Let's solve the problem step-by-step using the **integral method** for the **energy equation**, to determine the local Nusselt number for laminar flow over a vertical plate at constant surface temperature \( T_s \). --- ### **1. Problem Restatement** - **Flat vertical plate, laminar boundary layer.** - **Surface temperature \( T_s \), fluid temperature \( T_\infty \).** - **Velocity and temperature profiles are cubic polynomials:** \[ u(x, y) = a_0(x) + a_1(x) y + a_2(x) y^2 + a_3(x) y^3 \] \[ T(x, y) = b_0(x) + b_1(x) y + b_2(x) y^2 + b_3(x) y^3 \] - **Thermal boundary layer thickness: \( \delta_t(x) \).** - **We are to use the energy integral equation to find \( \delta_t(x) \), and hence the local Nusselt number.** --- ### **2. Energy Integral Equation** The **integral form of the energy equation** for boundary layer flow over a flat plate is: \[ \frac{d}{dx} \int_0^{\delta_t(x)} u (T - T_\infty) \, dy = \alpha \left. \frac{\partial T}{\partial y} \right|_{y=0} \] where \( \alpha \) is the thermal diffusivity. --- ### **3. Boundary Conditions** For temperature: - At \( y = 0 \): \( T = T_s \) - At \( y = \delta_t(x) \): \( T = T_\infty \) - At \( y = \delta_t(x) \): \( \frac{\partial T}{\partial y} = 0 \) (optional, for cubic profiles). For velocity: - At \( y = 0 \): \( u = 0 \) - At \( y = \delta(x) \): \( u = U_\infty \) - At \( y = \delta(x) \): \( \frac{\partial u}{\partial y} = 0 \) (optional). But since we're asked to use the **energy equation only** (i.e., only \( \delta_t(x) \)), we can choose a **velocity profile** that's known, for example: \[ u = U_\infty \cdot f\left(\frac{y}{\delta}\right) \] But the actual functional form isn't crucial for the energy equation unless we need to match the velocity and thermal boundary layers. For simplicity, let's assume \( u = U_\infty \) (i.e., neglecting the growth of the velocity boundary layer compared to the thermal BL, which is a standard assumption for very large Prandtl numbers; for generality, we can keep \( u = U_\infty \) in the integral). --- ### **4. Assume a Cubic Temperature Profile** Let: \[ \eta = \frac{y}{\delta_t(x)} \] Propose a cubic temperature profile: \[ \theta(\eta) = \frac{T(x, y) - T_\infty}{T_s - T_\infty} = 1 + b_1 \eta + b_2 \eta^2 + b_3 \eta^3 \] Apply BCs: - \( \eta = 0: \theta = 1 \implies b_0 = 1 \) - \( \eta = 1: \theta = 0 \implies 1 + b_1 + b_2 + b_3 = 0 \) - \( \eta = 1: \frac{d\theta}{d\eta} = 0 \implies b_1 + 2b_2 + 3b_3 = 0 \) We have three equations, and three unknowns (\( b_1, b_2, b_3 \)): 1. \( b_0 = 1 \) 2. \( 1 + b_1 + b_2 + b_3 = 0 \) 3. \( b_1 + 2b_2 + 3b_3 = 0 \) Let’s choose another condition, say, smoothness at the wall (just two BCs suffice for a cubic, but let's proceed as usual: sometimes \( d^2\theta/d\eta^2|_{\eta=0} = 0 \) is imposed, but typically two BCs are enough). Let's solve for \( b_1, b_2, b_3 \): From (2): \( b_1 + b_2 + b_3 = -1 \) From (3): \( b_1 + 2b_2 + 3b_3 = 0 \) Subtract (2) from (3): \[ (b_1 + 2b_2 + 3b_3) - (b_1 + b_2 + b_3) = 0 - (-1) \] \[ (2b_2 + 2b_3) = 1 \implies b_2 + b_3 = 0.5 \] Let’s let \( b_3 = a \), then \( b_2 = 0.5 - a \). Now, \( b_1 + b_2 + b_3 = -1 \implies b_1 + 0.5 - a + a = -1 \implies b_1 = -1.5 \). Now, check with the other equation: \( b_1 + 2b_2 + 3b_3 = 0 \implies -1.5 + 2(0.5 - a) + 3a = 0 \) \[ -1.5 + 1 - 2a + 3a = 0 \] \[ -0.5 + a = 0 \implies a = 0.5 \] So: \[ b_3 = 0.5,\quad b_2 = 0.5 - 0.5 = 0,\quad b_1 = -1.5 \] Thus, the temperature profile becomes: \[ \theta(\eta) = 1 - 1.5 \eta + 0.5 \eta^3 \] --- ### **5. Evaluate