Concept

Control volume — where it appears

A region drawn in the fluid across whose faces mass and momentum are counted. It allows a force to be found without going near the surface it acts on, which is why it settles arguments that a surface integral leaves open.

Named by 37 essays across 7 fields — each of them below, with the objects they name alongside it.

The tube widens because the air slows. The streamtube through an actuator disc at an induction factor of 0.333. The three radii are not drawn to taste: each is fixed by requiring the same mass to pass every station, and the slowest station is therefore the widest. The tube widening in front of a wind turbine is why some of the wind goes round it rather than through it, and it is the whole reason a disc cannot take everything.

The most a disc can take

A wind turbine cannot extract more than sixteen twenty-sevenths of the energy passing through the circle its blades sweep. That is not a limit on turbines — it is a limit on anything at all, and it follows from three conservation laws and no engineering.

applied · Actuator disc
Efficiency is decided before the engine is chosen. Froude's propulsive efficiency against the ratio of jet speed to flight speed. It is 2/(1+σ) and nothing else — no engine, no fuel, no combustion. A turbojet with a jet at three times flight speed cannot exceed 50% however good its core is, and a propeller moving a great deal of air slowly is above 90% before anybody has designed anything.

A big slow push

The same thrust can be had from a lot of air moved a little or a little air moved a lot, and the two are not equivalent. One number decides which, it contains no engine, and it is why every airliner built since 1970 has a fan far larger than the machine driving it.

applied · Actuator disc
A control volume round an aerofoil. A rectangle drawn in the fluid around a lifting section. The arrows on the right-hand face show the downward velocity of the air leaving it, drawn to scale. Adding the momentum carried through all four faces to the pressure acting on them gives the force on whatever is inside, without the calculation ever going near the surface.

Air must be pushed down, and the usual sum is wrong

The momentum explanation of lift is the one physicists reach for, and it is right — a wing does hold itself up by throwing air downwards. The version usually given then does the accounting badly, and the face of the control volume it keeps turns out to carry the least of it.

misconceptions · Momentum lift
Betz's ceiling, and the rotor that cannot reach it. The power coefficient of Glauert's optimum rotor against tip-speed ratio, with Betz's 16/27 drawn as the ceiling it is. The gap is wake rotation: a rotor that extracts power applies a torque, a torque leaves the wake spinning, and that rotational energy never reaches the shaft. It falls as the rotor is geared up and is never zero — which is why large wind turbines turn so slowly and yet have such fast tips.

The wake that has to spin

A rotor that takes power out of the wind must apply a torque to it, and a torque applied to air is angular momentum left behind. The axial theory has nowhere to put that energy, so Betz's ceiling is unreachable at every finite tip-speed ratio — and the gap is computable.

applied · Actuator disc
The box, and the one thing assumed about it. The control volume across a sudden enlargement. Mass and momentum crossing the two ends are known exactly. The only modelling statement in the whole derivation is written on the annular step: the pressure there is taken to be the upstream pressure, because the fluid in the corner is nearly stationary. Measurement supports it well. Nothing else is assumed, and in particular nothing at all is assumed about the eddy that lives in that corner — which this figure therefore does not draw.

A loss with no viscosity in it

Where a pipe suddenly widens, energy is destroyed. The amount is exact, it has been known since 1766, and the derivation never mentions viscosity, Reynolds number or roughness — because momentum does not care where the energy went, only that it left.

applied · Internal flow
Two depths, and a gap the model will not describe. The surface either side of a hydraulic jump at an arriving Froude number of 5.05. Both depths are exact consequences of the momentum balance. The distance between them is not: the shallow-water model has no length scale in it and cannot say how far the transition takes, so the region between the two levels is left blank and the six-depth rule of thumb beside it is somebody's measurement rather than this site's result.

