91Ó°ÊÓ

For the given problem, we can use the simplified form of the energy equation, as we have a steady-state problem, and we can neglect viscous dissipation. The simplified energy equation for the boundary layer is: \[ \rho u \frac{\partial T}{\partial x} + \rho v \frac{\partial T}{\partial y} = k \frac{\partial^2 T}{\partial y^2} \] Considering that \(u=u_{\infty}\) throughout the thermal boundary layer, we get: \[ \rho u_\infty \frac{\partial T}{\partial x} + \rho v \frac{\partial T}{\partial y} = k \frac{\partial^2 T}{\partial y^2} \]

For the initial and boundary conditions, we have: Initial condition: at \(x=0\), \(T(x,0) = T_\infty\) Boundary conditions: - at \(y=0\), \(T(x,0) = T_s\), - as \(y \rightarrow \infty\), \(T(x,\infty) = T_\infty\).

Let's introduce a dimensionless temperature \(\theta\) as follows: \[ \theta(x, y) = \frac{T(x, y) - T_\infty}{T_s - T_\infty} \] The boundary conditions become: - at \(y=0\), \(\theta(x,0) = 1\), - as \(y \rightarrow \infty\), \(\theta(x,\infty) = 0\). Substituting for \(T\) in the energy equation, we get: \[ \rho u_\infty \frac{\partial\theta}{\partial x} + \rho v \frac{\partial\theta}{\partial y} = k \frac{\partial^2 \theta}{\partial y^2} \] Now, let's follow the hint and find the analogy to one-dimensional heat transfer in a semi-infinite medium with a sudden change in surface temperature. A known solution for the problem in one-dimension is: \[ \theta (y) = \operatorname{erfc}\left(\frac{y}{2\sqrt{\alpha x}}\right) \] where \(\alpha\) is the thermal diffusivity, and \(\operatorname{erfc}\) is the complementary error function. With a little adjustment, we can write the boundary layer temperature field: \[ \theta (x, y) = \operatorname{erfc}\left(\frac{y}{2\sqrt{\alpha' x}}\right) \] where \(\alpha' = \operatorname{Pr}\cdot\nu\) with \(\nu\) being the kinematic viscosity of the liquid metal. Now we have the boundary layer temperature field: \[ T(x, y) = T_\infty + (T_s - T_\infty)\cdot\operatorname{erfc}\left(\frac{y}{2\sqrt{\operatorname{Pr}\cdot\nu x}}\right) \]

The local Nusselt number is defined as: \[ \mathrm{Nu}_x = \frac{h_x x}{k} \] where \(h_x\) is the local heat transfer coefficient, and we recall that the heat flux is given by: \[ q_x = -k \frac{\partial T}{\partial y} \Big|_{y=0} \] Substituting the expression for the temperature field, we get: \[ q_x = -k \frac{\partial}{\partial y}\left[T_\infty + (T_s - T_\infty)\cdot\operatorname{erfc}\left(\frac{y}{2\sqrt{\operatorname{Pr}\cdot\nu x}}\right)\right] \Big|_{y=0} \] After calculating the derivative and plugging in \(y=0\): \[ q_x = -k \frac{T_s - T_\infty}{2\sqrt{\operatorname{Pr}\cdot\nu x}} \cdot -\frac{1}{\sqrt{\pi}} = \frac{k(T_s - T_\infty)}{\sqrt{\pi\operatorname{Pr}\nu x}} \] Using the definition of the heat transfer coefficient: \[ h_x = \frac{q_x}{T_s - T_\infty} = \frac{k}{\sqrt{\pi\operatorname{Pr}\nu x}} \] Finally, we get the expression for the local Nusselt number: \[ \mathrm{Nu}_x = \frac{h_x x}{k} = \frac{kx}{\sqrt{\pi\operatorname{Pr}\nu x} \cdot k} = \sqrt{\frac{x}{\pi\operatorname{Pr}\nu}} \]