6 replies to this topic

[Renegade]

Hello everyone,

I have been on this forum previously and was fortunate to benefit from the experience of Art Montemayor in particular.

I now have a very fundamental question to pose to the engineers here. I have known for a long time that pressure drop in a pipe is directly proportional to the velocity of fluid flowing through it, i.e. at greater velocities we have greater pressure drop.

I am in a job designing heat exchangers and the rules of thumb are that greater velocities provide greater heat transfer coefficients (U), hence decreasing the area of the exchanger required. The caveat: at the expense of pressure drop resulting in potentially greater pumping costs.

My question is this: Aside from the Darcy-Weisbach equations and such which so very tastefully articulate the interdependence of deltaP and V, is there a physical explanation to the phenomena observed (perhaps on a molecular level)? I mean, why does increasing the fluid flow velocity, increase the pressure drop? I understand that a potential difference is needed for flow, but I'm not able to picture the exact mechanisms in my head.

Thank you for your input in advance!

Best regards,
Anirudh

[acer_asd]

Dear Renegade

I try to explain you this phenomenon based on my little knowledge on fluid dynamics.

When any fluid moves in a pipe, there is a collision between molecules because of the random movements of molecules which results in the decrease in the kinetic energy. This loss of kinetic energy is converted into some other form of energy (Internal energy, I think). As we know from continuity equation(A1*v1=A2*v2) that if cross section of flow is not changing then velocity of fluid has to be same. Due to this reason, pressure energy is converted into kinetic energy to keep the velocity same. Hence we observe drop in pressure.

When you increase the velocity of fluid, the collision between molecules increases resulting into greater loss of kinetic energy, in turn, greater pressure drop.

The explanation i gave you what i thought about this phenomenon when i was studying. I request other members to confirm if it is correct or not.

Regards

[katmar]

It is a question of conservation of energy. The sum of the individual energy components remains the same. Most of this is explained by the classic Bernoulli Equation - although he did not include friction. Almost any fluid mechanics book will have the explanation of the complete energy balance which extends the Bernoulli Equation to include the shaft work and friction terms.

[kkala]

Let us try to "feel" frictional pressure drop of an incompressible fluid, moving along an horizontal pipe of constant diameter. Continuity law advises that (mean) velocity v does not change along the pipe, on the condition that volumetric flow rate Q is constant. Q=(π*d^2/4)*v (d=internal diameter).
Since the fluid touches the pipe internal walls, a shear stress R (frictional force per unit area)is developed due to friction. Considering an internal cylindrical part of the pipe, length=l, lateral surface=π*d*l, frictional force is F=π*d*l*R against flow direction.
Horizontal forces on mentioned liquid cylinder are in equilibrium, so P1*π*d^2/4=F+P2*π*d^2/4, or (P1-P2)*π*d^2/4 = π*d*l*R. Hence ΔP= P1-P2= 4*R*l/d, or ΔP/l=4*R/d.
R represents the frictional force opposing to flow. It is according to our experience to assume that this resistance to flow increases with velocity, e.g. compare to drug resistance of objects in a moving fluid. We had learned in elementary physics that friction is independent of velocity, being an oversimplification (compare to oscillations, where friction is assumed proportional to velocity), especially for fluids in motion. So resistance increases with velocity, causing higher frictional pressure drop as a consequence. Experiment has verified it, R being a function of Reynolds Number, velocity and pipe roughness.
This is a simplified view, more on text books.

[kkala]

