Calculating required HP from drag

In the past few days I've staggered through teaching myself the basics of Free!ship and Michlet (with some key pointers from forum members - thank you oh so much!). Freeship for designing the 3D hull model, and Michlet for calculating the drag. Very cool tools - and FREE!

I've rough-designed a 40 ft power catamaran hull. Goal is to have very good fuel economy, good speed and range with reasonable payload capability for coastal cruising/liveaboard/fishing in SE Alaska and maybe points south (and north).

Target cruise speed is 17-20 kts. L/B ratio is 12.5/1. Loaded displacement is about 20,000 lbs. Max prop diameters is about 17 inches.

Michlet tells me that drag at 19 knots will be about 2,800 lbs. Two props - need to produce 1,400 lbs of thrust. Do those numbers sound reasonable?

Given that info how do I estimate what sort of horsepower I need? An 85% solution is fine at this point so I can determine fuel tankage etc. Getting to specific motors/props will come later.

 

2800 lbs ~ 1500kg ~ 150N
17-20 knots ~ 35-39 km/h ~ 10m/s
150N x 10m/s = 1500Nm/s = 1500J/s = 1500W = 1.5kW ~ 2HP
Theoretically, with 100% efficiency, of course

 

Thankks Raggi, but I think there was a loss in translation between imperial and metric. I don't think a 10 ton displacememt boat will go 20 knots with 1.5KW (2 HP), regardless of efficiency... (damn imperial system!)

Converting from lbs-force to KN, divide by 224. Gives about 12.5 KN. Using your method yields a theoretical power requirement of 120 KW, or 160 HP. That's 80 HP from each prop per side.

So if I need 80 HP output from the prop, what's a reasonable guess on the inefficiencies to work back to motor size? I understand there are lots of variables here...

 

Sorry about that
Quite embarrassing, but I often di things a bit too fast...
I am sure there are some motor boat experts around here whi can answer more accurately. I would guess 30% extra power in the engine, but thats just a wild guess.

 

the highest efficiency you should expect from a prop is 70% so a good rule of thumb is to design for 50% which includes a margin and 85% mcr... don't wanna run at full throttle all the time... if it's a cruising speed you're talking about... Also more power required for wind resistance and to get through waves...

 

Thanks Dave.

So when I look at a published diesel engine power graph, there's the shaft power curve and the propeller power curve. The propeller curve is well below the shaft curve until you get to the full rated power.

Am I correct that that's because a prop can't absorb all of the engine's shaft power across the whole engine power curve? Would that be the "inefficiency" of the prop? Or do I need to factor ANOTHER 50% inefficiency on top of that?

 

The absorbed HP by the propeller can't be given by the engine manufacturer. It depends on many things including hull design. I recomend you read "Propeller Handbook" by Dave Gerr. It has a good explanation of theory and all the necessary formulae.

 

Thanks Gonzo - the book is already on order from Amazon...

I thought the same as you - after you look at the power curves in my other posting on this subject you'll see my confusion.

 

Good luck. The book requires a bit of math, but only basic algebra. All you need is a common calculator.