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Module 1 · Chapter 2

The Three Pillars of State: Density, Pressure, and Viscosity

Reading 10 minutes

Where this fits: Having established the continuum assumption, we now define the macroscopic properties we will track at every point in the flow field.

By the end of this chapter you will be able to:

  • Define density and pressure mathematically.
  • Understand the difference between dynamic and kinematic viscosity.
  • Apply Sutherland's Law to predict gas viscosity changes with temperature.

Density and Pressure

At every point in a continuous fluid field, we can define properties.

Density (ρ\rho) is mass per unit volume:

ρ=lim⁡δV→δV∗δmδV\rho = \lim_{\delta \mathcal{V} \to \delta \mathcal{V}^*} \frac{\delta m}{\delta \mathcal{V}}

where δV∗\delta \mathcal{V}^* is the smallest volume where the continuum assumption still holds.

Pressure (pp) is the normal force per unit area exerted by the fluid:

p=lim⁡δA→0δFnδAp = \lim_{\delta A \to 0} \frac{\delta F_n}{\delta A}

Viscosity

Viscosity is a measure of a fluid’s resistance to deformation (its “thickness” or internal friction).

  • Dynamic Viscosity (μ\mu): The proportionality constant relating shear stress (τ\tau) to the velocity gradient (du/dydu/dy) in a Newtonian fluid: τ=μdudy\tau = \mu \frac{du}{dy}.
  • Kinematic Viscosity (ν\nu): The ratio of dynamic viscosity to density:
ν=μρ\nu = \frac{\mu}{\rho}

Kinematic viscosity is often interpreted as the diffusivity of momentum.

Temperature Dependence (Sutherland’s Law)

Watch Out — Liquids vs. Gases

In liquids, viscosity decreases as temperature increases (think of heating up syrup). In gases (like air), viscosity increases as temperature increases because higher thermal energy means more molecular collisions transferring momentum between layers.

For air, we calculate dynamic viscosity at a given temperature TT using Sutherland’s Law:

μ=μref(TTref)3/2Tref+ST+S\mu = \mu_{ref} \left( \frac{T}{T_{ref}} \right)^{3/2} \frac{T_{ref} + S}{T + S}

Where for air:

  • μref=1.716×10−5 kg/(m⋅s)\mu_{ref} = 1.716 \times 10^{-5} \text{ kg/(m}\cdot\text{s)}
  • Tref=273.15 KT_{ref} = 273.15 \text{ K}
  • S=110.4 KS = 110.4 \text{ K}

Worked Example — Viscosity at High Temp

Calculate the dynamic viscosity of air at a stagnation temperature of 400 K400 \text{ K}.

Using Sutherland’s law:

μ=1.716×10−5(400273.15)3/2273.15+110.4400+110.4\mu = 1.716 \times 10^{-5} \left( \frac{400}{273.15} \right)^{3/2} \frac{273.15 + 110.4}{400 + 110.4}μ≈2.3×10−5 kg/(m⋅s)\mu \approx 2.3 \times 10^{-5} \text{ kg/(m}\cdot\text{s)}

Recap in one minute

  • Density (ρ\rho) is mass per volume.
  • Pressure (pp) is normal force per area.
  • Dynamic viscosity (μ\mu) relates shear stress to velocity gradients.
  • Kinematic viscosity (ν=μ/ρ\nu = \mu/\rho) relates to momentum diffusion.
  • Gas viscosity increases with temperature (Sutherland’s Law), while liquid viscosity decreases.

Check your understanding

  1. What is the difference between dynamic and kinematic viscosity?

    Show answer

    Dynamic viscosity (μ\mu) represents internal friction. Kinematic viscosity (ν=μ/ρ\nu = \mu/\rho) also incorporates density and represents the diffusion rate of momentum in the fluid.

  2. If you heat up a tank of water and a tank of air, what happens to their respective viscosities?

    Show answer

    The water’s viscosity will decrease, while the air’s viscosity will increase.

Your action step

Write a simple Python script to plot Sutherland’s Law for air from 200 K to 1000 K.

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