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FLUID FLOWSection Page

TWO-PHASE (VAPOR-LIQUID) FLOWXIV-D 1 of 52

DESIGN PRACTICES August, 2004

EXXONMOBILMOBIL RESEARCH AND ENGINEERING COMPANY - FAIRFAX, VA

CONTENTSSection Page

SCOPE...............................................................................................................................................................3

REFERENCES...................................................................................................................................................3DESIGN PRACTICES (SUBSECTION A, B, AND C OF THIS SECTION).................................................3OTHER LITERATURE ...............................................................................................................................3

DEFINITIONS.....................................................................................................................................................3TWO-PHASE FLOW DEFINITIONS..........................................................................................................3

BASIC DESIGN CONSIDERATIONS ................................................................................................................4FLOW REGIMES IN HORIZONTAL OR SLIGHTLY INCLINED PIPEs .....................................................4FLOW REGIMES IN VERTICAL PIPEs .....................................................................................................5EFFECT OF FITTINGS ON TWO-PHASE FLOW .....................................................................................8PRESSURE DROP IN STRAIGHT PIPE ...................................................................................................8OTHER PRESSURE DROP ......................................................................................................................8PERFORATED-PIPE DISTRIBUTORS......................................................................................................8FLASHING CRITICAL FLOW ....................................................................................................................9

CALCULATION PROCEDURE..........................................................................................................................9DETERMINING FLOW REGIME ...............................................................................................................9PRESSURE DROP IN STRAIGHT PIPE .................................................................................................12METHOD A - MODIFIED HOMOGENEOUS METHOD ...........................................................................13METHOD B - SIMILARITY METHOD (LIQUID HOLDUP INCLUDED) ....................................................16FLOW RATE IN STRAIGHT PIPE ...........................................................................................................18PRESSURE DROP IN SINGLE PIPING COMPONENTS........................................................................19PERFORATED-PIPE DISTRIBUTORS....................................................................................................21DISTRIBUTION MANIFOLDS..................................................................................................................21INTEGRATED PRESSURE DROP CALCULATION FOR PIPING SYSTEMS.........................................21FLASHING CRITICAL FLOW ..................................................................................................................23

SAMPLE PROBLEMS .....................................................................................................................................23PROBLEM 1 - PRESSURE DROP ..........................................................................................................24PROBLEM 2 - FLOW PATTERN IN HORIZONTAL PIPE .......................................................................31PROBLEM 3 - FLOW PATTERN IN VERTICAL RISER ..........................................................................32PROBLEM 4 - HOLDUP AND PRESSURE DROP IN VERTICAL RISER ...............................................33PROBLEM 5 - CRITICAL MASS FLUX....................................................................................................38

NOMENCLATURE ...........................................................................................................................................40

Changes shown by �


Section Page FLUID FLOWXIV-D 2 of 52 TWO-PHASE (VAPOR-LIQUID) FLOW

August, 2004 DESIGN PRACTICES

ExxonMobilMobil Research and Engineering Company – Fairfax, VA

FIGURES

Figure 1 Two-Phase Flow Regimes In Horizontal Pipe ....................................................................................42

Figure 2A Two-Phase Flow Regimes In Vertical Pipe (Customary) .................................................................43

Figure 2B Two-Phase Flow Regimes In Vertical Pipe (Metric) .........................................................................46

Figure 3A Elevation Head Factor For Two-Phase Pressure Drop (Customary) ...............................................47

Figure 3B Elevation Head Factor For Two-Phase Pressure Drop (Metric).......................................................48

Figure 4 Liquid Holdup Correlation For Horizontal Pipe ...................................................................................49

Figure 5 Friction Factors In Two-Phase Flow...................................................................................................50

Figure 6 Liquid Holdup For Vertical Upflow Pipe..............................................................................................51

Figure 7 Manifold Designs For Distributing Two-Phase Flow...........................................................................52

Figure 8 Critical Flow Of Vapor - Liquid Mixtures .............................................................................................53

Revision Memo

08/04 Highlights of revisions:

� Equation changes on figure 1






SCOPEThis subsection covers the use of flow maps for determining two-phase (vapor-liquid) flow regimes and calculation methods fordetermining pressure drop and liquid holdup for isothermal two-phase flow in horizontal and vertical pipes. Methods forcalculating pressure drop across orifices, valves, pipe fittings, as well as sudden enlargements and contractions, are alsoincluded. Guidelines for designing manifolds and perforated pipes for distributing vapor-liquid mixtures are also given. Thesemethods are considered to be among the best currently available for general use by simple hand calculations.

REFERENCES

DESIGN PRACTICES (SUBSECTION A, B, AND C OF THIS SECTION)

OTHER LITERATURE1. O. Baker, “Multiphase Flow in Pipelines,” Oil and Gas Journal, 56, 156-67 (November 10, 1958).2. A. E. Dukler, et. al., “Frictional Pressure Drop in Two-Phase Flow,” AIChE Journal, 10, 38-51 (1964).3. AGA/API Monograph on Project NX-28, “Gas-Liquid Flow in Pipelines.”

a. Vol. I, Research Results, A. E. Dukler (May, 1969).b. Vol. II, Design Manual, O. Baker, et. al (October, 1970).

4. O. Flanigan, “Effect of Uphill Flow on Pressure Drop in Design of Two-Phase Gathering Systems,” Oil and Gas Journal, 56,132-141 (March 10, 1958).

