F_g = 21.4755 \cdot 760 \approx 16320.38 \, \text{N} \] So, the gross thrust is: \[ \boxed{F_g = 16320.38 \, \text{N}} \] --- ### **(e) Net thrust (\( F_n \)):** Net thrust accounts for the momentum of incoming air: \[ F_n = F_g - m_0 V_0 \] Substitute the values: \[ F_n = 16320.38 - (20.85 \cdot 242.59) \approx 16320.38 - 5057.91 \approx 11262.47 \, \text{N} \] So, the net thrust is: \[ \boxed{F_n = 11262.47 \, \text{N}} \] --- ### **(f) Thrust power (\( P_t \)):** Thrust power is the product of net thrust and flight velocity: \[ P_t = F_n V_0 \] Substitute the values: \[ P_t = 11262.47 \cdot 242.59 \approx 2730000 \, \text{W} \, \text{or} \, 2730 \, \text{kW} \] So, the thrust power is: \[ \boxed{P_t = 2730 \, \text{kW}} \] --- ### **(g) Propulsive efficiency (\( \eta_p \)):** Propulsive efficiency is given by: \[ \eta_p = \frac{2 V_0}{V_0 + V_e} \] Substitute the values: \[ \eta_p = \frac{2 \cdot 242.59}{242.59 + 760} = \frac{485.18}{1002.59} \approx 0.484 \] So, the propulsive efficiency is: \[ \boxed{\eta_p = 48.4\%} \] --- ### **(h) Thermal efficiency (\( \eta_{th} \)):** Thermal efficiency is given by: \[ \eta_{th} = \frac{F_n V_0}{m_f Q_r} \] Here, \( Q_r \) is the heat released per unit mass of fuel. Since \( Q_r \) is not provided in the problem, we cannot calculate \( \eta_{th} \) without additional information. --- ### **(i) Specific fuel consumption (\( SFC \)):** Specific fuel consumption is given by: \[ SFC = \frac{m_f}{F_n} \] Substitute the values: \[ SFC = \frac{0.6255}{11262.47} \approx 0.0000555 \, \text{kg/N·s} \, \text{or} \, 55.5 \, \text{mg/N·s} \] So, the specific fuel consumption is: \[ \boxed{SFC = 55.5 \, \text{mg/N·s}} \] --- ### **(j) Overall efficiency (\( \eta_0 \)):** Overall efficiency is the product of propulsive efficiency and thermal efficiency: \[ \eta_0 = \eta_p \cdot \eta_{th} \] Since \( \eta_{th} \) is not calculated due to missing \( Q_r \), \( \eta_0 \) cannot be determined without additional information. --- This concludes the solution.

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