The shock in a river

Shallow water is a gas whose ratio of specific heats is two. The white water below a weir is a shock wave, momentum is conserved across it exactly, energy is not, and one direction is forbidden for the same reason an expansion shock is forbidden — which makes the analogy exact to first order and wrong at the second.

applied · Open-channel
Two depths for the same energy, and one for the least. Specific energy against depth for a discharge of 0.5 square metres per second per metre of width. Every energy above the minimum is carried by two different depths — one fast and shallow, one slow and deep — and the minimum is carried by exactly one. That depth is the critical depth, the Froude number there is one, and the least energy is three halves of it; all three are found here by search and checked against their closed forms.

The depth that costs least

For a given flow there are two depths that carry it at any energy above a floor, and exactly one at the floor. That one depth is where the Froude number is one, the least energy is exactly three halves of it, and a bump in the bed that asks for more than the flow has does not thin the water — it backs it up.

applied · Open-channel
A normal shock at Mach 2.00, and what crosses it unchanged. The state in front of the shock and the state behind it. Every ratio was computed from the standard jump relations and then substituted back into mass, momentum and energy, which is an independent route — a mistyped exponent in the total-pressure expression cannot survive a momentum balance it never appeared in. The residuals are printed below because a check nobody can see is a check nobody can audit.

The jump the equations allow

A shock is a discontinuity in a fluid, which sounds like a breakdown of the description rather than a solution of it. It is a solution: mass, momentum and energy can all be satisfied across a jump, and every ratio across one follows from that alone.

compressible · Shock
Three times the wind, on a reach. The polar diagram: boat speed in every direction, as a multiple of the true wind speed, for drag angles of 14 and 6 degrees. The shaded wedge at the top is the no-go zone, whose half-angle is exactly the sum of the two drag angles. Everywhere outside about twice that angle the boat is faster than the wind, and the maximum is 2.92 times the wind at 110 degrees — which is 1/sin λ at 90° + λ, both checked.

Faster than the wind that drives it

An ice yacht in a fifteen-knot breeze does forty. That is not a trick and it does not need a special sail — it follows from two drag angles and a triangle, and the best speed a boat can reach is one over the sine of their sum.

applied · Sailing
Two triangles, and the work is the difference between them. The velocity triangles at inlet and outlet of a rotor at constant blade speed and constant axial velocity. The horizontal arrow is the blade speed; the arrow from the origin is the absolute velocity of the fluid; the arrow closing the triangle is what the blade sees. Euler's equation says the work is the blade speed times the change in the swirl component alone — the horizontal distance between the two upper corners, times U — and nothing else in the picture appears in it.

Work out of a change of swirl

The work a rotor does per unit mass is the blade speed times the change in swirl, and that is all of it — no blade shape, no pressure, no efficiency, no gas properties. It is the same equation for a pump, a compressor, a turbine and a fan, and it follows from angular momentum on a box with nothing assumed about the inside.

applied · Turbomachine
The jet is 61.1% of the hole. Flow out of a slot in a plane wall, solved by Kirchhoff's free-streamline method. The outer curve is not a wall and not a guess: it is the streamline on which the pressure is ambient, and where it goes is part of the solution. It leaves the edge of the slot travelling straight down the wall and turns through ninety degrees, settling to a jet whose width is π/(π+2) = 0.6110 of the opening. Every streamline drawn is a level set of the streamfunction the conformal map supplies.

The hole that halves the flow

A jet leaving a sharp-edged hole is narrower than the hole, and by an amount that is not measured but computed. One geometry gives exactly one half from momentum alone; another gives exactly π/(π+2) from a conformal map in which the shape of the free surface is part of the answer.

applied · Jet
Three contours, one force, three different accounts of it. The same vortex, and the same total force on it, computed by a momentum balance over three contours of the same area. All three give ρUΓ to eight figures. What differs is the bookkeeping: the tall box gets 16 per cent of it from pressure and the rest from momentum flux, the wide box gets 84 per cent from pressure, and the circle gets exactly half. Neither part converges on its own as the contour is enlarged — each falls off like 1/r while the contour grows like r — so the split is a property of the shape of the limit rather than of the flow.

