Fluid Flow & Aerodynamics
why things fly, and why the usual explanation is wrong
Concept

Method of characteristics — where it appears

A solution method for hyperbolic equations that follows the curves along which information actually travels. On each characteristic a combination of the variables is carried unchanged, which turns a wave problem into arithmetic along a grid of arrival times.

Named by 8 essays across 3 fields — each of them below, with the objects they name alongside it.

The wall is what cancels the waves it made. The characteristic net of a minimum-length nozzle designed for Mach 2.4, with 18 waves. The pale lines run from the sharp throat down to the axis, reflect there by symmetry, and run back up to the wall; the wall turns through exactly the angle needed to cancel each one as it arrives, so nothing reflects back into the flow and the exit is uniform at Mach 2.400 and parallel to 0.0 degrees. The area ratio decides the Mach number and this net decides the shape, and the two agree on the exit height to 0.51 per cent at this resolution.

The wall that cancels its own waves

The area ratio of a supersonic nozzle fixes its exit Mach number and says nothing whatever about its shape. What fixes the shape is a wave-by-wave construction in which the wall turns through exactly the angle needed to absorb each expansion as it arrives — and getting it wrong leaves a stream full of oblique shocks at precisely the right Mach number.

compressible · Characteristics How much the peak cares about the valve at each instant. A small bump is added to the valve's opening at one moment and the peak is recomputed; the difference, divided by the bump, is plotted against the moment. It is flat and zero for the first 0.45 seconds — that part of the closure is invisible to the peak — rises through the reflection arrivals, and falls to nothing after the peak has happened, which is the one part of the shape that needs no fluid mechanics.

The part of the closure a pipe cannot see

A surge specification says how long the valve takes to shut. The line does not integrate the duration, it integrates the shape — and it is measurably blind to the first half-second of an eight-second closure, while three closures of identical length give peaks 97 per cent apart.

applied · Water hammer A 204 m hammer traded for a 8.57 m swing over 299 seconds. The water level in a 10 m surge tank at the end of a 2 km tunnel 3 m across, carrying 2 m/s, after the turbine is shut off at once, with the level measured from the reservoir's. Without friction it rises to V₀√(L Aₜ/g Aₛ) = 8.57 m and swings with a period 2π√(L Aₛ/g Aₜ) = 299.1 s, the integration agreeing with both closed forms. With the tunnel's 5 m of friction the level starts 5 m below the reservoir, peaks at 5.61 m after 98 s, falls to −3.70 m, and decays. The same tunnel shut at its end with no tank would take the Joukowsky rise of 204 m. The tank does not remove the column's momentum; it gives it a free surface to push against, slowly.

A tank that turns a hammer into a swing

Shut a turbine at the end of a two-kilometre tunnel in two seconds and the valve takes a rise of 256 metres of head. Put a shaft open to the air beside it and the rise is 51, the tunnel never carries the closure as a wave at all, and its water slows instead against a level that climbs for a minute and a half — to a height that is a closed form with the tank's area under a square root.

applied · Water hammer 121 m after the cavity closes, against 69 m from the closure. The head at a valve shut instantly on water flowing at 0.36 m/s through 600 m of 100 mm pipe, a = 1200 m/s, with a steady head of 25 m. The closure raises it to 69.1 m, the Joukowsky head; the reflection returns at one round trip, 1.00 s, and takes the head down to the vapour head, −10.1 m, where a cavity opens (shaded). It closes 2.146 round trips after the closure, and the first pulse after it reaches 121.3 m — 52.3 m above the Joukowsky head — for 146 ms. The step line is the exact solution between events; the thin line is a 240-reach grid solver that was told nothing about it and agrees with its first pulse to better than a millimetre.

Twice the margin, on top of the hammer

Shut a valve on a line whose pressure is low and the returning wave boils the water beside it. When that cavity closes, the head at the valve can pass the Joukowsky rise — by up to twice the margin that let the water boil, in a sawtooth that jumps each time one more round trip fits into the cavity's life, and for a time that is shortest exactly when the pulse is tallest.

applied · Water hammer The throat holds the hammer back only while its cavity lasts. Left, pressure at the closing valve (red) and at the upstream face of the venturi (gold); right, the throat's cavity volume; after the valve shuts with a 5 mL cavity in the throat. The valve sees the full Joukowsky rise of 14.8 bar at once. The wave reaches the venturi 33.3 ms later, and for the next 7.9 ms the upstream pipe hears nothing: its pressure stays at 5 bar while the cavity is squeezed. When the cavity closes at 41.3 ms the surge passes into the upstream pipe at 13.8 bar above its steady pressure.

