Topic 05 · 12 min · 5 figures

Fluid kinematics

A solid has a shape you can track. A fluid does not — there is no marked point on it to follow with your eye, and the substance that was here a second ago has already gone somewhere else. Kinematics is the language built to describe motion without ever naming a piece of the fluid, and almost every later topic borrows a word from it.

A rigid body has a position, a velocity and an orientation, and statics gets a long way describing whole objects with three or six numbers. A fluid will not sit still for that. Squeeze it, and different parts of it move at different rates in different directions at the same instant; there is no single velocity that describes the water in a river, only a velocity at a point. Kinematics is the study of that motion before any force has been mentioned — no pressure, no viscosity, nothing about why the fluid moves the way it does, only a careful vocabulary for describing that it does. The vocabulary matters because the equations of the rest of the course are written in it, and a shaky grip on what a streamline or a material derivative actually means is where a great deal of confusion in later topics quietly starts.

The velocity field

There are two entirely different ways to describe a moving fluid, and the choice between them is made once, early, and then lived with for the rest of the calculation. The first is to pick a fixed point in space and ask what is happening there right now:

\[\mathbf{V} = \mathbf{V}(x, y, z, t)\]

This is the Eulerian description. It says nothing about which particle of fluid is at \((x, y, z)\) at time \(t\) — only what its velocity is, given that it is there. The point in space is the fixed thing; the occupant changes continuously. Almost every instrument you can build works this way too: a pressure tap, a thermocouple, a fixed anemometer all report on a location, not on a piece of matter, which is one reason the Eulerian view dominates engineering practice.

The second way is to pick a particle and follow it:

\[\mathbf{V} = \mathbf{V}(a, b, c, t)\]

where \((a, b, c)\) labels the particle by the position it happened to occupy at some reference time, usually \(t = 0\). This is the Lagrangian description, and it is the one you already use without thinking in particle mechanics, where a body keeps its identity and you write down where it goes. Applied to a fluid, it means tagging an enormous number of infinitesimal parcels and tracking each one — which is exactly what weather balloons, dye tracers and drifting ocean buoys do, and exactly why a single flow can be described by either method without contradiction.

Neither description is more correct than the other; they are different bookkeeping schemes for the same velocity field, and the field itself does not care which one you use to read it off. What changes is the mathematics: the Eulerian view turns every physical law into a partial differential equation in space and time, evaluated at fixed locations, while the Lagrangian view turns it into an ordinary differential equation followed along a trajectory. The Eulerian route is usually the tractable one for a flow field with a fixed geometry — a pipe, a duct, an aerofoil — and it is the convention this course follows from here on, with the Lagrangian picture kept in reserve for the one calculation that genuinely needs it: the acceleration of a particle, in the next section.

Two bookkeeping schemes — step 1 of 4

A field assigns a velocity to every point in space, at every instant.

\(\mathbf{V} = \mathbf{V}(x, y, z, t)\)
\(\mathbf{V}(x_0, y_0, z_0, t)\)
\(\mathbf{V} = \mathbf{V}(a, b, c, t)\)
\(\text{Eulerian, not Lagrangian}\)
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Streamlines, pathlines and streaklines

Three different curves get drawn through a flow field, they are often drawn identically in a textbook sketch, and conflating them is one of the most common ways students lose marks in this topic. Each answers a different question.

A streamline is a curve that is tangent to the velocity vector everywhere along it, at one fixed instant. It answers: if I could freeze the flow right now, which way would the fluid be heading at every point? In two dimensions it satisfies

\[\frac{dx}{u} = \frac{dy}{v}\]

and no fluid can cross it at that instant, because crossing it would mean having a velocity component normal to your own velocity vector. A pathline is the trajectory a single tagged particle actually traces out over some interval of time — the Lagrangian picture from the previous section, drawn as a curve. A streakline is different again: fix a point in space, release dye there continuously, and at some later instant look at the shape traced by every particle of dye released so far. It is what a smoke trail from a chimney shows you, or the streak of ink from a fixed injector in a wind-tunnel photograph.

In a steady flow — one where the velocity field itself does not change with time — all three coincide, and this is the case drawn in almost every introductory figure, which is exactly why the distinction seems pointless until it suddenly is not. The fluid passing through a fixed point always has the same velocity there, so the tangent curve frozen at any instant, the trail a particle leaves, and the accumulated dye streak from that point are all the same curve retraced over and over.

Make the flow unsteady and the three separate immediately. A jet from a nozzle that is being swung around traces this out cleanly. The instantaneous streamline is whatever direction the jet is pointing right now. The pathline of a particle released earlier is a straight run in whatever direction the jet was pointing when that particle left, since once released it keeps going the way it was sent — the nozzle turning afterwards does not reach back and bend a particle already in flight. The streakline is neither: it is the curve joining the current positions of every particle released between then and now, each one carrying the direction it was launched at and having travelled for a different length of time, and the result curves continuously from the oldest, furthest particle back to the nozzle itself. Three genuinely different curves, from one nozzle, at one instant.

