point source paradox?

[DBruceG]

Forgive me for showing my ignorance, but I'm a structural ...

If pressure drops with velocity, then it would seem that the pressure surrounding a point source in an infinite medium must be less than the pressure at an infinite distance from the source, regardless of the sign of that source. That would seem to imply that if one wishes to inject fluid into a reservoir, he needs a suction pump! What am I missing here?

DBruceG

 

[25362]

When you say "pressure drops with velocity" do you mean pressure drops with velocity increases ? Because this is what the famous Bernoulli equation tells us for incompressible, inviscid (frictionless) fluids when running at the same elevation (aka potential energy). Please clarify.

 

[PUMPDESIGNER]

Did you mean to say "pressure falls as velocity increases"?

PUMPDESIGNER

 

[smckennz]

The pressure drop you are referring to is the static pressure. There is also the velocity pressure, and a few others we need no concern ourselves with here. Inside a converging pipe, as the velocity rises so does the velocity pressure;the static pressure experiences a corresponding decrease in accordance with the principle of conservation of energy. Pressure is energy (per unit volume).
Bernoulli is the chap who did the hard yards.
Ignoring friction, the total energy (static + velocity) remains the same.
This concept drove me nuts at first; it was my understanding of static pressure that caused me the grief.
Static pressure and velocity(pressure) are interconvertible, but in such a way that there is no change in total energy.
Whatever that means............
You will get it; aint that hard.

Cheers

Steve

 

[DBruceG]

Thanks for the responses. Yes, I did mean that "pressure drops as velocity increases," though that's certainly not obvious upon my own reread. Special thanks to Steve. Conservation of energy, huh? I'll have to think about that, but it doesn't appear to wash at the moment. If the fluid is under compression at the point of injection, as it seems it must be, then you would lose both strain energy AND kinetic energy as you move away from the source.

Must be the Altzheimer's.

DBruceG

 

[IRstuff]

If the mass is moving at the same velocity, then there is no energy lost.

Conversely, if the mass is moving slower, the energy was lost, likewise, if energy was lost, the mass moves slower.

TTFN

 

[davefitz]

The fluid moves from a position of higher total pressure to a position of lower total pressure. Gravity head, static pressure and velocity head all contribute to the total pressure. The bottom line is that for a process to move in a certain direction, the entropy must increase.

 

[lilliput1]

The most important formula to remenber is pump bhp

pump bhp = (gpm x FT TDH)/(3960 x eff)

and fan bhp

fan bhp = (CFM x Inches wg)/ ((6344 x eff)

where eff is fan or pump efficiency in decimal. A value of 0.65 is good for initial estimate that should be refined with actual catalog data.

The above formulas are important because with them you can easily estimate motor size required.

 

[DBruceG]

Thanks again for the responses, but I'm beginning to think I asked the wrong question. If there is a spherically-symmetric flow emanating from a source at the origin in an infinite body of fluid, what is the pressure as a function of radius if it is p0 at infinite radius?

DBruceG

 

[IRstuff]

Wouldn't the pressure have to drop as a function of increasing radius simply from the fact that the mass is spreading and therefore must slow down?

TTFN

 

[DBruceG]

You're beginning to see my problem, IRstuff. But if the pressure DROPS as the radius increases, then the pressure GRADIENT exerts a force OUTWARD on a differential element of fluid, seemingly contradicting Newton's Second Law. If the mass is slowing down, and it certainly must be, the pressure gradient should exert a force INWARD.

DBruceG

 

[smckennz]

In your case, the velocity decreases with distance from source, assuming a constant fluid "thickness". This will be associated with a corresponding rise in static pressure and the velocity energy is converted to potential (static) energy, so the total (static + velocity) pressure remains constant.
I have ignored friction; davefitz covers this.
When allowing for friction, fluid flows from a higher total energy(pressure) point to a lower energy point; the energy gradient slopes downhill in the direction of flow. The less friction, the flatter the gradient. To increase total energy we need a pump.
It is not true to say flow always travels from higher to lower static pressure. It is true to say flow always travels from higher to lower total pressure, ignoring pumps and a few "special" cases.
I think The understanding problem comes about because in many real-world cases (e.g. pumps and pipes), the static pressure is far higher than the velocity pressure, so we can often ignore the latter without causing a serious error.
Your example appears to have the static pressure gradient going the wrong way. There is nothing wrong with this provided the total energy gradient is either zero (ideal frictionless case) or slopes downwards in the direction of flow. This effect is common to divergent flow and is used to good effect in a venturi pump.

Cheers

Steve

 

[DBruceG]

Steve, I think you've almost gotten to me. Forget the friction, forget the pumps, forget the "special cases." You're saying that in the ideal positive point source case, the energy gradient is zero and the total pressure (i.e., that which compresses the fluid) is constant. Right? I don't quite see the nature of the potential energy increase with radius, but I might in time.

DBruceG

 

[DBruceG]

Steve: I think I've got it well enough to use it. I was laboring under the misconception that the energy balance was strictly between strain energy (i'm working with a gas) and kinetic energy. It becomes clearer if you look at an INCOMPRESSIBLE flow, in which the strain energy is zero. Since there is no energy loss mechanism, the energy DENSITY gradient must be zero, so if kinetic energy density decreases with radius, SOME form of energy must increase. Clearly, whatever that form of energy is, I was ignoring it.

YOU tell me that whatever that form of energy density is, it appears as a pressure. I have to buy that, because if I don't, I arrive at the paradox - you need a suction pump to inject fluid. The necessary conclusion is that since both the kinetic energy and the mysterious (potential?) energy densities appear as pressures, then the pressure must be constant throughout the system.

Thanks!

DBruceG

 

[smckennz]

Sorry; for some reason I assumed you were dealing with a liquid. Must have been the reference to a reservoir.
Pressure is just mechanical energy per unit volume.
However when dealing with compressibles, things get a bit more tricky because of heating and cooling effects (gases cool when they expand), so conservation of thermal energy also needs to be considered. If your density changes are moderate - say 10% of the absolute pressure, then you probably dont need to go there.


Cheers

Steve