Restraint forces from head loss 1

[WARose]

Pardon me for such a fundamental question.....but I don't do much in terms of pipe design (just the supports since I am structural)....I got a set of calculations from this guy designing some pipe. What I don't see is any forces at the supports due to the head loss from friction.

As I recall (and from looking in my old fluids book) these forces aren't significant until the velocity of flow gets high. But I cannot recall if this produces a net force to be resisted by a support......so my bottom line question is: should I be asking/concerned about this? Thanks.

 

[MJCronin]

But I cannot recall if this produces a net force to be resisted by a support......so my bottom line question is: should I be asking/concerned about this? Thanks.

Do not be concerned

A piping system undergoing just fluid pressure drop does not exert any significant loadings on the pipe supports or the terminations....


MJCronin
Sr. Process Engineer

 

[WARose]

Thanks MJ. A quick follow up question (just for future reference): if one were to quantify this force.....it would be equivalent to the head loss.....correct?

 

[LittleInch]

What ever it is, and I'm having some difficulty thinking what that would be, it will be dwarfed by the axial force/ movement from even 1C thermal expansion.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[bimr]

Thanks MJ. A quick follow up question (just for future reference): if one were to quantify this force.....it would be equivalent to the head loss.....correct?

No, the forces are probably associated with velocity head (V²/2g).

For further information, refer to:

Link

Having said that it is velocity head, the forces depend on the type of piping. Welded piping doesn't exert any loading on the piping supports as noted by MJCronin. However, if you using other piping designs such as gravity flow, there may be some support loading in addition to weight of the pipe and pumped contents.

 

[1503-44]

Pressure loss axial stresses (change in hoop stress x Poison ratio between adjacent points of unequal pressure) are usually very small and are well distributed along the pipe length. You might keep an eye out for possible concentrations of pressure loss. A closed end cap, a closed valve, or a control and relief valves for example, can have high pressure on one side and no to low pressure on the other side, so a force of (P_hi-P_lo)*A results. Those are usually not all that much to worry about in welded or fusion jointed pipe, but pipe with push-in joints will need some special attention.

Change of momentum forces can occur at bends, tees and reducers as bimr noted and when waterhammer forces from rapid velocity changes occur.

The most problematic are Relief valves, especially those from high pressure gas to atmosphere, which can have large dynamic forces. There have been some instances of uncontrolled launches of poorly supported relief valves.

 

[WARose]

No, the forces are probably associated with velocity head (V2/2g).

That's part of the head loss equation.

What ever it is, and I'm having some difficulty thinking what that would be, it will be dwarfed by the axial force/ movement from even 1C thermal expansion.

Yeah I know that. The focus here is on something else.

 

[bimr]

That's part of the head loss equation.

Velocity head is part of the total hydraulic energy, but not part of the head loss.

 

[WARose]

Velocity head is part of the total hydraulic energy, but not part of the head loss.

According to my old fluids book it is:

Head loss= h_L=f(L/D)(V²/2g)

Where: f=Darcy friction factor

-From 'Fundamentals of Fluid Mechanics' (2nd ed.), by: Gerhart, et al. (1992), p.473

 

[1503-44]

Let's say there are two issues here, fluid mechanics and fluid dynamics.
Fluid Mechanics
The kinetic energy of a moving fluid, "velocity head" = V^2/2/g.
It is in terms of energy per unit mass of fluid, hence it doesn't look exactly like the KE = 1/2 M V^2 definition you may be used to seeing, because if you define it as a kinetic energy per unit mass basis, mass is eliminated from the standard KE equation.

fL/D is simply the portion of that kinetic energy which is being consumed by friction of the moving fluid within the pipe.
f(L/D) x (V^2/2g) is calculating just the portion of kinetic energy lost to friction as it moves a length L down the pipe.

Fluid Dynamics
Force = Mass x Acceleration
Acceleration = change in velocity over time = (V1-V0)/T
F = Mass * (V1-V0)/T
F = (Mass * V1 - Mass * V0)/T
Momentum, Mv = Mass x Velocity
So force at elbows, etc comes from change of momentum
F = (Mv1-Mv0)/T
FT = (Mv1-Mv0)
Impulse = Force x T, so the force being applied all the time is
F = Mv1-Mv0

That can happen due to a change of mass, adding flow, or splitting a stream at a tee, a change of velocity, at a closing valve (waterhammer), or at a reducer, or by a change in the velocity vector, at an elbow.

 

[bimr]

Addressing the title of your original post, there should be no restraint forces associated with the fluid headloss.

The momentum equation is used to find the required strength of tie-downs, anchors, and thrust blocks to restrain piping at elbows, tees, etc. If you have a small diameter piping with moderate fluid velocities, one doesn't have to be concerned. If you have large diameter piping, then you have to evaluate the stresses.