Hello HydroGuy,
Right away you can tell your equation is not quite right for two reasons:
1. It doesn't have dimensions of power. Compare:
A unit of power
1 W = 1 \frac{J}{s} = 1 \frac{N \cdot m}{s} = 1 \frac{kg\cdot m \cdot m}{s \cdot s^2} = 1 \frac{kg\cdot m^2}{s^3}
A unit of whatever quantity is being calculated by your equation:
\textrm{mass} \cdot \textrm{area} \cdot \textrm{velocity}^3
= \frac {kg \cdot m^2 \cdot m^3}{s^3} = \frac {kg \cdot m^5}{s^3}
2. The presence of "mass" doesn't makes sense in this context.
"Mass" of what? The entire river? Or the amount of water impinging on the blade per unit time? It is my sneaking suspicion that the m should actually be a rho \rho meaning density. This makes more sense because density is a intrinsic property of the fluid that doesn't depend on the amount you have. The fact that we are off by length to the third power in our dimensional analysis above also greatly strengthens this supposition (since density = mass/volume).
To give you a sense of why this might be true, I'll give you a sketchy outline/hand-waving derivation that appeals to our mutual knowledge of first year physics, nothing more:
1. Definition of Power
The instantaneous work done on the blade is just the total force on it times the instantaneous displacement, s:
dW = Fds
Of course, power is just the RATE at which work is done on the blade with time:
P = \frac{dW}{dt} = F\frac{ds}{dt} = Fv
where v is the flow rate of the water. This is just the good old "power = force * velocity" relation.
So what is the force, F, on the blade? Before I answer that question, I want to derive a relation for the mass flow rate, because it will come in handy later. The mass flow rate is just the mass of water arriving at the blade per unit area, per unit time.
2. Mass Flow Rate
To derive this, imagine a volume element, i.e. an infinitesimal cylinder of water with cross sectional area dA and length ds. Then obviously this volume element has volume:
dV = dAds
Now, if we choose the length so that ds = vdt, then in time interval dt, all of the water in the cylinder will flow out of it. So the rate of flow of water through area dA will just be:
dV = dAds = vdAdt
\frac{dV}{dt} = vdA
This is the volume flow rate (cubic metres per second). To get the mass flow rate, (kilograms per second), we must figure out how much mass is in volume dV. Obviously this is just the density of water multiplied by that volume:
dm = \rho dV
And so the mass flowing per unit area per unit time is just given by:
\frac{dm}{dtdA} = \rho v
Remember this result for later.
3. So what is the force?
Recall that force is the rate of change of momentum (Newton's 2nd).
F = \frac{dp}{dt}
Now, let's talk about the momentum "arriving" per unit AREA per unit time. This would be the "flow of momentum" or momentum FLUX. Obviously it would have units of pressure, because it would just be a force per unit area. Since we've already used p, P, AND rho, I'm going to use little f for force per unit area (i.e the fluid pressure):
\frac{dp}{dtdA} = f
But what is differential amount of momentum arriving, dp? This amount of momentum, given that the velocity is constant, is just the differential amount of mass, dm, multipied by v:
v\frac{dm}{dtdA} = f
But this is just the velocity times the mass flow rate that we derived earlier! Now we make an assumption that the fluid pressure, f, is the the same across the entire blade. N.B. This assumption is WRONG! So we have that the total force on the blade is just equal to f*A, where A is the area of the blade (force = pressure *area):
v\frac{dm}{dtdA}A = fA = F
Now substitute the result for the mass flow rate from part 2:
v \rho v A = fA = F
Substitute this expression for the force into the expression for the power from part 1:
P = Fv = v \rho v A v = \rho A v^3
So that gives you some idea of where this equation might have come from, assuming the mass was supposed to be, in fact, a density. Now, you might be asking, what about the factor of 0.5? Well, we made an erroneous assumption that certain quantities that are defined at a single point in space were in fact constant everywhere. This assumption was wrong. The fluid pressure, for instance, could have varied in space. That would have necessitated some sort of integration (which is probably how the random factors of 1/2 turn up).
I KNOW for a fact that in a static situation (water is not flowing), the pressure would vary with depth, h, by the relation:
f = f_{\textrm{initial}} + \rho g h
In a dynamical situation, I'm not sure how that changes. It's been a long time since I took fluid mechanics, and I was never very fond of it. I would encourage you to Google fluid statics and fluid dynamics for more details. I hope this helps!
