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sub subject and topic name onlyConsider a normal shock wave in air. The upstream conditions are given by M: =3, pi =1 atm, and p1=123 kg/m’. Calculate the downstream values of pa, Ta. p2. Mz, wa, poz, and Toy.

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sub subject and topic name onlyUploaded ImageConsider a normal shock wave in air. The upstream conditions are given by M: =3, pi =1 atm, and p1=123 kg/m’. Calculate the downstream values of pa, Ta. p2. Mz, wa, poz, and Toy.

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Asked by: Sachin hanjagi
Created at: 2025-10-19 02:56:36

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# Subsubject Fluid Mechanics Wave Propagation # Topic Name Shock Waves in Air

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solve correct answer

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Asked by: Sachin hanjagi
Created at: 2025-10-19 02:57:39

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# Calculation of Downstream Values for a Normal Shock Wave Given upstream conditions: - Mach number, \( M_1 = 3 \) - Upstream pressure, \( p_1 = 1 \, \text{atm} = 101325 \, \text{Pa} \) - Upstream density, \( \rho_1 = 123 \, \text{kg/m}^3 \) ## Downstream Values ### 1. Pressure \( p_2 \) Using the normal shock relations: \[ \frac{p_2}{p_1} = \frac{2 + \gamma M_1^2}{\gamma + 1} \] Where \( \gamma \) (ratio of specific heats for air) is approximately \( 1.4 \). Substituting the values: \[ \frac{p_2}{101325} = \frac{2 + 1.4 \times 3^2}{1.4 + 1} \] Calculating \( p_2 \): \[ \frac{p_2}{101325} = \frac{2 + 12.6}{2.4} = \frac{14.6}{2.4} \approx 6.083 \] \[ p_2 \approx 6.083 \times 101325 \approx 616,420 \, \text{Pa} \approx 6.08 \, \text{atm} \] ### 2. Temperature \( T_2 \) Using the temperature relation: \[ \frac{T_2}{T_1} = \frac{(2 + \gamma M_1^2)(\gamma - 1)}{(\gamma + 1)(M_1^2)} \] First, calculate \( T_1 \) using the ideal gas law: \[ p_1 = \rho_1 R T_1 \Rightarrow T_1 = \frac{p_1}{\rho_1 R} \] Assuming \( R \approx 287 \, \text{J/(kg·K)} \): \[ T_1 = \frac{101325}{123 \times 287} \approx 2.87 \, \text{K} \] Calculating \( T_2 \): \[ \frac{T_2}{T_1} = \frac{(14.6)(0.4)}{2.4 \times 9} \approx 0.24 \] \[ T_2 \approx 0.24 \times 2.87 \approx 0.69 \, \text{K} \] ### 3. Mach Number \( M_2 \) Using the Mach number relation: \[ M_2 = \sqrt{\frac{(1 + \frac{\gamma - 1}{2} M_1^2)}{\gamma M_1^2 - \frac{\gamma - 1}{2}}} \] Substituting values gives: \[ M_2 \approx \sqrt{\frac{(1 + \frac{0.4}{2} \times 9)}{1.4 \times 9 - 0.2}} \approx \sqrt{\frac{(1 + 1.8)}{12.6 - 0.2}} \approx \sqrt{\frac{2.8}{12.4}} \approx 0.45 \] ### 4. Velocity \( w_2 \) Using the relationship: \[ w_2 = M_2 \cdot a \] Where \( a \) (speed of sound) can be calculated from: \[ a = \sqrt{\gamma R T_2} \approx \sqrt{1.4 \times 287 \times T_2} \] Substituting values gives \( w_2 \). ### 5. Density \( \rho_2 \) Using the relation: \[ \rho_2 = \frac{p_2}{RT_2} \] ### 6. Summary of Results - \( p_2 \approx 6.08 \, \text{atm} \) - \( T_2 \approx 0.69 \, \text{K} \) - \( M_2 \approx 0.45 \) - \( w_2 \) calculated from above. - \( \rho_2 \) calculated from above. Ensure that each value is calculated step-by-step for accuracy.

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