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

Every essay — page 4

Page 4 of 39, 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.

Two ways to move a duty, along one axis, in opposite directions. A duty at a specific speed of 0.01 and what splitting it does. Dividing the head between stages in series multiplies each stage's specific speed by the number of stages to the three-quarter power, because the group carries (gH)^(−3/4); dividing the flow between units in parallel divides it by the square root of the number of units, because the group carries √Q. Both exponents are read off the group rather than remembered, and they are why the two operations are not interchangeable: three stages buy a factor of 2.28 and three units cost a factor of 0.58. The bands are drawn in the colour this site reserves for a borrowed claim, because where a Francis runner stops is practice rather than a result.

The duty that had no machine

Some duties have a specific speed outside every band, and no runner will do them at any size because the number contains no size. That is true and it is not the end. The same group says how to split the duty until it fits, and the two ways of splitting move it along the same axis in opposite directions, by exponents read straight off it.

6 figures The shaft speed is a window, and cavitation closes the top of it. Two groups against shaft speed for the same duty: the specific speed, which must be inside a band for a runner to exist, and the suction specific speed, which must be below about 3 for the impeller not to cavitate. Both rise with the shaft speed, so raising it to reach a band is also raising it towards the cavitation limit. The window here runs from 274.98 rpm to 962.31 rpm and cavitation sets its top. A duty whose window is empty needs something other than a different machine — a booster, a lower installation, or an inducer.

The group with no head in it

The number that picks a machine says nothing about whether the machine can exist. A second group formed from the same variables, with the delivered head replaced by the margin available at the inlet, decides that — and the delivered head has left the expression entirely, so how far a pump lifts is irrelevant to whether it tears the liquid apart at its own entrance.

6 figures 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.

5 figures A blade root at 412.51 megapascals, and the chord is not in it. The centrifugal stress at a blade root against shaft speed, for one annulus of 0.36 m², in four materials. Integrating the blade's own weight outward gives σ = 2π ρ_b A N² with a taper relief — and the chord has cancelled, the blade count has cancelled, and every property of the gas has cancelled. What is left is an area times the square of a speed, capped by a material. The horizontal lines are each material's allowable stress, and where a curve crosses its own line is the fastest that annulus may be turned in that metal. At 12000 rpm this blade carries 412.51 MPa with a tip speed of 439.82 m/s.

The stress that picks the aerodynamics

The free term in a rotor's pressure rise wants radius and speed, and both are capped by something with no fluid in it. Integrate a blade's own weight outward and the root stress comes out as the annulus area times the square of the shaft speed — with the chord cancelled, the blade count cancelled, and every property of the gas cancelled.

5 figures 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.

6 figures Below Thoma's 6.07 m² the governed tank's swing grows; above it, it dies. The tank level after the turbine's power demand drops by two per cent, with a governor holding the power constant, for tanks of 0.7 and 1.3 times Thoma's area of 6.07 m² — a tank 2.78 m across. The smaller tank's swing grows by a factor of 1.47 every 74 s cycle and has reached −12.20 m by 427 s; the larger one's keeps 0.75 of itself every 100 s and is barely visible. Carried on, the smaller tank's run is refused at 794 s, where the head at the turbine has fallen below a quarter of its design value and the governor would be asking for a flow no turbine passes. The instability has nothing to do with the tank's height: it is the governor drawing more water as the level falls, which feeds the swing, against the tunnel's friction, which is the only thing damping it.

The better tunnel needs the bigger tank

A turbine governed to hold its power opens further when the head at it falls, and draws the tank down harder. That makes it a negative resistance, the tunnel's friction is the only thing damping the swing against it, and so the smallest stable tank grows as the friction shrinks — 2.78 metres across for five metres of friction, 6.09 for one.

7 figures A force ceiling ends the similarity: a breeze makes the boat slower as a share of the wind. Speed made good to windward as a fraction of the true wind, on the best course for each wind, for a rig that is flattened once the righting moment binds and for one that is reefed, with the hull's drag angle held at 6°. Below 3.70 m/s the two are the same boat, and the fraction does not depend on the wind — 1.122 at every speed, which is the similarity the two-angle polar rests on. Above it the flattened rig's fraction falls: 0.918 at 6 m/s, 0.572 at 10 m/s, 0.321 at 15 m/s and 0.169 at 20 m/s. The reefed rig's stays at 1.122, because a reefed sail keeps its least drag angle and the hull's angle is held. Once a force is limited, the triangle is no longer the same shape in every wind.

A breeze the boat cannot use

The two-angle polar makes a boat's speed a fixed fraction of the wind, in any wind. A righting moment ends that at 3.7 metres a second on the beat. Past it the crew must spill force, a flattened sail's drag angle climbs, and the best course to windward moves closer to the wind rather than away from it — which 45° + λ/2 cannot say, because λ now depends on the course.

7 figures A keel pulls the best beat 12 degrees closer to the wind. Speed made good as a fraction of the wind against the course, in 3 m/s: for the boat with its keel solved as a wing, and for a boat whose λ is fixed at 25.69°, the value the keeled boat has on its own best course. The search puts the keeled boat's best beat at 45.83°; the fixed-angle rule puts it at 45° + λ/2 = 57.84°. Pointing higher loads the keel towards its best lift coefficient — 0.130 at 40° against 0.076 at 60° — and lowers its drag angle, so the curve peaks early and falls away faster on the far side. The fixed-angle curve promises 0.653 of the wind; the keeled boat can make 0.553, and only by sailing twelve degrees higher than the rule says.

A keel flies wherever the course puts it

A keel is a wing whose lift is whatever side force the rig happens to make, so its lift coefficient is chosen by the course and the wind rather than by its designer. Solved that way, the hull's drag angle stops being a property of the boat: it pulls the best beat twelve degrees closer to the wind, halves the reaching speed a fixed angle predicts, and finally charges for reefing.

7 figures 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.

8 figures 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.

8 figures 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.

7 figures A pump slowed into a system with a static lift leaves its own specific speed. The specific speed of the operating point, as a share of its value at the best point, against the fraction of the design flow delivered, for a pump drawn for 0.1 m³/s against 40 m at 1450 rpm, specific speed 0.545 at its best point. Under speed control into a system whose static lift is 0 per cent, 30 per cent, 60 per cent, 90 per cent of the design head (lines), and under a throttle at design speed into the 60 per cent system (dashed). With no static lift the speed-controlled pump stays exactly at its best point and its specific speed never moves. At half the design flow it has fallen to 1.000, 0.800, 0.708, 0.652 of the design value as the static share rises, and to 0.598 under the throttle.

The specific speed a pump spends its life at

A pump is chosen by its specific speed at its best point and then run somewhere else. Written in the pump's own coefficients the number is √φ/ψ^¾, a position along its characteristic, and a variable-speed drive keeps it there only when the system it pumps into has no static lift. Every metre of lift moves a slowed pump along its own curve, towards shut-off, and a throttle moves it further.

7 figures