5. G. W. Govier and K. Aziz, “The Flow of Complex Mixtures in Pipes,” Krieger, New York (1977).6. J. L. Greene, “Symmetrical Piping Arrangement Solves Two-Phase Flow Distribution Problems,” Hydrocarbon Processing,

46, 141-43 (1967).7. K. C. Hong, “Two-Phase Flow Splitting at a Pipe Tee,” J. Petroleum Technology, 290-296 (February, 1978).8. G. A. Hughmark and B. S. Pressburg, “Holdup and Pressure Drop With Gas-Liquid Flow in a Vertical Pipe,” AIChE Journal,

7, 677-82 (1961).9. J. C. Leung, “A Generalized Correlation for One-Component Homogeneous Equilibrium Flashing Choked Flow,” AIChE

Journal, 32, 1743-46 (1986).10. F. N. Nazario, “Sizing Pressure Relief Valves in Flashing and Two-Phase Service - an Alternative Procedure,” EE.28E.90

(April, 1990).11. L. L. Simpson, “Sizing Piping for Process Plants,” Chemical Engineering, 75, 192-214 (June 17, 1968).12. Y. W. Wong and R. D. Patel, “Improved Vapor-Liquid Manifolds,” EE.23E.88 (February, 1988).13. Y. W. Wong, “Single Component Critical Flashing Flow,” 89CET355 (April 11, 1989).

DEFINITIONSMany general fluid dynamics terms defined in Section XIV-A are applicable to two-phase flow. Additional definitions specific totwo-phase flows are provided below.

TWO-PHASE FLOW DEFINITIONSCRITICAL POINTThe location along the flow path that experiences the steepest pressure gradient in critical flow. For a valve or a nozzle, thecritical point is located at the narrowest point of the device and for a crack or a short pipe, it is usually at the exit.FLOW MALDISTRIBUTIONIn branching flows, such as tee junctions or perforated pipe distributors, the flow maldistribution measures the preferentialdistribution of flow among alternative paths. As a percentage, it is calculated as follows.

%1001branchany inflow mass Smallestbranchany inflow mass Largest x��

�

��

��

In the preceding relation, the branch flows refer to branches or flow paths downstream of the split and not to the main brainch. QUALITYMass fraction of vapor in vapor-liquid flow.





STAGNATIONLocation upstream of the critical point where flow is stagnant or moving very slowly.VELOCITY SLIPIn actual two-phase flow there is slip between vapor and liquid, with vapor flowing at a higher average velocity.

BASIC DESIGN CONSIDERATIONSThe considerations discussed below affect the basis for the calculation procedures given later in this subsection.

FLOW REGIMES IN HORIZONTAL OR SLIGHTLY INCLINED PIPESIn two-phase flow, interactions between liquid and vapor phases, as influenced by their physical properties and flow rates and bythe size, roughness and orientation of the pipe, cause the fluids to flow in various types of patterns. These patterns are calledflow regimes. Only one type of flow exists at a given point in a line at any given time. However, as flow conditions change, theflow regime may change from one type to another.Seven principal flow regimes have been defined to describe flow found in horizontal or slightly inclined pipes. These flowregimes are described below, in order of increasing vapor velocity. In the accompanying sketches, the direction of flow is fromleft to right.

BUBBLE FLOW - liquid occupies the bulk of the cross-section and vapor flows in the formof bubbles along the top of the pipe. Vapor and liquid velocities are approximately equal.If the bubbles become dispersed throughout the liquid, then this is sometimes called frothflow. In uphill flow bubbles retain their identity over a wider range of conditions. Indownhill flow the behavior is displaced in the direction of plug flow.

PLUG FLOW - as the vapor rate increases, the bubbles coalesce, and alternating plugs ofvapor and liquid flow along the top of the pipe with liquid remaining the continuous phasealong the bottom. In an uphill orientation, the behavior is displaced in the direction ofbubble flow; downhill, stratified flow is favored.

STRATIFIED FLOW - as the vapor rate continues to increase, the plugs become acontinuous phase. Vapor flows along the top of the pipe and liquid flows along the bottom.The interface between phases is relatively smooth and the fraction occupied by eachphase remains constant. In uphill flow, stratified flow rarely occurs with wavy flow beingfavored. Downhill, stratified flow is somewhat enhanced, as long as the inclination is nottoo steep.

WAVY FLOW - as the vapor rate increases still further, the vapor moves appreciably fasterthan the liquid, and the resulting friction at the interface forms liquid waves. The waveamplitude increases with increasing vapor rate. Wavy flow can occur uphill, but over anarrower range of conditions than in a horizontal pipe. Downhill, the waves are milder fora given vapor rate and the transition to slug flow, if it occurs at all, takes place at highervapor rates than in horizontal pipes.

SLUG FLOW - when the vapor rate reaches a certain critical value, the crests of the liquidwaves touch the top of the pipe and form frothy slugs. The velocity of these slugs, andthat of the alternating vapor slugs, is greater than the average liquid velocity. In the bodyof a vapor slug the liquid level is depressed so that vapor occupies a large part of the flowarea at that point. Uphill, slug flow is initiated at lower vapor rates than in horizontal pipe.Downhill, it takes higher vapor rates to establish slug flow than in horizontal pipe, and thebehavior is displaced in the direction of annular flow. SINCE SLUG FLOW MAY LEAD TOPULSATION AND VIBRATION IN BENDS, VALVES and other flow restrictions, itshould be avoided where possible.

DP14DFS1

ANNULAR FLOW - the liquid flows as an annular film of varying thickness along the wall,while the vapor flows as a high-speed core down the middle. There is a great deal of slipbetween phases. Part of the liquid is sheared off from the film by the vapor and is carriedalong in the core as entrained droplets. At the same time, turbulent eddies in the vapordeposit droplets on the liquid film. The annular film on the wall is thicker at the bottom ofthe pipe than at the top, the difference decreasing with distance from slug flow conditions.Downstream of bends, most of the liquid will be at the outer wall.

