riaology
Unit 03 · Lesson 03 of 09

How do we describe fluid motion?

A steady flow can accelerate its particles.

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

What you’ll be able to do
  • Eulerian and material descriptions.
  • Continuity fixes the mean speed.
  • Acceleration has two contributions.
Before you touch the controls

Can the speed at each location stay constant while a fluid parcel speeds up?

An experiment you can change

Track a parcel through a contracting channel

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

Contraction: inlet speed 2, outlet speed 4 meters per second2 m/s4 m/sArea fixes speed through continuityTrack a parcel through a contracting channel. Position along channel (m) versus Mean speed (m/s). Selected result 4 m/s. Steady means velocity at each fixed point does not change in time. A moving parcel can nevertheless accelerate as it reaches narrower sections.0020.541Mean speed (m/s)Position along channel (m)
V2=V1A2/A1=20.5=4 m/s,a2=16 m/s2V_2=\frac{V_1}{A_2/A_1}=\frac{2}{0.5}=4\,\mathrm{m/s},\quad a_2=16\,\mathrm{m/s^2}

Steady means velocity at each fixed point does not change in time. A moving parcel can nevertheless accelerate as it reaches narrower sections.

Model assumptions and units

Steady, incompressible, one-dimensional mean flow; area decreases linearly over 1 m. Particle paths are not the same concept as instantaneous streamlines in unsteady flow.

m˙=ρAV,V(x)=A1V1A(x),ax=∂V∂t+V∂V∂x\dot m=\rho AV,\quad V(x)=\frac{A_1V_1}{A(x)},\quad a_x=\frac{\partial V}{\partial t}+V\frac{\partial V}{\partial x}
01

Eulerian and material descriptions.

An Eulerian description reports velocity at fixed locations. A material description follows individual fluid particles. In a steady field, a fixed probe sees no time change, yet a particle can reach locations with different speeds.

V=V(x,t)\mathbf V=\mathbf V(\mathbf x,t)
02

Continuity fixes the mean speed.

For steady incompressible flow through a channel, volume rate is constant. A smaller cross-section requires a higher mean normal velocity. This relation uses section averages; it does not specify the entire velocity profile.

A1V1=A2V2A_1V_1=A_2V_2
03

Acceleration has two contributions.

A particle experiences both local time variation and spatial variation as it moves. In a steady one-dimensional flow only the convective contribution remains. This is why steady does not imply zero acceleration.

DVDt=∂V∂t+V∂V∂x\frac{DV}{Dt}=\frac{\partial V}{\partial t}+V\frac{\partial V}{\partial x}
04

Streamlines and pathlines.

A streamline is tangent to the instantaneous velocity field; a pathline is the history of one particle. They coincide for steady flow, but can differ when the field changes with time. The contracting channel below is a steady mean-flow illustration.

A worked example

A channel halves its area.

  1. A steady incompressible flow enters at V1=2 m/sV_1=2\,\mathrm{m/s} and exits with A2=0.5A1A_2=0.5A_1.
  2. Continuity gives V2=2/0.5=4 m/sV_2=2/0.5=4\,\mathrm{m/s}.
  3. For the displayed linear area contraction over 1 m, V(x)=2/(1−0.5x)V(x)=2/(1-0.5x) and ax=V dV/dxa_x=V\,dV/dx; exit acceleration is 16 m/s216\,\mathrm{m/s^2}.
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

Halve the area

A channel has outlet/inlet area ratio 0.50 and inlet mean speed 2 m/s. Find outlet speed.

Hint

Continuity keeps A V constant for steady incompressible flow.

Explore the visual solution
Contraction: inlet speed 2, outlet speed 4 meters per second2 m/s4 m/sArea fixes speed through continuityTrack a parcel through a contracting channel. Position along channel (m) versus Mean speed (m/s). Selected result 4 m/s. Steady means velocity at each fixed point does not change in time. A moving parcel can nevertheless accelerate as it reaches narrower sections.0020.541Mean speed (m/s)Position along channel (m)
V2=V1A2/A1=20.5=4 m/s,a2=16 m/s2V_2=\frac{V_1}{A_2/A_1}=\frac{2}{0.5}=4\,\mathrm{m/s},\quad a_2=16\,\mathrm{m/s^2}

Steady means velocity at each fixed point does not change in time. A moving parcel can nevertheless accelerate as it reaches narrower sections.