Let us try to "feel" frictional pressure drop of an incompressible fluid, moving along an horizontal pipe of constant diameter. Continuity law advises that (mean) velocity v does not change along the pipe, on the condition that volumetric flow rate Q is constant. Q=(π*d^2/4)*v (d=internal diameter).
Since the fluid touches the pipe internal walls, a shear stress R (frictional force per unit area)is developed due to friction. Considering an internal cylindrical part of the pipe, length=l, lateral surface=π*d*l, frictional force is F=π*d*l*R against flow direction.
Horizontal forces on mentioned liquid cylinder are in equilibrium, so P1*π*d^2/4=F+P2*π*d^2/4, or (P1-P2)*π*d^2/4 = π*d*l*R. Hence ΔP= P1-P2= 4*R*l/d, or ΔP/l=4*R/d.
R represents the frictional force opposing to flow. It is according to our experience to assume that this resistance to flow increases with velocity, e.g. compare to drug resistance of objects in a moving fluid. We had learned in elementary physics that friction is independent of velocity, being an oversimplification (compare to oscillations, where friction is assumed proportional to velocity), especially for fluids in motion. So resistance increases with velocity, causing higher frictional pressure drop as a consequence. Experiment has verified it, R being a function of Reynolds Number, velocity and pipe roughness.
This is a simplified view, more on text books.

Another way to indicate it can be as follows.
Mentioned liquid cylinder (of dia=d & length=l) moves along the pipe by a travel L. Its average pressure (e.g. at its center) is p1 at the start and p2 at the end. Frictional force opposing its motion is F, as before.
Ambient temperature is same as liquid temperature, which is assumed not changing (this is according to our senses for incompressible fluids, but is is a weak point here). Liquid mass of the cylinder is m=(π*d^2/4)*l*ρ
An energy balance on the liquid cylinder would give:
kinetic energy at start & end: 0.5*m*v^2 & 0.5*m*v^2 (same).
pressure energy at start & end : p1*(π*d^2/4)*l & p2*(π*d^2/4)*l
Work of frictional force between start & end: F*L=π*d*l*R*L.
Energy change equals frictional work: (p1-p2)*(π*d^2/4)*l = π*d*l*R*L, or ΔP/L = 4*R/d, similar as before. The only difference is than now we speak of energy (instead of force); p1*(π*d^2/4)*l must be higher than p2*(π*d^2/4)*l to overbalance frictional work, so p1 must be higher than p2. So higher upstream pressure is needed to counterweight friction.The fact that frictional ΔP increases with velocity is indicated as before (not in a more clear way).

Edited by kkala, 17 July 2011 - 11:32 AM.

[sheiko]

I have known for a long time that pressure drop in a pipe is directly proportional to the velocity of fluid flowing through it, i.e. at greater velocities we have greater pressure drop.

Laminar (Re<2100): ΔP is proportional to v
Turbulent (Re>4000): ΔP is proportional to v²

[kkala]

Indeed, dependence of frictional ΔP from velocity should have been pointed out from the beginning of this thread. We should not neglect the two basic relations, for laminar or tubular flow, that clarify the query based on experimental data; theories have been developed to support them. And ΔP is proportional to v only in case of laminar flow.
Moreover the meaning of the query seems to go beyond the mathematical expressions, in my understanding. It must be something like this: how can we "feel" that frictional ΔP increases with v, neglecting the empirical equations? Is there a "mechanism" to explain it, based on simple data apparent to our experience? Answer may not be so easy. Some of us are convinced by mathematical procedures starting from valid experimental data. Others are doubtful, searching for evidence & interpretations that they can feel as real, especially in case that the path from experimental data to mathematical expressions is long (mathematics merely being a way of expression).
One can say that the engineer has to take empirical equations for granted, and often has to, due to schedule restraints. Even so, a better understanding based on simple principles caught by our senses would help to apply these equations better. I have seen that (at least locally) physical experiments in schools and universities are quite limited, even though we are reluctant to apply a formula without having got its physical feeling.
In the last year of high school, our late teacher of astronomy told us. "Do you think that you know more than ancient astronomers? Prove that the earth rotates about the sun". The effort improved our judgment & knowledge, but it was not easy; we found little evidence based on real observations and much based on accepted theories and data (not 100% understood by us, or even not taught).
Probably a practical way to see that frictional ΔP increases with velocity is to put a rubber hose to a tap, with a manometer located at the beginning. Then vary flow rate through the tap and read manometer, actually showing ΔP. Our empirical feeling may tell us, even before the experiment, that manometer reading increases with flow rate; for friction between moving water and internal wall of hose increases with flow rate.

Edited by kkala, 18 July 2011 - 12:42 AM.