# Source Panel Problem: Solution Let's address each part, step by step. --- ## **Given** - Source strength per unit length: \(\gamma(s) = \gamma_\) (constant) - The source panel is parameterized by \(s\) from to 1- The panel center (origin) is at \((, )\). - We want quantities at point \((x, y)\), which is at distance \(r_m\) and angle \(\phi_m\) from the center. --- ## **(a) Potential Function, \(\phi(x, y)\)** For a line of sources, the potential function at \((x, y)\) due to an infinitesimal element at \(s\) is: \[ d\phi = \frac{\gamma_}{2\pi} \ln(r(s))\, ds \] where \(r(s)\) is the distance from the element at \(s\) to the field point \((x, y)\). Parameterize by the angle \(\phi(s)\) (from the panel center to \(s\)), and integrate from \(\phi_1\) to \(\phi_2\): \[ \phi(x, y) = \frac{\gamma_}{2\pi} \int_{\phi_1}^{\phi_2} \ln(r(s))\, ds \] But more commonly, we express this in terms of the panel geometry. For a straight panel of length \(L\), integrating along the panel gives: \[ \phi(x, y) = \frac{\gamma_}{2\pi} \int_{\phi_1}^{\phi_2} \ln(r_m) \frac{d\phi}{ds}\, ds \] But since \(r(s)\) varies, the exact form (for a straight panel) is: \[ \phi(x, y) = \frac{\gamma_}{2\pi} \left[ (\phi_2 - \phi_1) \ln(r_m) + (\phi_2 - \phi_1) \right] \] But for a general position: \[ \boxed{ \phi(x, y) = \frac{\gamma_}{2\pi} (\phi_2 - \phi_1) \ln(r_m) } \] *(Check your panel's geometry for precise limits and distances, but this is the standard result for a straight panel with constant strength)* --- ## **(b) Stream Function, \(\psi(x, y)\)** For a source panel, the stream function due to an infinitesimal element is: \[ d\psi = \frac{\gamma_}{2\pi} \theta(s)\, ds \] where \(\theta(s)\) is the angle subtended from the source point to the field point. For a straight panel: \[ \psi(x, y) = \frac{\gamma_}{2\pi} \int_{\phi_1}^{\phi_2} d\phi = \frac{\gamma_}{2\pi} (\phi_2 - \phi_1) \] \[ \boxed{ \psi(x, y) = \frac{\gamma_}{2\pi} (\phi_2 - \phi_1) } \] --- ## **(c) Velocities at \((x, y)\)** The velocity components are determined from the potential and stream function: \[ u = \frac{\partial \phi}{\partial x} = \frac{\partial \psi}{\partial y} \] \[ v = \frac{\partial \phi}{\partial y} = -\frac{\partial \psi}{\partial x} \] But for a source panel, the induced velocity at \((x, y)\) can be written as: \[ u = \frac{\gamma_}{2\pi} [\cos(\phi_2) - \cos(\phi_1)] \frac{1}{r_m} \] \[ v = \frac{\gamma_}{2\pi} [\sin(\phi_2) - \sin(\phi_1)] \frac{1}{r_m} \] Or, more generally, the velocity potential gradient gives: \[ \vec{V}(x, y) = \nabla \phi(x, y) \] But for the purposes of this problem, the answer is typically given in terms of the difference in angles and the geometry: \[ \boxed{ u = \frac{\gamma_}{2\pi r_m} (\sin \phi_2 - \sin \phi_1) } \] \[ \boxed{ v = -\frac{\gamma_}{2\pi r_m} (\cos \phi_2 - \cos \phi_1) } \] --- ## **Summary Table** | Quantity | Expression (in terms of \(r_m, \phi_1, \phi_2, \phi_m, \gamma_\)) | |--------------------------|--------------------------------------------------------------------| | **Potential, \(\phi\)** | \(\displaystyle \frac{\gamma_}{2\pi} (\phi_2 - \phi_1) \ln(r_m)\) | | **Stream function, \(\psi\)** | \(\displaystyle \frac{\gamma_}{2\pi} (\phi_2 - \phi_1)\) | | **Velocity, \(u\)** | \(\displaystyle \frac{\gamma_}{2\pi r_m} (\sin \phi_2 - \sin \phi_1)\) | | **Velocity, \(v\)** | \(\displaystyle -\frac{\gamma_}{2\pi r_m} (\cos \phi_2 - \cos \phi_1)\) | --- ## **Notes** - The above assumes the panel is straight and the angles are measured from the panel center. - For non-straight panels or varying source strength, the integrals would need to be evaluated accordingly. If you need more detail or a derivation, let me know!