Model assumptions and units

Steady, incompressible, one-dimensional mean flow; area decreases linearly over 1 m. Particle paths are not the same concept as instantaneous streamlines in unsteady flow.

m˙=ρAV,V(x)=A1V1A(x),ax=∂V∂t+V∂V∂x\dot m=\rho AV,\quad V(x)=\frac{A_1V_1}{A(x)},\quad a_x=\frac{\partial V}{\partial t}+V\frac{\partial V}{\partial x}

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

Steady, incompressible, one-dimensional mean flow; area decreases linearly over 1 m. Particle paths are not the same concept as instantaneous streamlines in unsteady flow.

m˙=ρAV,V(x)=A1V1A(x),ax=∂V∂t+V∂V∂x\dot m=\rho AV,\quad V(x)=\frac{A_1V_1}{A(x)},\quad a_x=\frac{\partial V}{\partial t}+V\frac{\partial V}{\partial x}
2 · Compare

Quarter the area

Use area ratio 0.25 and inlet speed 1 m/s. Find exit mean speed, then inspect convective acceleration.

Hint

The area ratio is already an area ratio, so do not square it again.

Explore the visual solution
Contraction: inlet speed 1, outlet speed 4 meters per second1 m/s4 m/sArea fixes speed through continuityTrack a parcel through a contracting channel. Position along channel (m) versus Mean speed (m/s). Selected result 4 m/s. Steady means velocity at each fixed point does not change in time. A moving parcel can nevertheless accelerate as it reaches narrower sections.0020.541Mean speed (m/s)Position along channel (m)
V2=V1A2/A1=10.25=4 m/s,a2=48 m/s2V_2=\frac{V_1}{A_2/A_1}=\frac{1}{0.25}=4\,\mathrm{m/s},\quad a_2=48\,\mathrm{m/s^2}

Steady means velocity at each fixed point does not change in time. A moving parcel can nevertheless accelerate as it reaches narrower sections.

Model assumptions and units

Steady, incompressible, one-dimensional mean flow; area decreases linearly over 1 m. Particle paths are not the same concept as instantaneous streamlines in unsteady flow.

m˙=ρAV,V(x)=A1V1A(x),ax=∂V∂t+V∂V∂x\dot m=\rho AV,\quad V(x)=\frac{A_1V_1}{A(x)},\quad a_x=\frac{\partial V}{\partial t}+V\frac{\partial V}{\partial x}

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

Steady, incompressible, one-dimensional mean flow; area decreases linearly over 1 m. Particle paths are not the same concept as instantaneous streamlines in unsteady flow.

m˙=ρAV,V(x)=A1V1A(x),ax=∂V∂t+V∂V∂x\dot m=\rho AV,\quad V(x)=\frac{A_1V_1}{A(x)},\quad a_x=\frac{\partial V}{\partial t}+V\frac{\partial V}{\partial x}
3 · Challenge the model

Remove the contraction

Set outlet/inlet area ratio to 1 and inlet speed 3 m/s. Find exit speed and the acceleration in the modeled channel.

Hint

A uniform area gives a spatially uniform mean speed.

Explore the visual solution
Contraction: inlet speed 3, outlet speed 3 meters per second3 m/s3 m/sArea fixes speed through continuityTrack a parcel through a contracting channel. Position along channel (m) versus Mean speed (m/s). Selected result 3 m/s. Steady means velocity at each fixed point does not change in time. A moving parcel can nevertheless accelerate as it reaches narrower sections.001.50.531Mean speed (m/s)Position along channel (m)
V2=V1A2/A1=31=3 m/s,a2=0 m/s2V_2=\frac{V_1}{A_2/A_1}=\frac{3}{1}=3\,\mathrm{m/s},\quad a_2=0\,\mathrm{m/s^2}

Steady means velocity at each fixed point does not change in time. A moving parcel can nevertheless accelerate as it reaches narrower sections.

Model assumptions and units

Steady, incompressible, one-dimensional mean flow; area decreases linearly over 1 m. Particle paths are not the same concept as instantaneous streamlines in unsteady flow.

m˙=ρAV,V(x)=A1V1A(x),ax=∂V∂t+V∂V∂x\dot m=\rho AV,\quad V(x)=\frac{A_1V_1}{A(x)},\quad a_x=\frac{\partial V}{\partial t}+V\frac{\partial V}{\partial x}

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

Steady, incompressible, one-dimensional mean flow; area decreases linearly over 1 m. Particle paths are not the same concept as instantaneous streamlines in unsteady flow.

m˙=ρAV,V(x)=A1V1A(x),ax=∂V∂t+V∂V∂x\dot m=\rho AV,\quad V(x)=\frac{A_1V_1}{A(x)},\quad a_x=\frac{\partial V}{\partial t}+V\frac{\partial V}{\partial x}
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