Question

A vertical steel rod has radius $a$. The rod has a coat of a liquid film on it. The liquid slides under gravity. It was found that the speed of liquid layer at radius $r$ is given by

$v = \frac{\rho g b^2}{2\eta} \ln(\frac{r}{a}) - \frac{\rho g}{4\eta}(r^2 - a^2)$

Where $b$ is the outer radius of liquid film, $\eta$ is coefficient of viscosity and $\rho$ is density of the liquid.

(i) Calculate the force on unit length of the rod due to the viscous liquid?

(ii) Set up the integral to calculate the volume flow rate of the liquid down the rod. [you may not evaluate the integral]

Answer 1

(i) Force per unit length = $\pi \rho g (b^2 - a^2)$ (downwards)

(ii) Volume flow rate = $\int_{a}^{b} \left[ \frac{\rho g b^2}{2\eta} \ln\left(\frac{r}{a}\right) - \frac{\rho g}{4\eta}(r^2 - a^2) \right] (2\pi r) dr$

Answer 2

The problem involves the flow of a viscous liquid film down a vertical rod. We are given the velocity profile of the liquid and asked to calculate the force on the rod and set up an integral for the volume flow rate.

Part (i): Calculate the force on unit length of the rod due to the viscous liquid.

The force exerted by a viscous fluid is related to the shear stress ($\tau$) at the surface. According to Newton's law of viscosity, the shear stress is given by: $\tau = \eta \frac{dv}{dr}$ where $\eta$ is the coefficient of viscosity and $\frac{dv}{dr}$ is the velocity gradient.

The given velocity profile is: $v = \frac{\rho g b^2}{2\eta} \ln\left(\frac{r}{a}\right) - \frac{\rho g}{4\eta}(r^2 - a^2)$

First, we need to find the velocity gradient $\frac{dv}{dr}$: $\frac{dv}{dr} = \frac{d}{dr} \left[ \frac{\rho g b^2}{2\eta} (\ln r - \ln a) - \frac{\rho g}{4\eta}(r^2 - a^2) \right]$ $\frac{dv}{dr} = \frac{\rho g b^2}{2\eta} \left(\frac{1}{r}\right) - \frac{\rho g}{4\eta}(2r)$ $\frac{dv}{dr} = \frac{\rho g b^2}{2\eta r} - \frac{\rho g r}{2\eta}$ $\frac{dv}{dr} = \frac{\rho g}{2\eta} \left( \frac{b^2}{r} - r \right)$

The force on the rod is due to the viscous liquid at its surface, which is at radius $r=a$. So, we evaluate the velocity gradient at $r=a$: $\left(\frac{dv}{dr}\right)_{r=a} = \frac{\rho g}{2\eta} \left( \frac{b^2}{a} - a \right) = \frac{\rho g}{2\eta a} (b^2 - a^2)$

Now, calculate the shear stress at the rod's surface ($r=a$): $\tau_a = \eta \left(\frac{dv}{dr}\right)_{r=a} = \eta \left[ \frac{\rho g}{2\eta a} (b^2 - a^2) \right]$ $\tau_a = \frac{\rho g}{2a} (b^2 - a^2)$

The force on unit length of the rod ($F'$) is the shear stress multiplied by the surface area per unit length. The surface area per unit length of the rod is its circumference, $2\pi a$. $F' = \tau_a \times (2\pi a)$ $F' = \left[ \frac{\rho g}{2a} (b^2 - a^2) \right] \times (2\pi a)$ $F' = \pi \rho g (b^2 - a^2)$ Since the liquid is flowing downwards, and the rod is stationary, the liquid exerts a downward viscous force on the rod.

Part (ii): Set up the integral to calculate the volume flow rate of the liquid down the rod.

The volume flow rate ($Q$) through a cross-section is given by the integral of the velocity over the area of flow. The liquid flows in an annular region between radius $a$ and $b$. Consider an elemental annular ring of radius $r$ and thickness $dr$. The area of this elemental ring is $dA = 2\pi r dr$. The volume flow rate through this elemental ring is $dQ = v \cdot dA = v (2\pi r) dr$.

To find the total volume flow rate, we integrate $dQ$ from the inner radius $a$ to the outer radius $b$: $Q = \int_{a}^{b} v (2\pi r) dr$ Substitute the given expression for $v$: $Q = \int_{a}^{b} \left[ \frac{\rho g b^2}{2\eta} \ln\left(\frac{r}{a}\right) - \frac{\rho g}{4\eta}(r^2 - a^2) \right] (2\pi r) dr$

This is the required integral setup for the volume flow rate.