The practical consequence is a matter of reading instruments correctly. A tuft or a smoke-wire experiment gives you streaklines. A particle-tracking measurement gives you pathlines. A computed flow field, snapshotted, gives you streamlines. Treating any one of the three as if it were another is only harmless in the steady case, and the moment a flow starts up, a valve throttles, or a wake sheds vortices, it stops being harmless.

A swinging jet — step 1 of 4

Steady flow: the tangent curve, one particle's trail and the dye streak all agree.

\(\frac{dx}{u} = \frac{dy}{v} = \frac{dz}{w}\)
\(\mathbf{V} = \mathbf{V}(x, y, z, t)\)
\(\text{streamline} \ne \text{pathline} \ne \text{streakline}\)
\(\text{equal only if } \partial \mathbf{V}/\partial t = 0\)
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The material derivative

Newton's second law is written for a particle, so applying it to a fluid means finding the acceleration of a specific parcel — the Lagrangian quantity — while still working, for convenience, in the Eulerian field. The material derivative is the bridge between the two, and it is worth sitting with because it produces a result that surprises almost everyone the first time: a particle can accelerate in a flow that is not changing with time at all.

Consider a particle's velocity as a function of the coordinates it currently occupies, and differentiate with respect to time using the chain rule, remembering that its position is itself a function of time:

\[\frac{D\mathbf{V}}{Dt} = \frac{\partial \mathbf{V}}{\partial t} + u\frac{\partial \mathbf{V}}{\partial x} + v\frac{\partial \mathbf{V}}{\partial y} + w\frac{\partial \mathbf{V}}{\partial z} = \frac{\partial \mathbf{V}}{\partial t} + (\mathbf{V}\cdot\nabla)\mathbf{V}\]

The first term, \(\partial \mathbf{V}/\partial t\), is the local acceleration: how the velocity at a fixed point is changing with time, exactly what an Eulerian sensor bolted in place would read. The remaining terms are the convective acceleration: the change a particle experiences simply by moving to a new location where the field has a different value, even if that field is frozen in time. It is the second term that is easy to underrate, because nothing at any fixed point is changing, and the intuition trained on solid mechanics — no change means no acceleration — is simply wrong here.

A nozzle makes the point concretely. Run water through a converging duct at a constant flow rate — steady, by definition, so \(\partial u/\partial t = 0\) everywhere. A particle entering slowly at the wide end is unmistakably faster by the time it reaches the narrow end, which means it accelerated, which means the material derivative was not zero. The resolution is entirely in the convective term: the duct narrows with distance, so \(\partial u/\partial x\) is positive, and \(u\,\partial u/\partial x\) is the whole story. Open a valve upstream instead, so the flow rate itself climbs with time, and the local term switches on too — the particle now feels both a squeeze from the geometry and a general rise from the unsteadiness, added together as the material derivative says they should be.

This is not a bookkeeping curiosity. It is the acceleration that appears on the right-hand side of Newton's second law applied to a fluid element, and the whole of the momentum equation — Euler's equation, the Navier–Stokes equations, and the derivation of the Bernoulli equation a few topics on — is built by substituting the material derivative for \(a\) in \(F = ma\) and then working out what forces produce it.

Two ways to accelerate — step 1 of 4

A steady nozzle: nothing at any fixed point changes with time.

\(\frac{\partial u}{\partial t} = 0\)
\(u\,\frac{\partial u}{\partial x} \ne 0\)
\(\frac{\partial u}{\partial t} \ne 0\)
\(\frac{Du}{Dt} = \frac{\partial u}{\partial t} + u\,\frac{\partial u}{\partial x}\)
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Rate of deformation

A solid element resists a change of shape; a fluid element does not, and the velocity gradients around it are exactly what determine how fast it deforms. Take a small parcel and look at how the velocity varies across it. Part of that variation stretches the parcel, part of it shears it, and part of it simply spins it without changing its shape at all — three effects tangled together in a single gradient, and separating them is the point of this section.

The stretching is the linear strain rate. Along the direction of flow it is

\[\varepsilon_{xx} = \frac{\partial u}{\partial x}\]

with matching terms \(\varepsilon_{yy}\) and \(\varepsilon_{zz}\) in the other two directions. It measures the fractional rate at which a line of fluid aligned with that axis is getting longer or shorter — positive in the nozzle of the previous section, where fluid speeds up along \(x\) and a marked line of particles pulls apart.