DP14DFS2

In annular flow, the effects of friction pressure drop and momentum outweigh the effect of






gravity, so that pipe orientation and direction of flow have less influence than in theprevious flow regimes. Annular flow is a very stable flow regime. For this reason andbecause vapor-liquid mass transfer is favored, this flow regime is advantageous for somechemical reactions.SPRAY FLOW (ALSO KNOWN AS MIST FLOW OR DISPERSED FLOW) - when thevapor velocity in annular flow becomes high enough, all of the liquid film is torn away fromthe wall and is carried by the vapor as entrained droplets. This flow regime is almostcompletely independent of pipe orientation or direction of flow.

DP14DFS3

FLOW REGIMES IN VERTICAL PIPESFlow behavior in vertical pipes, where gravity plays an important role, has been less extensively investigated than has flow inhorizontal pipes. Most of the available information on vertical flow pertains to upflow.Conditions under which certain flow regimes exist depend largely on the orientation of the pipe and the direction of flow. In asituation where stratified or wavy flow would exist in a horizontal pipe, tilting the pipe downward increases the relative velocity ofthe liquid, making a larger part of the flow area available for the vapor. On the other hand, tilting the pipe upward causes theliquid to drain back downhill until enough has accumulated to block off the entire cross-section. The vapor can then no longerget past the liquid, and therefore pushes a slug of liquid through the inclined section of the line.





Five principal flow regimes have been defined to describe vertical flow. These flow regimes aredescribed below, in order of increasing vapor velocity. In the accompanying sketches, the direction offlow is upward.BUBBLE FLOW - upward flowing liquid is the continuous phase, with dispersed bubblesof vapor rising through it. The velocity of the bubbles exceeds that of the liquid, because of buoyancy.As vapor flow rate is increased, the sizes, number and velocity of the bubbles increase. The bubblesretain their identity, without coalescing into slugs, at higher vapor rates than in a horizontal pipe.

SLUG FLOW - as the vapor rate increases, bubbles coalesce into slugs which occupy the bulk of thecross-sectional area. Alternating slugs of vapor and liquid move up the pipe with some bubbles of vaporentrained in the liquid slugs. Surrounding each vapor slug is a laminar film of liquid which flows towardthe bottom of the slug. As the vapor rate is increased, the lengths and velocity of the vapor slugsincrease.Slug flow can occur in the downward direction, but is usually not initiated in that orientation. However, ifslug flow is well established in an upward leg of a coil, it will persist in a following downward leg, providedthat other conditions remain the same.IN DESIGNING FOR TWO-PHASE FLOW IT IS NORMAL PRACTICE TO TRY TO AVOID SLUGFLOW, since this regime can lead to serious pressure fluctuations and vibration, especially at vesselinlets and in bends, valves and other flow restrictions. This could lead to serious equipment deteriorationor operating problems. When slug flow cannot be avoided (for instance, in thermosyphon reboilers), oneshould avoid flow restrictions and use long-radius bends to make turns as smooth as possible.

DP14DFS4






FROTH FLOW - as the vapor rate increases further, the laminar liquid film is destroyed by vaporturbulence and the vapor slugs become more irregular. Mixing of vapor bubbles with the liquidincreases and a turbulent, disordered pattern is formed with ever shortening liquid slugsseparating successive vapor slugs. The transition to annular flow is the point at which liquidseparation between vapor slugs disappears and the vapor slugs coalesce into a continuous,central core of vapor. Since froth flow has much in common with slug flow, the two regimes areoften lumped together and called slug flow. In the downward direction, froth flow behaves muchthe same as slug flow does, except that the former is more easily initiated in this orientation,particularly if conditions are bordering on those for annular flow.

ANNULAR FLOW - this flow regime is similar to annular flow in horizontal pipe, except that theslip between phases is affected by gravity. In upflow, the annular liquid film is slowed down bygravity, which increases the difference in velocities between vapor and liquid. In downflow, thereverse is true, with gravity speeding up the liquid and reducing the difference in velocitiesbetween vapor and liquid. On the other hand, the liquid film thickness is more uniform around thecircumference of the pipe than in horizontal flow. Annular flow tends to be the dominant regime invertical downflow.

MIST FLOW - this flow regime is essentially the same as spray flow in horizontal pipe. The veryhigh vapor rates required to completely disperse the liquid essentially eliminate the effects oforientation and direction of flow. In identification of vertical two-phase flow regimes, annular andmist flow are often considered together (and called annular-mist).

DP14DFS5





EFFECT OF FITTINGS ON TWO-PHASE FLOWFittings may strongly affect the flow of vapor-liquid mixtures. Bends tend to cause phase segregation. Because of differences inmomenta, liquid tends to move to the outer wall while vapor tends to stay close to the inner wall. The segregation can beminimized by the use of blanked-off tees instead of elbows. The fluid should enter through one straight-through end and exitthrough the branch. Orifices, valves, and other flow restrictions tend to disperse the two phases into each other. For example, ifthe flow upstream of an orifice is in a separated flow regime such as stratified or slug the flow regime immediately downstreamof the orifice will be dispersed flow in most cases. It may take 100 pipe diameters downstream of the orifice for the flow to re-establish the separated flow pattern.

