Does the pressure drop in a liquid line depend on the operating pressure of the line? 5

[jesahe]

I'm studying the pressure drop of oil flow in a single phase liquid line by utilizing the Fanning equation [ ΔP = (0.00115f*Q[sup]2[/sup]*S)/d[sup]5[/sup] ]. Where f = Moody friction factor, Q = liquid flow rate (barrels/day), S = liquid specific gravity, and d = pipe inside diameter (inches).

I can see very clearly from this equation that the answer to my question is 'NO'. However, I'm having a hard time justifying this to myself in physical terms. Why would a line operating at 100psig behave the same as another operating at 2000psig given all other variables are the same.

Thank you for taking the time to answer this question!

 

[MortenA]

There is no change in velocity for an incompressible fluid when moveing from one operating pressure to another. That is the reason why.

 

[jesahe]

Thank you! So, would it be correct to say that the pressure drop of single phase, incompressible liquid flow is only a result of the energy loss due to friction? Furthermore, the energy loss due to friction will always be the same regardless of operating pressure.

If I'm grasping this I still can't reconcile the fact that friction losses will always be the same at different pressures. Why wouldn't it act differently?

 

[LittleInch]

Whilst no fluid is actually incomprssible, the difference in density for a single phase liquid at the same temperature between your 2000 psig and 100 psig is negligible and therefore can be ignored in the flow calculations. If you search for it you can find liquid compressibility - e.g. for water 1% compression requires about 3,000 psig.

Any gas or two phase flid however is completetly different....



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Also: There's usually a good reason why everyone does it that way

 

[cvg]

pressure drop is due to friction caused by movement of the fluid. higher velocity equals more friction. pressure is irrelevant. turbulence, expansion and contraction also cause pressure drop, but again it is due to change in inertia which is related to velocity, not pressure.

 

[Oldhydroman]

For many fluids the viscosity increases with absolute pressure. This will alter the Reynolds number and change your friction factor.

DOL

 

[BigInch]

I can appreciate that thought and it is true only to some very small extent in my experience, so I will say that viscosity is about a thousand times more sensitive to temperature than pressure and the temp effect will most always be observed first and about a thousand times more pronounced. I think it is probably way more than difficult, if not a total waste of time, to concentrate on any pressure effect on viscosity in almost any typical case. Have I calculated such effect before? Yes. Has the pressure effect on viscosity ever made a difference to any of my pipeline designs? No. Not even with super-heavy crude. Two phase, or gas, or entrained gas, yes. That's a completely different problem.


Independent events are seldomly independent.

 

[Oldhydroman]

Mineral oils certainly do exhibit a significant change of viscosity with pressure.

The Barus equation gives this relationship as:

V[sub]P[/sub] = V[sub]atm[/sub] * e^(alpha * P)
Where “V[sub]p[/sub]” is the absolute viscosity at the new pressure, “V[sub]atm[/sub]” is the absolute viscosity at atmospheric pressure, “P” is the new pressure (in N/mm[sup]2[/sup]) and “alpha” is the pressure-viscosity coefficient (in mm[sup]2[/sup]/N ... it has to have the reciprocal units of the pressure) . A typical value for alpha for a mineral oil at 50 deg C would be 0.02 mm[sup]2[/sup]/N.

Disregarding the particular viscosity of the oil, the term “e^(alpha *P)” gives a measure of the relative viscosity increase. At a pressure of 100 bar (= 10 MPa = 10[sup]7[/sup] N/m[sup]2[/sup] = 10 N/mm[sup]2[/sup]) the relative increase is e^0.2 , i.e. 1.22. It’s not much but I would be of the opinion that a 22% increase in viscosity it is more than insignificant.

Although the pressure related increase in viscosity is usually considered in terms of its effect on fluid film thickness in heavily loaded bearings, the effect is also present in the bulk volume of the oil. For example, on subsea hydraulic equipment, such as ROV’s, the various speed control valves are set up on the deck and everything works fine but when at depth the equipment is found to behave a little sluggishly even though there's no significant difference in temperature.

 

[BigInch]

I can also appreciate that you would not want an ROV to behave slugishly, whereas for a pipeline, slugish behavior would hardly be noticable.

Independent events are seldomly independent.

 

[katmar]

I'm probably answering a question that the OP did not ask, but the system pressure does have an effect with rubber hoses used for fire fighting. The ratings for these hoses usually include the system pressure. In turbulent flow the pressure drop varies with approximately the 5th power of the diameter so a relatively small stretching of the hose can have a measurable impact on the flow.

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[Oldhydroman]

I only mentioned the ROV application as an example where the physical effect of a general increase in absolute pressure causes a general increase of bulk fluid viscosity.

Jesahe's original question was all about the Fanning equation indicating that the absolute pressure was not a consideration in the calculation of pressure drop and the difficulty in justifying this in the light of everyday experience.

If the fluid behaves as a mineral oil with a pressure-viscosity coefficient of 0.02 mm[sup]2[/sup]/N across the whole of the pressure range and at the particular operating temperature, then the actual viscosity of the fluid at 2000 psi will be about 30% higher than the viscosity of the fluid at 100 psi. This will reduce the Reynolds number by 30% which, for laminar flow, will increase the friction factor by 30%. So, all other things being equal (which they rarely are), the 2000 psi flow line will have a 30% greater pressure drop per unit length. With fully turbulent flow the effect is very much reduced.

Note, however, that the increased pressure drop is an increase in energy losses, i.e. a heating effect on the fluid. As BigInch noted, the temperature change has a very significant effect on viscosity. It is possible that, in practice, the viscosity increase consequent on any increase in absolute pressure causes an increase in pressure drop per unit length which will cause an increase in fluid temperature which then decreases the viscosity which reduces the pressure drop. A new steady state condition will arise where you have a slightly warmer fluid and a slightly higher pressure drop than before and I can quite appreciate BigInch's conclusion that, at the end of the day, it wasn't worth calculating.