Unit 03 · Lesson 01 of 09

What makes a fluid a fluid?

Viscosity connects motion to shear stress.

Explore the model. Derive the result. Check its limits.

What you’ll be able to do

Fluids deform under shear.
No slip supplies boundary values.
Find stress and force.

Before you touch the controls

If the plate speed stays fixed and the gap halves, what happens to the shear force?

An experiment you can change

Drag a plate over a Newtonian fluid

Predict the direction of change, move one control, then explain the response using the equation.

Plate speed1 m/sDynamic viscosity0.02 Pa sGap5 mm

\tau=0.02\frac{1}{5\times10^{-3}}=4\,\mathrm{Pa}

The linear profile comes from constant shear stress. A smaller gap steepens the velocity gradient even when plate speed is unchanged.

Shear force on 0.1 m² plate0.4 N

Model assumptions and units

Steady laminar Newtonian Couette flow; no pressure gradient, no slip, fixed lower plate and moving upper plate. Edge effects neglected.

u(y)=U\frac yH,\quad \tau=\mu\frac{du}{dy}=\mu\frac UH,\quad \nu=\frac\mu\rho

Fluids deform under shear.

A fluid keeps deforming under a sustained shear stress. Density relates mass to volume; dynamic viscosity relates a velocity gradient to shear stress in a Newtonian fluid. A fluid at rest has no viscous shear from motion, but still exerts normal pressure.

\rho=\frac m{\mathcal V},\quad\tau=\mu\frac{du}{dy}

No slip supplies boundary values.

Fluid touching a solid surface takes that surface velocity in this model. Between a stationary lower plate and a moving upper plate, steady motion with no pressure gradient has a linear velocity profile.

u(0)=0,\quad u(H)=U,\quad u(y)=U\frac yH

Find stress and force.

The gradient is U/H, so stress is constant across the gap. Multiply by wetted plate area for the required tangential force. The plate applies the force needed to balance viscous resistance.

\tau=\mu\frac UH,\quad F=\tau A

Kinematic viscosity is a ratio.

Kinematic viscosity is dynamic viscosity divided by density, with units of m²/s. It compares viscous momentum diffusion to inertial transport and appears in Reynolds number. Newtonian behavior is a model; some fluids have nonlinear stress-gradient relations.

\nu=\frac\mu\rho

A worked example

A plate dragged over oil.

Let $\mu=0.020\,\mathrm{Pa\,s}$ , $U=1\,\mathrm{m/s}$ , and $H=0.005\,\mathrm m$ .
The gradient is $200\,\mathrm{s^{-1}}$ , so $\tau=4\,\mathrm{Pa}$ .
For area $0.1\,\mathrm{m^2}$ , $F=0.4\,\mathrm N$ and plate power is $FU=0.4\,\mathrm W$ .

Make the idea your own

Calculate. Compare. Test the assumptions.

Work out your answer, check it, then explore the visual solution. The answer checker uses a 1.5% rounding tolerance; for model-limit questions it checks the displayed model result.

1 · Calculate

Find viscous stress

A plate moves at 1 m/s over a 5 mm gap with μ = 0.020 Pa s. Find shear stress.

Hint

Convert millimeters to meters, then use μU/H.

Explore the visual solution

Plate speed1 m/sDynamic viscosity0.02 Pa sGap5 mm

\tau=0.02\frac{1}{5\times10^{-3}}=4\,\mathrm{Pa}

The linear profile comes from constant shear stress. A smaller gap steepens the velocity gradient even when plate speed is unchanged.

Shear force on 0.1 m² plate0.4 N

Model assumptions and units

Steady laminar Newtonian Couette flow; no pressure gradient, no slip, fixed lower plate and moving upper plate. Edge effects neglected.

u(y)=U\frac yH,\quad \tau=\mu\frac{du}{dy}=\mu\frac UH,\quad \nu=\frac\mu\rho

Controls explore variations. The steps below retain the original question’s values.

Steady laminar Newtonian Couette flow; no pressure gradient, no slip, fixed lower plate and moving upper plate. Edge effects neglected.

u(y)=U\frac yH,\quad \tau=\mu\frac{du}{dy}=\mu\frac UH,\quad \nu=\frac\mu\rho

2 · Compare

Halve the gap

Keep speed 1 m/s and viscosity 0.020 Pa s but use a 2 mm gap. Find stress and compare with a 4 mm gap.

Hint

The gradient is inversely proportional to the gap.

Explore the visual solution

Plate speed1 m/sDynamic viscosity0.02 Pa sGap2 mm

\tau=0.02\frac{1}{2\times10^{-3}}=10\,\mathrm{Pa}

The linear profile comes from constant shear stress. A smaller gap steepens the velocity gradient even when plate speed is unchanged.

Shear force on 0.1 m² plate1 N

Model assumptions and units

Steady laminar Newtonian Couette flow; no pressure gradient, no slip, fixed lower plate and moving upper plate. Edge effects neglected.

u(y)=U\frac yH,\quad \tau=\mu\frac{du}{dy}=\mu\frac UH,\quad \nu=\frac\mu\rho

Controls explore variations. The steps below retain the original question’s values.

Steady laminar Newtonian Couette flow; no pressure gradient, no slip, fixed lower plate and moving upper plate. Edge effects neglected.

u(y)=U\frac yH,\quad \tau=\mu\frac{du}{dy}=\mu\frac UH,\quad \nu=\frac\mu\rho

3 · Challenge the model

Stop the moving plate

With μ = 0.080 Pa s and gap 10 mm, set plate speed to zero. Find viscous shear under this model. Does pressure have to be zero?

Hint

No velocity gradient means no Newtonian viscous shear; normal pressure is separate.

Explore the visual solution

Plate speed0 m/sDynamic viscosity0.08 Pa sGap10 mm

\tau=0.08\frac{0}{10\times10^{-3}}=0\,\mathrm{Pa}

The linear profile comes from constant shear stress. A smaller gap steepens the velocity gradient even when plate speed is unchanged.

Shear force on 0.1 m² plate0 N

Model assumptions and units

Steady laminar Newtonian Couette flow; no pressure gradient, no slip, fixed lower plate and moving upper plate. Edge effects neglected.

u(y)=U\frac yH,\quad \tau=\mu\frac{du}{dy}=\mu\frac UH,\quad \nu=\frac\mu\rho

Controls explore variations. The steps below retain the original question’s values.

Steady laminar Newtonian Couette flow; no pressure gradient, no slip, fixed lower plate and moving upper plate. Edge effects neglected.

u(y)=U\frac yH,\quad \tau=\mu\frac{du}{dy}=\mu\frac UH,\quad \nu=\frac\mu\rho

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