COURSE NUMBER: ME 423COURSE NUMBER: ME 423

Fluids Engineering

Gas Pipeline Hydraulics: Pressure Drop

Course teacher

Dr. M. Mahbubur Razzaque

Professor

Department of Mechanical Engineering

BUET

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FLOW EQUATIONS

Several equations are available that relate the gas flow rate with gasproperties, pipe diameter and length, and upstream and downstreampressures. These equations are listed as follows:

1. General Flow equation2. Colebrook-White equation3. Modified Colebrook-White equation4 AGA ti4. AGA equation5. Weymouth equation6. Panhandle A equation7 Panhandle B equation7. Panhandle B equation8. IGT equation9. Spitzglass equation10. Mueller equation10. Mueller equation11. Fritzsche equation

A comparison of these equations will also be discussed using an examplep q g ppipeline.

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The General Flow equation, also called the Fundamental Flow equation, forthe steady-state isothermal flow in a gas pipeline is the basic equation forrelating the pressure drop with flow rate. The most common form of thisequation in the U.S. Customary System (USCS) of units is given in terms ofthe pipe diameter, gas properties, pressures, temperatures, and flow rate asf ll It t b t d th t f th i t f ti 1 t tifollows. It must be noted that for the pipe segment from section 1 to section2, the gas temperature Tf is assumed to be constant (isothermal flow).

In SI units, the General Flow equation is stated as follows:

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Sometimes the General Flow equation is represented in terms of thetransmission factor F instead of the friction factor f This form of thetransmission factor F instead of the friction factor f. This form of theequation is as follows.

where the transmission factor F and friction factor f are related by

and in SI units

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When there is elevation difference between the ends of a pipe segment, theGeneral Flow equation needs further modification.

AVERAGE PIPE SEGMENT PRESSUREIn the General Flow equation, the compressibility factor Z is used. This

t b l l t d t th fl i t t d imust be calculated at the gas flowing temperature and average pressure inthe pipe segment. Therefore, it is important to first calculate the averagepressure in a pipe segment.

Consider a pipe segment with upstream pressure P1 and downstreampressure P2. An average pressure for this segment must be used to calculatethe compressibility factor of gas at the average gas temperature Tf. As a firstthe compressibility factor of gas at the average gas temperature Tf. As a firstapproximation, we may use an arithmetic average of (P1 + P2)/2. However,it has been found that a more accurate value of the average gas pressure in apipe segment isp p g

Another form of the average pressure in a pipe segment is

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VELOCITY OF GAS IN A PIPELINEUnlike a liquid pipeline, due to compressibility, the gas velocity dependsupon the pressure and hence will vary along the pipeline even if the pipeupon the pressure and, hence, will vary along the pipeline even if the pipediameter is constant. The highest velocity will be at the downstream end,where the pressure is the least. Correspondingly, the least velocity will be atthe upstream end where the pressure is higherthe upstream end, where the pressure is higher.

Consider a pipe transporting gas from point A to point B. Under steady stateflow, at A, the mass flow rate of gas is designated as M and will be the same, , g gas the mass flow rate at point B, if between A and B there is no injection ordelivery of gas. We can write the following relationship for point A:

The volume rate Q can be expressed in terms of the flow velocity u andpipe cross sectional area A as follows:

Th f bi i E ti 2 16 d E ti 2 17 d l i thTherefore, combining Equation 2.16 and Equation 2.17 and applying theconservation of mass to points A and B, we get

If the pipe is of uniform cross section between A and B the velocities atIf the pipe is of uniform cross section between A and B, the velocities atA and B are related by the following equation:

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Since the flow of gas in a pipe can result in variation of temperature frompoint A to point B, the gas density will also vary with temperature andpressure. If the density and velocity at one point are known, thecorresponding velocity at the other point can be calculated using Equation2.19.

If inlet conditions are represented by point A and the volume flow rate Q atstandard conditions (60°F and 14.7 psia) are known, we can calculate thevelocity at any point along the pipeline at which the pressure andvelocity at any point along the pipeline at which the pressure andtemperature of the gas are P and T, respectively.

The mass flow rate M at section 1 and 2 is the same for steady-state flowThe mass flow rate M at section 1 and 2 is the same for steady state flow.Therefore,

where Qb is the gas flow rate at standard conditions and ρb is theQb g ρbcorresponding gas density. Therefore, simplifying Equation 2.20,

Applying the gas law, we getor

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where P1 and T1 are the pressure and temperature at pipe section 1.Similarly, at standard conditions,

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EROSIONAL VELOCITY

As flow rate increases, so does the gas velocity. How high can the gas velocity be ina pipeline? As the velocity increases, vibration and noise are evident. In addition,higher velocities will cause erosion of the pipe interior over a long period of time.The upper limit of the gas velocity is usually calculated approximately from theThe upper limit of the gas velocity is usually calculated approximately from thefollowing equation:

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In the preceding Examples 1 and 2, we have assumed the value ofibilit f t Z t th t t A t l ti ill b tcompressibility factor Z to the constant. A more accurate solution will be to

calculate the value of Z using CNGA or Standing-Katz method.

