Let's proceed step by step as requested: --- ## **Given Data** - **Flight Mach number (\(M_\))**: 2. - **Burner exit (station 3) temperature (\(T_3\))**: 170 K - **Flight altitude**: 10 km - **Assume air is an ideal gas:** \( \gamma = 1.4 \), \( R = 287 \) J/kg·K --- ## **Step 1: Find Atmospheric Properties at 10 km** From standard atmosphere tables: - \( T_ = 223.3 \) K - \( P_ = 26.5 \) kPa --- ## **Step 2: Stagnation Properties at Inlet (Station )** ### **Total (Stagnation) Temperature at Inlet (\(T_{,}\))** \[ T_{,} = T_ \left[ 1 + \frac{\gamma-1}{2} M_^2 \right] \] \[ T_{,} = 223.3 \left[ 1 + \frac{.4}{2} \times (2.)^2 \right] = 223.3 \left[ 1 + .2 \times 4 \right] = 223.3 \left[ 1 + .8 \right] = 223.3 \times 1.8 = 401.94~\text{K} \] ### **Total (Stagnation) Pressure at Inlet (\(P_{,}\))** \[ P_{,} = P_ \left[ 1 + \frac{\gamma-1}{2} M_^2 \right]^{\frac{\gamma}{\gamma-1}} \] \[ P_{,} = 26.5 \left[ 1.8 \right]^{\frac{1.4}{.4}} = 26.5 \times (1.8)^{3.5} \] Calculate \(1.8^{3.5}\): \[ \ln(1.8^{3.5}) = 3.5 \ln(1.8) = 3.5 \times .5878 = 2.0573 \] \[ e^{2.0573} \approx 7.82 \] So: \[ P_{,} = 26.5 \times 7.82 = 207.3~\text{kPa} \] --- ## **Step 3: Conditions after Diffuser (Station 2)** **Assume a normal shock at the inlet, then subsonic flow isentropically to the combustor.** ### **a) Behind Normal Shock** #### **Calculate Downstream Mach Number (\(M_1\))** For a normal shock: \[ M_1 = \sqrt{\frac{(\gamma-1)M_^2 + 2}{2\gamma M_^2 - (\gamma-1)}} \] \[ M_1 = \sqrt{\frac{(.4)(4) + 2}{2.8 \times 4 - .4}} = \sqrt{\frac{1.6 + 2}{11.2 - .4}} = \sqrt{\frac{3.6}{10.8}} \approx \sqrt{.333} \approx .577 \] #### **Static Temperature behind Shock (\(T_1\)):** \[ \frac{T_1}{T_} = \frac{[2\gamma M_^2 - (\gamma-1)] [(\gamma-1)M_^2 + 2]}{(\gamma+1)^2 M_^2} \] But simpler: Use energy: \[ \frac{T_1}{T_} = \frac{(1 + \frac{\gamma-1}{2}M_^2)}{(1 + \frac{\gamma-1}{2}M_1^2)} \cdot \frac{P_1}{P_} \] But let's use a shortcut: First, find \(P_1/P_\): #### **Pressure Ratio across Shock (\(P_1/P_\)):** \[ \frac{P_1}{P_} = 1 + \frac{2\gamma}{\gamma+1}(M_^2 - 1) = 1 + \frac{2 \times 1.4}{2.4}(4-1) = 1 + \frac{2.8}{2.4} \times 3 = 1 + 1.167 \times 3 = 1 + 3.5 = 4.5 \] #### **Temperature Ratio across Shock (\(T_1/T_\)):** \[ \frac{T_1}{T_} = \left[ \frac{P_1}{P_} \right] \left[ \frac{\rho_}{\rho_1} \right] \] But, \[ \frac{\rho_1}{\rho_} = \frac{(\gamma+1) M_^2}{(\gamma-1) M_^2 + 2} = \frac{2.4 \times 4}{.4 \times 4 + 2} = \frac{9.6}{1.6 + 2} = \frac{9.6}{3.6} = 2.67 \] So, \[ \frac{\rho_}{\rho_1} = \frac{1}{2.67} = .375 \] So, \[ \frac{T_1}{T_} = 4.5 \times .375 = 1.6875 \] \[ T_1 = 1.6875 \times 223.3 = 377.15~\text{K} \] --- ## **Step 4: Diffuser (1 → 2): Isentropic to Mach ** Since the combustor entry Mach is zero, all kinetic energy converts to pressure (isentropic deceleration). \[ T_{,1} = T_1 \left[ 1 + \frac{\gamma-1}{2} M_1^2 \right] = 377.15 \left[ 1 + .2 \times (.577)^2 \right] = 377.15 \left[ 1 + .2 \times .333 \right] = 377.15 \left[ 1 + .0666 \right] = 377.15 \times 1.0666 = 402.8~\text{K} \] But this must equal the total temperature after the shock, which should be less than the inlet stagnation temperature (loss through shock): \[ T_{,1} = T_ \left[ 1 + \frac{\gamma-1}{2} M_^2 \right] \cdot \text{total temperature loss factor} \] But for simplicity, let's proceed: --- ## **Step 5: Burner Exit (Station 3) Pressure (\(P_3\))** Burner exit Mach = , so stagnation pressure = static pressure. \[ \text{From station 2 to 3:} \] \[ P_3 = P_2 \] \[ T_3 = \text{given} = 170~K \] To find \(P_3\), use isentropic relation from 1 to 2 (Mach ): \[ P_{02} = P_1 \left[ 1 + \frac{\gamma-1}{2} M_1^2 \right]^{\frac{\gamma}{\gamma-1}} \] \[ = P_1 \left[ 1 + .2 \times (.577)^2 \right]^{3.5} = P_1 \left[ 1 + .2 \times .333 \right]^{3.5} = P_1 (1.0666)^{3.5} \] \[ \ln(1.0666^{3.5}) = 3.5 \times \ln(1.0666) = 3.5 \times .0645 = .2256 \] \[ e^{.2256} = 1.253 \] \[ P_{02} = P_1 \times 1.253 \] Recall \(P_1 = 4.5 \times 26.5 = 119.25~\text{kPa}\): \[ P_{02} = 119.25 \times 1.253 = 149.5~\text{kPa} \] **Thus:** \[ \boxed{P_3 = 149.5~\text{kPa}} \] --- ## **Step 6: Nozzle Exit Velocity (\(V_e\))** At nozzle exit, pressure matched to atmosphere (\(P_e = P_\)), and isentropic expansion from \(T_3, P_3\) to \(P_\). Use isentropic relation for exit temperature: \[ \frac{T_e}{T_3} = \left( \frac{P_e}{P_3} \right)^{\frac{\gamma-1}{\gamma}} \] \[ T_e = T_3 \left( \frac{P_}{P_3} \right)^{.286} \] \[ T_e = 170 \left( \frac{26.5}{149.5} \right)^{.286} \] \[ \frac{26.5}{149.5} = .1774 \] \[ \ln(.1774) = -1.729 \] \[ -1.729 \times .286 = -.4948 \] \[ e^{-.4948} = .610 \] \[ T_e = 170 \times .610 = 1037~\text{K} \] Now, from energy balance: \[ V_e = \sqrt{2 C_p (T_3 - T_e)} \] For air (\(C_p = \frac{\gamma R}{\gamma-1} = 1005~\text{J/kg·K}\)) \[ V_e = \sqrt{2 \times 1005 \times (170 - 1037)} = \sqrt{2 \times 1005 \times 663} = \sqrt{2 \times 1005 \times 663} = \sqrt{133146} = 1154~\text{m/s} \] --- ## **Final Answers** ### **a) Nozzle entrance (burner exit) pressure:** \[ \boxed{P_3 = 149.5~\text{kPa}} \] ### **b) Nozzle exit velocity:** \[ \boxed{V_e = 1154~\text{m/s}} \] --- **Let me know if you want the full process or a summary for your report!**

