Let's break down the solution step by step: --- ## Given Data - **Freestream Mach number:** \( M_ = 1.5 \) - **Mach number at cowl lip:** \( M_1 = .62 \) - **Normal-shock inlet, subcritical mode** - **Neglect viscous losses** --- ## (a) Inlet Spillage Mass Fraction \(\frac{\dot{m}_{spillage}}{\dot{m}_}\) ### **Step 1: Understand Key Terms** - **\(\dot{m}_\):** Mass flow that would enter the inlet if there was *no spillage* (i.e., if the inlet captured all the air in its geometric area at \( M_ \)). - **\(\dot{m}_1\):** Actual mass flow entering the inlet at the cowl lip (\( M_1 \)). - **\(\dot{m}_{spillage}\):** Mass flow that is "spilled" (not captured by the inlet). \[ \dot{m}_ = \rho_ U_ A_{lip} \] \[ \dot{m}_1 = \rho_1 U_1 A_{lip} \] But in subcritical mode, the actual mass flow entering the inlet at the cowl is less than the freestream mass flow that would have entered. \[ \dot{m}_ = (\rho_ U_) A_{lip} \] \[ \dot{m}_1 = (\rho_1 U_1) A_{lip} \] So, \[ \frac{\dot{m}_{spillage}}{\dot{m}_} = 1 - \frac{\dot{m}_1}{\dot{m}_} \] --- ### **Step 2: Express in Terms of Mach Numbers** \[ \frac{\dot{m}_1}{\dot{m}_} = \frac{\rho_1 U_1}{\rho_ U_} \] For perfect gas: \[ \rho = \frac{P}{RT}, \quad U = M a = M\sqrt{\gamma R T} \] \[ \frac{\rho_1}{\rho_} = \frac{P_1/T_1}{P_/T_} = \frac{P_1}{P_} \cdot \frac{T_}{T_1} \] \[ \frac{U_1}{U_} = \frac{M_1 a_1}{M_ a_} = \frac{M_1}{M_} \cdot \frac{a_1}{a_} = \frac{M_1}{M_} \cdot \sqrt{\frac{T_1}{T_}} \] So, \[ \frac{\dot{m}_1}{\dot{m}_} = \left(\frac{P_1}{P_}\right) \left(\frac{T_}{T_1}\right) \left(\frac{M_1}{M_}\right) \left(\sqrt{\frac{T_1}{T_}}\right) \] \[ = \frac{P_1}{P_} \cdot \frac{M_1}{M_} \cdot \frac{1}{\sqrt{\frac{T_1}{T_}}} \] --- ### **Step 3: Isentropic Relations** \[ \frac{T_1}{T_} = \frac{1 + \frac{\gamma-1}{2} M_^2}{1 + \frac{\gamma-1}{2} M_1^2} \] \[ \frac{P_1}{P_} = \left(\frac{T_1}{T_}\right)^{\frac{\gamma}{\gamma-1}} \] But since \( M_1 \) is downstream of the cowl, and \( M_ \) is freestream, use isentropic relations between them. #### **Calculate \( T_1/T_ \) and \( P_1/P_ \):** For air, \( \gamma = 1.4 \). \[ \frac{T_1}{T_} = \frac{1 + .2 \times 1.5^2}{1 + .2 \times .62^2} = \frac{1 + .45}{1 + .2 \times .3844} = \frac{1.45}{1.07688} \approx 1.347 \] \[ \left(\frac{P_1}{P_}\right) = \left(\frac{T_1}{T_}\right)^{\frac{\gamma}{\gamma-1}} = (1.347)^{3.5} \approx 2.85 \] **But this is for isentropic, and we have a normal shock at the cowl; however, for subcritical mode, the mass flux is determined by the ratio of Mach numbers:** \[ \frac{\dot{m}_1}{\dot{m}_} = \frac{M_1}{M_} \cdot \frac{1 + \frac{\gamma-1}{2} M_^2}{1 + \frac{\gamma-1}{2} M_1^2}^{1/2} \] \[ = \frac{.62}{1.5} \cdot \left( \frac{1 + .2 \cdot 1.5^2}{1 + .2 \cdot .62^2} \right)^{1/2} \] \[ = .413 \cdot \left( \frac{1.45}{1.07688} \right)^{1/2} \] \[ = .413 \cdot (1.3467)^{.5} \] \[ = .413 \cdot 1.161 \] \[ = .480 \] So, \[ \frac{\dot{m}_{spillage}}{\dot{m}_} = 1 - .480 = .520 \] --- ### **Final Answers for (a):** \[ \boxed{\frac{\dot{m}_{spillage}}{\dot{m}_} = .52} \] --- ## (b) Inlet Total Pressure Recovery, \( \pi_d \) \[ \pi_d = \frac{P_{,2}}{P_{,}} \] Where \( P_{,2} \) is the stagnation (total) pressure after the normal shock at the inlet throat. ### **Step 1: Isentropic Total Pressure Ratio to Shock** From freestream to just before the shock (isentropic): \[ \frac{P_{,1}}{P_{,}} = 1 \] (If the inlet is at the cowl, otherwise see the notes.) ### **Step 2: Total Pressure Drop Across Shock** For a normal shock: \[ \frac{P_{,2}}{P_{,1}} = \left[ \frac{(1 + \frac{\gamma-1}{2} M_1^2)}{(1 + \frac{\gamma-1}{2} M_2^2)} \right]^{\frac{\gamma}{\gamma-1}} \cdot \frac{(1+\gamma)M_1^2}{(2+\frac{\gamma-1}{2}M_1^2)}^{1/(\gamma-1)} \] But it's easier to use the normal shock tables for \( M_1 = 1.5 \): - Upstream Mach number: \( M_{shock,up} = 1.5 \) - Downstream Mach number: \( M_{shock,down} \) (find using normal shock relations) From normal shock tables for \( M_1 = 1.5 \): - \( M_2 = .701 \) - \( P_{02}/P_{01} = .855 \) ### **Step 3: Isentropic Compression from \( M_2 \) to \( M_1 \) (if necessary)** But in this problem, the total pressure recovery is just the normal shock recovery, since we are neglecting viscous losses. So, \[ \pi_d = \frac{P_{02}}{P_{00}} = \frac{P_{02}}{P_{01}} \cdot \frac{P_{01}}{P_{00}} \] But \( P_{01} = P_{00} \) since the inlet at the cowl lip is at freestream, so: \[ \pi_d = \frac{P_{02}}{P_{01}} = .855 \] --- ### **Final Answer for (b):** \[ \boxed{\pi_d = .855} \] --- ## **Summary Table** | Quantity | Value | |-------------------------------------------|-----------| | Inlet spillage mass fraction | .52 | | Inlet total pressure recovery (\( \pi_d \)) | .855 | --- ### **Assumptions** - Neglected viscous losses. - Used normal shock tables for \( M_1 = 1.5 \). - Subcritical mode, so spillage is calculated as above. --- Let me know if you'd like more detail or derivation!

