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# Empirical Equations For Pressure Drop In Crude Oil And White Oil Pipelines Most of the experienced chemical engineers know about the Darcy-Weisbach equation for frictional pressure drop using the Darcy friction factor values from Moody's chart. The Moody chart provides the Darcy friction factor values for a given Reynold's number and a given relative roughness for conduits with a circular cross-section (circular pipes).

For newcomers to pressure drop calculations the references provided for Darcy-Weisbach and Moody's chart below could be great starting points:

http://en.wikipedia....isbach_equation

http://en.wikipedia....iki/Moody_chart

Today's blog entry is related to some empirical equations for pressure drop calculations in Crude Oil & Petroleum products (Gasoline, Naphtha, Kerosene) pipelines. Let us get to the equations straight away:

Shell-MIT Equation - Pressure Drop in Heavy Crude Oil & Heated Liquid Pipelines (USC Units)

R = 92.24*Q / (v*Di)

where:

R = Reynolds no, dimensionless

Q = Flow Rate, bbl / day

v = Kinematic Viscosity, cSt

Di = Pipe internal diameter, inches

Rm = R / 7742

where:

Rm = modified Reynolds number, dimensionless

Pressure Drop

Pm = 0.241*f*Sg*Q2 / Di5

where:

Pm = Frictional Pressure Drop, psi / mile

Sg = Liquid Specific Gravity, dimensionless (water = 1.0)

f = friction factor, dimensionless (calculated as below)

Laminar Flow:

f = 0.00207 / Rm

Turbulent Flow:

f = 0.0018 + 0.00662*(1 / Rm)0.355

Shell-MIT Equation - Pressure Drop in Heavy Crude Oil & Heated Liquid Pipelines (Metric Units)

R = 353,678*Q / (v*Di)

where:

R = Reynolds no, dimensionless

Q = Flow Rate, m3 / h

v = Kinematic Viscosity, cSt

Di = Pipe internal diameter, mm

Rm = R / 7742

where:

Rm = modified Reynolds number, dimensionless

Pressure Drop

Pm = 6.2191E10*f*Sg*Q2 / Di5

where:

Pm = Frictional Pressure Drop, kPa / km

Sg = Liquid Specific Gravity, dimensionless (water = 1.0)

f = friction factor, dimensionless (calculated as below)

Laminar Flow:

f = 0.00207 / Rm

Turbulent Flow:

f = 0.0018 + 0.00662*(1 / Rm)0.355

T R Aude Equation - Pressure Drop in Refined Petroleum Products (Gasoline, Naphtha, Kerosene) Pipelines (6" and 8" line size only) (USC units)

Pm = (Q*µ0.104*Sg0.448 / (0.871*K*Di2.656))1.812

where:

Pm = Pressure drop due to friction, psi / mile

Q = flow rate, bbl / hr

Di = pipe internal diameter, inches

Sg = Liquid specific gravity, dimensionless

µ = Liquid Viscosity, cP

K = pipe roughness / efficiency factor (usually 0.9 to 0.95)

T R Aude Equation - Pressure Drop in Refined Petroleum Products (Gasoline, Naphtha, Kerosene) Pipelines (6" & 8" line size only) (Metric units)

Pm = 8.888E8*(Q*µ0.104*Sg0.448 / (0.871*K*Di2.656))1.812

where:

Pm = Pressure drop due to friction, kPa / km

Q = flow rate, m3 / h

Di = pipe internal diameter, mm

Sg = Liquid specific gravity, dimensionless

µ = Liquid Viscosity, cP

K = pipe roughness / efficiency factor (usually 0.9 to 0.95)

Reference: Liquid Pipeline Hydraulics by E. Shashi Menon

It is important to note that these equations cannot be generalized for any liquid flow and are specifically meant for the liquids mentioned in the titles of these equations. Also note that the T. R. Aude equation is targeted for 6" & 8" pipelines for white oil products. Apparently these equations were developed for specific products, line sizes and for long distance pipelines and would provide more accurate pressure drop values than the Darcy-Weisbach equation for pipe friction loss. The interpretation behind these equations being more accurate would be based on their being developed based on actual field tests for the mentioned liquids and line sizes. An important thing to note for liquid transmission in long-distance pipelines is that, the liquid density and viscosity is not constant along the entire length of the pipeline. Another aspect of long-distance pipeline transmission of liquids is that, liquids generally considered to be incompressible, are not so over the long transmission distances encountered in pipelines. In fact for pipeline transmission of liquids, liquids are considered to be partially compressible which would explain the development of these empirical equations.

If high accuracy is not a pre-requisite for pressure drop calculations for the mentioned liquids and line sizes then my advice to all would be to use the Darcy-Weisbach equation which should provide conservative pressure drop values.

Hope to get quite a few comments on this blog entry from the readers of my blog.

Regards,
Ankur. sheiko
Thanks Ankur!
This is useful! JAVilleg
Thank so much Mr. Ankur, your posts are really useful, specially for Excel sheets. Greetings from Mexico. Thanks Ankur darwish
There is one thing I did notice Ankur,

The shell-MIT equation does not need the length of the pipeline in question, where conventional approaches (Darcy-Weir) require a length parameter. I made a quick spreadsheet comparing the two equations and it seems that the pressure drop calculation results vary significantly when compared to the Darcy-Weirbash approach.

The pressure drops are not in agreement unless the length parameter in the Darcy equation is above 9000 m or 9 km.

I was just wondering if this is to be expected?

I appreciate the input ankur2061
Darwish,

Th Shell-MIT equation is for a unit length of either 1 mile (USC units) or 1 km (Metric Units), hence there is no length parameter in the equation.

While the Darcy-Wiesbach is a generalized pressure drop equation with friction factor derived from the Moody's chart for circular conduits the Shell-MIT equation is specifically developed for heavy crude oil (high viscosity) long-distance pipelines.

As I mentioned if you are looking for accurate pressure drop over long-distance heavy crude oil pipelines than Shell-MIT should give the more accurate results.

Regards,
Ankur. greengeek
Thank very much Mr. Ankur.

Can you post some correlations for Bitumen and Fuel oil too? boyleT

Can anyone help...The MIT equations seems to have three options for calculating Pm. The friction factor is calculated from f = 0.0018 + 0.00662 1/Rm^0.355 for turbulent flow. If you want to calculate Pm in SI units and so, Pm = 6.2191 + 1^10(f Sg Q^2)/D^5, is f in this equation calculated from the above equation and is it fanning of darcy ?

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