The Loss Budget: Where a Propeller's Energy Actually Goes

Our reference ship is a Panamax bulk carrier of about 75,000 tonnes deadweight, making 14.0 knots. Her propeller absorbs 6,386 kW and hands the ship 3,447 kW of useful thrust power. The other 2,939 kW, 46 per cent of everything the propeller is given, goes into the water and never comes back.

That single number is the whole of this course. Every duct, stator, boss cap, end plate, rudder bulb and thrust fin ever sold is competing for a share of those 2,939 kW. None of them can take more than the channel it attacks contains, and once you know how big each channel is, most vendor claims answer themselves.

Building the budget in kilowatts

The ship's propeller is 7.00 m in diameter, four bladed, turning at 84.0 rpm, which is 1.400 rev/s. Her wake fraction is 0.32, so the speed of advance, the mean axial velocity the propeller actually sees, is 0.68 × 7.202 = 4.897 m/s against a ship speed of 7.202 m/s. That gives an advance ratio J of 0.500, at which the open-water curves read KT = 0.146 and KQ = 0.0215. Our course Open-Water Performance derives those coefficients; here we only spend them.

Torque and thrust follow directly: Q = 726 kN·m and T = 704 kN. The power the propeller absorbs in the open-water condition is

P = 2 π n Q = 2 π × 1.400 × 726,000 = 6,386 kW

where n is the rotation rate in revolutions per second and Q the open-water torque in newton metres. The thrust power, the only part of that the ship gets any benefit from, is thrust times speed of advance:

PT = T · Va = 704,000 × 4.897 = 3,447 kW

The ratio is 3,447 / 6,386 = 0.540, which is precisely the open-water efficiency on the propeller's data sheet. Nothing has been assumed yet. The delivered power at the shaft flange behind the hull is about 6,320 kW rather than 6,386 kW, because a relative rotative efficiency of 1.01 says the propeller runs marginally better in the ship's wake than in uniform flow.

Two denominators are now in play and they differ by that same 1 per cent. This lesson sizes the four loss channels against the 6,386 kW the propeller absorbs, because that is the power they come out of. From lesson 2 onwards every device claim is measured against the 6,320 kW delivered, because that is the number a vendor writes his percentage on. Name the one you are using and the arithmetic stays honest; leave it unsaid and two people can spend an hour arguing over a difference of 1 per cent.

A caution on these numbers. Our reference Panamax is a teaching hull with round figures, and her 704 kN of thrust comes from a resistance of 577 kN that is light for her size: a real Panamax at 14.0 knots needs nearer 680 kN of resistance, and so nearer 830 kN of thrust. On a real hull every kilowatt in the budget below is about a fifth larger, and the loss channels more than that, because the propeller is also more heavily loaded. The teaching hull therefore understates the pot rather than flattering it, and since a vendor's percentage grows with the same power, the verdicts this lesson reaches come out the same on a real ship.

So 2,939 kW is lost. Where? The honest answer is that the split is an engineering estimate, not a measurement. The published review literature on stern energy-saving devices agrees on the ranking of the channels and on the mechanisms, but no source fixes the percentages for a given hull, and you should treat any table that claims to, including this one, as indicative. For this ship at this operating point a defensible split is: axial kinetic energy left in the slipstream about 1,810 kW, blade section friction about 660 kW, rotational swirl about 320 kW, and hub and tip vortices together about 150 kW. The swirl figure is the one lesson 2 works out from the propeller's own torque, rounded; the axial term takes what the other three leave.