The shearing is the rate of angular deformation, built from the cross-derivatives:

\[\varepsilon_{xy} = \frac{1}{2}\left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right)\]

Picture a small square dropped into a shear flow, one where \(u\) increases with \(y\) but \(v\) stays zero — the profile next to a wall, or between two plates. The top edge of the square, sitting in faster fluid, is dragged further along than the bottom edge in the same interval, and the square leans over into a parallelogram. This quantity is not a curiosity either: multiplied by viscosity, it is precisely the shear stress that Newton's law of viscosity describes, and it is the mechanism by which a moving fluid drags on a stationary wall.

What is left over once the pure stretching and the pure shearing are accounted for is a rigid rotation — the parcel spinning about its own centre without changing shape at all. Its rate is

\[\zeta = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\]

the vorticity, equal to twice the angular velocity of the fluid element. A common and costly confusion is to assume that because a flow curves — round a bend, through a vortex — it must be rotational, and the reverse: that a flow travelling in a straight line must have zero vorticity. Neither follows. A straight shear flow next to a wall has vorticity, because the differential speed between adjacent layers spins each element even though the element's centre travels in a straight line, while a free vortex away from its core can curve every streamline into a circle and still have zero vorticity almost everywhere, because the shearing and the curving exactly cancel the spin. Whether a flow is irrotational — vorticity zero throughout — turns out to control which regions later simplifications, including Bernoulli's equation applied across streamlines rather than along just one, are allowed to reach.

One gradient, three effects — step 1 of 4

Drop an undistorted square of fluid into a sheared stream.

\(\varepsilon_{xx} = \varepsilon_{xy} = \zeta = 0\)
\(\varepsilon_{xx} = \frac{\partial u}{\partial x}\)
\(\varepsilon_{xy} = \tfrac{1}{2}\left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right)\)
\(\zeta = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\)
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The Reynolds transport theorem

Every conservation law you already know — mass is conserved, momentum changes only under a force, energy is conserved — was proved for a system: a fixed, identifiable quantity of matter, the Lagrangian object. That is a poor description of a turbine, a nozzle or a length of pipe, because the matter inside those never stays the same from one instant to the next. What engineers actually want to write balances for is a control volume — a fixed region of space, the box drawn around the hardware — and the Reynolds transport theorem is the formula that translates a law proved for the first into a law usable for the second, at the cost of exactly one extra term.

Let \(B\) be any extensive property of the system — mass, momentum, energy — and \(b = B/m\) the corresponding amount per unit mass. The theorem states

\[\frac{dB_{\text{sys}}}{dt} = \frac{d}{dt}\int_{CV} \rho\, b \, dV + \int_{CS} \rho\, b\, (\mathbf{V}\cdot\mathbf{n})\, dA\]

The left-hand side is the familiar system rate: whatever the conservation law for \(B\) says it should equal — zero for mass, the net force for momentum, the net heat and work for energy. The first term on the right is the rate of change of \(B\) stored inside the fixed box, exactly the kind of term an instrument sitting still could report. The second term is the net rate at which \(B\) is carried across the box's surface by the flow, with \(\mathbf{n}\) the outward normal, so the dot product is positive where fluid leaves and negative where it enters.

The derivation is a matter of accounting, not new physics. At some instant, let the system exactly fill the control volume. An instant later, the system — being matter, and having moved — no longer coincides with the box: a sliver has left through one part of the control surface and a sliver of new matter has entered through another, while the box itself has not moved at all. The change in \(B\) for the system, tracked the honest Lagrangian way, is exactly the change stored inside the box plus what left minus what arrived — which is precisely the right-hand side above, arranged so both slivers appear with the same sign convention through \(\mathbf{V}\cdot\mathbf{n}\).

What makes the theorem worth learning once, carefully, rather than re-deriving for each new quantity, is that every major balance in the rest of this course is the same theorem with a different \(b\) substituted in. Conservation of mass is \(b = 1\) with a left-hand side of zero. Linear momentum is \(b = \mathbf{V}\) with a left-hand side equal to the net force. The steady-flow energy equation met in thermodynamics is the same theorem again, with \(b\) equal to the specific energy. None of that is a coincidence worth marvelling at for long; it is the entire reason the theorem is taught before any of the balances that use it.

System to control volume — step 1 of 4

At this instant the system — a fixed lump of matter — exactly fills the box.

\(B_{\text{sys}} = \int_{CV} \rho\, b \, dV\)
\(\frac{dB_{\text{sys}}}{dt}\)
\(\int_{CS} \rho\, b\, (\mathbf{V}\!\cdot\!\mathbf{n})\, dA\)
\(\frac{dB_{\text{sys}}}{dt} = \frac{d}{dt}\int_{CV} \rho b\, dV + \int_{CS} \rho b (\mathbf{V}\!\cdot\!\mathbf{n})\, dA\)
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