6.5.4 Drag at Low Reynolds Number

A $10 -μ m-diameter$ dust particle settles to the ground at about 20 mm/s. It’s not hard to calculate that the Reynolds number for the falling particle is $R e \approx 0.01$ . This is an example of motion at low Reynolds number, which we define to be $R e < 1$ . The drag force for low Reynolds number is almost entirely a viscous force.

For low Reynolds number, the drag force for motion through a fluid at speed v is

\vec{F}_{drag} = (b v, direction opposite the motion)

(6.17)

where b is a constant that depends on the size and shape of the object and on the viscosity of the fluid. For low Reynolds number, the magnitude of the drag force is proportional to the object’s speed. If the speed doubles, drag also increases by a factor of 2. This model of viscous drag is called linear drag.

NOTE Drag is more complex for intermediate values of Reynolds number, $1 < R e < 1000$ . We don’t consider those situations in this text.

Viscous drag is mostly applicable to very small objects, many of which can be reasonably modeled as spheres. That’s fortunate because only for spheres does theory give us any guidance for the coefficient b. For a spherical object of radius r moving in a fluid with viscosity $η$ , it can be shown that $b = 6 π η r$ . Thus

(\vec{F}_{drag})_{s p h e r e} = (6 π η r v, direction opposite the motion)

(6.18)

This expression for the viscous drag on a sphere is called Stokes’ law. Notice that linear drag at low Reynolds number depends on the fluid’s viscosity, whereas quadratic drag at high Reynolds number depends on the fluid’s density.

As before, an object reaches its terminal speed when the drag force exactly balances an applied force. At low Reynolds number, unlike high Re, this happens almost instantly, as you’ll see, so motion at low Reynolds number is almost entirely constant-speed motion at the terminal speed. The applied force needed to propel a sphere at its terminal speed is

F_{applied} = 6 π η r v_{term}

(6.19)

You can find the terminal speed of a small sphere falling in air by using $F_{applied} = m g$ , as we did for high Reynolds number, but this approach doesn’t work for objects falling in liquids. The reason is that the upward buoyant force of the liquid—a topic we’ll study in Chapter 14—cannot be neglected. We’ll return to this after we introduce buoyancy.

EXAMPLE 6.9 Measuring the mass of a pollen grain

Pollen grains are very light. In one experiment to determine their mass, researchers dropped grains inside a clear glass cylinder and then watched their motion with a microscope. A $40 -μ m-diameter$ pollen grain was observed to fall at a rate of $5.3 cm/s$ . What is the mass in nanograms of this grain?

MODEL We will model the grain as a sphere. At terminal speed, the force of gravity on the grain is equal to the drag force. This is a very small object moving quite slowly, so its Reynolds number is very low and the drag force is linear drag. We’ll assume the air in the cylinder was at a room temperature of 20°C and use the value for air viscosity in Table 6.2. In SI units, the terminal speed is $v_{term} = 0.053 m/s$ .

SOLVE We set the force of gravity equal to the linear drag force to get

m g = 6 π η r v_{term}

We can then solve for the mass as

\begin{matrix} m & = & \frac{6 π η r v_{term}}{g} \\ = & \frac{6 π (1.8 \times 10^{- 5} Pa s) (20 \times 10^{- 6} m) (0.053 m / s)}{9.8 m / s^{2}} \\ = & 3.7 \times 10^{- 11} kg \end{matrix}

Converting this result to nanograms (ng) gives

m = 3.7 \times 10^{- 11} kg \times \frac{1000 g}{kg} \times \frac{1 ng}{10^{- 9} g} = 37 ng

REVIEW This extremely tiny mass is hard to assess. It’s likely that a pollen grain’s density is near that of water, and the mass of a 40- $μ m$ -diameter sphere of water can be calculated—volume times density—to be about 30 ng. So our answer seems reasonable.