# Step-by-Step Solution for Ideal Ramjet Engine Parameters ## Given Data - **Flight Mach number (\(M_\))**: 2.0 - **Burner exit temperature (\(T_3\))**: 1700 K - **Flight altitude**: 10 km - **Assumptions**: - Mach number in the combustor is zero (\(M_2 = M_3 = 0\)). - Pressure losses in the combustor are ignored (\(P_2 = P_3\)). - Flow in the diffuser and nozzle is isentropic. - Air as an ideal gas: \( \gamma = 1.4 \), \( R = 287 \) J/kg·K ## Step 1: Atmospheric Properties at 10 km From standard atmosphere tables: - **Temperature (\(T_a\))**: 223.3 K - **Pressure (\(P_a\))**: 26.5 kPa ## Step 2: Stagnation Properties at Inlet (Station 1) ### Total Temperature at Inlet (\(T_{01}\)) \[ T_{01} = T_a \left( 1 + \frac{\gamma - 1}{2} M^2 \right) \] \[ T_{01} = 223.3 \left( 1 + \frac{0.4}{2} \cdot 2^2 \right) = 223.3 \left( 1 + 0.4 \right) = 223.3 \times 1.8 = 401.94~\text{K} \] ### Total Pressure at Inlet (\(P_{01}\)) \[ P_{01} = P_a \left( 1 + \frac{\gamma - 1}{2} M^2 \right)^{\frac{\gamma}{\gamma - 1}} \] \[ P_{01} = 26.5 \left( 1.8 \right)^{\frac{1.4}{0.4}} = 26.5 \times (1.8)^{3.5} \approx 207.3~\text{kPa} \] ## Step 3: Conditions after Diffuser (Station 2) ### Downstream Mach Number (\(M_1\)) behind Normal Shock \[ M_1 = \sqrt{\frac{(\gamma - 1) M^2 + 2}{2 \gamma M^2 - (\gamma - 1)}} \] \[ M_1 = \sqrt{\frac{(0.4)(4) + 2}{(2.8)(4) - 0.4}} \approx 0.577 \] ### Static Temperature behind Shock (\(T_1\)) Using the shock relations, we find: \[ \frac{T_1}{T_a} = \frac{(1 + \frac{\gamma - 1}{2} M^2)}{(1 + \frac{\gamma - 1}{2} M_1^2)} \approx 1.6875 \] \[ T_1 \approx 1.6875 \times 223.3 = 377.15~\text{K} \] ### Pressure behind Shock (\(P_1\)) Using pressure ratio across shock: \[ \frac{P_1}{P_a} = 4.5 \implies P_1 = 4.5 \times 26.5 = 119.25~\text{kPa} \] ## Step 4: Burner Exit Pressure (\(P_3\)) Since \(P_2 = P_3\), we can use isentropic relations: \[ P_3 \approx 149.5~\text{kPa} \] ## Step 5: Nozzle Exit Velocity (\(V_e\)) ### Exit Temperature (\(T_e\)) Using the isentropic expansion: \[ \frac{T_e}{T_3} = \left( \frac{P_a}{P_3} \right)^{\frac{\gamma - 1}{\gamma}} \] \[ T_e \approx 1037~\text{K} \] ### Exit Velocity Calculation Using energy balance: \[ V_e = \sqrt{2 C_p (T_3 - T_e)} \] \[ V_e \approx 1154~\text{m/s} \] ## Final Answers ### a) Nozzle entrance (burner exit) pressure: \[ \boxed{P_3 = 149.5~\text{kPa}} \] ### b) Nozzle exit velocity: \[ \boxed{V_e = 1154~\text{m/s}} \]