A racing boat planes over the water at 80 MPH. The water temperature is 75F. The wetted surface of the boat, i.e., the part of the boat in contact with the water, is 3 ft wide and 8 ft long. Estimate the power required to overcome the surface resistance generated by the water against the wetted surface of the boat.

Sigma(75F) = 4.93E-3 lbf/ft http://www.engineeringtoolbox.com/water-surface-tension-d_597.html

Surface tension, by definition, is force per unit length of contact, so the length of contact is either the width or the length (I'm not 100% sure so I'm going with length)

Not having the equation in front of me, we should be able to just do this:

Power is in units ft-lbf/s so we need to multiply by 2 units of length and divide by one unit of time

Power = Sigma * L * v which has the same units
Power = (4.93E-3)(8)(80)(88/60) ft-lbf/s
Power = 4.6276 ft-lbf/s = .008413 hp

If for some reason it is width and not length, then it would be that power * 3/8

.008413 seems extremely small. I got an answer of 156.3 HP

It did seem extremely small to me as well but I couldn't find an equation for it so I had to derive a guess.

Do you have an equation?

Yea,

*Reynolds Number RE = (V x L) / Kinematic Viscosity

Drag Force (Fd) = (Shear-Stress Coefficient x Density x Speed x Area)/ 2

Power = (Drag Force x Speed) / 550

Reynolds equation is right, power equation is right if drag force is in lbf and power is in hp

Your drag force equation's units don't line up at all, so I'm inclined to believe it's a bad equation for this scenario. Density*Area*Speed is a unit of mass flow (mass/second). "Shear-Stress Coefficient" is not the same thing as surface tension and that's all you're given, so you're missing an equation somewhere.

This post was edited by Dontrunaway on Nov 13 2014 04:04pm

force is a function of v^2

That would even out the units, but we still need a way to get the shear-stress coefficient (which is unitless) of water at 75F

This post was edited by Dontrunaway on Nov 13 2014 04:25pm