Where the reaction to a wing's lift is

The force on a body can be computed on any contour drawn round it, and the answer is the same every time. How much of that answer is pressure and how much is momentum flux is not — it runs from three per cent to ninety-seven, and the difference is the shape of the contour.

inviscid · Far field
The jet divides 75% to 25%. A jet striking a plate at 60 degrees. Both sheets leave at the jet's own speed, because their surfaces are at ambient pressure and Bernoulli allows nothing else, and the plate can exert no force along itself because the fluid has no viscosity. Momentum along the plate then fixes the split at (1 + cos β)/2 = 0.7500, and the normal force at ṁV sin β = 0.8660. Nothing about the plate's material, size or roughness enters either.

What a jet cannot push sideways

A jet striking a plate divides in two, and how it divides is fixed by a single sentence — an inviscid fluid exerts no force along a surface. That one statement, plus mass, gives the split exactly — and the same sentence turns a flat plate into a bucket worth twice as much.

applied · Jet
Most power at exactly half the jet speed, found by search at 0.5000. The power a bucket takes from a jet, against how fast the bucket runs, for four deflection angles. Every curve is a parabola with roots at zero — where the force is greatest and the bucket is not moving — and at the jet speed, where the bucket is running away and there is no force at all. The peak is halfway between, at U = V/2, and it is there for every angle and every flow rate. A golden-section search that knows none of the algebra puts it at 0.500000.

Half the jet speed takes everything

A bucket standing still feels the largest force and does no work; a bucket running with the jet does no work either. Between them the power peaks at exactly half the jet speed, for every bucket shape and every flow rate — and at that speed a perfect bucket leaves the water motionless.

applied · Jet
The wall's condition, on its way to the middle. Five profiles across the half-channel, from just inside the entrance to fully developed, each drawn at the station where it occurs. The march starts from a slab of uniform flow and never assumes a shape: what arrives at the far end is a parabola, with a centre-line speed of 1.4979 times the mean against the exact 3/2 and a momentum flux of 1.1995 against 6/5. Notice what the middle does while the edges are being slowed: it speeds up, because the flow rate is held, and that acceleration is what the entrance's extra pressure drop pays for.

How far before a duct forgets what was fed into it

A pipe is always drawn with its answer already in place. Getting there takes a distance proportional to the Reynolds number, which means a more viscous fluid is done sooner — and the entrance costs a fixed number of dynamic pressures however long the pipe is.

viscous · Entrance
The same reading, and only one of them gives it back. A Venturi and an orifice plate at the same diameter ratio, with the pressure along the axis drawn beneath each. Both narrow the flow by the same amount, both read the same difference between the pipe and the narrowest section, and both infer the same flow rate from it. Downstream they part company: the Venturi's diffuser turns the throat's speed back into pressure, and the orifice's jet expands into the pipe and destroys 73% of the reading. The picture is a section rather than a solved field: nothing here computes the jet, and the recirculating corner is not drawn.

The price of knowing the flow rate

Two flowmeters can narrow a pipe by the same amount, read the same pressure difference and infer the same flow rate, and cost pressures that differ by an order of magnitude. What separates them is not viscosity, and not workmanship — it is whether the flow is decelerated or abandoned.

applied · Metering
A material region, and the dye that stays inside it. The same fluid at four times, carried by an unsteady straining flow whose strain rate oscillates. The outline is a circle of the fluid at the first instant, tracked by integrating the velocity field; the shading is a blob of passive dye. The region is stretched to nearly seven to one and its area is unchanged to fifteen decimal places, because the flow is incompressible. The amount of dye inside it is unchanged to thirteen, because the dye is carried by the same fluid.

