Let's solve the problem step by step, addressing each part: --- # Q2 | Compressor Cascade Choking Analysis ## **Given Data** - Inlet Mach number, \( M_1 = .9 \) - Inlet flow angle, \( \alpha_1 = 52^\circ \) - Throat to frontal area ratio, \( \frac{A^*}{A_{1s}} = .625 \) - Choked at these conditions We'll assume **air** (\( \gamma = 1.4 \)), and the flow is **steady and adiabatic**. --- ## **a. One-dimensional compressible flow and Mach at the throat** ### **Assumptions:** - 1D steady flow - Perfect gas (air) - Adiabatic, no heat transfer - Neglecting body forces and shaft work - Flow is choked at the throat: \( M^* = 1 \) at the throat ### **Verification of Choking** For choked flow, the Mach number at the throat is 1. The area relation for isentropic nozzles is: \[ \frac{A^*}{A} = \frac{1}{M} \left[ \frac{2}{\gamma+1} \left(1 + \frac{\gamma-1}{2} M^2 \right) \right]^{\frac{\gamma+1}{2(\gamma-1)}} \] Here, \( A_{1s} \) is the **frontal area corresponding to the inlet Mach number**. ### **Calculate \( \frac{A^*}{A_{1s}} \) for \( M_1 = .9 \):** Plug in values: - \( \gamma = 1.4 \) - \( M_1 = .9 \) \[ \frac{A^*}{A_{1s}} = \frac{1}{.9} \left[ \frac{2}{2.4} \left(1 + \frac{.4}{2} \cdot .9^2 \right) \right]^{\frac{2.4}{.8}} \] Calculate step by step: - \( \frac{2}{2.4} = .833 \) - \( \frac{.4}{2} = .2 \) - \( .9^2 = .81 \) - \( 1 + .2 \times .81 = 1 + .162 = 1.162 \) Now: \[ \frac{A^*}{A_{1s}} = \frac{1}{.9} \left[ .833 \times 1.162 \right]^{3} \] \[ .833 \times 1.162 = .968 \] \[ .968^{3} = .907 \] \[ \frac{1}{.9} \times .907 = 1.008 \] But the given \( \frac{A^*}{A_{1s}} = .625 \), so **either the reference area \(A_{1s}\) is not exactly the upstream area, or there is a loss** (as expected). ### **Verification:** - The throat is choked, \( M^* = 1 \) at the throat. --- ## **b. Isentropic Area-Mach Relation and Stagnation Pressure Loss Coefficient** ### **1. Isentropic Stagnation Pressure Ratio (Ideal):** The isentropic stagnation pressure ratio from upstream to throat is: \[ \left( \frac{P_{,2}}{P_{,1}} \right)_{\text{isentropic}} = 1 \] (because for isentropic, no loss: stagnation pressure is conserved) But with area change, the actual loss is: \[ \text{Loss coefficient:} \quad \sigma = \frac{P_{,2}}{P_{,1}} \] ### **2. Compute Isentropic Area Ratio for \( M_1 = .9 \):** As above: \[ \left( \frac{A^*}{A_{1s}} \right)_{\text{isentropic}} = 1.008 \] But measured is .625, so the ratio of actual to isentropic area is: \[ \frac{.625}{1.008} = .62 \] ### **3. Stagnation Pressure Loss Calculation** For a given area ratio, the actual stagnation pressure loss follows: \[ \frac{A^*}{A_{1s}} = \frac{A^*}{A_{1s}} \bigg|_{\text{isentropic}} \cdot \sigma \] Thus, \[ \sigma = \frac{\text{actual } (A^*/A_{1s})}{\text{isentropic } (A^*/A_{1s})} \] \[ \sigma = \frac{.625}{1.008} = .62 \] **So, the stagnation pressure loss coefficient is \( \sigma = .62 \), i.e., 38% loss.** --- **_Comment on Possible Causes:_** - **Shock waves:** Sudden compression, non-isentropic. - **Viscous effects:** Boundary layer growth, separation, turbulence, and wake mixing. - **Leakage and secondary flows:** In cascades, tip leakage and endwall effects can cause extra loss. --- ## **c. Estimate Static Pressure and Temperature at Throat, and Effects of Non-idealities** ### **At the Throat (\(M^* = 1\)):** From isentropic relations: \[ \frac{T^*}{T_} = \frac{2}{\gamma+1} = \frac{2}{2.4} = .833 \] \[ \frac{P^*}{P_} = \left( \frac{2}{\gamma+1} \right)^{\frac{\gamma}{\gamma-1}} \] \[ \left( \frac{2}{2.4} \right)^{3.5} = .833^{3.5} \approx .486 \] So, - Static T at throat: \( T^* = .833 T_ \) - Static P at throat: \( P^* = .486 P_ \) **Effect of Shock Waves and Viscous Effects:** - **Shock waves:** Cause sudden increase in entropy, larger loss in stagnation pressure. - **Viscous effects:** Boundary layer and wake mixing reduce total pressure further compared to ideal estimates. --- ## **d. If Inlet Mach is Reduced Below .9 (at Same Flow Angle):** ### **Qualitative Effects:** - **Choking Condition:** No longer choked, since \( M_1 < .9 \) and area is fixed. - **Mass Flow Rate:** Decreases, since choking mass flow is maximum for given geometry. - **Stagnation Pressure Loss:** Loss is less dramatic, as no shock forms; but viscous losses still exist. - **Cascade Choking:** No, the cascade would NOT choke at inlet Mach below .9. --- # **Summary Table** | Quantity | Value | |-----------------------------------|--------------| | Isentropic \(A^*/A_{1s}\) | 1.008 | | Actual \(A^*/A_{1s}\) | .625 | | Stagnation Pressure Loss (\(\sigma\)) | .62 (i.e., 38% loss) | | Throat \(T^*/T_\) | .833 | | Throat \(P^*/P_\) | .486 | --- # **Final Answers (Summary)** **a.** The throat is choked (\(M^* = 1\)) at the given inlet Mach number using 1D compressible flow relations. **b.** The stagnation pressure loss coefficient is \(\sigma = .62\) (i.e., 38% loss), caused by shocks, viscous effects, and secondary flows. **c.** At the throat: \(T^* = .833 T_\), \(P^* = .486 P_\). Shock and viscous effects increase losses above isentropic predictions. **d.** Reducing inlet Mach below .9 prevents choking, decreases mass flow rate, and reduces stagnation pressure loss; cascade will not choke. --- Let me know if you want detailed calculations or further explanation on any step!