Does the pressure drop in a liquid line depend on the operating pressure of the line?
Does the pressure drop in a liquid line depend on the operating pressure of the line?
(OP)
I'm studying the pressure drop of oil flow in a single phase liquid line by utilizing the Fanning equation [ ΔP = (0.00115f*Q2*S)/d5 ]. 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!
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!





RE: Does the pressure drop in a liquid line depend on the operating pressure of the line?
RE: Does the pressure drop in a liquid line depend on the operating pressure of the line?
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?
RE: Does the pressure drop in a liquid line depend on the operating pressure of the line?
Any gas or two phase flid however is completetly different....
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
RE: Does the pressure drop in a liquid line depend on the operating pressure of the line?
RE: Does the pressure drop in a liquid line depend on the operating pressure of the line?
DOL
RE: Does the pressure drop in a liquid line depend on the operating pressure of the line?
Independent events are seldomly independent.
RE: Does the pressure drop in a liquid line depend on the operating pressure of the line?
The Barus equation gives this relationship as:
VP = Vatm * e^(alpha * P)
Where “Vp” is the absolute viscosity at the new pressure, “Vatm” is the absolute viscosity at atmospheric pressure, “P” is the new pressure (in N/mm2) and “alpha” is the pressure-viscosity coefficient (in mm2/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 mm2/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 = 107 N/m2 = 10 N/mm2) 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.
RE: Does the pressure drop in a liquid line depend on the operating pressure of the line?
Independent events are seldomly independent.
RE: Does the pressure drop in a liquid line depend on the operating pressure of the line?
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RE: Does the pressure drop in a liquid line depend on the operating pressure of the line?
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 mm2/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.