A rate of change that will not hold still

Three boxes drawn in one flow at one instant give three different answers to how fast the dye inside them is changing — one falling, one falling twice as fast, one rising. All three reconcile with a single material rate, and that rate is zero.

kinematics · Transport theorem
The same number, by two integrals that share no arithmetic. Three flows whose dissipation is in closed form both ways. The volume route integrates the dissipation function over the fluid; the boundary route multiplies a force or a torque by the speed of whatever is applying it. Neither calculation contains the other, and the residual column is what is left when they are subtracted.

The price of a gradient

Viscosity does not charge for motion. It charges for the rate at which a parcel is being deformed, and a fluid in solid-body rotation at any speed whatever destroys nothing at all. What is charged for is a sum of squares, which is why the bill can be computed twice.

viscous · Dissipation
12 bar from stopping one metre per second. The head at the valve after it shuts, computed by the method of characteristics on a 600 m pipe. The rise is 122.4 m of water, which is ρaΔV/ρg to 1.4e-14 m — and the scheme was told neither ρaΔV nor anything else about the answer. The wave then runs to the reservoir and back every 2.000 s, and with no friction in the model it never decays: a real pipe damps this out in a few tens of cycles.

Stopping water costs more than moving it

Shut a valve on water running at one metre per second and the pressure that appears is twelve bar — not because the water was pushing hard, but because the only way to stop a column of fluid is to send a message back along it, and the message travels at the speed of sound in the pipe.

applied · Water hammer
How far downstream the heat is still being made. The dissipation accumulated from a station ahead of a cylinder to a station behind it, as a fraction of the whole of what is made inside the frame, at five Reynolds numbers. At Reynolds number 1 the fluid has finished paying by about a diameter behind the body. At 100 it has not finished at five, and the curve is still climbing at the edge of the picture — the drag is a force on the body, and the heat it stands for is somewhere else.

Where the heat of a drag is made

The power it takes to tow a body through a fluid becomes heat, all of it, eventually. None of the interesting words in that sentence are the first four. It is the "eventually" that decides how a wake behaves, how far a disturbance reaches, and why no box drawn round a body contains its own bill.

viscous · Dissipation
Nothing can deliver more than h/H, and here that is 10.0%. The fraction of the supply a ram can deliver, against the height it is asked to deliver to, from a supply falling 2 m. The upper curve is the exact ceiling h/H, which follows from the energy audit with every loss set to zero and can be reached by no real machine; the lower one is what a ram at 65% efficiency actually sends. Asking for twice the height halves the delivery, exactly, and there is no design that escapes it.

A pump with no engine

A hydraulic ram lifts water uphill using nothing but the water that is already falling. It has one moving part and no power supply, and everything it can and cannot do follows from an energy audit that fits on one line — including a ceiling nothing about its design can move.

applied · Water hammer
A pump curve out of the momentum theorem. The pressure an ejector delivers, against how much it is entraining, at a fixed nozzle. It has the shape of every pump characteristic ever measured — a shut-off pressure with no flow, falling to no pressure at free delivery — and it was obtained from a momentum balance on a tube with nothing in it. The shut-off value here is 36.0 kPa and the machine at its best power runs at 4.56 times its own motive flow.

Mixing is a pump

Two streams at different speeds mixing in a tube destroy energy — the same Borda–Carnot expression a handbook prints beside a sudden enlargement, with two streams in it instead of one. And while they destroy it the pressure rises, which makes the loss the mechanism of a machine with no moving parts.

applied · Ejector
The momentum of the fluid, against the shape of the region it is added up over. Momentum of the fluid around a cylinder moving through it, divided by the body's hydrodynamic impulse, against the aspect ratio of the rectangle the integral was taken over. Every rectangle has the same area and contains the same body. A tall region gives minus the impulse, a long one gives plus it, a square gives exactly zero, and the limit of a large region is whichever of those the region was shaped like. The momentum of an unbounded ideal flow is not a number.