The inlet and outlet gas velocities then will be modified

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The inlet and outlet gas velocities then will be modified.

REYNOLDS NUMBER OF FLOWThe Reynolds number is a function of the gas flow rate, pipe insidedi t d th d it d i it d i l l t d f thdiameter, and the gas density and viscosity and is calculated from thefollowing equation:

In gas pipeline hydraulics, using customary units, a more suitable equationfor the Reynolds number is as follows:

In SI units, the Reynolds number is

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COMPARISON OF FLOW EQUATIONS

I Fi 2 5 id i li 100 il l NPS 16 ith 0 250 iIn Figure 2.5, we consider a pipeline 100 mile long, NPS 16 with 0.250 in.wall thickness, operating at a flow rate of 100 MMSCFD. The gas flowingtemperature is 80°F. With the upstream pressure fixed at 1400 psig, thedownstream pressure was calculated using the different flow equations Bydownstream pressure was calculated using the different flow equations. Byexamining Figure 2.5, it is clear that the highest pressure drop is predictedby the Weymouth equation and the lowest pressure drop is predicted by thePanhandle B equation. It must be noted that we used a pipe roughness of 700Panhandle B equation. It must be noted that we used a pipe roughness of 700µin. for both the AGA and Colebrook equations, whereas a pipelineefficiency of 0.95 was used in the Panhandle and Weymouth equations.

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Figure 2.6 shows a comparison of the flow equations from a differentperspective. In this case, we calculated the upstream pressure required for anNPS 30 i li 100 il l h ldi th d li t t tNPS 30 pipeline, 100 miles long, holding the delivery pressure constant at800 psig. The upstream pressure required for various flow rates, rangingfrom 200 to 600 MMSCFD, was calculated using the five flow equations.Again it can be seen that the Weymouth equation predicts the highestAgain it can be seen that the Weymouth equation predicts the highestupstream pressure at any flow rate, whereas the Panhandle A equationcalculates the least pressure. We therefore conclude that the mostconservative flow equation that predicts the highest pressure drop is theconservative flow equation that predicts the highest pressure drop is theWeymouth equation and the least conservative flow equation is PanhandleA.

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The wall thickness required for this pipe diameter and pressure will be dictated by the pipe material

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EQUIVALENT LENGTH METHOD: Pipes in Series

Th l l h h d b l d h h f flThe equivalent length method can be applied when the same uniform flowexists throughout the pipeline consisting of pipe segments of differentdiameter, with no intermediate deliveries or injections.

Consider the same flow rate Q through all pipe segments. The first pipesegment has an inside diameter D1 and length L1, followed by the secondsegment of inside diameter D2 and length L2 and so on We calculate thesegment of inside diameter D2 and length L2 and so on. We calculate theequivalent length of the second pipe segment based on the diameter D1such that the pressure drop in the equivalent length matches that in theoriginal pipe segment of diameter D2. The pressure drop in diameter D2g p p g 2 p p 2and length L2 equals the pressure drop in diameter D1 and equivalent lengthLe2. Thus, the second segment can be replaced with a piece of pipe oflength Le2 and diameter D1. Similarly, the third pipe segment with diameterD3 and length L3 will be replaced with a piece of pipe of Le3 and diameterD1. Thus, we have converted the three segments of pipe in terms ofdiameter D1 as follows:

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We now have the series piping system reduced to one constant-diameter(D1) pipe of total equivalent length given by

For the same flow rate and gas properties, neglecting elevation effects, thepressure difference (P12 – P22) is inversely proportional to the fifth power ofthe pipe diameter and directly proportional to the pipe length Thereforethe pipe diameter and directly proportional to the pipe length. Therefore,we can state that, approximately,

From Equation 3.2 we conclude that the equivalent length for the samepressure drop is proportional to the fifth power of the diameter. Therefore,in the series piping discussed in the foregoing, the equivalent length of thesecond pipe segment of diameter D2 and length L2 is

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or

Similarly, for the third pipe segment of diameter D3 and length L3, theequivalent length is

Therefore, the total equivalent length Le for all three pipe segments interms of diameter D isterms of diameter D1 is

It can be seen from Equation 3.6 that if D1 = D2 = D3, the total equivalentlength reduces to (L1 + L2 + L3), as expected.

