# Step-by-Step Solution Given: - Inlet pressure, \( P_1 = .5 \) MPa = \( 500 \) kPa - Inlet temperature, \( T_1 = 420 \ K - Inlet velocity, V_1 110 \) m/s - Ratio of specific heats, \( k 1. \) - Specific heat at constant pressure, \( c_p =1.005 \) kJ/kg·K - Gas constant, \( R = .287 \) kJ/kg·K We are to determine at the **sonic condition** (\( V_2 = a_2 \)): 1. The pressure \( P_2 \) 2. The temperature \( T_2 \) 3. The speed of sound at the inlet \( a_1 \) 4. The Mach number at the inlet \( Ma_1 \) --- ## 1. Find Speed of Sound at the Inlet The speed of sound in a perfect gas: \[ a_1 = \sqrt{kRT_1} \] Plug in the values: \[ a_1 = \sqrt{1.4 \times .287 \times 420} \] \[ a_1 = \sqrt{1.4 \times 120.54} \] \[ a_1 = \sqrt{168.756} \] \[ a_1 \approx 12.99 \text{ m/s} \quad \text{(incorrect units!)} \] **Note:** \( R \) is in kJ/kg·K, \( 1\, \text{kJ} = 100\, \text{J} \). \[ R = .287\, \text{kJ/kg·K} = 287\, \text{J/kg·K} \] So, \[ a_1 = \sqrt{1.4 \times 287 \times 420} \] \[ = \sqrt{1.4 \times 120,540} \] \[ = \sqrt{168,756} \] \[ a_1 \approx 410.7 \text{ m/s} \] --- ## 2. Find Mach Number at the Inlet \[ Ma_1 = \frac{V_1}{a_1} \] \[ Ma_1 = \frac{110}{410.7} \approx .2677 \] --- ## 3. Isentropic Relations for Sonic Condition We know at the exit, \( V_2 = a_2 \) (sonic). ### a) Use Energy Equation For isentropic flow, the total (stagnation) enthalpy is constant: \[ h_ = h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2} \] For an ideal gas, \( h = c_p T \): \[ c_p T_1 + \frac{V_1^2}{2} = c_p T_2 + \frac{V_2^2}{2} \] At \( V_2 = a_2 \): \[ c_p T_1 + \frac{V_1^2}{2} = c_p T_2 + \frac{a_2^2}{2} \] But \( a_2^2 = kRT_2 \): \[ c_p T_1 + \frac{V_1^2}{2} = c_p T_2 + \frac{1}{2}kRT_2 \] \[ c_p T_1 + \frac{V_1^2}{2} = T_2 \left[ c_p + \frac{1}{2}kR \right] \] \[ T_2 = \frac{c_p T_1 + \frac{V_1^2}{2}}{c_p + \frac{1}{2}kR} \] #### Plug in the values Convert \( c_p \) and \( R \) to J/kg·K: - \( c_p = 1.005\, \text{kJ/kg·K} = 1005\, \text{J/kg·K} \) - \( R = 287\, \text{J/kg·K} \) Calculate denominator: \[ c_p + \frac{1}{2}kR = 1005 + .5 \times 1.4 \times 287 \] \[ = 1005 + .7 \times 287 \] \[ = 1005 + 200.9 \] \[ = 1205.9 \] Calculate numerator: \[ c_p T_1 + \frac{V_1^2}{2} = 1005 \times 420 + \frac{110^2}{2} \] \[ = 1005 \times 420 + 605 \] \[ = 422100 + 605 \] \[ = 428150 \] So, \[ T_2 = \frac{428150}{1205.9} \approx 355.2\, \text{K} \] --- ### b) Find Pressure at Sonic Condition (\( P_2 \)) Isentropic relation: \[ \frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}} \] \[ \frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^{\frac{k}{k-1}} \] \[ \frac{P_2}{500} = \left( \frac{355.2}{420} \right)^{\frac{1.4}{.4}} \] \[ \left( \frac{355.2}{420} \right) = .846 \] \[ \left( .846 \right)^{3.5} = \exp(3.5 \times \ln(.846)) \] \[ = \exp(3.5 \times (-.167)) = \exp(-.5845) \approx .558 \] \[ P_2 = 500 \times .558 = 279\, \text{kPa} \] --- ## 4. Speed of Sound and Mach Number at the Inlet **Already calculated above:** - \( a_1 = 410.7\, \text{m/s} \) - \( Ma_1 = .2677 \) --- # **Final Answers** ## At the location where air velocity equals the speed of sound: - **Pressure:** \( P_2 = 279\, \text{kPa} \) - **Temperature:** \( T_2 = 355.2\, \text{K} \) - **Speed of sound at the inlet:** \( a_1 = 410.7\, \text{m/s} \) - **Mach number at the inlet:** \( Ma_1 = .268 \) --- ## **Summary Table** | Location | Pressure (kPa) | Temp (K) | Velocity (m/s) | Speed of Sound (m/s) | Mach Number | |-----------|:--------------:|:--------:|:--------------:|:--------------------:|:-----------:| | Inlet | 500 | 420 | 110 | 410.7 | .268 | | Sonic Pt. | 279 | 355.2 | 398.2* | 398.2 | 1 | \* At the sonic point, \( V_2 = a_2 = \sqrt{1.4 \times 287 \times 355.2} \approx 398.2\, \text{m/s} \)