BASIC DESIGN CONSIDERATIONS It has been shown that the behavior of a two-phase mixture in a tee is highly unpredictable, i.e., Liquid or vapor may prefer onebranch over the other resulting in uneven splitting, so that the volume fraction of vapor (or liquid) differs in the two split streams.However, it is generally true that if the flow entering the tee is a dispersed flow the split will be more even than when the enteringstream is in a separated flow regime. The splitting of a two-phase stream in a pipe tee is also dependent on the configuration ofthe flow. In general the splitting is more even if it is done in a more symmetrical fashion. For example, splitting is more even inan impacting tee (a) than in a straight-through tee (b) as shown below.

(a) (b)In order to maintain symmetry, elbows immediately upstream of an impacting tee should be mounted perpendicular to the planeof the tee. If this is not possible a blanked-off tee should be used instead of the elbow.

PRESSURE DROP IN STRAIGHT PIPETwo methods are described in this subsection for calculating two-phase pressure drop in straight pipe. Method a assumes ahomogeneous mixture of vapor and liquid, with no slip between the phases. Method b allows for slip between the two phases,but assumes that the local ratio of liquid velocity to vapor velocity is constant independent of position. Neither method requiresthe prediction of the two-phase flow regime. Method b is more accurate than method a, especially for vertical upflow butrequires a trial-and-error procedure.

OTHER PRESSURE DROPFor two-phase pressure drop through other than straight pipe, the fluid is treated as a single (“liquid") phase with the averageproperties of the mixture, and the method given in section xiv-b for liquid flow is used. This applies to valves and other fittings;orifices, nozzles and venturis; sudden contractions and expansions; and combining and dividing of streams. An exception is thatfor orifices, nozzles and venturis: the pressure recovery factor is not used.

PERFORATED-PIPE DISTRIBUTORSTHE DESIGN OF PERFORATED-PIPE DISTRIBUTORS FOR SINGLE-PHASE FLUID AS DOCUMENTED IN SECTION XIV-Bmay be applied to pseudo-liquid (vapor volume flow rate � 5% of the total flow) or pseudo-vapor (liquid volume flow rate � 5% ofthe total flow) two-phase mixtures. With two-phase flow there is the potential complication that liquid may flow preferentiallythrough some of the holes, and vapor through the others. For pseudo-liquids and pseudo-vapors, this effect will be small.However, for mixed-phase fluids (vapor volume flow rate > 5% or liquid volume flow rate > 5% of the total flow) it may be quitesevere. As a result the section xiv-b procedure is not recommended for mixed-phase flow. Currently no general procedures areavailable for mixed-phase fluids, and the design should be considered on an individual basis. ExxonMobil engineering shouldbe consulted for such a design.






FLASHING CRITICAL FLOWWhen a liquid or a vapor-liquid mixture under pressure is released to the surroundings through a pressure relief valve or a crackin a pipe or vessel, flashing can occur along the flow path due to the reduction in pressure. If the pressure drop is severeenough, a situation can arise in which the release rate becomes independent of the surrounding pressure. In other words, therelease rate cannot be increased by further reduction in the surrounding pressure. This condition is called critical flow.If the two-phase mixture can be considered as a homogeneous mixture, i.e., There is no velocity slip between the vapor and theliquid, and equilibrium between the phases is maintained during the expansion, the critical mass flux and the critical pressureratio can be calculated when the stagnation conditions, i.e., Pressure, temperature, and quality, are given. The critical pressureratio is the ratio of the pressure at the critical point to the stagnation pressure.In reality, phase equilibrium is seldom achieved during the expansion. This is especially true for liquid systems and for two-phase mixtures with low quality (< 0.1). This normally results in a higher critical mass flux than that predicted by assumingphase equilibrium. Methods that include the non-equilibrium effects are available and usually require computer solution.ExxonMobil engineering should be consulted for these more sophisticated calculation methods.

CALCULATION PROCEDUREThe following design methods, equations, and guidelines should be used with the “basic design considerations" given above.The first part gives procedures for determining the flow regime. The succeeding parts deal with methods for calculatingpressure drop and liquid holdup in straight pipes, pressure drop in piping components such as valves, orifices and pipe fittings,integrated pressure drop calculation for piping systems, guidelines for designing perforated-pipe distributors and manifolds fordistributing two-phase flow to parallel equipment, and finally a procedure for calculating critical flashing flow.

DETERMINING FLOW REGIMEThe flow regime is a very important design consideration for two-phase flow since it has significant effects on mechanicaldesign, heat transfer, flow splitting, etc. For example, slug flow can lead to pipe vibration and should be avoided wheneverpossible. Depending on the flow regime, a two-phase fluid can have quite different heat transfer characteristics. As mentionedin the basic design considerations section, the splitting of a two-phase mixture in a tee is very sensitive to its flow pattern.Knowledge of the flow regime is thus essential to design of equipment involving two-phase flow.Flow regime maps for the most common and significant pipe orientations: horizontal and vertical upflow are given in Figures 1and 2, respectively, as functions of the flow rates and system properties. It should be noted that these maps predict flowregimes for fully-developed flow, which may require up to 100 pipe diameters downstream of an obstruction or fitting toestablish. Further, these maps are mainly based on air-water data obtained from small diameter pipes [typically 1/2 in. (13 mm)to 3 in. (76 mm) dia.)]. Therefore, the maps are more reliable for systems with properties similar to the air-water system flowingin pipes not much larger than 3 in. (76 mm) dia. For systems with vastly different properties than air-water or for flow in largepipes ( > > 3 in. (76 mm) dia.), The maps should be considered only as qualitative guides to the flow pattern. Also, when using aflow map it must be recognized that the boundaries between flow regimes do not mean that there is a sharp transition from oneregime to the other.





