Let's solve the problem step by step: Given Data: - \( M_1 = .4 \) - \( M_2 = .1 \) - \( p_1 = 100 \) kPa - \( T_1 = 15^\circ \text{C} = 288.15 \) K - \( C_{PR} = .75 \) - \( \gamma = 1.4 \) - \( R = 287 \) J/kg·K --- ## (a) Inlet Dynamic Pressure, \( q_1 \) Dynamic pressure: \[ q_1 = \frac{1}{2} \rho_1 V_1^2 \] But, \[ V_1 = M_1 \cdot a_1 \] \[ a_1 = \sqrt{\gamma R T_1} \] Calculate \( a_1 \): \[ a_1 = \sqrt{1.4 \times 287 \times 288.15} = \sqrt{1.4 \times 287 \times 288.15} \approx 340.3 \text{ m/s} \] So, \[ V_1 = .4 \times 340.3 = 136.12 \text{ m/s} \] Now, calculate \( \rho_1 \): \[ \rho_1 = \frac{p_1}{RT_1} = \frac{100,000}{287 \times 288.15} \approx 1.21 \text{ kg/m}^3 \] Calculate \( q_1 \): \[ q_1 = \frac{1}{2} \times 1.21 \times (136.12)^2 \approx 11,210 \text{ Pa} = 11.2 \text{ kPa} \] --- ## (b) Exit Static Pressure, \( p_2 \) Diffuser static pressure recovery: \[ C_{PR} = \frac{p_2 - p_1}{p_{t1} - p_1} \] \[ p_2 = p_1 + C_{PR}(p_{t1} - p_1) \] First, find inlet total (stagnation) pressure \( p_{t1} \): \[ \frac{p_{t1}}{p_1} = \left[1 + \frac{\gamma-1}{2}M_1^2 \right]^{\frac{\gamma}{\gamma-1}} \] \[ = [1 + .2 \times .16]^{3.5} = [1 + .032]^{3.5} = 1.032^{3.5} \] \[ = e^{3.5 \ln(1.032)} \approx e^{3.5 \times .03147} = e^{.1101} \approx 1.116 \] \[ p_{t1} = 1.116 \times 100 = 111.6 \text{ kPa} \] Now, solve for \( p_2 \): \[ p_2 = 100 + .75 \times (111.6 - 100) = 100 + .75 \times 11.6 = 100 + 8.7 = 108.7 \text{ kPa} \] --- ## (c) Exit Total Pressure, \( p_{t2} \) \[ \frac{p_{t2}}{p_2} = \left[1 + \frac{\gamma-1}{2}M_2^2 \right]^{\frac{\gamma}{\gamma-1}} \] \[ [1 + .2 \times .01]^{3.5} = [1 + .002]^{3.5} = 1.002^{3.5} \] \[ = e^{3.5 \times .001998} = e^{.00699} \approx 1.007 \] \[ p_{t2} = 1.007 \times 108.7 = 109.46 \text{ kPa} \] --- ## (d) Diffuser Area Ratio \( \frac{A_2}{A_1} \) \[ \frac{A_2}{A_1} = \frac{M_1}{M_2} \left( \frac{1 + \frac{\gamma-1}{2}M_2^2}{1 + \frac{\gamma-1}{2}M_1^2} \right)^{\frac{\gamma+1}{2(\gamma-1)}} \] First, compute the ratio inside the exponent: - \( 1 + .2 \cdot .01 = 1.002 \) - \( 1 + .2 \cdot .16 = 1.032 \) - Exponent: \( \frac{1.4+1}{2 \times .4} = \frac{2.4}{.8} = 3 \) So, \[ \frac{A_2}{A_1} = \frac{.4}{.1} \left( \frac{1.002}{1.032} \right)^3 = 4 \times (.970)^3 \approx 4 \times .913 = 3.65 \] --- ## **Final Answers** **(a)** Inlet dynamic pressure, \( q_1 = \boxed{11.2 \text{ kPa}} \) **(b)** Exit static pressure, \( p_2 = \boxed{108.7 \text{ kPa}} \) **(c)** Exit total pressure, \( p_{t2} = \boxed{109.5 \text{ kPa}} \) **(d)** Diffuser area ratio, \( \frac{A_2}{A_1} = \boxed{3.65} \)