The momentum with no value

A cylinder is pushed from rest to a steady speed. Work was done, energy went into the fluid, something was pushed. How much momentum does the fluid carry? The integral converges, the answer is finite, and it is a different finite number for every shape of region it is summed over.

inviscid · Impulse
The pressure drop stops rising at 0.213 m/s. The pressure drop across a bed of 500 µm sand, against the velocity through it, in units of the fluidisation velocity. The rising branch is Ergun's resistance and the flat one is the bed's buoyant weight, which the flow cannot exceed however hard it is pushed: past the corner the bed expands rather than resisting more. That flat line is the reason fluidisation is unmistakable in practice — the corner is a crossing of two curves rather than a gradual departure, and it can be read off a gauge.

The bed that weighs itself

Blow hard enough through a pile of sand and the pressure drop stops rising. It cannot rise — a control volume round the bed says the drop can never exceed the buoyant weight of the solid in it, and at the velocity where the two meet the bed stops being a structure and starts being a fluid.

applied · Porous
And the net drift, which is where they disagree. The Stokes drift plus each return flow. All three drift forward at the surface, because the Stokes drift there swamps any return current of the right size. Below that they part company completely: the uniform current has most of the column moving upstream, and the two that satisfy no slip have almost none of it. The reversal depths span 77 per cent of the water column.

The drift a closed box will not allow

A wave in a wave tank carries mass forward, and the tank has nowhere to put it. So a return current appears carrying exactly the opposite transport — exactly, from mass conservation and nothing else. Which fixes a total and leaves the answer anybody wants entirely open.

kinematics · Stokes drift
A uniform scalar in a fluid at rest, under two face rules. Nothing is flowing and the scalar starts at one everywhere. The swept-volume rule leaves it at one to the last bit, at every step of the two time units. The midpoint rule moves it by two parts in ten thousand, on a mesh motion that begins and ends in the same place, and the excursion looks exactly like a physical transient.

The mesh that makes its own mass

The transport theorem holds for a region moving at any velocity, which is what makes a moving-mesh calculation possible. Discretised carelessly it is not an identity but an approximation, and a fluid at rest with a uniform density then gains density from the motion of a grid — smoothly, plausibly, and looking exactly like a physical transient.

kinematics · Transport theorem
A vortex has no energy. The kinetic energy of a Lamb–Oseen vortex inside a circle, per metre of its length, against the logarithm of that circle's radius. It is a straight line and it does not stop: outside the core the swirl is Γ/2πr, the energy density falls as 1/r², and the area grows as r², so every decade of radius adds the same amount. There is no such thing as the energy of a line vortex without a stated cutoff, and no cutoff is physical.

The energy a vortex cannot have

A line vortex has infinite kinetic energy. Not a large amount — infinite, growing without limit as the logarithm of however far out the counting stops. And it is losing that energy at a rate that is finite, exactly known, and contains no cutoff at all.

viscous · Diffusion
Five approach profiles a meter might be looking at. The velocity across the pipe upstream of a contraction, for a uniform flow, fully developed laminar flow, two turbulent power laws and an annular jet of the kind a bend or a partly open valve leaves. All five carry the same volume flow. The meter reads a pressure difference and cannot see any of this.

The profile a meter cannot see

A differential-pressure flowmeter measures a force balance and reports a flow rate. The step between them needs two integrals of a velocity profile the instrument has no access to — and two profiles differing by half the mean velocity across the pipe give identical readings, which is why the standards specify straight pipe rather than a correction.

applied · Metering
Two integrands, and the wrong one claims 9.58 per cent more drag. The two things that get integrated across a wake, each scaled to its own peak so the shapes can be compared. The momentum integrand u(U − u)/U² is the drag; the mass integrand (U − u)/U is the displacement thickness, and it is not a drag at all. They differ by a factor of u/U inside them, so the mass one is fatter wherever the deficit is deep — and its integral here is 1.1 times the momentum one's. That ratio is decided by how deep the wake is rather than by how wide: at a twentieth of this momentum thickness it falls to 1, and at twice it rises to 1.25. The error is smallest exactly where a survey is properly done, far downstream where the wake has spread and shallowed.