HORIZONTAL FLOW - FIGURE 1 is a flow map for horizontal flow. Caution should be used in applying it to slightly inclinedupflow or downflow since the effect of gravity on the two-phase mixture can be quite strong. For example, stratified flow isfavored in inclined pipes with downflow. For flow regime estimation in inclined pipes contact ExxonMobil engineering.THE FOLLOWING PROCEDURE IS RECOMMENDED FOR DETERMINING THE FLOW REGIME WHEN THE PIPE SIZE ANDFLOW CONDITIONS ARE GIVEN:1. Calculate the parameter X in Figure 1 from the following equation:

X F xx

G L

L�

��

��

��

�

��

1

1 2 1 3

1 61 � �

� �

/ /

/ Eq. (1)

where:X = Abscissa of Figure 1

F1 = 530.7 (Customary)

= 2100 (Metric)

�G = Vapor density, lbm/ft3 (kg/m3)

�L = Liquid density, lbm/ft3 (kg/m3)

� = Liquid surface tension, dynes/cm (mN/m)

x = Mass fraction vapor (quality), dimensionless

�L = Liquid viscosity, cp (Pa�s)

2. Calculate the parameter Y in Figure 1 from the following equation:

Y Fx W

d G L� 2 2 1 2 1 2

� �/ / Eq. (2)

where:Y = Ordinate of Figure 1

F2 = 396 (Customary)

= 3.24 x 107 (Metric)

W = Total mass flow rate, klbm/h (kg/s)

d = Inside diameter of pipe, in. (mm)

3. Using X and Y, determine the flow regime from Figure 1.WHEN A PARTICULAR FLOW REGIME IS DESIRED, THE FOLLOWING PROCEDURE IS RECOMMENDED FORCALCULATING THE PIPE SIZE TO ACHIEVE THE FLOW PATTERN:

1. Calculate the parameter X from Eq. (1).

2. Using this value of X and Figure 1 pick a value of Y in the desired flow regime. Keep in mind that the boundaries betweenflow regimes are not sharp as indicated in the map, so that it is prudent to pick a value of Y away from the boundaries.






3. Use Y to calculate the pipe diameter d from the following equation:

d F x WYG L

��

��

�

��

��3 1 4 1 4

1 21� �/ /

/Eq. (3)

where:


= 5695 (Metric)

VERTICAL FLOW - Figure 2 is a flow map for vertical upflow. Maps for vertical downflow are not usually needed since in mostcases such flows are in the annular/mist flow regime. As a result a vertical downflow map is not given. ExxonMobil engineeringshould be contacted if such a map is needed.When the pipe size and flow conditions are given the flow regime can be found by the following procedure:

1. Calculate the parameter � from the following equation:

��

��

��

�F L

4

1 4/Eq. (4)

where:


= 0.52 (Metric)

2. Calculate Vsl from the following equation:

V F x Wdsl

L�

�

5 21( )�

Eq. (5)

where:

Vsl = Superficial velocity of liquid, ft/s (m/s)


= 1.273 x 106 (Metric)

3. Calculate the parameter � from the following equation:

� ��

��

��

�F L

G6

1 41 3

// Eq. (6)

where:


= 0.486 (Metric)





4. Calculate Vsg from the following equation:

V Fx W

dsgG

� 5 2�

Eq. (7)

where:

Vsg = Superficial velocity of vapor, ft/s (m/s)


= 1.273 x 106 (Metric)

5. Calculate �Vsl and �Vsg and determine the flow regime from Figure 2 (A for customary units and B for metric units).For calculating the pipe size which will give a desired flow regime in vertical upflow, the following procedure is recommended:

1. Calculate the parameter � from Eq. (6).

2. Pick a value of �Vsg from Figure 2 that will give the desired flow regime. For example pick �Vsg > 70 ft/s (21.3 m/s) for anannular flow regime and �Vsg < 0.4 ft/s (0.122 m/s) for a bubble flow regime.

3. Calculate Vsg based on the value of �Vsg picked in Step 2 and � calculated in Step 1.

4. Calculate the pipe diameter from the following equation:

d x WVG sg

��

��

�

12 0 354

1 2.

/

�(Customary) Eq. (8a)

d x WVG sg

��

��

�

��1000 1273

1 2.

/

�(Metric) Eq. (8b)

5. Calculate Vsl from Eq. (5) using the d calculated in Step 4.

6. Calculate �Vsl where � is calculated from Eq. (4).

7. Determine the flow regime based on �Vsl from Step 6 and �Vsg from Step 2 using Figure 2 and check whether it is thedesired one. If not, repeat Steps 2 to 7 until the desired flow regime is obtained.

PRESSURE DROP IN STRAIGHT PIPEThe methods given below for calculating pressure drop in circular pipes are independent of the flow regime. For non-circularconduits, these methods may be used with the equivalent hydraulic diameter deq defined by the following relation. It should bepointed out that the equivalent hydraulic diameter is a measure of the size of the cross sectional area available for flow, butcannot be used in place of the pipe diameter for flow rate calculations (see the note following the definition in subsection a).

d Cross AreaConduit Perimetereq �

��

��

�

��4 section Eq. (9)

The modified homogeneous method (method a) is the simplest since it assumes no velocity slip between the vapor and theliquid and is independent of liquid holdup. Both methods a and b are included in the two-phase pressure drop section ofpegasys (pc engineering application system). If liquid holdup is needed method b, which is based on dukler's similarity method,should be used. Method b is slightly more accurate than method a when applied to horizontal flow and may be expected to beaccurate to ± 30%. For vertical flow in which the pressure drop due to gravity may dominate, method b is appreciably morereliable. This is because the two-phase density calculated by method b is based on liquid holdup and as a result is moreaccurate. For more SOPHISTICATED METHODS SUCH AS THOSE DEPENDENT ON FLOW REGIME AND METHODS FORINCLINED PIPES, EXXONMOBIL ENGINEERING SHOULD BE CONSULTED.






