The Two Propulsion Problems: Direct and Inverse

Take an illustration, quoted rather than worked here: a twin-screw ship, 1,800 metric horsepower a shaft, a five-bladed propeller 1.83 m across. Turn it at 375 rpm with a pitch ratio of 1.1 and the ship makes 17.0 knots. Gear the same power down to 550 rpm, where the pitch must drop to 0.63 to absorb it, and the same hull makes 16.5 knots. Half a knot, three per cent of the ship's speed, given away to a convenient gear ratio. Nothing was wrong with the propeller. The wrong question had been asked of it.

Nine variables, three equations, six choices

The direct problem is analysis: ship and propeller are fully specified and you ask how the system behaves. Put a 7.00 m four-bladed propeller of pitch ratio 0.715 on our reference Panamax, a bulk carrier of about 75,000 tonnes deadweight, turn it at 84.0 rpm and ask what speed she makes and what it costs. A usable model returns one answer, stably.

The inverse problem is design: the ship, the required speed and the available engine are given, and you are asked what the propeller must be. It does not have one answer, and pretending otherwise is how a competent propeller ends up solving somebody else's problem.

Count the unknowns. Inside a systematic series, a propeller behind a hull is described by nine quantities: four geometric (pitch as the ratio P/D, diameter D, expanded area ratio AE/A0, blade number z), two kinematic (rotation rate n, ship speed V) and three of loading (thrust T, torque Q, and any external tow or pull force F). Three equations tie them together:

T = ρ n² D⁴ · KT(J, P/D, AE/A0, z) with J = Va/(nD) = V(1 - w)/(nD) Q = ρ n² D⁵ · KQ(J, P/D, AE/A0, z) / ηR T = (R0(V) + F) / (1 - t)

Here ρ is the water density, 1,025 kg/m³ in seawater; KT and KQ the open-water thrust and torque coefficients of the series; J the advance ratio and Va the speed of advance; w the Taylor wake fraction, t the thrust deduction fraction, ηR the relative rotative efficiency, and R0(V) the resistance our course Ship Resistance Fundamentals builds. F is zero for a conventional ship and non-zero for a tug on the hook.

Nine unknowns less three equations leaves six quantities that must be specified before the answer is unique. In the direct problem the geometry hands over four of them at a stroke and F is known, which is five; leave the shaft speed free and what remains is a one-parameter family, which is why a direct problem fits on one flat chart with rpm along the bottom. Name the shaft speed as well and the sixth is spent, and the answer is a single point. In the inverse problem four of the unknowns are the geometry itself, and finding six numbers to hold still is your job, not the physics'.

Shaft power is not a tenth variable: SHP = 2πnQ is a function of two quantities already on the list, so it may be specified in place of n or of Q, never as well as one of them.

What closes the design space

You rarely choose the six freely: the ship closes most of them before the propeller itself is designed.

Diameter is capped from above. A screening rule for bulkers and tankers is that the propeller should not exceed about 0.65 of the design draught, which on our Panamax's 12.80 m gives 8.3 m. Lower still is the aperture: the stern has to find the tip clearance the yard and class guidance ask for above the blade, and find it without spoiling the wake field the disc is fed from, and on this ship that caps the diameter near 7.6 m. The yard settled on 7.00 m, 0.6 m inside even that, and what put it there was the shaft speed: at the engine's 84 rpm the best diameter behind this hull is about 7.0 m. Immersion in ballast never came into it on this ship. Lesson 2 takes that apart.

Revolutions are capped by the machinery. On a directly coupled two-stroke the propeller speed is the engine speed, and engines exist only at the speeds their makers build; a gearbox buys that freedom back for weight, cost and three or four points of transmission efficiency (MAN Energy Solutions, now Everllence, puts the shaft efficiency in its propulsion guide near 0.99 on a short, directly coupled shaft and at 0.95 to 0.96 through a reduction gear).

Blade area has a floor set by cavitation, because the thrust must be spread over enough surface to keep the suction peaks clear of vapour pressure; lessons 3 and 4 put numbers on it. Blade number is a vibration decision, not an efficiency one, since the blade-rate excitation z·n must miss the hull's own natural frequencies.

Pitch is the one quantity with no external limit. It is the free variable and takes up whatever slack the others leave, which is why so many propulsion problems reduce to finding a pitch.