Weighing what is missing

A control volume drawn round a wing gets the lift out of it, on a flow that was solved exactly. The wake survey asks the same box for the drag on a flow nobody has solved, and it is how a real aerofoil's drag is known — with three assumptions, all of which are checkable, and one integral standing next to it that is wrong.

misconceptions · Momentum lift
A duct that does not change, and a parcel that does. A contraction of area ratio four, with the parcel marked at five equal intervals of time. The walls do not move, the field at every point is the same at every instant, and the spacing of the markers grows because the parcel is carrying its own history through the duct.

How long the fluid has been in there

Age is the simplest thing a flow can remember. It obeys the shortest transport equation in the subject — its material derivative is one — and no instrument pointed at a steady flow can read it, because a steady flow's every field is constant and its fluid is getting older all the time.

kinematics · Acceleration
Where a rotor's pressure rise comes from, as the radius moves. The static pressure rise across a rotor, split into the two terms rothalpy gives it. The diffusion term is held at the de Haller limit throughout — the blade is being asked to slow the relative flow as hard as a boundary layer will allow — so it is a flat 11558.4 Pa at every radius ratio. Everything above that line is the centrifugal term, which costs no diffusion and has no limit of its own. At a radius ratio of 2 it supplies 49.92 per cent of the rise and at 3, 72.66 per cent. An axial machine, at a ratio of exactly one, gets none of it.

What a turning frame keeps

Euler's equation prices the work and says nothing about where the pressure comes from. In the frame turning with the blades — which is accelerating, and carries two fictitious forces — a Bernoulli-like quantity survives both of them, and it splits the pressure rise into a term a boundary layer limits and a term that is free if the radius moves.

applied · Turbomachine
Four wakes carrying exactly the same drag. Four velocity-deficit profiles behind a body, each normalised so that the integral of the deficit across the wake is exactly the same. That integral is the drag: the far-wake momentum balance says so with no assumption about the shape of anything. The four are a narrow Gaussian, a wide top hat, the two-lobed wake a body with a splitter plate leaves, and a profile with heavy tails.

Exact in the total, free in the profile

A constraint is one number imposed on a function. Four wakes built to carry exactly the same drag differ by a factor of four and a half in their peak deficit, and the general statement behind that is a question about angles: how much of the wanted answer survives being projected off the constraints, and how much does not.

regimes · Dimensional
Four fluids under the same stress, and the four profiles that result. Each fluid carries the same linear stress and answers it differently: a Newtonian fluid with a parabola, a shear-thinning one with a blunter profile, a shear-thickening one with a sharper one, and a Bingham plastic with a plug in the middle where the stress is below its yield point.

The stress a pipe knows

A capillary viscometer measures a pressure drop and a flow rate and reports a viscosity. The first half of that inference is a force balance and is exact for any fluid there is; the second needs the slope of a whole flow curve, which is the experiment the instrument was bought to avoid.

viscous · Non-newtonian
A cone sends 1.7 per cent of its jet's momentum sideways at 15°. A conical divergent section of 15° half-angle and exit area ratio 25, drawn to scale from its throat to its lip, with its virtual apex to the left. The gas leaves as a source flow: straight streamlines from the apex, each at its own angle, and a Mach number uniform on spheres centred there. On the spherical cap through the lip the flow is normal to the surface and uniform, at Mach 3.925 for a gas with γ = 1.2; its axial momentum is ρV² times the cap's projection onto the exit disc, while the mass crossing it is ρV times the cap itself. The ratio of the two areas is (1 + cos α)/2 = 0.9830, and it is the only thing the cone's shape does to the momentum. On the flat exit plane the flow is not uniform: its edge is further from the apex than its centre, and the Mach number there is higher.

