## Example 1 Solution (Step-by-step) ### **Step 1: Find the upstream Mach number (M₁)** : - **₁ = 250 m/s** (velocity before shock) - **T₁ = 300 K** (temperature before shock) - **γ = 5/3** - **R = 2077 J/(kg·K)** First, calculate the speed of sound upstream: \[ a_1 = \sqrt{\gamma R T_1} \] \[ a_1 = \sqrt{\frac{5}{3} \times 2077 \times 300} \] \[ a_1 = \sqrt{1.6667 \times 2077 \times 300} \] \[ a_1 = \sqrt{1.6667 \times 623100} \] \[ a_1 = \sqrt{1,038,500} \] \[ a_1 \approx 1019 \text{ m/s} \] Now, calculate the upstream Mach number: \[ M_1 = \frac{V_1}{a_1} = \frac{250}{1019} \approx 2.45 \] **Explanation:** We calculated the Mach number to describe the flow speed relative to the speed of sound before the shock. --- ### **Step 2: Find the downstream (after shock) properties** #### **2a. Downstream Mach number (M₂) for normal shock:** \[ M_2^2 = \frac{(\gamma - 1)M_1^2 + 2}{2\gamma M_1^2 - (\gamma - 1)} \] \[ M_2^2 = \frac{(5/3-1) \cdot (2.45)^2 + 2}{2 \cdot (5/3) \cdot (2.45)^2 - (5/3 - 1)} \] \[ = \frac{(.6667) \cdot 6.0025 + 2}{2 \cdot 1.6667 \cdot 6.0025 - .6667} \] \[ = \frac{4.0017 + 2}{20.0067 - .6667} \] \[ = \frac{6.0017}{19.34} \] \[ M_2^2 \approx .3104 \implies M_2 \approx .557 \] #### **2b. Downstream velocity (V₂):** \[ a_2 = \sqrt{\gamma R T_2} \] But first, let's use the Mach number ratios and conservation laws. It is easier to use the following normal shock relations for velocity ratio: \[ \frac{V_2}{V_1} = \frac{(\gamma - 1)M_1^2 + 2}{(\gamma + 1)M_1^2} \] \[ = \frac{(.6667) \cdot 6.0025 + 2}{(2.6667) \cdot 6.0025} \] \[ = \frac{4.0017 + 2}{16.007} \] \[ = \frac{6.0017}{16.007} \] \[ \approx .375 \] \[ V_2 = .375 \times 250 = 938 \text{ m/s} \] **Explanation:** We used normal shock relations to find the Mach number and velocity after the shock. --- #### **2c. Downstream static temperature (T₂):** \[ \frac{T_2}{T_1} = \frac{[1 + \frac{\gamma-1}{2}M_1^2][2\gamma M_1^2 - (\gamma-1)]}{(\gamma+1)^2M_1^2} \] Alternatively, for practical use, you can relate it via pressure and density ratios, but let's use another direct formula: \[ \frac{T_2}{T_1} = \frac{(1 + \frac{\gamma-1}{2}M_1^2) \cdot [2 + (\gamma-1)M_1^2]}{(\gamma+1)^2 M_1^2 / (\gamma-1)} \] But it's easier to use the following: \[ \frac{T_2}{T_1} = \frac{[1 + \frac{\gamma-1}{2}M_1^2][2\gamma M_1^2 - (\gamma-1)]}{(\gamma+1)^2 M_1^2 / (\gamma-1)} \] Given the complexity, let's use standard tables or the usual temperature ratio for a normal shock: \[ \frac{T_2}{T_1} = \frac{[1 + \frac{\gamma-1}{2}M_1^2][2 + (\gamma-1)M_1^2]}{(\gamma+1)^2 M_1^2 / (\gamma-1)} \] But for quick purposes, \[ \frac{T_2}{T_1} = \frac{[1 + \frac{\gamma-1}{2} M_1^2] \cdot [2\gamma M_1^2 - (\gamma-1)]}{(\gamma+1)^2 M_1^2} \] Plug in the numbers: \[ 1 + \frac{.6667}{2} \cdot 6.0025 = 1 + .3333 \cdot 6.0025 = 1 + 2.0008 = 3.0008 \] \[ 2\gamma M_1^2 - (\gamma-1) = 2 \times 1.6667 \times 6.0025 - .6667 = 20.0067 - .6667 = 19.34 \] \[ (\gamma+1)^2 M_1^2 = (2.6667)^2 \cdot 6.0025 = 7.1111 \cdot 6.0025 = 42.677 \] So, \[ \frac{T_2}{T_1} = \frac{3.0008 \cdot 19.34}{42.677} \] \[ = \frac{58.03}{42.677} \approx 1.36 \] \[ T_2 = 1.36 \times 300 = 408 \text{ K} \] --- #### **2d. Stagnation temperature** For a normal shock, the stagnation temperature is constant (no heat transfer): \[ T_{_2} = T_{_1} \] \[ T_{_1} = T_1 \left[1 + \frac{\gamma - 1}{2} M_1^2 \right] \] \[ = 300 \left[1 + \frac{.6667}{2} \cdot 6.0025 \right] \] \[ = 300 \times 3.0008 \approx 900.24 \text{ K} \] --- ## **Final Answers** - **Downstream velocity \(V_2\):** **938 m/s** - **Downstream static temperature \(T_2\):** **408 K** - **Stagnation temperature:** **900 K** --- **Summary:** 1. Calculated Mach number and velocity after the shock using normal shock relations. 2. Determined static and stagnation temperatures after the shock.