the Energy Integral** Recall the energy integral equation: \[ \frac{d}{dx} \int_0^{\delta_t} u (T - T_\infty) dy = \alpha \left. \frac{\partial T}{\partial y} \right|_{y=0} \] Let’s assume \( u = U_\infty \) (for simplicity; you can use a more accurate profile if desired). Change variables: \( y = \eta \delta_t \), \( dy = \delta_t d\eta \), \( \eta \) from 0 to 1. So, \[ \int_0^{\delta_t} u (T-T_\infty) dy = U_\infty (T_s-T_\infty) \int_0^{\delta_t} \theta(y) dy = U_\infty (T_s-T_\infty) \delta_t \int_0^1 \theta(\eta) d\eta \] Compute the integral: \[ \int_0^1 \left[1 - 1.5\eta + 0.5\eta^3\right] d\eta = [\eta - 0.75\eta^2 + 0.125\eta^4]_0^1 = 1 - 0.75 + 0.125 = 0.375 \] So, \[ \int_0^{\delta_t} u (T-T_\infty) dy = U_\infty (T_s-T_\infty) \delta_t \times 0.375 \] Therefore, \[ \frac{d}{dx} \left[U_\infty (T_s-T_\infty) \delta_t \times 0.375 \right] = \alpha \left. \frac{\partial T}{\partial y} \right|_{y=0} \] \[ U_\infty (T_s-T_\infty) \times 0.375 \frac{d \delta_t}{dx} = \alpha \left. \frac{\partial T}{\partial y} \right|_{y=0} \] --- ### **6. Evaluate the Wall Temperature Gradient** Recall: \[ T(x, y) = T_\infty + (T_s-T_\infty) [1 - 1.5\eta + 0.5\eta^3] \] \[ \theta(\eta) = 1 - 1.5\eta + 0.5\eta^3 \] So, \[ \frac{\partial T}{\partial y} = (T_s - T_\infty) \frac{d \theta}{d\eta} \cdot \frac{d \eta}{d y} = (T_s - T_\infty) \theta'(\eta) \cdot \frac{1}{\delta_t} \] At \( y=0 \), \( \eta = 0 \): \[ \theta'(\eta) = -1.5 + 3 \times 0.5 \eta^2 = -1.5 \text{ at } \eta = 0 \] So, \[ \left. \frac{\partial T}{\partial y} \right|_{y=0} = (T_s - T_\infty) \times (-1.5) \frac{1}{\delta_t} \] --- ### **7. Plug Back into the Energy Integral** \[ U_\infty (T_s - T_\infty) \times 0.375 \frac{d \delta_t}{dx} = \alpha (T_s - T_\infty) \times (-1.5) \frac{1}{\delta_t} \] Divide both sides by \( (T_s - T_\infty) \): \[ U_\infty \times 0.375 \frac{d \delta_t}{dx} = -1.5 \frac{\alpha}{\delta_t} \] Multiply both sides by \( \delta_t \): \[ U_\infty \times 0.375 \delta_t \frac{d \delta_t}{dx} = -1.5 \alpha \] \[ 0.375 U_\infty \delta_t \frac{d \delta_t}{dx} = -1.5 \alpha \] We can write \( \delta_t \frac{d \delta_t}{dx} = \frac{1}{2} \frac{d}{dx} (\delta_t^2) \): \[ 0.375 U_\infty \times \frac{1}{2} \frac{d}{dx} (\delta_t^2) = -1.5 \alpha \] \[ 0.1875 U_\infty \frac{d}{dx} (\delta_t^2) = -1.5 \alpha \] \[ \frac{d}{dx} (\delta_t^2) = -\frac{1.5}{0.1875} \frac{\alpha}{U_\infty} = -8 \frac{\alpha}{U_\infty} \] Integrate both sides from 0 to \( x \): \[ \delta_t^2(x) = -8 \frac{\alpha}{U_\infty} x + C \] At \( x = 0 \), \( \delta_t = 0 \implies C = 0 \): \[ \delta_t^2(x) = 8 \frac{\alpha x}{U_\infty} \implies \boxed{ \delta_t(x) = \sqrt{8 \frac{\alpha x}{U_\infty}} } \] --- ### **8. Local Nusselt Number** The local heat transfer coefficient is: \[ h_x = -\frac{k}{T_s - T_\infty} \left. \frac{\partial T}{\partial y} \right|_{y=0} \] So, \[ Nu_x = \frac{h_x x}{k} \] From above, \[ \left. \frac{\partial T}{\partial y} \right|_{y=0} = -1.5 (T_s - T_\infty)/\delta_t \] So, \[ h_x = \frac{1.5 k}{\delta_t} \] Therefore, \[ Nu_x = \frac{h_x x}{k} = \frac{1.5 x}{\delta_t} \] Now substitute \( \delta_t \): \[ Nu_x = \frac{1.5 x}{\sqrt{8 \alpha x / U_\infty}} = \frac{1.5 x}{\sqrt{8 (\alpha x/U_\infty)}} = \frac{1.5 x}{\sqrt{8} \sqrt{\alpha x / U_\infty}} = \frac{1.5 x}{2.828 \sqrt{\alpha x / U_\infty}} \] \[ = \frac{1.5 x U_\infty^{1/2}}{2.828 \alpha^{1/2} x^{1/2}} = \frac{1.5}{2.828} \left( \frac{U_\infty x}{\alpha} \right)^{1/2} \] But \( \frac{U_\infty x}{\nu} = Re_x \), and \( \frac{\nu}{\alpha} = Pr \