A choked throat buys time, not silence

A venturi whose throat has reached vapour pressure passes a flow the downstream pressure cannot change, and it is tempting to read that as isolation: whatever happens downstream, the upstream pipe will not hear it. Slam a valve downstream and it hears it. The cavity at the throat holds the surge back only for as long as it takes to fill, and then lets 93 per cent of it through.

misconceptions · Venturi Gas takes the top off a tall, brief pulse. The head at a closed valve against time, for a line whose margin to vapour pressure is 0.49 Joukowsky rises: the exact vapour-cavity history, whose first pulse after the collapse reaches 1.94 rises for only 0.028 of a round trip, and the same line with a pocket of free gas at the valve of a millionth, a hundred-thousandth and a ten-thousandth of the pipe's volume. A hundred-thousandth brings the pulse down to 1.62: the pocket cannot be squeezed fast enough to follow a pulse that short.

Air at a valve softens the hammer only in quantity

A vapour cavity at a closed valve holds the head at one value, and that is why the pressure after it collapses climbs a staircase of equal steps. Put a pocket of free air there instead and the valve becomes a spring. A millionth of the pipe's volume trims only the tallest, briefest pulses; a hundred-thousandth takes a third off them and moves them; a ten-thousandth can make the pulse taller than it was, even on a line that never cavitates. The air that reliably removes the hammer is a thousandth of the pipe, which is a deliberate air vessel rather than a little dissolved gas.

applied · Water hammer Tailored, the reservoir waits for the driver's own expansion. Distance against time in a helium-driven air tube at the tailored Mach number, 3.41, with a driver half as long as the driven tube: the incident shock, the contact surface, the reflected shock and the shock it transmits into the driver gas, and the driver's expansion — its head running back to the driver's closed end, and the head reflected from there. Nothing returns from the contact, and the reservoir at the end wall holds from 0.293 until the reflected head arrives at 0.555 L/a₁: a test time of 0.261.

A tailored tube buys its test time with its driver

Tailoring a shock tube removes the wave the contact surface would send back to the reservoir. What ends the reservoir then is slower: the driver's own expansion, which runs back to the driver's closed end, reflects, and has to cross the whole tube to reach the end wall. Its arrival is exact in one dimension, because the reflected head crosses the incident fan as a simple wave, and the answer is that test time is bought with driver length — about seven-tenths of a driven-tube crossing time per driver length for helium — and that tailoring is worth nothing with a driver shorter than a quarter of the tube.

compressible · Shock tube Air that comes out fast makes the first pulse taller. The head at the closed valve of a 600 m main, in Joukowsky rises above its steady head, for eight round trips after the valve shuts, with the water saturated with air at atmospheric pressure. With no release the cavity is vapour and the first pulse is 1.3 rises. If all the water gives up its air slowly, over ten round trips, the gas cushions the collapse and the pulse is 1.16. If it comes out fast, in three-hundredths of one, the cavity holds the head near atmospheric pressure rather than the vapour pressure, it lasts a round trip longer, and the pulse is 1.46 — the air-cavity limit's 1.45.

The air a cavity releases is capped by the cavity

When a shut valve pulls the pressure behind it down, the air dissolved in the water starts to come out, and a little gas was already known to soften the hammer. But air leaves the water only while its pressure in the cavity is below the pressure the water was saturated at, so a cavity can fill with at most its own volume of air at that pressure — a seven-thousandth of the pipe here, however much water gives up its air and however fast. That is less than a third of what removes the hammer. Released quickly, the air does not cushion the collapse at all: it turns the vapour cavity into an air cavity, and the pulse becomes the column-separation pulse with the margin measured to the atmosphere — taller, on this main.

applied · Water hammer

Named alongside it

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

Model limitWater hammerColumn-separationWave speedCavitationVapour pressureReflectionRiemann invariantsCharacteristicsComplianceExpansion fanSurge tank

All concepts