METHOD A - MODIFIED HOMOGENEOUS METHOD

1. Assume the downstream pressure if upstream pressure is given or vice versa and calculate an average line pressure.

2. Calculate the two-phase mixture density, �ns, based on homogeneous flow assumption:� � � � �ns L G� � �( )1 Eq. (10)

and

�

QQ Q

L

L GEq. (11)

where:�ns = Two-phase, no slip density, lb/ft3 (kg/m3)

� = Volume fraction liquid, dimensionless

QL = Liquid volumetric flow rate, ft3/s (m3/s)

QG = Gas volumetric flow rate, ft3/s (m3/s)

3. Assume that the average mixture viscosity, �2�, is equal to the liquid viscosity:�2� = �L Eq. (12)

where:�2� = Two-phase viscosity, cP (Pa�s)

4. Calculate the average mixture velocity, V2�:

V FQ Q

dL G

2 7 2� �

�( )Eq. (13)

where:V2� = Average mixture velocity, ft/s


= 1.273 x 106 (Metric)

5. Calculate the two-phase Reynolds Number, assuming homogeneous flow (no slip):

RensnsF

d V� 8

2

2

�

�

�

�Eq. (14)

where:Rens = No-slip Reynolds Number, dimensionless


= 10-3 (Metric)

6. Find the Fanning friction factor, f, from Figure 1 or 2 in Subsection B, using Rens calculated from Eq. (14):





7. Calculate the pressure drop due to friction (P)f:

( )�Pf V L

g dfns

c�

22

6� � (Customary) Eq. (15a)

( )�Pf V L

dfns

�2 2

2� � (Metric) Eq. (15b)

where:L = Pipe length, ft (m)

(P)f= Frictional pressure drop, psi (kPa)

gc = Gravitational conversion factor, 32.174 ft lbm/lbf s2

8. For calculating the pressure drop due to elevation changes, one assumes no pressure recovery in the downhill legs. Whenthe pressure gain on a negative leg must be known, consult ExxonMobil Engineering.a. Calculate the superficial vapor velocity, Vsg. This is the velocity the vapor would have if there were no liquid in the line.

V FQdsg

G� 7 2 Eq. (16)

where:

Vsg = Superficial vapor velocity, ft/s (m/s)


= 1.273 x 106 (Metric

b. Calculate the pressure drop due to elevation changes (P)e:

( )��

PE H

eH L��

144(Customary) Eq. (17a)

( ) .� �P x E He H L� �9 8 10 3 � (Metric) Eq. (17b)

where:

(P)e = Pressure drop due to elevation change, psi (kPa)

EH = Elevation head factor from Figure 3, using Vsg calculated from Eq. (16)

H = Summation of all elevation uplegs measured in the vertical direction, ft. Note: This is not the same as the net elevation change between pipe inlet and outlet.






9. Pressure drop due to acceleration is usually small and can normally be neglected. However, it should be checked, asindicated below:a. Calculate the acceleration group, J:

J W W W P

gc d P P

L G G

G

��

�

184

1 2

( )_

�

(Customary) Eq. (18a)

J x W W W P

d P P

L G G

G

��

�

162 109

41 2

. ( )_

�

(Metric) Eq. (18b)

where:

J = Acceleration group, dimensionless

WL = Liquid mass flow rate, klbm/h (kg/s)

WG = Vapor mass flow rate, klbm/h (kg/s)

gc = Dimensional constant, 32.174 ft lbm/lbf s2

d = Inside diameter of pipe, in. (mm)

P1 = Upstream pressure, psia (kPa)

P2 = Downstream pressure, psia (kPa)

P�

= Average pressure, P P1 22� , psia (kPa)

�

�

G = Average vapor density, � �G G1 2

2�

, lbm/ft3 (kg/m3)

b. If J � 0.1, acceleration may be neglected, and the total pressure drop can be calculated from Eq. (19). See Step 10a.If J > 0.1, a trial-and-error procedure is required (Step 10b).

10. Calculate the total pressure drop:a. If the acceleration as determined in Step 9 is negligible, calculate the total pressure drop (P)t from Eq. (19):

(P)t = (P)f + (P)e Eq. (19)

b. If the acceleration is not negligible, use the values (P)f and (P)e from Eqs. (15) and (17) in Eq. (20):

( ) ( ) ( )�

� �P P PJt

f e�

�1Eq. (20)

where:

(P)t= Total pressure drop due to friction, elevation, and acceleration, psi (kPa)

11. Check the assumed average pressure (Step 1) and repeat the procedure if more accuracy is required, particularly whenacceleration (Step 10b) must be accounted for:





METHOD B - SIMILARITY METHOD (LIQUID HOLDUP INCLUDED)Use the following procedure for calculating pressure drop and liquid holdup in HORIZONTAL FLOW when flow rate and eitherupstream or downstream pressure are given.

1. Assume the downstream pressure if the upstream pressure is given or vice versa and calculate the average line pressure.

