# Viscosity

#### Prerequisites

When an object moves in a fluid -- a liquid or gas -- it drags bits of the fluid with it along its surface. This results in a layer of fluid sliding over a neighboring layer of fluid. The interactions of the molecules in the fluid result in a kind of internal friction that acts to slow the relative motion of neighboring layers of fluid. The full description of the mechanism of viscosity is rather complex, so we will treat it phenomenologically. Let's start by looking at the simplest possible example.

Imagine a liquid that is sandwiched between two plates. If the top plate is moved while the bottom is held fixed, the liquid will be *sheared* -- pulled so that the amount of deformation of the fluid changes perpendicular to the direction of flow. Some of the fluid sticks to and moves with the top plate, while the parts of the liquid next to the bottom plate remain at rest. This sets up a **gradient** (a rate of change with respect to space) in velocity across the liquid. This requires the layers of the fluid to move past one another as shown in the lower picture. The friction between the layers of fluid moving with different speeds is the source of the viscous resistive force.

## An equation for the viscous force

To get an equation for the resistive force that the fluid exerts on the plates as a result of the internal sliding, consider the example shown in the figure. Suppose the two plates have an area A and are separated by a thickness y of liquid. Suppose we are holding the bottom plate fixed and are dragging the top plate with a constant velocity $v$. If there were no internal resistance, any force on the top plate would continue to speed it up. But if we apply a constant force, $F^{app}_{P}$* *on the plate, we discover that it will speed up, but then its velocity will increase more slowly until it approaches a constant velocity, which we will call $u$. What's happening is that the acceleration of the plate, P, is determined by Newton's second law for the plate:

$$m_P a_P = F^{net}_P = F^{app}_{P} + F^{viscous}_{fluid \rightarrow P}$$

When the plate reaches a steady speed, the acceleration is 0, so the force we are applying is equal (and opposite) to the viscous force.

What we discover from our experiments is:

- the viscous force is proportional to the speed of the plate
- the viscous force is proportional to the area of the plate
- the viscous force is inversely proportional to the distance between the moving plate and the fixed plate.

The proportionality constant, *μ*, is called *the viscosity of the fluid *and is defined by:

$$\frac{F^{viscosity}_P}{A} = \mu \frac{u}{y}$$

We're sorry, but we've done it again. We're using the same symbol to mean two different things. We've already used $\mu$ to mean the coefficient of friction and here we are using it for viscosity. And what's worse is that they're not even the same kind of quantities. The coefficient of friction is a dimensionless number (the ratio of two forces) but the viscosity has dimensions as you'll see below. You'll have to be careful. When we're talking about solids $\mu$ means the friction coefficient while for objects moving in liquids it's the viscosity coefficient.

## Using dimensional analysis to find the viscous force on a moving sphere

While the two-plate experiment is a good way to figure out what's going on in the phenomenon of viscosity, it's a bit tricky to apply to an object moving in a fluid. Mostly, we'll want to be considering small objects like bacteria moving inside a fluid. Let's consider the motion of a sphere in a fluid and see what we can get from a dimensional analysis.

From our experiments with the plates, we see that our viscosity coefficient has the dimensionality of

[$\mu$] = $\left[ \frac{Fy}{Av} \right]$

[$Fy$] = (ML/T^{2})L = ML^{2}/T^{2}

[$Av$] = (L^{2})(L/T) = L^{3}/T

so

[$\mu$] = (ML^{2}/T^{2}) / (L^{3}/T) = ML^{2}T/L^{3}T^{2} = M/LT

Now suppose we have a small sphere of radius $R$ moving through a fluid with a velocity $v$. We expect there to be a viscous force from the fluid on the moving object, and we expect it to depend on the viscosity coefficient, $μ$, the velocity of the object, $v$, and the size of the object — some function of $R$.

We want to construct a force, which has dimensionality the same as "$ma$", or ML/T^{2}. Using our intuition from the plates example, we expect the force to be proportional to $μ$, which has dimensionality, M/LT. We expect it to be proportional to velocity, $v$, which has dimensionality L/T. So $μv$ has dimensionality (M/LT) * (L/T) = M/T^{2}.

To get a force we're only missing a factor with dimensionality "L". The only quantity with dimensionality L we have is the radius. So we expect our force to look something like $μvR$. This turns out to be right — but there's a dimensionless factor of $6 \pi$ — something we couldn't know by dimensional analysis. This factor depends on the fact that we have a sphere and that the boundaries of the fluid (the walls) are very far away compared to $R$ (otherwise we'd have to worry about the fluid stuck to the walls as well as to our sphere). The result for a sphere of radius $R$ moving at a velocity $v$ through a fluid with viscosity $\mu$ is

$$F^{viscosity}_{fluid \rightarrow sphere} = 6 \pi \mu Rv$$

## Units and Scales

Since viscosity has dimensions of M/LT, it will have units (in the SI system) of kg/m-s. Sometimes it's convenient to express this unit in different forms.

For example, since we will typically be building a force with it, we might want to rearrange this so it looks like Newtons: 1 N = 1 kg-m/s^{2}. So we can make the units of viscosity include a Newton by multiplying by m^{2}-s and dividing by the same factor. Looking at the dimensions of $mu$ and pulling out a force (ML/T^{2}), the result is

M/LT = (ML/T^{2}) (T/L^{2}).

So the units of viscosity will be Newtons x seconds / meters^{2}. The unit "Newton/meter^{2}" is a unit of pressure called a *Pascal*. So the units you will see for viscosity are typically "Pascal-seconds (Pa-s)".

The measured viscosities for air and water at standard temperature and pressures are:

$\mu_{air}$ | 1.81 x 10^{-5} Pa-s |

$\mu_{water}$ | 1.00 x 10^{-3} Pa-s |

$\mu_{sea water}$ | 1.07 x 10-^{3} Pa-s |

Karen Carleton & Joe Redish 9/29/11

#### Follow-ons

Last Modified: May 22, 2019