# Step-by-Step Solution: Axial Stress in Nozzle Throat Given data: - **Gas constant, \( R \) = 693 J/kg·K** - **Throat (subscript \( t \)):** - Pressure, \( p_t = 2.8\, \text{MPa} = 2.8 \times 10^6\, \text{Pa} \) - Temperature, \( T_t = 269\, \text{K} \) - Velocity, \( V_t = 1514\, \text{m/s} \) - Area, \( A_t = .05\, \text{m}^2 \) - **Exit (subscript \( e \)):** - Pressure, \( p_e = .08\, \text{MPa} = 8. \times 10^4\, \text{Pa} \) - Velocity, \( V_e = 3325\, \text{m/s} \) - Area, \( A_e = .40\, \text{m}^2 \) - **Ambient pressure, \( p_a = .1\, \text{MPa} = 1. \times 10^5\, \text{Pa} \)** - **Wall thickness at throat, \( t = 5\, \text{mm} = .005\, \text{m} \)** - **Nozzle is circular in cross-section.** We are to **calculate the axial (longitudinal) stress experienced by the nozzle ring at the throat.** --- ## 1. **Find Inner Radius at Throat** The area at the throat: \[ A_t = \pi r_t^2 \implies r_t = \sqrt{\frac{A_t}{\pi}} \] \[ r_t = \sqrt{\frac{.05}{\pi}} \approx \sqrt{.01592} \approx .1262\, \text{m} \] --- ## 2. **Determine Axial Force on Throat Section** The axial force is the net force due to pressure difference and gas momentum acting on the cross-section. The **axial force (F)** on the nozzle wall at the throat is: \[ F = [p_t - p_a]\, A_t + \dot{m} V_t \] But, for a control surface at the throat, the typical stress is from the pressure difference across the (cut) section. For the **axial stress in the ring wall**, we typically use the pressure difference across the wall (\( p_t - p_a \)), as the wall must support the pressure inside against the ambient outside. Thus, \[ F_{\text{axial}} = (p_t - p_a) A_t \] --- ## 3. **Calculate Axial Stress** The **axial stress** in the wall: \[ \sigma_{\text{axial}} = \frac{F_{\text{axial}}}{A_{\text{wall}}} \] Where: - \( A_{\text{wall}} \) = cross-sectional area of the wall at the throat = area of the ring For a thin ring: \[ A_{\text{wall}} = \text{circumference} \times \text{thickness} = 2\pi r_t \times t \] ### Plug in the values: - \( p_t = 2.8 \times 10^6\, \text{Pa} \) - \( p_a = 1. \times 10^5\, \text{Pa} \) - \( A_t = .05\, \text{m}^2 \) - \( r_t = .1262\, \text{m} \) - \( t = .005\, \text{m} \) **Axial force:** \[ F_{\text{axial}} = (2.8 \times 10^6 - 1. \times 10^5) \times .05 \] \[ F_{\text{axial}} = (2.7 \times 10^6) \times .05 = 1.35 \times 10^5\, \text{N} \] **Wall area:** \[ A_{\text{wall}} = 2\pi \times .1262 \times .005 \] \[ A_{\text{wall}} = 2 \times 3.1416 \times .1262 \times .005 \approx .00397\, \text{m}^2 \] **Axial stress:** \[ \sigma_{\text{axial}} = \frac{1.35 \times 10^5}{.00397} \approx 3.40 \times 10^7\, \text{Pa} \] \[ \sigma_{\text{axial}} \approx 34.\, \text{MPa} \] --- ## 4. **Final Answer** **The axial stress experienced by the nozzle ring at the throat is:** \[ \boxed{34.~\text{MPa}} \] --- ### **Summary Table** | Quantity | Value | |-------------------------|--------------| | Throat radius | .126 m | | Wall area at throat | .00397 m² | | Axial force on ring | 135,000 N | | **Axial stress** | **34. MPa** | --- **Note:** - If the wall is not thin, for thick-walled cylinders, more advanced stress analysis (Lame's equations) may be needed, but for most nozzle walls, this thin-wall approximation is common. - The stress calculated is the average axial (longitudinal) stress due to the net pressure across the wall at the throat. --- **Nozzle throat stress diagram**  *Diagram of nozzle throat showing internal pressure \(p_t\), ambient pressure \(p_a\), area \(A_t\), and wall thickness \(t\).*