2. Calculate �, the volume fraction liquid, from Eq. (11).

3. Calculate the mixture viscosity, �2� = �L� + �G(1 - �). Eq. (21)

4. Calculate the average mixture velocity, V2� from Eq. (13).5. Calculate the two-phase Reynolds Number, Re2�. This is a trial-and-error procedure consisting of the following steps:

a. Estimate a value of RL, the liquid holdup. (Use � and Figure 4 for estimate.)b. Calculate Re2� using �, �2�, and V2� from:

Re2 82 2

2�

� �

�

�

� F

d VEq. (22)

��

�2

2 211

� �

�L

L

G

LR R( ) Eq. (23)

where:

Re2�= Two-phase Reynolds Number, dimensionless


= 10-3 (Metric)

�2� = Two-phase density, lbm/ft3 (kg/m3)

c. Use Figure 4 with � and Re2� to obtain a new value of RL. If the assumed and calculated values of RL agree within 5%,the accuracy is sufficient. If not, Step 5b must be repeated with the new value of RL.

d. When the assumed and calculated values of RL agree within 5%, use the last calculated value of RL to calculate Re2�from Eq. (22).

6. Calculate f1�, the single-phase Fanning friction factor from Eq. (24):

f12

0 320 0014 0125�

�

� . .(Re ) . Eq. (24)

7. Look up f2�/f1� from Figure 5 and calculate f2� from this ratio and the value of f1� calculated in Step 6.

8. Calculate pressure drop due to friction, (P)f, from Eq. (25):

( )�Pf V L

g dfc

�2 2

22

6� � ��

(Customary) Eq. (25a)

( )�Pf V L

df �2 2 2

22� � ��

(Metric) Eq. (25b)






9. Pressure drop due to acceleration is usually small and can normally be neglected. However, it should be checked, asindicated below:a. Calculate (P)a, the acceleration pressure drop, from Eq. (26):

( ) .�Pd

Q Qd

Q Qa

G G L L G G L L��

��

�

�

��

�

��

�

��

�

��

��

��

�

�

��

�

��

�

��

�

��

��

��

��

��7 25 1

11

14

2 2

24

2 2

1

� � � �

R R R R L L L L(Customary) Eq. (26a)

( ) .�P xd

QR

QR d

QR

QRa

G G

L

L L

L

G G

L

L L

L�

��

��

�

��

�

��

�

�

�

��

��

��

�

��

�

��

�

�

�

��

��

��

��

��162 10 1

11

19

4

2 2

24

2 2

1

� � � �(Metric) Eq. (26b)

where the subscripts 1 and 2 refer to upstream and downstream conditions, respectively, and all other terms are aspreviously defined. In Eq. (26), the unknown pressure is either at 1 or 2, and is obtained using the known pressure and(P)f.If two or more lines join together, the upstream term should be calculated separately for each line and added togetherto get the total upstream acceleration term. In Eq. (26) �G, QG, and RL vary with position. RL can be obtained fromFigure 4 using � at upstream and downstream conditions.

b. If ( )( ) ( )

�

� �

PP P

a

f e � 0.1, the value of (P)a calculated above is close enough (or acceleration may be neglected).

If ( )( ) ( )

�

� �

PP P

a

f e > 0.1, a trial-and-error procedure is required, involving Eqs. (22), (23), (25), (17), and (26). The

procedure involves recalculating the unknown pressure using the new value of P = (P)a + (P)f. The calculations inEqs. (22) through (26) are then repeated using the new value of the unknown pressure and the new value of �g at thatpoint until convergence on (P)a is obtained.Note that (P)e = 0 for horizontal flow.

10. Calculate the total pressure drop (P)t from Eq. (27):(P)t = (P)f + (P)a Eq. (27)

11. Check the assumed average pressure(Step 1) and repeat the procedure if more accuracy is required.Use the following procedure for calculating pressure drop and liquid holdup in VERTICAL UPFLOW when flow rate and eitherupstream or downstream pressure are given.

1. Assume the downstream pressure if the upstream pressure is given or vice versa and calculate the average line pressure.

2. Calculate �, the volume fraction liquid, from Eq. (11).

3. Calculate the mixture viscosity, �2�, from Eq. (21).

4. Calculate the average mixture velocity, V2�, from Eq. (13).





5. Determine the two-phase Reynolds Number Re2� by the following steps:a. Calculate the parameter � from the following equation:

��

��

�

��

�F W

W GL

G

L G G

t L9

0 9 0 19 0 205 0 7 2 75

0 435 0 72

. . . . .

. . Eq. (28)

where:

Gt = Total mass velocity, lbm/ft2-s (kg/m2-s)

If Gt > 50 lbm/ft2-s (244 kg/m2-s), set Gt = 50 lbm/ft2-s (244 kg/m2-s)

F9 = 6.28 x 104 (Customary)

= 8.75 x 1013 (Metric)

b. Determine the liquid holdup RL from Figure 6.c. Calculate �2� from Eq. (23).d. Calculate Re2� from Eq. (22).

6. Calculate f1� from Eq. (24).

7. Look up f2�/f1� from Figure 5 and calculate f2� from this ratio and the value of f1� calculated in Step 6.

8. Calculate pressure drop due to friction, (P)f, from Eq. (25).

9. Calculate the pressure drop due to elevation change from the following equation:

� �( )( )

�PR R

LeL L G L

�� 1

144(Customary) Eq. (29a)

� �( ) . ( )�P x R R Le L L G L� ��9 8 10 13 � � (Metric) Eq. (29b)

10. Same as Step 9 in horizontal flow for calculating (P)a except that RL is obtained from Figure 6 with � determined atupstream and downstream conditions.

