Let's break down the problem and solve it step by step. --- # **Problematement** **Given:** - A vertical wetted-wall tower with a thin liquid film (thickness = δ) flowing down the wall in the _z_-direction under gravity. - The film thickness, δ = 5 m (though this is unusually thick for a film, but we'll use it as given). - Viscosity = μ, Density = ρ (both constant) - \( x = \): wall surface; \( x = \delta \): liquid-gas interface - Find: 1. **Velocity distribution** \( V_z(x) \) 2. **Average velocity** 3. **Force exerted by fluid in the _z_ direction** on the surface perpendicular to the _x_ direction --- # **1. Velocity Distribution \( V_z(x) \)** **Assumptions:** - Steady, incompressible, laminar flow - Flow in the \( z \)-direction only - Gravity acts downward (in \( z \)-direction) - No pressure gradient in \( z \)-direction (open to atmosphere at interface) **Governing Equation (Navier-Stokes for this scenario):** \[ \mu \frac{d^2 V_z}{dx^2} = \rho g \] Where: - \( V_z \): Velocity in the \( z \)-direction - \( x \): Distance from wall **Boundary Conditions:** 1. **No-slip at wall:** \( V_z() = \) 2. **Zero shear at interface:** \( \left. \frac{dV_z}{dx} \right|_{x=\delta} = \) --- ## **Solving the Differential Equation** Integrate once: \[ \mu \frac{dV_z}{dx} = \rho g x + C_1 \] Integrate again: \[ \mu V_z = \frac{\rho g}{2} x^2 + C_1 x + C_2 \] \[ V_z(x) = \frac{\rho g}{2\mu} x^2 + \frac{C_1}{\mu} x + \frac{C_2}{\mu} \] Apply boundary conditions: 1. **At \( x= \): \( V_z() = \)** \[ = + + \frac{C_2}{\mu} \implies C_2 = \] 2. **At \( x = \delta \): \( \left. \frac{dV_z}{dx} \right|_{x=\delta} = \)** \[ \frac{dV_z}{dx} = \frac{\rho g}{\mu} x + \frac{C_1}{\mu} \] Set \( x = \delta \): \[ = \frac{\rho g}{\mu} \delta + \frac{C_1}{\mu} \implies C_1 = -\rho g \delta \] So, \[ V_z(x) = \frac{\rho g}{2\mu} x^2 - \frac{\rho g \delta}{\mu} x \] Or, more neatly: \[ \boxed{ V_z(x) = \frac{\rho g}{2\mu} \left[ x^2 - 2\delta x \right] } \] or \[ \boxed{ V_z(x) = \frac{\rho g}{2\mu} \left[ x^2 - 2\delta x \right] } \] --- # **2. Average Velocity** The average velocity across the film thickness (\( \le x \le \delta \)) is: \[ V_{avg} = \frac{1}{\delta} \int_^\delta V_z(x) dx \] Plug in the expression for \( V_z(x) \): \[ V_{avg} = \frac{1}{\delta} \int_^\delta \frac{\rho g}{2\mu} (x^2 - 2\delta x) dx \] \[ = \frac{\rho g}{2\mu \delta} \int_^\delta (x^2 - 2\delta x) dx \] Integrate: \[ \int_^\delta x^2 dx = \frac{\delta^3}{3} \] \[ \int_^\delta 2\delta x dx = 2\delta \cdot \frac{x^2}{2} \Big|_^\delta = \delta^3 \] So, \[ V_{avg} = \frac{\rho g}{2\mu \delta} \left( \frac{\delta^3}{3} - \delta^3 \right) = \frac{\rho g}{2\mu \delta} \left( -\frac{2}{3}\delta^3 \right) = -\frac{\rho g \delta^2}{3\mu} \] But the negative sign here simply reflects the direction (downwards). So, the magnitude is: \[ \boxed{ V_{avg} = \frac{\rho g \delta^2}{3\mu} } \] --- # **3. Force Exerted by the Fluid in the \(z\)-direction** This is the **shear force** exerted by the fluid on the wall (\( x = \)) in the \(z\)-direction. Shear stress at the wall: \[ \tau_w = \mu \left. \frac{dV_z}{dx} \right|_{x=} \] From earlier: \[ \frac{dV_z}{dx} = \frac{\rho g}{\mu} x - \frac{\rho g \delta}{\mu} \] At \( x = \): \[ \left. \frac{dV_z}{dx} \right|_{x=} = -\frac{\rho g \delta}{\mu} \] So, \[ \tau_w = \mu \left( -\frac{\rho g \delta}{\mu} \right) = -\rho g \delta \] The negative sign indicates the direction (opposite to flow). The force per unit area is: \[ \boxed{ \tau_w = \rho g \delta } \] (magnitude, in the \( z \)-direction) If you want **total force** on an area \( A \): \[ F = \tau_w \cdot A = (\rho g \delta) \cdot A \] --- ## **Summary Table** | Quantity | Expression | |----------|------------| | **Velocity profile** \( V_z(x) \) | \( \frac{\rho g}{2\mu}(x^2 - 2\delta x) \) | | **Average velocity** \( V_{avg} \) | \( \frac{\rho g \delta^2}{3\mu} \) | | **Wall shear stress** \( \tau_w \) | \( \rho g \delta \) | --- If you need a step-by-step example with numbers or further explanation, just ask!