91Ó°ÊÓ

To estimate the boundary layer thickness at a location \(0.25\,m\) from the lower edge, we can use the similarity solution provided by Equation \(7.41\) for the laminar boundary layer: \[ \delta = 5x \left(\frac{\nu}{v_s}\right)^{1/2} \] where - \(\delta\) is the boundary layer thickness - \(x = 0.25\,m\) is the distance from the lower edge - \(\nu\) is the kinematic viscosity of air - \(v_s\) is the free stream velocity Now, we need the free stream velocity. Since the flow is natural convection, we can use the Rayleigh number to find it: \[ Ra_x = \frac{g \beta (T_s - T_\infty) x^3}{\nu \alpha} \] where - \(g = 9.81\,m/s^2\) is the acceleration due to gravity - \(\beta = 1/T_\infty\) is the volumetric thermal expansion coefficient - \(T_s = 130^\circ C\) is the surface temperature - \(T_\infty = 25^\circ C\) is the ambient temperature - \(\alpha\) is the thermal diffusivity of air From the Rayleigh number, we can find the Grashof number: \[ Gr_x = \frac{Ra_x}{Pr} \] where - \(Pr = \frac{\nu}{\alpha}\) is the Prandtl number For natural convection, the laminar flow holds when \(Gr_x < Gr_{crit} = 10^9\). Since we are given that the flow is laminar, we can assume this condition holds. Now, we can find the free stream velocity (\(v_s\)) using the relation between Grashof number and the Reynolds number: \[ Re_x = \frac{Gr_x^{1/2}}{Pr^{1/2}} = \frac{v_s x}{\nu} \] With \(v_s\), we can find \(\delta\) using the above equation for the boundary layer thickness.

To find the maximum velocity in the boundary layer at the given location and its position, we can use the similarity solution given by Equation \(9.7\): \[ u(x,y) = v_s \frac{\eta^2}{2} \frac{d^2F(\eta)}{d\eta^2} \] where - \(u(x,y)\) is the local velocity in the horizontal direction - \(\eta = \frac{y}{\delta}\) is the similarity variable - \(F(\eta)\) is the stream function The maximum velocity occurs when \(\frac{d^2F(\eta)}{d\eta^2} = 0\), which means: \[ \frac{d^2F(\eta)}{d\eta^2} = 0 \Rightarrow \eta_{max} = \sqrt{2} \] Plugging this into the equation for the maximum velocity and the position in the boundary layer, we have: \[ u_{max} = v_s \frac{2}{2} \frac{d^2F(\eta_{max})}{d\eta^2} = v_s\] \[ y_{max} = \eta_{max} \delta = \sqrt{2} \delta \]

To find the heat transfer coefficient at \(0.25\,m\) from the lower edge, we can use the similarity solution result, Equation \(9.19\) for the Nusselt number: \[ Nu_x = \frac{h x}{k} = 0.664 Re_x^{1/2} Pr^{1/3} \] where - \(h\) is the heat transfer coefficient - \(k\) is the thermal conductivity of air Now, we can find the heat transfer coefficient: \[ h = \frac{Nu_x k}{x} \]

To find the location where the boundary layer becomes turbulent, we need to find the critical Reynolds number \(Re_{crit}\). For the flow over a flat plate, the critical Reynolds number is around \(5 \times 10^5\). Using the Reynolds number definition: \[ Re_{crit} = \frac{v_s x_{crit}}{\nu} \] We can solve for the transition location \(x_{crit}\): \[ x_{crit} = \frac{Re_{crit} \nu}{v_s} \]