Chapter 04 of 12 · GATE NM

Propulsion & Propellers

A chain of efficiencies from the towrope to the engine. Most questions come down to knowing which link is which, and to remembering that the propeller sees the speed of advance, not the ship's speed.

Worked examples3, fully stepped
Read time≈ 15 min
PrerequisiteResistance

1. The hull and the propeller interfere with each other

A propeller in open water is a well-understood machine. Put it behind a hull and two things change, in opposite directions, and both have names worth knowing precisely.

Wake. The hull drags a layer of water along with it, so the water arriving at the propeller disc is moving slower than the ship. The propeller therefore sees a reduced speed, the speed of advance.

V_a = V(1 − w) w = (V − V_a)/V

Wake comes from three sources: the boundary layer (viscous wake), the orbital motion of the ship's own wave system (wave wake), and the streamline flow closing in around the stern (potential wake). Taylor's rough approximation for a single-screw ship is w ≈ 0.5CB − 0.05.

Thrust deduction. A working propeller accelerates water past the stern, which lowers the pressure there and effectively increases the hull's resistance. The thrust the propeller must produce therefore exceeds the resistance measured when towing.

R = T(1 − t) t = (T − R)/T

It is important to see that this is a hull effect, not a propeller loss. The propeller has not become less efficient; the hull has become harder to push.

The key idea

Wake helps — the propeller works in slower water, so it needs less power. Thrust deduction hurts — more thrust is needed than the bare resistance. Their combination is hull efficiency.

2. The efficiency chain, link by link

Start at the towrope and work back to the engine. Each link divides, so the power grows at every step.

P_E effective power = R_T · V ÷ η_H hull efficiency ÷ η_O open-water efficiency ÷ η_R relative rotative efficiency P_D delivered power at the propeller ÷ η_S shaft transmission P_B brake power at the engine coupling
  • ηH = (1−t)/(1−w) — typically 1.0 to 1.15 on a single-screw ship. Greater than one, which surprises people the first time.
  • ηO — the propeller's own efficiency in uniform flow, read from an open-water chart. Typically 0.55 to 0.70.
  • ηR — a correction for the fact that the flow behind a hull is non-uniform and rotational, unlike the open-water test. Usually between 0.98 and 1.05.
  • ηD = ηO·ηH·ηR — the quasi-propulsive coefficient, typically 0.65 to 0.75.

A design instinct worth carrying: for a given power, a larger, slower-turning propeller is more efficient, because it accelerates a larger mass of water by a smaller amount. Draught and hull clearance are what stop designers from making it larger still.

3. The open-water diagram

A propeller's performance in uniform flow is captured by three non-dimensional numbers, and the whole of propeller selection rests on them.

J = V_a/(n·D) advance coefficient K_T = T/(ρ·n²·D⁴) thrust coefficient K_Q = Q/(ρ·n²·D⁵) torque coefficient

Plot KT, 10KQ and ηO against J and you have the open-water chart. KT and KQ fall as J rises; efficiency rises to a peak and then collapses as thrust runs out.

Open-water efficiency is thrust power out divided by shaft power in, and writing that ratio in coefficient form gives:

η_O = (T·V_a)/(2π·n·Q) = J·K_T/(2π·K_Q)

Two unit conventions cause most of the lost marks here, and both are worth writing at the top of the page before you start: n is in revolutions per second, and Va, not V, appears in J. Note also the different powers of D — fourth for thrust, fifth for torque.

Propeller geometry sits alongside: pitch ratio P/D sets how coarse the blade is, blade area ratio AD/AO sets how much area carries the thrust, and skew staggers the blade so that it enters the wake peak gradually rather than all at once — which is a vibration measure, not an efficiency one.

4. Slip: two definitions, one useful

If a propeller were a screw turning in a solid nut, one revolution would advance the ship by exactly the pitch. It does not, because water yields — and it must yield, or no thrust would be produced at all.

Theoretical speed = P × n Apparent slip = (P·n − V)/(P·n) uses ship speed Real slip = (P·n − V_a)/(P·n) uses speed of advance

Apparent slip is what a bridge log reports. It is contaminated by current: a following current raises the observed speed and can even drive apparent slip negative, which looks impossible until you remember the ship is being carried along by the water.

