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Old Dec 18, 2012, 10:18 AM
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So Crossplot, I think there’s value in trying to “tie up” the discussion about the connection between the normal acceleration (flow curvature) experienced by a fluid element and the pressure variation in the fluid. I think we are in complete agreement about how the radial pressure gradient at any point in the fluid is related to the acceleration normal to the flow streamlines:

dp = density *( v^2 / r) * dr

At any instant in time, a fluid element moves along a path that can be described by a circle tangent to the path that has the same radius of curvature. The above equation says that if you move out from the center of the circle by a very small distance dr, the pressure will increase by a small amount equal to the fluid density times the velocity squared times the distance you moved divided by the radius of curvature.

The remaining question is: how do you turn this expression into one that relates pressure changes in the fluid to velocity changes in the fluid? Your approach has been to multiply the right hand side by r/2, where r/2 represents an “integration factor” to account for the effects of pressure gradients elsewhere in the field. Multiplying the right hand side by r/2 and replacing the left hand side with the change in pressure gives the familiar Bernoulli equation:

Change in pressure = (density/2) * change in (v^2)

I don’t think you have any real support for this approach (multiplication by r/2) other than “it works”. There is an alternative approach that doesn’t involve any mysterious integration factors. If the fluid is irrotational, then you can relate changes in location (radius) within the fluid to changes in velocity:

dr = -(r/v)dv or equivalently: dv = (-v/r)dr

This equation says that if you move a very small radial distance dr in an irrotational flow, the velocity will decrease by a small amount equal to the fluid velocity times the distance you moved divided by the radius of curvature.

If you combine the equation relating pressure changes to radius changes and the equation relating velocity changes, you end up with an equation relating pressure changes to velocity changes:

dp = density *( v^2 / r) * dr = density *( v^2 / r) * (-r/v)*dv

This simplifies to:

dp = - density * v * dv

You can easily show that if the change in velocity (dv) is very small compared to v, then v*dv is almost exactly equal to (1/2)*[(v+dv)*(v+dv) – v*v]. In other words, the change in the value of (1/2)v^2 is equal to v times the change in v. If you need convincing, just compare these two expressions using a variety of values for v and dv (keeping dv<<v). This lets you write:

dp = -density * d[(1/2)v^2]

where d[(1/2)v^2] represents the change in the quantity (1/2)v^2. If the density stays the same throughout the fluid, then you can also write this:

dp = -d[(1/2)*density*v^2]

where d[(1/2)*density*v^2] represents the change in the quantity (1/2)*density*v^2. This equation tells you that the change in pressure as you move from one location to another is equal to the change in (1/2)*density*v^2 (with a minus sign). This is true whether the changes are big or small. This gets you to the steady, uniform density, irrotational form of the Bernoulli equation without having to resort to a mysterious and unsupportable r/2 “integration factor". Psst... post # 821 is total nonsense.
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