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PARALLEL PIPING

Sometimes two or more pipes are connected such that the gas flow splitsamong the branch pipes and eventually combines downstream into a singlepipe, as illustrated in Figure 3.7. The reason for installing parallel pipes orl i d d i i i f h i li dloops is to reduce pressure drop in a certain section of the pipeline due topipe pressure limitation or for increasing the flow rate in a bottlenecksection.

By installing a pipe loop from B to E, in Figure 3.7 we are effectivelyreducing the overall pressure drop in the pipeline from A to F, sincebetween B and E the flow is split through two pipes.p g p p

Applying the principle of flow conservation, at junction B, the incomingflow into B must exactly equal the total outflow at B through the parallel

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pipes. Therefore, at junction B,

Where, Q = inlet flow at A, Q1 = flow through pipe branch BCE and Q2 =flow through pipe branch BDE.

Both pipe branches have a common starting point (B) and common endingpoint (E). Therefore, the pressure drop in the branch pipe BCE and branchi BDE h l (P P ) h P d P hpipe BDE are each equal to (PB – PE), where PB and PE are the pressures atjunctions B and E, respectively. Therefore, we can write

The pressure drop due to friction in branch BCE can be calculated from

Where, K1 = a parameter that depends on gas properties, gas temperature,etc., L1 = length of pipe branch BCE, D1 = inside diameter of pipe branchBCE and Q1 = flow rate through pipe branch BCE.

Similarly, the pressure drop due to friction in branch BDE is calculated from

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In Eq. 3.10 and Eq. 3.11, the constants K1 and K2 are equal, since they donot depend on the diameter or length of the branch pipes BCE and BDE.

Combining both equations, we can state the following for common pressuredrop through each branch:

Simplifying further, we get the following relationship between the two flowQ d Qrates Q1 and Q2:

In equivalent diameter method, we replace the pipe loops BCE and BDEwith a certain length of an equivalent diameter pipe that has the samepressure drop as one of the branch pipes Since the pressure drop in thepressure drop as one of the branch pipes. Since the pressure drop in theequivalent diameter pipe, which flows the full volume Q, is the same as thatin any of the branch pipes, from Eq. 3.10, we can state the following:

where Q = Q1 + Q2 from Equation 3.7 and Ke represents the constant for

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the equivalent diameter pipe of length Le flowing the full volume Q.We get,using Eq. 3.10, Eq. 3.11, and Eq. 3.14:

Setting K1 = K2 = Ke and Le = L1, we simplify Equation 3.15 as follows:

Using Equation 3.16 in conjunction with Equation 3.7, we solve for theequivalent diameter De as

Where

and the individual flow rates Q1 and Q2 are calculated fromand the individual flow rates Q1 and Q2 are calculated from

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and

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Solving for the outlet pressure at E, we get P2 = 1145.60 psia, which is almostthe same as what we calculated before.

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Therefore, using the equivalent diameter method, the parallel pipes BCE andBDE can be replaced with a single pipe 24 mi long, having an inside diameterof 17.67 in.

LOCATING PIPE LOOP

How do we determine where a loop should be placed for optimum results?How do we determine where a loop should be placed for optimum results?Should it be located upstream, downstream, or in a midsection of the pipe?Three looping scenarios are presented in Figure 3.8.

In case (a), a pipeline of length L is shown looped with X miles of pipe,beginning at the upstream end A In case (b) the same length X of pipe is

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beginning at the upstream end A. In case (b), the same length X of pipe islooped, but it is located on the downstream end B. Case (c) shows themidsection of the pipeline being looped.

For most practical purposes, we can say that the cost of all three loops willbe the same as long as the loop length is the same.

Which of these cases is optimum,? It is found that if the gas temperature isconstant throughout, at locations near the upstream end, the pressuredrops at a slower rate than at the downstream end Therefore there isdrops at a slower rate than at the downstream end. Therefore, there ismore pressure drop in the downstream section compared to that in theupstream section. Hence, to reduce the overall pressure drop, the loopmust be installed toward the downstream end of the pipe. This argument ismust be installed toward the downstream end of the pipe. This argument isvalid only if the gas temperature is constant throughout the pipeline.

In reality, due to heat transfer between the flowing gas and the surroundingsoil (buried pipe) or the outside air (above-ground pipe), the gastemperature will change along the length of the pipeline. If the gastemperature at the pipe inlet is higher than that of the surrounding soil(buried pipe), the gas will lose heat to the soil and the temperature willdrop from the pipe inlet to the pipe outlet.

If the as is c m ressed at the inlet sin a c m ress r then the as

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If the gas is compressed at the inlet using a compressor, then the gastemperature will be much higher than that of the soil immediatelydownstream of the compressor.The hotter gas will cause higher pressure

drops. Hence, in this case the upstream segment will have a larger pressuredrop compared to the downstream segment. Therefore, considering heattransfer effects the pipe loop should be installed in the upstream portiontransfer effects, the pipe loop should be installed in the upstream portionfor maximum benefit.

