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

Every essay — page 5

Page 5 of 61, continuing through the fields in the same order.

Flows and fields Ideal flow Circulation and lift Viscosity Regimes and numbers Compressible flow Transition and turbulence Fluids at work What is taught wrongly Series Concepts Regimes Refutations Search

Fluids at work

Turbines, pipes, weirs, balls, sails, arteries and blades. What the conservation laws say about machines, which is more than a designer expects and less than a brochure claims.

Six nozzles, six pump curves, and where each one is best. The head ratio a water jet pump delivers against the flow ratio it entrains, for six area ratios from a narrow nozzle to one filling four-fifths of the throat. A wide nozzle makes a tall, steep curve that is finished at a small flow; a narrow one makes a low, long curve. The dots are each curve's best-efficiency point. Nozzle, suction, throat-friction and diffuser losses are included at borrowed representative values, and the mixing loss is computed.

The nozzle that is best at one thing

Put the four losses back into a jet pump and three questions get three answers. The most head comes from a nozzle four-fifths of its throat, in closed form; the best efficiency from one a quarter of it; and the most flow from whichever nozzle is smallest, because flow has no optimum at all.

8 figures The characteristic gets a wall, and the wall does not care about the discharge. One jet pump's characteristic, at an area ratio of 0.275, with the flow ratio at which its throat entry reaches vapour pressure drawn for three values of the cavitation parameter σ = (Pₛ − pᵥ)/(Pₘ − Pₛ). Left of a wall the machine runs on its curve. At the wall no lower discharge pressure raises the flow: the head ratio can fall to zero along the vertical and the flow ratio stays where it is. The wall's position contains the nozzle and suction losses and nothing downstream of the throat entry.

The wall the suction puts in the curve

A liquid jet pump's lowest pressure is where the entrained stream enters the throat, and when that reaches vapour pressure the pump curve stops being a curve. The flow ratio freezes at a value no lower discharge pressure can move — and raising the motive pressure, the obvious cure, brings the wall closer.

7 figures Flat until the back pressure reaches a value, then gone. Entrainment ratio against back pressure, as a multiple of the suction pressure, for three mixing-section sizes of one steam ejector driven from a motive supply a hundred times its suction pressure. Each is exactly flat while the entrained stream is choked beside the jet, up to its critical back pressure. The dashed lines join that point to the back pressure at which the entrainment has fallen to nothing, 1.4 per cent higher for the middle machine. The model fixes those two ends and not the path between them. A larger mixing section entrains more and breaks at a lower back pressure.

A choke that belongs to two streams

A steam ejector entrains a fixed amount of gas whatever its back pressure, up to a pressure where it stops. The flat part is a choke, and the entrained gas is not at Mach one when it happens: it is at 0.886, because the supersonic jet beside it is part of the same throat. One area ratio then trades that entrainment for compression, and the trade decides how a vacuum train is built.

7 figures The current that runs across the river. The transverse velocity through the depth in a bend, beside the downstream velocity that causes it. The surface water is thrown outward and the bed water is driven inward, the two cross at about half the depth, and the whole profile carries no net discharge — which it cannot, because the cross-section has to pass what it is given. What it does carry is the bed load, inward.

The current that runs across the river

The standard channel calculation is one-dimensional, and a river is not. Put a bend in it and the surface tilts by a few centimetres, which is invisible — and a helix forms, a few per cent of the main flow, which decides where the river will be in a hundred years.

6 figures The depth at the edge, against how much pressure is left there. The brink depth as a fraction of the critical depth, against the share of the hydrostatic pressure the brink is taken to retain. A free jet — no pressure at all — gives exactly two thirds, to 1e-16. The measured ratio of 0.715 is met at 40 per cent of hydrostatic, which is more residual pressure than a falling nappe looks as though it should have.

The depth at the edge is not the critical one

Every open-channel calculation assumes the pressure is hydrostatic, and usually says so nowhere. At a free overfall it is not, the assumption's failure can be measured, and the measurement is one number — a brink depth that is 0.715 of the critical depth where a hydrostatic reading would make it one.

6 figures The vapour fraction that would stop a cavity, against temperature. How much vapour a cavity would have to make, as a multiple of the liquid volume it cools, before the cooling had used up the whole driving tension. In cold water it is six hundred and the effect is unreachable; at the boiling point it is below one and the effect is the dominant thing in the problem. Four decades across the range a pump might see, out of a vapour density that rises by a factor of two hundred.

A cavity that cools the water it came from

The usual account takes the vapour pressure as a property of the liquid, looked up once. It is a property of the liquid at whatever temperature the cavity has left it — and making vapour costs latent heat, so a cavity suppresses itself. In cold water that is unmeasurable and in liquid hydrogen it is the dominant term.