Real slip is the physically meaningful one and is always positive. Trended over weeks at similar draught and weather, a rising real slip is genuine evidence of hull or propeller fouling.

5. Cavitation

Water boils when the local pressure drops to its vapour pressure — at ambient temperature, if the pressure falls far enough. On the back of a propeller blade, where the flow accelerates and pressure drops, that is exactly what can happen.

The vapour cavities collapse violently when they reach a region of higher pressure. Three consequences follow: thrust breakdown as the blade loses its low-pressure side, erosion as repeated implosions pit the metal, and noise and vibration.

σ = (p₀ + ρgh − p_v)/(½ρV_R²)

Here h is the depth of the blade section — which is why cavitation is worst at the top of the disc, where immersion and hence pressure are least, and why a lightly ballasted ship in heavy weather is the classic condition for it.

The defence is blade area. Spreading the same thrust over more area lowers the mean thrust loading and raises the pressure on the back of the blade. Burrill's diagram plots thrust loading against cavitation number with limit lines for acceptable cavitation, and increasing the developed area ratio is how a designer moves back below the line.

Note the cost: more blade area means more frictional drag, so efficiency falls slightly. Blade area ratio is always a compromise between cavitation margin and efficiency.

6. Worked examples

The first walks the whole efficiency chain. The second is the open-water chart question that appears in some form almost every year. The third is the slip question that tests whether you understand what you are measuring.

Worked example 1

Working down the efficiency chain

A ship requires an effective power of 4000 kW at 16 knots. The wake fraction is 0.25, the thrust deduction fraction 0.18, the relative rotative efficiency 1.02 and the open-water efficiency 0.62. Shaft transmission efficiency is 0.98. Find the speed of advance, the hull efficiency, the delivered power and the brake power required.

Given

P_E = 4000 kW, V = 16 kn w = 0.25, t = 0.18, η_R = 1.02, η_O = 0.62, η_S = 0.98

Required

Find the speed of advance, the hull efficiency, the delivered power and the brake power required

  1. Speed of advance first.

    V=16 × 0.5144 = 8.230 m/s V_a=V(1 − w) = 8.230 × 0.75 =6.173 m/s

    The hull drags water along with it, so the propeller works in a stream moving slower than the ship.

  2. Hull efficiency.

    η_H=(1 − t)/(1 − w) =(1 − 0.18)/(1 − 0.25) =0.82/0.75 = 1.093

    It compares the useful work done on the hull with the work done by the propeller on the water it actually sees.

  3. Note that it exceeds 1.

    That is not an error and it is a favourite conceptual question: the propeller recovers some of the energy already lost into the wake, so the hull-and-propeller combination does better than the propeller alone would in open water.

  4. Quasi-propulsive coefficient.

    η_D=η_O × η_H × η_R =0.62 × 1.093 × 1.02 =0.691

    The product of the three efficiencies between towrope and propeller shaft.

  5. Delivered power at the propeller.

    P_D=P_E/η_D = 4000/0.691 =5789 kW
  6. Brake power at the engine.

    P_B=P_D/η_S = 5789/0.98 =5907 kW
    The power chain, towrope to engine
    StageDivided byPower (kW)
    P_E effective4000
    P_D deliveredη_D = 0.6915789
    P_B brakeη_S = 0.985907

    After transmission losses in shafting and bearings.

AnswerV_a = 6.17 m/s, η_H = 1.093, P_D = 5789 kW, P_B ≈ 5910 kW

The trap: answering with the wrong power. P_E → P_D → P_S → P_B, each larger than the last. Read the question twice and note which one it wants before you start.

Worked example 2

Reading an open-water diagram

A propeller of 5.5 m diameter turns at 110 rpm behind a hull whose wake fraction is 0.28 at a ship speed of 15 knots. From the open-water chart at the resulting advance coefficient, K_T = 0.185 and K_Q = 0.0290. Find the thrust, the torque, the delivered power and the open-water efficiency.

Given

D = 5.5 m, N = 110 rpm, V = 15 kn, w = 0.28 K_T = 0.185, K_Q = 0.0290, ρ = 1025 kg/m³

Required

Find the thrust, the torque, the delivered power and the open-water efficiency

  1. Get the units right before anything else.

    n=110/60 = 1.833 rev/s V=15 × 0.5144 = 7.716 m/s V_a=7.716 × (1 − 0.28) = 5.556 m/s

    The coefficients are defined with n in revolutions per second, not per minute, and with speed of advance, not ship speed.

