Let's break this down **step by step**! --- ## **Step 1: Understanding the Problem** The stream function, \(\psi(x, y)\), is given at the four corners of a square cell (\(.7\, \text{m} \times .7\, \text{m}\)). You need to estimate the **velocity magnitude** and **direction** at the cell center. --- ## **Step 2: Stream Function to Velocity** For **2D incompressible flow**, the velocity components are: \[ u = \frac{\partial \psi}{\partial y} \] \[ v = -\frac{\partial \psi}{\partial x} \] where: - \(u\): velocity in \(x\) direction - \(v\): velocity in \(y\) direction --- ## **Step 3: Assign the Values** Let's assign the corners: | Corner | Coordinates | \(\psi\) value \((\text{m}^2/\text{s})\) | |------------------|-----------------|-------------------------------------------| | Bottom left | (, ) | 1.7308 | | Bottom right | (.7, ) | 1.7978 | | Top left | (, .7) | 1.9552 | | Top right | (.7, .7) | 2.0206 | --- ## **Step 4: Estimate Derivatives at Center** Use **central differences** at the center \((.35, .35)\): \[ u = \frac{\partial \psi}{\partial y} \approx \frac{\psi_{\text{top avg}} - \psi_{\text{bottom avg}}}{\Delta y} \] Where: - \(\psi_{\text{top avg}} = \frac{1.9552 + 2.0206}{2}\) - \(\psi_{\text{bottom avg}} = \frac{1.7308 + 1.7978}{2}\) - \(\Delta y = .7\, \text{m}\) Calculate: \[ \psi_{\text{top avg}} = \frac{1.9552 + 2.0206}{2} = 1.9879 \] \[ \psi_{\text{bottom avg}} = \frac{1.7308 + 1.7978}{2} = 1.7643 \] \[ u = \frac{1.9879 - 1.7643}{.7} = \frac{.2236}{.7} = .3194\, \text{m/s} \] --- \[ v = -\frac{\partial \psi}{\partial x} \approx -\frac{\psi_{\text{right avg}} - \psi_{\text{left avg}}}{\Delta x} \] Where: - \(\psi_{\text{right avg}} = \frac{1.7978 + 2.0206}{2} = 1.9092\) - \(\psi_{\text{left avg}} = \frac{1.7308 + 1.9552}{2} = 1.843\) - \(\Delta x = .7\, \text{m}\) Calculate: \[ v = -\frac{1.9092 - 1.843}{.7} = -\frac{.0662}{.7} = -.0946\, \text{m/s} \] --- ## **Step 5: Resultant Velocity Magnitude** \[ V = \sqrt{u^2 + v^2} \] \[ V = \sqrt{(.3194)^2 + (-.0946)^2} \] \[ V = \sqrt{.102 + .0089} = \sqrt{.1109} = .333\, \text{m/s} \] --- ## **Step 6: Angle with Respect to x-axis** \[ \alpha = \tan^{-1}\left(\frac{v}{u}\right) \] \[ \alpha = \tan^{-1}\left(\frac{-.0946}{.3194}\right) = \tan^{-1}(-.2962) = -16.5^\circ \] --- ## **Final Answers** ### **Resultant Velocity at Center:** - **Magnitude \(V\):** \(\boxed{.33\, \text{m/s}}\) - **Angle \(\alpha\) (from x-axis):** \(\boxed{-16.5^\circ}\) (below the x-axis) --- ### **Summary Table** | Quantity | Value | |-------------------------|------------------| | \(u\) | \(.32\, \text{m/s}\) | | \(v\) | \(-.095\, \text{m/s}\) | | \(V\) | \(.33\, \text{m/s}\) | | \(\alpha\) | \(-16.5^\circ\) | --- **Let me know if you'd like a diagram or further explanation of any step!**