Viscosity and Turbulent flow

Viscosity and turbulent flow

Viscosity characterizes the degree of internal friction in fluids. This internal friction, or viscous force, is associated with the resistance experienced by two adjacent layers of fluid moving relative to each other. Viscosity causes part of the kinetic energy of a fluid to be converted to internal energy. The viscosity mechanism is similar to the one by which an object sliding on a rough horizontal surface loses kinetic energy. In previous sections, the discussion focused only on an ideal fluid (steady flow non-viscous and incompressible fluid). In this section, you will learn about the characteristics of viscous fluid and turbulent flow, specifically, the Newton's law of viscosity, Poiseuille's formula, Stokes' law and applications of viscosity in daily life.

Coefficient of viscosity

Coefficient of viscosity $\eta$ of a fluid is a measure of the degree to which the fluid exhibits viscous effects. This effect is described by Newton's law of viscosity which states that, "The frictional force $F$ between the layers is directly proportional to the area $A$ of the layers and the velocity gradient $\frac{dv}{dh}$."

Consider a pipe that contains a fluid flowing steadily. There are so many cylindrical fluid layers located at different positions within the pipe. Layers have varying speed ranging from zero at the wall of the pipe to the maximum speed at the center of the pipe. Fluid layers between C and B have velocities which are less than that of C but greater than that at P. If A is the surface area of layers in contact, $\frac{v_1 - v_2}{h}$ is the velocity gradient where $h$ is the distance of separation of the two layers with velocities $v_1$ and $v_2$. Then according to Newton's law of viscosity: F \propto A \left( \frac{v_1 - v_2}{h} \right) \tag{5.17} Introducing the constant of proportionality into equation (5.17), gives, F = \eta A \left( \frac{v_1 - v_2}{h} \right) \tag{5.18} Making $\eta$ the subject of the formula gives: \eta = \frac{Fh}{A(v_1 - v_2)} \tag{5.19} Hence, from equation (5.19) the coefficient of viscosity force is defined as the frictional force per unit area per unit velocity gradient. $[\eta] = \frac{[F]}{[A] \cdot [v_G]}$ But, $[F] = MLT^{-2},\quad [A] = L^2,\quad [v_G] = T^{-1}$ hence the dimensions of viscosity, $[\eta] = ML^{-1}T^{-1}$ Therefore, the unit of coefficient of viscosity is $\text{kg} \cdot \text{m}^{-1} \cdot \text{s}^{-1}$ or $\text{N} \cdot \text{s} \cdot \text{m}^{-2}$.

Note that, viscosity of an ideal fluid is zero. The coefficient of viscosity of a liquid decreases with an increase in temperature. But for gases, the coefficient of viscosity increases with increase in temperature.

Viscosity (particularly of oil and grease) is utilized in lubricants for various parts of machines. Viscosity is related to internal friction and hence it affects heat generation in bearings, cylinders and gear sets, therefore, various parts of machines require specific density of lubricants. The knowledge of viscosity is therefore important in measuring and choosing related lubricants for machinery parts. This means that the viscosity of an oil is foremost to be considered when selecting lubricating oil for a specific application.

Poiseuille's formula

Poiseuille studied the flow of a liquid through a horizontal pipe and found that the volume of liquid flowing out per second $\frac{V}{t}$, depends on the coefficient of viscosity $\eta$, the pressure gradient $P$ and the radius $a$ of the tube. Thus, $\frac{V}{t} \propto \frac{Pa^4}{\eta l}$ where $P$ is the pressure difference between the two ends of the pipe of length $l$.

By method of dimensional analysis: $\frac{V}{t} = k \cdot \eta^x a^y \left( \frac{P}{l} \right)^z$ Equating dimensions, $M^0L^3T^{-1} = (ML^{-1}T^{-1})^x \cdot L^y (ML^{-2}T^{-2})^z$ Equating powers of like terms: $M: x + z = 0$ $L: -x + y - 2z = 3$ $T: -x - 2z = -1$ Solving for $x, y, z$: $x = -1,\quad y = 4,\quad z = 1$ Substituting the values of $x, y, z$ into the formula: $\frac{V}{t} = k \cdot \frac{Pa^4}{\eta l}$ The constant of proportionality $k$ was experimentally found to be $\frac{\pi}{8}$. Hence: $\frac{V}{t} = \frac{\pi Pa^4}{8 \eta l}$ This is called the Poiseuille's formula. Through Poiseuille's formula, the coefficient of viscosity of a liquid can be determined. Consider the liquid flowing steadily from a tank of height $h$ through a capillary tube of length $l$ at constant pressure. A volume $V$ of the liquid to be collected at time interval $t$ is obtained by applying Poiseuille's formula, $\frac{V}{t} = \frac{\pi Pa^4}{8 \eta l}$ but $P = \rho gh$ where $h$ is the height of the liquid column from the capillary tube to the top of the liquid level and $\rho$ is the density of the liquid.

Hence the coefficient of viscosity can be obtained from the formula: $\frac{V}{t} = \pi \rho g h a^4 \div (8 \eta l)$

Stokes' law and terminal velocity

When a small solid sphere is dropped into a viscous liquid, the ball will accelerate and eventually reach a point where it moves with a constant velocity known as terminal velocity.

There are three forces that act upon the ball namely, the weight of the ball, the upthrust and the viscous force (Figure 5.13(b)). At terminal velocity, the net force is zero, since the acceleration is zero, then $F + U = W$ According to Stokes' law, "The viscous force is proportional to the radius $a$ of the ball, velocity $v$ of the ball, and coefficient of viscosity $\eta$ of the liquid".

By dimensional analysis, $F = k a^x \eta^y v^z$ Equating the dimensions, $[F] = MLT^{-2}, \quad [a^x] = L^x, \quad [\eta^y] = (ML^{-1}T^{-1})^y, \quad [v^z] = (LT^{-1})^z$ $MLT^{-2} = L^x M^y L^{-y} T^{-y} L^z T^{-z}$ Combining the exponents:

• Mass (M): $y = 1$
• Length (L): $x - y + z = 1$
• Time (T): $-y - z = -2$

Solving gives: $x = 1, y = 1, z = 1$

Substitution of these values into the equation gives: $F = k a \eta v$ From mathematical analysis, the proportional constant $k = 6\pi$

Therefore, the viscous force $F$ is expressed as: F = 6 \pi a \eta v \tag{5.22}

Suppose $\rho$ is the density of the sphere and $\sigma$ is the density of a fluid, then, the weight $W$ of the sphere: $W = mg, \quad m = \frac{4}{3} \pi \rho a^3$ $W = \frac{4}{3} \pi \rho a^3 g$ Similarly, upthrust $U$: $U = \frac{4}{3} \pi \sigma a^3 g$ At terminal velocity $v_t$: $F + U = W$ $6 \pi a \eta v_t + \frac{4}{3} \pi \sigma a^3 g = \frac{4}{3} \pi \rho a^3 g$ Simplifying for $v_t$: v_t = \frac{2 a^2 (\rho - \sigma) g}{9 \eta} \tag{5.23} Thus, equation (5.23) represents the expression of terminal velocity which is a constant velocity attained by a spherical body when falling through a viscous fluid.

A graph of velocity against time for motion of a ball falling in a viscous fluid and attaining terminal velocity is shown in Figure