Originally Posted by UpTheChinas
...why its Half Density in the formula for coefficient of lift
another way of putting what people have well said is that it's not Half-Density, it's Half-V-Squared (Density x 1/2 x V^2), coming from the integration in the derivation of Bernoulli's (actually Euler's, so I understand) equation.
I find Genghis's link too hard to understand. I think it's a lot simpler, as follows.
Euler's equation is
(dp) = -Rho x V x (dV)
where (dp) is differential increase in pressure, Rho is density, V is (local) velocity, and (dV) is incremental increase in velocity. (We mathematicians balk at these infinitesimals, but what's good enough for Newton is surely good enough for us).
We treat Rho as constant. That is the assumption of incompressibility.
Integrate left and right sides of Euler's equation between arbitrary points (arbitrary pressures) p1 and p2. I write the indefinite integrals "[..]" and the limits afterwards "(between .....)":
[p](between p1 and p2) = -Rho x [1/2 x V^2](between p1 and p2)
Let V1 be the velocity at the point at which pressure is p1. Similarly V2. Substitute the limits:
(p2 - p1) = Rho x 1/2 x (V1^2 - V2^2)
(notice I have absorbed the "minus" sign). Rearrange:
p1 + 1/2 x Rho x V1^2 = p2 + 1/2 x Rho x V2^2
This 1/2 x Rho x V^2 term is dubbed "dynamic pressure", because it plus static pressure (the p term) is by this result constant throughout the flow (p2 and p1 were arbitrary), and so this looks like the energy equation (for much the same formal reasons). Dynamic pressure + static pressure is called total pressure, so another way of expressing Bernoulli's result is to say that total pressure is constant throughout an incompressible flow.
The way one derives Euler's equation is by considering an infinitesimal element of volume in the flow. I don't know whether this will work in words, but if words were good enough for Newton it's worth a try. Probably best to draw things on paper as one reads.
Let the flow be 2D (i.e. constant cross-section in the third dimension, so you can draw this cross section on paper). Assume a unit distance in the 3rd dimension (orthogonal to the 2D cross section). Let the flow be from left to right. Take two stream lines very close to each other. They are going to be parallel or converging or diverging, and since we are considering an "infinitesimal" volume, we can take the stream lines to be straight lines, not curves (the beauty of infinitesimals!). Now chop off the flow geometrically with a plane orthogonal to the flow, to the left, and another plane orthogonal to the flow to the right (these become vertical lines in the 2D cross section). Remember they are "infinitesimally" apart. Let the distance at the left "chop" line between the streamlines be S. To the right, they will be S+some increment dS, because the streamflows are not necessarily parallel. Similarly, let the flow enter the left plane with velocity V, pressure p, and exit the right plane with velocity V + some increment dV, pressure p + some increment dp. Any of these increments might be positive or negative, depending.
This is all just nomenclature so far. Now, a crucial step comes with the observation of the pressure orthogonal to the streamlines. The streamlines are straight. From the notion of "streamline", no particles are going in or out of it (idealisation of course on the molecular level), and the pressure orthogonal to the streamlines to keep this so must be the average pressure between the left and right faces, i.e.
( p + (p + dp) ) / 2
or p + 1/2 dp.
Assume a perfect fluid, so the viscosity is 0 and all forces arise from the pressures.
Newton's second law says that the force acting on a particle equals the time rate of change of its linear momentum (F = ma). This averaged momentum over all particles can be taken at the left "face" and right "face", and top and bottom faces contribute 0 since particles are flowing parallel to them (they are streamlines).
So Rho x S x V is the mass entering from the left in unit time, thus with momentum (Rho x S x V) x V. Same mass leaves on the right, but with incremented velocity, so momentum (Rho x S x V) x (V + dV). Momentum change is then
(Rho x S x V) x (V + dV) - (Rho x S x V) x V
which is Rho x S x V x dV. This must equal force, as just said. Force on left face is (p x S x 1), on right face is - (p + dp) x (S + dS) x 1, which when you multiply out, and ignoring the "vanishingly small" product dp x dS (an engineering move), and then you subtract the forces, the resultant is
- (p x dS) - (S x dp)
The net force from the sides is (p + 1/2 x dp) x dS (by geometry), which is
p x dS
when one ignores the product dp x dS. So total force is the sum of these:
- (p x dS) - (S x dp) + (p x dS) which is just - S x dp
and this equals rate of change of momentum, which we derived above:
- (S x dp) = Rho x S x V x dV
Divide by S:
- dp = Rho x V x dV
which is Euler's equation as promised.
Acknowledgement: This derivation is Shevell's: Richard S. Shevell, Fundamentals of Flight, Prentice-Hall, 1983 (there is a second edition, which I don't have).
PBL