The installation of the pipe loop in the midsection of the pipeline, as in caseThe installation of the pipe loop in the midsection of the pipeline, as in case(c) in Figure 3.8, will not be the optimum location, based on the precedingdiscussion. It can therefore be concluded that if the gas temperature isfairly constant along the pipeline, the loop should be installed toward they g p p pdownstream end, as in case (b). If heat transfer is taken into account andthe gas temperature varies along the pipeline, with the hotter gas beingupstream, the better location for the pipe loop will be on the upstream end,as in case (a).

HYDRAULIC PRESSURE GRADIENT

Since pressure in a gas pipeline is nonlinear compared to liquid pipelines,the hydraulic gradient for a gas pipeline appears to be a slightly curved lineinstead of a straight line The slope of the hydraulic gradient at any point

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instead of a straight line. The slope of the hydraulic gradient at any pointrepresents the pressure loss due to friction per unit length of pipe. Thisslope is more pronounced as we move toward the downstream end of the

pipeline, since the pressure drop is larger toward the end of the pipeline. Ifthere are intermediate deliveries or injections along the pipeline, thehydraulic gradient will be a series of broken lines as indicated in Figure 3 10hydraulic gradient will be a series of broken lines, as indicated in Figure 3.10.

A similar broken hydraulic gradient can also be seen in the case of apipeline with variable pipe diameters and wall thicknesses, even if the flowrate is constant. Unlike liquid pipelines, the breaks in hydraulic pressuregradient are not as conspicuous in gas pipelines. In a long-distance gaspipeline, due to limitations of pipe pressure, intermediate compressori ill b i ll d b h h i d l

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stations will be installed to boost the gas pressure to the required value sothe gas can be delivered at the contract delivery pressure at the end of thepipeline.

PRESSURE REGULATORSAND RELIEFVALVES

In a long distance gas pipeline with intermediate delivery points there mayIn a long-distance gas pipeline with intermediate delivery points, there maybe a need to regulate the gas pressure at certain delivery points in order tosatisfy the customer requirements. Suppose the pressure at a delivery pointis 800 psig, whereas the customer requirement is only 500 psig. Obviously,is 800 psig, whereas the customer requirement is only 500 psig. Obviously,some means of reducing the gas pressure must be provided so that thecustomer can utilize the gas for his or her requirements at the correctpressure. This is achieved by means of a pressure regulator that will ensurep y p ga constant pressure downstream of the delivery point, regardless of thepressure on the upstream side of the pressure regulator.

This concept is further illustrated using the above example. The mainpipeline from A to C has a branch pipe BE The flowrate from A to B is 100

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pipeline from A to C has a branch pipe BE. The flowrate from A to B is 100MMSCFD, with an inlet pressure of 1200 psig at A. At B, gas is deliveredinto a branch line BE at the rate of 30 MMSCFD.

The remaining volume of 70 MMSCFD is delivered tothe pipeline terminus C at a delivery pressure of 600psig Based on the delivery pressure requirement of 600psig. Based on the delivery pressure requirement of 600psig at C and a takeoff of 30 MMSCFD at point B, thecalculated pressure at B is 900 psig. Starting with 900psig on the branch line at B, at 30 MMSCFD, gas ispsig on the branch line at B, at 30 MMSCFD, gas isdelivered to point E at 600 psig. If the actualrequirement at E is only 400 psig, a pressure regulatorwill be installed at E to reduce the delivery pressure byy p y200 psig.

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LINE PACK

The quantity of gas contained within the pipeline under pressure, measuredq y g p p p ,at standard conditions (generally 14.7 psia and 60°F), is termed the linepack volume.

Consider a segment of pipe, of length L, with upstream pressure andtemperature of P1 and T1 and downstream values of P2 and T2, respectively.Suppose the inside diameter of the pipe is D; then the physical volume ofhthe pipe section is

This volume is the gas volume at pressures and temperatures ranging fromP1 T1 at the upstream end to P2 T2 at the downstream end of the pipeP1, T1 at the upstream end to P2, T2 at the downstream end of the pipelength L. In order to convert this volume to standard conditions ofpressure, Pb, and temperature,Tb, we apply the gas law Equation as follows:

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From Equation 3.31, solving for line pack Vb at standard conditions, we get

Substituting the value ofVp from Equation 3.30 and simplifying, we get

Equation 3.33 is modified in terms of commonly used units as follows:

3Where, Vb is in standard ft3, D in inch and L in mi. The correspondingequation in SI units is

55Where, Vb is in standard m3, D in mm and L in km.

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