The jet a cone sprays sideways

The one-dimensional nozzle sends all its gas straight out along the axis. A real divergent section is a cone, and the gas leaves it as a spray of straight lines from the cone's apex. Only the axial part of that momentum pushes, and the share that does is (1 + cos α)/2 — 98.3 per cent at fifteen degrees, 93.3 at thirty — whatever the gas, the Mach number or the area ratio, and it touches the momentum and never the pressure.

compressible · Area mach
At 70 per cent speed the first stage runs at 0.64 of its flow coefficient and the last at 1.17. The flow coefficient of each stage of a compressor of 8 stages at 300 m/s mean blade speed, drawn for a flow coefficient of 0.5 and a work coefficient of 0.35, as a share of the value its blades were cut for, along the operating line a choked exit nozzle sets, at 110 per cent, 100 per cent, 90 per cent, 80 per cent, 70 per cent of design speed. At design speed every stage is at exactly one. Below it the front stages fall towards the stall limit (shaded below 0.82) and the rear ones rise towards the choke limit (shaded above 1.3): at 70 per cent the first stage is at 0.642 and the eighth at 1.170. Above design speed the pattern reverses, the front stages rising and the rear falling.

Matched at one speed and at no other

Every stage of a compressor passes the same mass flow, and the annulus behind each one is cut for the density the air will have reached there at design speed. Slow the shaft and the air is less dense than the metal expects, so the rear stages carry more volume than they were drawn for while the front ones starve — and below a definite speed no throttle setting keeps all of them working at once.

applied · Turbomachine
The trailing sheet rolls up into two vortices, and nothing it carries is lost. The trailing vortex sheet behind an elliptically loaded wing, seen in a plane across the wake, at times 0, 0.05, 0.2, 0.6 in units of b²/Γ₀, represented by 160 point vortices with a smoothing length of 0.03 of the span. The tips curl up first and the sheet winds into two concentrated vortices while the whole system sinks under its own induced velocity; by t = 0.6 the pair's centroid has descended 0.122 of the span. Through all of it the crossflow energy — the induced drag — and the separation of the two halves' centroids, 0.7854 of the span, stay exactly what they were.

The drag a wake keeps however it rolls up

A plane drawn across the wake of a finite wing contains its induced drag as the kinetic energy of the swirling crossflow. The trailing sheet then rolls up into two vortices, and the energy does not change at all — roll-up moves the drag around the plane without spending any of it. What does spend it is viscosity, which turns crossflow energy into a total-pressure defect, so a plane farther back reads less induced drag, more profile drag, and the same total.

misconceptions · Momentum lift
Momentum theory answers every descent rate except those between hover and twice the hover inflow. The induced velocity at a rotor disc against its climb speed, both in units of the hover induced velocity √(T/2ρA), at fixed thrust. The climb branch (thick) solves v(V + v) = 1 and is a streamtube for every climb and for hover, where v = 1. Continued into descent (dashed) it still has a root, but the air it describes leaves the tube at both ends. The windmill-brake branch (thin) solves v(V + v) = −1 and is real only for descent faster than two hover inflows, where it meets v = 1 again. Between V = −2 and V = 0 (shaded) neither is a streamtube. The faint diagonal is v = −V, where the rotor would need no power: it crosses the band and touches neither valid branch.

Between hover and twice the hover inflow

A rotor's momentum balance has an answer for every climb and for every fast descent, and none for descending at anything between zero and twice its own hover inflow. There one root sends air out of both ends of its streamtube and the other root is not a real number, and at each edge of the band one end of the tube stops moving — which is where the vortex ring state lives, and where a wind turbine's thrust coefficient of one sits.

applied · Actuator disc

Named alongside it

The objects these essays reach for when they reach for this one.

Momentum theoremConservationModel limitMeasurementMomentum fluxEfficiencyKinetic energyBernoulli's equationMass conservationDissipationDragThrust

All concepts