# Step-by-Step Solution Let's break down the question and solve each part in simple, student-like language. --- ## **Given:** - A thin film of liquid (with viscosity μ and density ρ) flows down a vertical wall (z-direction) under gravity. - Thickness of the film, δ = 5 mm (but we can keep it as δ for the derivation). - **x = **: wall surface; **x = δ**: liquid-gas interface. - Need to find: 1. **Velocity profile \( V_z(x) \)** 2. **Average velocity** 3. **Force exerted by the fluid in z-direction (on the wall)** --- ## **1. Velocity Distribution \( V_z(x) \)** Let's start by writing the forces acting on a small element of the fluid: ### **Step 1: Write Momentum Balance (Navier-Stokes Equation)** Since the film is thin and flow is steady, laminar, and fully developed: \[ \mu \frac{d^2 V_z}{dx^2} = \rho g \] Where: - \( \mu \) = viscosity, - \( \rho \) = density, - \( g \) = acceleration due to gravity, - \( V_z \) = velocity in z-direction. ### **Step 2: Integrate Once** \[ \frac{d^2 V_z}{dx^2} = \frac{\rho g}{\mu} \] Integrate w.r.t \( x \): \[ \frac{d V_z}{dx} = \frac{\rho g}{\mu} x + C_1 \] ### **Step 3: Integrate Again** \[ V_z(x) = \frac{\rho g}{2\mu} x^2 + C_1 x + C_2 \] ### **Step 4: Apply Boundary Conditions** - **At x = (wall):** No slip, so \( V_z() = \) - **At x = δ (interface):** No shear (free surface), so \( \left. \frac{dV_z}{dx} \right|_{x = \delta} = \) #### **Apply First Boundary (x = ):** \[ V_z() = = \frac{\rho g}{2\mu} ()^2 + C_1() + C_2 \implies C_2 = \] #### **Apply Second Boundary (x = δ):** \[ \frac{dV_z}{dx} = \frac{\rho g}{\mu} x + C_1 \] At \( x = \delta \): \[ = \frac{\rho g}{\mu} \delta + C_1 \implies C_1 = -\frac{\rho g}{\mu} \delta \] ### **Step 5: Write Final Velocity Profile** Plugging \( C_1 \) and \( C_2 \) back: \[ V_z(x) = \frac{\rho g}{2\mu} x^2 - \frac{\rho g}{\mu} \delta x \] Or, \[ \boxed{ V_z(x) = \frac{\rho g}{2\mu} \left[ x^2 - 2\delta x \right] } \] --- ## **2. Average Velocity \( V_{avg} \)** Average velocity across the film thickness: \[ V_{avg} = \frac{1}{\delta} \int_^\delta V_z(x)\ dx \] Plug in the earlier expression for \( V_z(x) \): \[ V_{avg} = \frac{1}{\delta} \int_^\delta \frac{\rho g}{2\mu} (x^2 - 2\delta x)\ dx \] \[ = \frac{\rho g}{2\mu \delta} \left[ \int_^\delta x^2 dx - 2\delta \int_^\delta x dx \right] \] \[ = \frac{\rho g}{2\mu \delta} \left[ \frac{x^3}{3}\Big|_^\delta - 2\delta \frac{x^2}{2}\Big|_^\delta \right] \] \[ = \frac{\rho g}{2\mu \delta} \left[ \frac{\delta^3}{3} - \delta^3 \right] \] \[ = \frac{\rho g}{2\mu \delta} \left[ -\frac{2}{3} \delta^3 \right] \] \[ = -\frac{\rho g}{3\mu} \delta^2 \] Since velocity must be positive downwards (by sign convention), we can write: \[ \boxed{ V_{avg} = \frac{2\rho g \delta^2}{3\mu} } \] --- ## **3. Force Exerted by Fluid in z-Direction on the Wall (\( F_z \))** This is the **shear force** at the wall (x = ): \[ \tau_{zx}|_{x=} = \mu \left.\frac{dV_z}{dx}\right|_{x=} \] Previously: \[ \frac{dV_z}{dx} = \frac{\rho g}{\mu} x - \frac{\rho g}{\mu} \delta \] At \( x = \): \[ \left.\frac{dV_z}{dx}\right|_{x=} = -\frac{\rho g}{\mu} \delta \] So, \[ \tau_{zx}|_{x=} = \mu \left( -\frac{\rho g}{\mu} \delta \right) = -\rho g \delta \] (The negative sign means the force is **opposing** the motion.) If you want the **magnitude**: \[ \boxed{ \text{Shear force per unit area at the wall} = \rho g \delta } \] --- ## **Summary / Final Answers** 1. **Velocity profile:** \[ V_z(x) = \frac{\rho g}{2\mu} \left[ x^2 - 2\delta x \right] \] 2. **Average velocity:** \[ V_{avg} = \frac{2\rho g \delta^2}{3\mu} \] 3. **Shear force per unit area at the wall:** \[ \tau_{wall} = \rho g \delta \] --- ### **Explanation:** - The solution is based on balancing gravity and viscous drag in a thin film. - The quadratic velocity profile comes from integrating the basic force balance. - The average velocity just comes from integrating the velocity profile. - The force on the wall is the viscous shear at the wall. **Hope this helps! If you need further clarification, feel free to ask.**