  2. Advance coefficient.

    J=V_a/(n·D) = 5.556/(1.833 × 5.5) =5.556/10.08 = 0.551

    The number that located K_T and K_Q on the chart in the first place.

  3. Thrust from K_T.

    T=K_T·ρ·n²·D⁴ =0.185 × 1025 × 1.833² × 5.5⁴ =0.185 × 1025 × 3.361 × 915.06 =583 000 N ≈ 583 kN
  4. Torque from K_Q.

    Q=K_Q·ρ·n²·D⁵ =0.0290 × 1025 × 3.361 × 5033 =502 700 N·m ≈ 503 kN·m

    Note the fifth power of diameter — a common slip is to use the fourth.

  5. Delivered power is torque times angular velocity.

    P_D=2π·n·Q = 2π × 1.833 × 502 700 =5.79e6 W = 5790 kW
  6. Open-water efficiency — thrust power out over shaft power in.

    η_O=J·K_T/(2π·K_Q) =0.551 × 0.185/(2π × 0.0290) =0.1019/0.1822 =0.559

    The formula is just that ratio written in coefficient form.

  7. Cross-check it directly, which is always worth thirty seconds.

    Thrust power=T·V_a =583 000 × 5.556 =3.239e6 W η_O=3.239e6/5.79e6 = 0.559 ✓

AnswerT = 583 kN, Q = 503 kN·m, P_D = 5790 kW, η_O = 0.559

The trap: using ship speed in J. The wake fraction exists precisely because the propeller does not see the ship's speed, and substituting V for V_a shifts J by 28 % — which lands you on a completely different part of the chart.

Worked example 3

Apparent slip, real slip and a current

A ship's propeller has a pitch of 4.8 m and turns at 95 rpm. The ship's log shows 13.5 knots. The wake fraction is 0.30. Find the apparent slip and the real slip. The next day, in the same conditions, the apparent slip is reported as −2 %. What does that tell you?

Given

P = 4.8 m, N = 95 rpm, V = 13.5 kn, w = 0.30

Required

Find the apparent slip and the real slip
What does that tell you?

  1. Theoretical speed.

    n=95/60 = 1.583 rev/s P·n=4.8 × 1.583 = 7.60 m/s

    How fast the ship would advance if the propeller were a screw in a solid nut, losing nothing.

  2. Convert the observed speed and find apparent slip.

    V=13.5 × 0.5144 = 6.944 m/s Apparent slip=(P·n − V)/(P·n) =(7.60 − 6.944)/7.60 =0.0863 = 8.6 %

    Which uses ship speed.

  3. Real slip uses speed of advance instead.

    V_a=6.944 × (1 − 0.30) = 4.861 m/s Real slip=(P·n − V_a)/(P·n) =(7.60 − 4.861)/7.60 =0.3604 = 36.0 %

    The water the propeller is actually working in.

  4. Real slip is always the larger, and always positive.

    The propeller must slip relative to the water to generate thrust at all. Apparent slip is a navigational quantity contaminated by whatever the water itself is doing.

  5. Now the negative reading.

    Apparent slip<0 ⇒ ship advancing faster than P·n following current (or a log error)

    Apparent slip goes negative when the ship's observed speed over the ground exceeds the theoretical propeller speed — which cannot happen through the water, but happens easily with a following current.

  6. This is why hull fouling is trended on real slip.

    At similar draught and weather, over weeks. Apparent slip on any single day says more about the current than about the hull.

AnswerApparent slip = 8.6 %, real slip = 36.0 %; a negative apparent slip means a following current

The trap: treating a change in apparent slip as evidence about the hull. Only real slip, trended at comparable draught and weather, means anything about fouling.

Reference sheet
60-second recall
  1. V_a = V(1−w), and J uses V_a with n in rev/s.
  2. η_O = J·K_T/(2π·K_Q).
  3. η_D = η_O·η_H·η_R, and η_H can exceed 1.
  4. K_T uses D⁴, K_Q uses D⁵.
  5. Cavitation is a pressure problem, cured with area.