# Solution to Inlet Flow Problem ## Given Data - **Freestream Mach number:** \( M = 1.5 \) - **Mach number at cowl lip:** \( M_1 = 0.62 \) - **Normal-shock inlet, subcritical mode** - **Neglect viscous losses** ## (a) Inlet Spillage Mass Fraction \(\frac{\dot{m}_{spillage}}{\dot{m}}\) ### Step 1: Key Terms - **\(\dot{m}\):** Mass flow that would enter the inlet without spillage. - **\(\dot{m}_1\):** Actual mass flow entering the inlet at the cowl lip (\( M_1 \)). - **\(\dot{m}_{spillage}\):** Mass flow that is spilled. The spillage mass fraction can be expressed as: \[ \frac{\dot{m}_{spillage}}{\dot{m}} = 1 - \frac{\dot{m}_1}{\dot{m}} \] ### Step 2: Express in Terms of Mach Numbers For mass flow: \[ \frac{\dot{m}_1}{\dot{m}} = \frac{\rho_1 U_1}{\rho U} \] Using the relationships for density and velocity in terms of Mach number: \[ \rho = \frac{P}{RT}, \quad U = M a = M\sqrt{\gamma R T} \] Thus, \[ \frac{\dot{m}_1}{\dot{m}} = \frac{\rho_1 U_1}{\rho U} = \frac{P_1}{P} \cdot \frac{M_1}{M} \cdot \sqrt{\frac{T}{T_1}} \] ### Step 3: Isentropic Relations Using the isentropic relations, we have: \[ \frac{T_1}{T} = \frac{1 + \frac{\gamma - 1}{2} M^2}{1 + \frac{\gamma - 1}{2} M_1^2} \] Substituting values (using \( \gamma = 1.4 \)): \[ \frac{T_1}{T} = \frac{1 + 0.2 \times 1.5^2}{1 + 0.2 \times 0.62^2} = \frac{1.45}{1.07688} \approx 1.347 \] Next, calculate \( \frac{P_1}{P} \): \[ \frac{P_1}{P} = \left(\frac{T_1}{T}\right)^{\frac{\gamma}{\gamma - 1}} = (1.347)^{3.5} \approx 2.85 \] ### Step 4: Calculate Spillage Mass Fraction Using the earlier derived equation: \[ \frac{\dot{m}_1}{\dot{m}} = \frac{P_1}{P} \cdot \frac{M_1}{M} \cdot \sqrt{\frac{T}{T_1}} \] Calculate \( \frac{\dot{m}_1}{\dot{m}} \): \[ \frac{\dot{m}_1}{\dot{m}} = \frac{2.85 \cdot 0.62}{1.5} \cdot \sqrt{\frac{1}{1.347}} \approx 0.480 \] Thus, the spillage mass fraction is: \[ \frac{\dot{m}_{spillage}}{\dot{m}} = 1 - 0.480 = 0.520 \] ### Final Answer for (a) \[ \boxed{\frac{\dot{m}_{spillage}}{\dot{m}} = 0.52} \] --- ## (b) Inlet Total Pressure Recovery, \( \pi_d \) ### Step 1: Total Pressure Ratio The total pressure recovery is given by: \[ \pi_d = \frac{P_{02}}{P_{0}} \] Where \( P_{02} \) is the stagnation pressure after the normal shock. ### Step 2: Pressure Drop Across Shock Using known values from normal shock relations for \( M_1 = 1.5 \): - \( P_{02}/P_{01} = 0.855 \) (from normal shock tables) ### Step 3: Total Pressure Recovery Since \( P_{01} = P_0 \): \[ \pi_d = \frac{P_{02}}{P_{01}} = 0.855 \] ### Final Answer for (b) \[ \boxed{\pi_d = 0.855} \] --- ## Summary | Quantity | Value | |------------------------------------------|-----------| | Inlet spillage mass fraction | 0.52 | | Inlet total pressure recovery (\( \pi_d \)) | 0.855 | --- ### Assumptions - Viscous losses are neglected. - Normal shock relations are used for calculations.