The ceiling on the biggest slice

The axial channel is the largest, and it is also the one where physics has already taken most of the recovery off the table. A propeller makes thrust by throwing water aft, and thrown water carries kinetic energy the ship cannot get back. Momentum theory puts a hard floor under how much it must carry. Define the thrust-loading coefficient

CTh = T / (½ ρ A Va²)

where ρ is seawater density in kilograms per cubic metre, A the disc area in square metres and Va the speed of advance in metres per second. For our ship, A = π × 7.00² / 4 = 38.48 m², so ½ρAVa² = 473 kN and CTh = 704 / 473 = 1.49. The ideal axial efficiency, the best any device working by pushing water aft can reach at that loading, is

ηi = 2 / (1 + √(1 + CTh)) = 2 / (1 + 1.578) = 0.776

Turn that into kilowatts. A perfect actuator disc producing 3,447 kW of thrust power at this loading would still need 3,447 / 0.776 = 4,442 kW, so 995 kW of axial kinetic energy is irreducible. The real propeller's axial term is about 1,810 kW; roughly 800 kW of it comes from finite blade number, non-optimal radial loading and the non-uniform wake, and that 800 kW is the only part any device can chase.

This is why diameter is the strongest lever in the whole subject. Take the same thrust on a 7.70 m disc instead of a 7.00 m one. Area rises to 46.57 m², CTh falls to 1.23, ηi rises to 0.802, and the irreducible axial loss drops to 850 kW. That is 145 kW saved, 2.3 per cent of absorbed power, from geometry alone, with no appendage, no added wetted surface and nothing to foul. A ten per cent larger propeller outperforms most of the hardware in this course, which is worth remembering before you buy any of it. Hull clearance, cavitation and pressure pulses are what usually stop you, not the hydrodynamics. (On a real Panamax's nearer 830 kN of thrust the loading coefficient is near 1.75, the ceiling near 0.75 and the irreducible axial loss nearer 1,340 kW, and the same 7.70 m disc saves nearer 190 kW: the lever is stronger on a real ship, not weaker.)

What is genuinely on the table

Now the arithmetic that makes this course honest. The entire swirl channel is about 320 kW, 5.0 per cent of what the propeller absorbs. The best recovery anyone credibly reports from a pre-swirl or post-swirl device is 30 to 50 per cent of that channel, which is 96 to 160 kW, or 1.5 to 2.5 per cent. Add whatever share of the recoverable axial term a device can genuinely reach, which for a duct is disputed (lesson 8 grades it at no more than about 60 kW on this ship), and you arrive at the familiar band: credible single-device gains cluster at 2 to 6 per cent across the fleet, and on this ship nearer the bottom of it.

Set that against what is quoted. The ranges that circulate in the retrofit trade give pre-swirl devices 2 to 9 per cent, propeller measures 2 to 7 per cent and post-swirl devices 0 to 5 per cent, and the zero at the bottom of the post-swirl range is doing real work. Those are broker and vendor bands rather than a published measurement, and where a class society does publish, it is more cautious: a 2024 DNV article on decarbonisation options for bulk carriers puts each hydrodynamic measure at 1 to 5 per cent. One peer-reviewed review of stern devices records makers' claims as high as 15 per cent, and says plainly of the 2 to 8 per cent that manufacturers commonly guarantee that it is not well supported by experimental data. Fifteen per cent of 6,386 kW is 958 kW, three times the whole swirl channel. That is not a pessimistic reading or an optimistic one; it is an impossible number.

The budget also raises an awkward question this course has to answer. Hub and tip vortices together are only 150 kW here, 2.3 per cent, yet boss cap fins are routinely credited with 2 to 5 per cent, and one careful model-and-computation study of a finned contraction-type cap reports at most 1.28 per cent on a six-bladed container-ship propeller, and 0.69 per cent on a five-bladed bulk carrier's. Either the hub share is larger on some propellers than this estimate allows, or the higher claims include effects that are not hub-vortex recovery at all. Lesson 3 takes that apart.

Monday morning

Pull your ship's open-water curve and her model basin report, read off KT, KQ and J at the service condition, and do the two multiplications above. Ten minutes gives you the absorbed power, the thrust power and the loss, and you will never again read a device brochure without asking which of the four channels it proposes to spend. When the next quotation lands, ask the vendor to name the channel and the fraction of it he expects to recover; a supplier who cannot answer that has not designed anything. Lesson 2 starts at the front of the propeller, with the ducts and pre-swirl fins that go after the axial and the swirl slices at the same time.