What survives is not a point but an arc, and a one-constant sizing rule traces it closely enough to argue with:

n = C · (PD / D⁵)^(1/3)

with n in rpm, PD the power in kilowatts, D in metres, and C equal to 125, 115, 104 and 93 for three, four, five and six blades. Those constants are a fit to single-screw fixed-pitch merchant ships rather than anything published as a standard, so use the rule to open an argument and never to close one. Fed the reference ship's 7.00 m four-bladed propeller and 6,320 kW, it returns 115 × (6,320/16,807)^(1/3) = 115 × 0.7218 = 83.0 rpm, against the 84.0 rpm she turns. The shape matters more than that one per cent: at constant power the shaft speed falls as D to the power minus five thirds, so every diameter has its own shaft speed.

The optimum is a property of the question

The classical way to see the whole family at once is a grid: freeze four of the six quantities and discretise the other two. The common version freezes diameter, area ratio, blade number and tow force, sweeps pitch ratio and ship speed, and plots shaft power against rpm. Two overlaid families of curves appear, iso-pitch and iso-speed, with a cavitation limit across the top; for a tug the second becomes iso-pull, or constant thrust, at a fixed towing speed. You read it by entering with what you know and following both curves to their intersection.

Four different questions all get called optimisation: minimise power at a given ship speed, or maximise speed at a given power, each with either the diameter or the shaft speed held fixed. They do not share an answer. Fixing the twin-screw ship's diameter at its maximum 1.83 m returns 1,800 PS at 17.0 knots with a pitch ratio of 1.1 and 375 rpm; fixing the shaft speed at 400 rpm instead returns the same power at the same speed with a pitch ratio of 0.96 and a diameter of 1.89 m, 60 mm more than the maximum the first answer had just accepted. Same ship, same objective, two propellers, because a different quantity was held still, and the second answer quietly breaches a wall the first one respected. Nothing in the optimisation announces that; you have to check the answer against the constraint you did not hand it.

Minimising power at a speed and maximising speed at that power are duals; on the grid they are the same tangency point. Change the objective to pull and the gap widens. In an illustration of the same kind, fifteen tonnes (147 kN) of bollard pull at 1.83 m diameter wants about 1,690 PS at 320 rpm and a pitch ratio near 1.0 from the best open five-bladed propeller; the same pull from a Kaplan blade in a 19A nozzle takes about 1,130 PS.

Check any pull figure against the actuator disc before you repeat it. The ideal static thrust from P watts through a disc of area A is (2 ρ A P²)^(1/3), which for 1,690 PS through a 1.83 m disc is 203 kN, or 20.7 tonnes. Fifteen tonnes is 72 per cent of that frictionless, swirl-free ceiling, which is about what a good propeller reaches. A quoted pull above the ceiling is not an ambitious propeller; it is an arithmetic error, and tug specifications are full of them.

Two design points, one propeller

A trawler needs its best free-running speed to the grounds and its best pull while towing. A merchant ship has the same problem more quietly: it trades laden and in ballast, where the propeller runs about 2 per cent lighter, and it is often asked to work at speeds nobody designed it for.

In a similar illustration for a fishing vessel wanting 17 tonnes of pull at 4 knots and the best free-running speed a ducted 1.83 m propeller will give, the near-optimum regions of the two grids overlap between 300 and 370 rpm. Take 320 rpm and you get the 17 tonnes at a pitch ratio of 1.2 for 1,690 PS, and 15.1 knots free running for 1,100 PS. A controllable-pitch propeller would add about 1.2 knots, which its hub and hydraulics have to be worth.

A single design point is a convenient fiction, defensible only while you can say which condition it describes and how far the ship's real life sits from it.

Monday morning

Find your ship's propeller drawing and the engine's shop trial report and write down D, z, AE/A0, the pitch ratio at 0.7R and the rated power and speed. Ask which six quantities the yard froze when it fixed that geometry, and whether the condition it froze them in is the one the ship trades in now. If the propeller was drawn for a laden hull at 14 knots and the ship has spent three years at 11, you know something the design file does not. Lesson 2 takes diameter and revolutions, and shows how far apart the hydrodynamic optimum and the buildable one really sit.