0
$\begingroup$

Given:

  • speed 30 knots
  • engine power 70 hp at 5800 RPM
  • propeller 3 blades, diameter 14 inch, pitch 17 inch, at 2300 RPM
  • fluid density 1023 kg/m3

I know that each blade is basically one wing. Calculating lift is extremely complex; even Navier-Stokes equations will not give an accurate result.

But can I estimate the thrust this produces, for these values or in general?

Improve this question
$\endgroup$
11
  • 3
    $\begingroup$ power is basically speed times force but you can make your life harder by not using metric. $\endgroup$
    – user3528438
    Sep 11, 2020 at 18:55
  • $\begingroup$ I use a very rough rule of thumb of about 3-4 lbs of thrust per HP. A 150 HP Lyc O-320 makes about 600lbs with a typical prop. For small propellers like that, it would be in the lower end, somewhere around 3 lbs, so you should get about 200-250 lbs with 70 HP. $\endgroup$
    – John K
    Sep 12, 2020 at 3:22
  • $\begingroup$ @JohnK Propeller is small because it is boat propeller on outboard engine.But formula must be same just fluid density is different. $\endgroup$
    – user52248
    Sep 12, 2020 at 4:37
  • $\begingroup$ related: is there any equation to bind velocity, thrust and power? $\endgroup$
    – Peter Kämpf
    Sep 12, 2020 at 11:01
  • $\begingroup$ @PeterKämpf In your formula with power...Gasoline outboard engine efficiency is 35%,so why I need multiply power with efficiency (70HP x0.35) if I has 70HP at propeller axis?That mean I has only 24HP available for thrust,that cant be true..also I dont know hom much is my delta v $\endgroup$
    – user52248
    Sep 12, 2020 at 15:03

2 Answers 2

Reset to default
3
$\begingroup$

Explaining maritime propeller design in comments doesn't work well, so I give it a shot here.

Given: 70 hp at 5800 RPM engine output, geared down at the prop shaft to 2300 RPM. Propeller with 3 blades, 14" in diameter and 17" in pitch.

First let's write that in SI units: 52.2 kW, 0.3556 m diameter, 0.4318 m pitch. That is an advance ratio of 1.2143. Speed is 38.333 rev/sec. If prop efficiency is 100%, thrust is simply power divided by speed. If the prop has no slip (unrealistic, but provides us with a starting point), the boat speed would be 16.5 m/s. The advance ratio above neglects slip, so the real value is maybe 1 (more for a lightly loaded prop and vice versa). 20% slip would reduce speed to 13.8 m/s. This corresponds to a thrust between 3782 N for 100% efficiency and 1891 N for 50% efficiency. The truth will be somewhere between these two extremes.

How is efficiency calculated?

Yes, prop flow is complex and this doesn't get any less complex when the prop is mounted at the stern of a boat. Flow is not uniform there and if the propeller is too close to the surface, it will operate in two media. Since one is a liquid, it itself can turn into a gas if pressure is sufficiently low (cavitation).

In order to straighten the flow and reduce the impact of the boat on the prop, all kinds of things have been tried (Schneekluth duct, pre- and post swirl fins, different values for rake and skew and even winglets in the shape of Kappel propellers). While aircraft propeller blades can reasonably be approximated by wings, ship propeller blades have tight restrictions on diameter which makes them more similar to twisted disks than to wings. Consequently, much of the flow is influenced by the tip effects and the propeller has a much higher solidity (called the developed area ratio in maritime propellers) so mutual interference between the single blades is also more pronounced. This all has an influence on efficiency.

That should be enough to explain why there is no simple formula. Naval engineers are used to diagrams and coefficients, and in this case use a Kramer diagram (see below, picture source) to calculate the ideal efficiency of a propeller. Ideal means that this diagram does not include viscous losses and assumes uniform flow.

Kramer diagram

To use this diagram, you need to know the propeller area A$_O$, the inflow velocity v$_A$ (which is normally lower than the ship's speed due to the vicinity of the hull), the thrust T, the rotation rate n in rev/sec and the propeller diameter D. With a known advance coefficient you start at the bottom and move up on the diagonal lines until you hit the horizontal line for the right number of blades. From that point you go straight up until you hit the known ideal efficiency (and read the corresponding thrust coefficient on the right side) or until you hit the known thrust coefficient and read the ideal efficiency from the curves.

In your case both efficiency and thrust coefficient are unknown and we don't even know the boat's speed. So to proceed means a lot of guessing until you go out and measure the speed your propeller-engine combination is capable of, and then return to the Kramer diagram.

Improve this answer
$\endgroup$
2
  • $\begingroup$ If boat speed is 26knots(13.3m/s) when engine produce 52.5kW and "theoretical no-slip speed" is 32knots(16.5m/s),then propeller efficiency is 13.3/16.5=80% ....then....Thrust=52.5kWx0.8 / 13.3m/s= 3139 N Is this correct results? $\endgroup$
    – user52248
    Sep 13, 2020 at 13:41
  • $\begingroup$ @Сократ No, efficiency and slip are not related. $\endgroup$
    – Peter Kämpf
    Sep 13, 2020 at 13:58
3
$\begingroup$

Power P, stationary thrust T, propeller swept area A, and air density rho are linked by this expression, derived from first principles and momentum theory:

T^3 = 2 · P^2 · A · rho

That's the maximum power available assuming a 100% prop efficiency.

Things change if the prop is moving forward, as in a flying plane. With the forward speed component, the blades have their angle of attack reduced in proportion to that forward speed. The blades work as wings, and a wing profile has a coefficient of lift that changes with the AoA. Drag does also change with a growing AoA, Hence, for every section of a fixed-pitch prop's blade, there's an optimum L/D at a given combination of rotating and forward speeds that gives the best efficiency. For a fixed angular speed of the prop, there is a given forward airspeed at which the prop works with maximum efficiency, that efficiency –and hence the thrust– dropping asymptotically as the forward speed grows...

Concerning the power delivered by a propeller in forward motion, it's the product of the thrust and speed...

Improve this answer
$\endgroup$
11
  • 1
    $\begingroup$ This is valid for static thrust. Now add that power is thrust times speed, so propeller thrust varies inversely with speed, and you'll get an upvote. Bonus if you add something on propeller efficiency. $\endgroup$
    – Peter Kämpf
    Sep 12, 2020 at 10:58
  • $\begingroup$ @xxavier Concerning the power delivered by a propeller in forward motion, it's the product of the thrust and speed... speed is 30knots(15m/s) and power 70HP(51kw) ,so Thrust=51000W / 15m/s=3400N ?so I dont need any data from propeller ,only need speed and engine power to calculate thrust? $\endgroup$
    – user52248
    Sep 12, 2020 at 15:17
  • $\begingroup$ @Cokpat Any object moving at a constant speed v through a medium with a drag d will need a propulsive power v·d It's the same if it's a submarine, an airship or an airplane... And at a constant speed, thrust = drag. Any imbalance will mean an acceleration, and we've specified a constant speed... $\endgroup$
    – xxavier
    Sep 12, 2020 at 21:57
  • $\begingroup$ @xxavier I understand that but if I look at 5m boats with 5HP they all go approx 5knots at full throttle.With this basic formula I get thrust of 1430 N,this is way too much, 5HP engine cant produce 143kg of thrust!So I cant find where this huge error comes from. $\endgroup$
    – user52248
    Sep 13, 2020 at 5:54
  • $\begingroup$ @Сократ How do you know that 1430 N is too much thrust...? $\endgroup$
    – xxavier
    Sep 13, 2020 at 6:24

You must log in to answer this question.