Small displacement craft resistance

Hey all,

I am designing a flat bottom boat with following particulars:

Loa = 3.3 m
Lwl = 3.02 m
Bwl = 1.24
D = 0.177 m
Displ = 385 kg
LCB -4.2% of Lwl aft Lwl/2



For my own interest i did a CFD resistance analysis in Star CCM+. The analysis was performed in calm water, at a speed of 2.2 m/s (Fn = 0.4) and fixed draught, so no movement possible (translation and rotation). The analysis resulted in a total resistance of 140 N and thus a theoretical shaft power of R*v = 140 N * 2.2 m/s = 308 W. Although this was a very basic simulation (no trim, no sinkage, no engine modeled, probably coarse mesh, etc.) 308 W still sound a bit low for me to move a boat at it's theoretical hull speed. A 2.9 kW outboard would be mouch more than required as a propulsor.

What do you guys think?

 

Me thinks i need to change from dragging a model hull across the pond, and into the matrix. Sorry, beyond me, but i will be looking at that programme.

 

That is the computed propulsive power. Now you need to include all the losses to get to the powerhead power. Prop efficiency is probably about 50% . Shaft losses on tiny outboard legs are going to be at least 10% (higher at part throttle). So 308W*1/0.5 * 1/0.9 = 684W as a first guess. The friction, wave, and interference losses of the drive leg are also significant.

 

That's what I meant with "very basic simulation", so the bare hull resistance without any added resistance by the outboard itself nor I didn't count with efficiency of prop and outboard.

But even with your efficiency losses, do 700W make sense? I mean there are many boats like my design with a 3kW outboard - do they actually need the power or would a smaller engine be sufficient to reach hull speed? I'm gonna simulate higher speeds above the theoretical hull speed to see the resistance rising.

@skaraborgcraft I can make use of the software via my university but I have no chance to make real world tests on a pond although I would like to.

 

My dink planes quite nicely on 3 hp. It will push 500 pounds to about 12 knots. People row at about 60-80 Watts, and they can go at four knots at that power on a bit longer boat. Four people can paddle a 28' canoe weighing 2 tons at 4 knots. Displacement speed was usually achievable by 2 hp per ton, and that includes all losses - windage, fouling allowance, maneuvering allowance, everything. It was a practical level of power. I ran about 3 hp on average on the intercoastal waterway (3 gallons per day) on my 14,000#, 38' sailboat. Installed power was about 18hp and I ran a 14" two-bladed folding Martec prop that would fit in your pocket.

 

You are modeling half the hull. Did you remember to double your forces?
In addition since you are not leeting the boat heave your displacement is likely to be significantly too low

 

Yes, only half hull is modelled due to symmetry. For the half hull it was 70 N so yes I doubled the value.

I was waiting for someone to notice this ;D

 

Seat of my pants (and lots of experience with little boats)..... displacement of 385kg on a short very fat boat will need a lot more thrust than might be provided by 308 watts to achieve hull speed. At a displacement of 100 kg then I think 308 watts might work.