7 figures The rudder should carry an eighth of the side force. The drag of keel and rudder together, as a percentage of the keel carrying everything, against the rudder's share of the side force, for a rudder 0.809 of the keel's depth. The induced drag is least with the keel alone — a shallower rudder makes the combined loading less elliptic for any load it takes. The profile drag is least near equal lift coefficients. Their sum is least at 12.4 per cent, 0.97 per cent below the keel alone; for the yacht drawn here, a rudder 0.809 of the keel's depth, that is 12.4 per cent.

The rudder pays for the keel's wake

A rudder angled to carry side force feels drag out of all proportion to its share, and that is the sensation behind the rule that weather helm is slow. The pair's drag does not care who is billed. It is least with the rudder carrying an eighth of the load, at under a degree of helm, and that share does not change with how hard the boat is pressed.

6 figures An open rotor's slipstream shrinks; a duct holds it open. The radius of the slipstream behind a hovering rotor of radius R, against the distance behind the disc: for an open rotor, from the vortex-cylinder model, contracting towards R/√2 so that the wake ends with half the disc's area; and for a rotor in a straight duct, whose slipstream leaves at the full area, and in a duct that widens to 1.3 times it. The same thrust from a wider jet needs a slower one, and a slower jet wastes less energy.

A duct is worth the square root of two

An open rotor squeezes its slipstream to half its own area and pays for the fast jet that results. Put the same rotor in a straight duct and the slipstream leaves at the rotor's full area, the jet is slower, and hovering costs 29 per cent less power. The duct is not a passive guard: it carries half the thrust itself, on the suction round its inlet lip. In cruise almost all of the advantage disappears.

6 figures Time per litre rises in a straight line with the litres already filtered. The elapsed time divided by the filtrate collected, against the filtrate, for a constant-pressure filtration of a silica slurry through a cloth: integrated step by step, and Ruth's closed form. The line's slope is the cake's resistance, which grows with every litre, and its intercept is the cloth's, which does not. After a quarter of a cubic metre the cake is already the larger.

A filter is slowed by what it has caught

Push a slurry through a cloth and within minutes the cloth no longer matters: the particles it has caught form a cake, the cake is a packed bed, and every litre filtered makes the bed thicker for the next. The filtrate grows only as the square root of time. And if the cake is soft, it packs tighter against the cloth the harder it is pushed, until more pressure buys almost nothing.

6 figures A washed filter's best cycle is shorter, and still has one. The filtrate a batch filter produces per hour, averaged over filtering, washing and twenty minutes of emptying and refilling, against the filtrate collected in each cycle, for a slurry of 400 kilograms of silica per cubic metre and wash ratios of 0, 1, 2 and 5 pore volumes. Each curve has a best cycle, marked, and the wash both lowers it and moves it to a thinner cake: from 0.307 to 0.217 m³ per m² of cloth.

A washed filter works as long as it rests

A batch filter that has to be emptied between runs has a best cycle, and the rule for it is exact: the time the cake is responsible for equals the time the filter stands idle. Washing the cake does not break the rule, because a wash through a finished cake costs time in proportion to the square of its thickness, just as filtering it did. What a wash changes is the price, and the price is set by the slurry's concentration and by how far the wash front smears, not by the cake's softness.

6 figures One fast pump or two slow ones. Pump efficiency against the flow asked for, for two pumps into a system with 60 per cent static lift: one pump under speed control, up to the flow at which it reaches full speed; both under speed control; and the best number running with the pumps held at full speed and a valve taking up the difference. Below 0.601 of the design flow one pump is better than two, and above it two are better — before the single pump has run out of speed at 0.65. The throttle is far worse at every flow.

One fast pump or two slow ones

A pumping station with several identical pumps in parallel has a control that a single pump does not: how many of them are running. With no static lift the answer is all of them, always, slowed together. With lift it is a number that falls with the flow and is almost never whole — the flow times the number of pumps, over the square root of the head the system asks. Staging rounds it, and the right moment to round up comes before the running pumps have run out of speed.

6 figures Surge is a loop round the characteristic's peak. The two runs in the plane of flow coefficient and plenum pressure rise, over the characteristic continued to reversed flow. With B = 0.5 the state slides off the peak and settles on the stalled characteristic. With B = 2 it traces a large loop: the flow collapses at nearly constant pressure, reverses while the plenum empties, recovers to the right-hand branch at low pressure, and climbs back up it while the plenum fills, round and round.

The plenum decides whether a compressor surges

Throttle a compressor past the peak of its characteristic and it does one of two things. It settles into rotating stall, a steady state with a cell of dead flow running round the annulus, or it surges, the whole flow through the machine collapsing, reversing and recovering over and over. Which one is not decided by the blades. It is decided by the volume the compressor discharges into, against the inertia of the air in its duct — one number, Greitzer's B, which grows with the blade speed, so the same machine stalls at part speed and surges at full speed.

7 figures