# Convert kN to required kW

Discussion in 'Propulsion' started by Mat-C, Mar 1, 2012.

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### Mat-CSenior Member

I've quite often read hereabouts that a design requires a certain amount of thrust in order to move at a particular speed... typically given as kN.
Simple question, I'm sure, but how do you go about converting that to the required SHP required (or to keep the units constant... kW)?
Is it as simple as mulitplying by the prop efficiency... so around 0.5 for conventional shafts, or 0.6 for sterndrives...?

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The total drag/resistance of a hull is usually expressed kN. The trust required by prop/jet etc must be at least equal to this value, at each given speed.

You then multiply that drag by the speed (in m/s) you require to obtain the naked EHP; which is kN x m/s = kW.

If the hull has appendages, then you need to add a %'age allowance for that. Then mechanical losses, typically 3-4%. And then finally either actual values from supplier or estimated values of losses from previous sea trials, for the PC.

So the actual power required is the = EHP/PC

Note, the PC also depends upon the arrangement of the shafting prop/waterjet and how the flow of the water is effected into the propellers. Since a shaft angle of say 1 degree with a smooth ramp up aft shall be different from say a 10 degree and short abrupt hull change aft. So while the theoretical PC may be the same (from suppliers estimation of their eqpt only), the actual, installed, shall be different for said reasons.

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### gonzoSenior Member

As he noted shaft and propeller arrangement influence the calculations. If you look at the water flow at the stern of a boat, at some speeds it may be reverse from the movement of the boat. In other words, the speed and flow direction of water in relation with the hull is different in each area. That changes at different speeds too, particularly if a boat get up on a plane.

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### Mat-CSenior Member

Thanks

Ok... so for by way of an example...

Resistance calcs show the total resistance = say 20 kN
Required speed = 20 knots x 0.5144 = 10.3 m/sec

10.3 x 20 = 206 kW

so, if prop efficiency = 50%, you need 412 kW at the prop, plus allowance for appendage drag....

Correct?

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Kind of.

As i noted above and as Gonzo noted, the arrangement of the installation and how it is also affected by the hull shape play a significant part. As Gonzo noted, it is not uncommon to sometimes have a positive thrust deduction factor!

The prop efficiency, is that open water efficiency??..since this is where everyone goes wrong. They plug in their numbers into a fancy bit of software and out pops an answer with 70% or 80% efficiency etc. The theoretical and open water efficiency is just that. Very very rare shall you ever have a prop installation that approaches the open water efficiency of the prop.

It is significantly influenced by the hull shape immediately in front of it and above, the size of the boss and to a lesser degree the shaft angle.

So if you said the PC was 50% then yes your calculation is correct. But prop efficiency, then no. Since the other factors need to be taken into account to establish the kW per engine at the output shaft.

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### Mat-CSenior Member

Gotcha... thanks again.

So how would an NA go about taking all these things into account... is there a way of claculating them, or is it more from experience?

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Both...thus the more designs you do the more of a database you have for estimation.

It is easier with waterjets though, all you do is place the drag curve over the thrust curve of the jets.

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### gonzoSenior Member

I think that one of the reasons the VolvoPenta pods are so efficient, is that they are really close to ideal open water efficiency. When Regal Marine switched from inboards to pods, they managed the same speed with almost 30% less power. This is something I tested.

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### WillallisonSenior Member

I agree - there's absolutely no doubting the efficiency gain by going from shafts to pods, though published tests indicate that those gains vary considerably from boat to boat.
I still have a problem with their vulnerability to (very expensive) damage though. I know, I know...rip a pair of shafts out of the bottom of the boat and see how much it costs...
The gain over their (better protected) aft facing counterparts (ZF / Zeus) is less obvious. And equally, I haven't come across anything that suggests there's much of a gain to be had over (infinitely less expensive) sterndrives. (Though these are currently difficult to get in the higher HP range, of course)

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### gonzoSenior Member

The gain with pods is not only on thrust efficiency, but in maneuvering.

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### WillallisonSenior Member

Yes - that was certainly the case when they were 1st released. Now though, there are an increasing number of similarly effective "joystick" set up available for all manner of drivetrain installations. Of course, a number of those rely on thrusters to assist, which brings complexities and compromises of their own...
One aspect of pods that hasn't been widely publicised is the increased potential for broaching in a following sea. Because they rely on thrust almost entirely for their steering effect, they can be quite a handfull. One experienced skipper recently told me that the pod-driven boat he brought from Sydney to Hobart was the worst boat he'd ever driven. He cosidered it to be downright dangerous. The boat was a sportsfishing boat fitted with a tower, which exacerbated the problem. By the time they reached Hobart, the tower had ripped itself almost free from the the boat and all the bulkhead tabbing had pulled apart. Whether it's a problem on other IPS equippped boats, I have no idea...

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### AlikSenior Member

- reduction of appendage drag;
- higher efficiency of propellers;

So as share of appendage drag in total resistance depends on FnV and specs of craft, the total gain will differ depending on those factors.

With Zeus/ZF, one should add resistance of tunnels that might comprise 4% extra to total drag.

Basically pods (at least those from VP) have same advantages as sterndrives, but provide solutions for higher horsepower.

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### johneckSenior Member

There is also a thrust deduction factor that must be included in the powering calculation. For planing hulls it is probably quite small and can be assumed to be 1, but for larger displacement vessels is can 15-20% and must be included. Thus the required thrust is the resistance time the thrust deduction factor (1-t) in PNA. There are many books such as The Principles of Naval Architecture that have details on the method used to get from resistance data to powering estimates.

Pod propellers are not really more efficient than typical open propellers, but the vessel resistance is usually less due to the elimination of shafts, struts, rudders, etc.

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sure and when you fill the fuel tank and the esky on your new pod drive boat with small engines it wont get on the plane, ive had a demo of that in a regal

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### gonzoSenior Member

Which model was it? They all performed really well.

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