11. Calculate the total pressure drop, (P)t, from the following equation:(P)t = (P)f + (P)e + (P)a Eq. (30)

12. Check the assumed average pressure (Step 1) and repeat the procedure if more accuracy is required.For VERTICAL DOWNFLOW use Method A.

FLOW RATE IN STRAIGHT PIPEFor calculating the flow rate when upstream and downstream pressures are known, use the procedures given below. In two-phase flow there can be multiple solutions or values of vapor and liquid flow rates for a given pressure drop over a given pipeequivalent length. Some other condition must be fixed to give a unique solution, for example the volume fraction liquid, �, or thetotal volumetric or mass flow rate of the stream. The procedure given below assumes that the volume fraction liquid, �, is knownbut analogous procedures for other cases can be developed readily. First, for non-circular conduits, calculate the equivalenthydraulic diameter, deq, from Eq. (9).

If P PP

1 2

1

� � 0.2, the fluid can be treated as being incompressible, i.e., the acceleration pressure drop can be neglected. Either

Method A or Method B may be used, as given below. If P PP

1 2

1

� > 0.2, either Method A or Method B may be used, but the

respective acceleration pressure drop terms, Eq. (18) or Eq. (26), must be included in the trial-and-error procedure.






1. Method A

a. Calculate �ns and �2� from Eqs. (10) and (12) using � and �G evaluated at P P1 22� .

b. Assume f = 0.005 and calculate V2� from Eq. (15).c. Calculate Rens from Eq. (14) and obtain new value for f in Figure 1 or 2, Subsection B.d. Recalculate V2� from Eq. (15) with new f. Repeat until convergence is obtained.e. Calculate QL and QG from Eqs. (11) and (13).

2. Method B (for Horizontal Flow Only)

a. Calculate �2� from Eq. (21) using � evaluated at P P1 22� .

b. Obtain ff2

1

�

�

from Figure 5.

c. Assume value for f2� = 0.01 and calculate f1�.d. Calculate Re2� from Eq. (24).e. Use Figure 4 with � and Re2� to obtain value of RL.f. Calculate �2� from Eq. (23).g. Calculate V2� from Eq. (22).h. With �2� and V2�, calculate new f2� from Eq. (25).i. Repeat Steps e through h until procedure converges.j. Calculate QL and QG from Eqs. (11) and (13).

PRESSURE DROP IN SINGLE PIPING COMPONENTSBends - Use the following procedures:

1. Find the resistance coefficient K from Figure 5C, Figure 5D, or Figure 5E of Subsection B.

2. Calculate the frictional pressure drop, (P)f, from:

( )�P F K Wdf

ns�

�

��

�

��10

2

4Eq. (31)

�ns is obtained from Eq. (10).

where:


= 8.1 x 108 (Metric)

Valves - Use the same procedure as for bends. Values for the resistance coefficient, K, should be taken from Figure 5A orFigure 5B from Subsection B.





Tees and Y’s - For blanked-off tees, use the same procedure as for bends. For tees and Y’s in which streams are split orjoined, use the procedures given in Subsection B for the particular configuration. Values for the resistance coefficient, K, shouldbe taken from Figure 5C Or Figure 5D from Subsection B.ORIFICES - Use the following equation:

�P F WC do ns

��

��

�

��10

2

2 42

Eq. (32)

where:

C = Flow coefficient, dimensionless (see Figure 7A or 7B of Subsection B)

do = Orifice diameter, in. (mm)


= 8.1 x 108 (Metric)

Note that �ns2 is the downstream density of the vapor-liquid mixture, Eq. (10).

Flow Nozzles - Proceed as for orifices, but use flow coefficient C from Figure 8 in Subsection B.Venturis - Proceed as for orifices, but use flow coefficient C given by Eq. (33):

41)/(1

98.0

ddC

o�

� Eq. (33)

where:

d1 = Inside diameter of upstream pipe, in. (mm)

Contractions and Expansions - Use the following procedure:

1. Calculate the frictional pressure drop from Eq. (34):

( )�P F K Wdf

ns s�

�

��

�

��10

2

4Eq. (34)

where:

ds = Inside diameter or equivalent hydraulic diameter of the smaller-diameter pipe, in. (mm)

K = Resistance coefficient, dimensionless (see Figure 6 of Subsection B).


= 8.1 x 108 (Metric)

For the first trial, use either the upstream or downstream mixture density for �ns, whichever is known.

2. Calculate the pressure drop due to change in kinetic energy of the flow from Eq. (35):

( )�P F Wd dk

ns ns� �

�

��

�

�10

2

24

2 14

1

1 1

Eq. (35)

where:


= 8.1 x 108 (Metric)

For the first trial, use either the upstream or downstream mixture density, whichever is known, for both �ns1 and �ns2.






3. Calculate the total pressure drop by adding (P)f and (P)k.(P) = (P)f + (P)k Eq. (36)

4. Calculate the unknown pressure and mixture density, find a new value for the average mixture density, �ns, and repeatSteps 1 to 4 until adequate convergence is obtained.

PERFORATED-PIPE DISTRIBUTORSUse the following procedure:

1. Using the two-phase regime map (Figure 1 or 2), find out which flow regime exists in the pipe leading to the distributor:a. If the flow regime is spray, mist or bubble flow, proceed to Step 2, below.b. If the flow regime is annular or froth flow, either reduce the pipe diameter to give spray or mist flow for a length of 50

pipe diameters upstream of the distributor, or provide an orifice just upstream of the distributor, with a diameter of 0.7times the pipe diameter. Then proceed to Step 2.

c. If the flow regime is plug, stratified, wavy or slug flow, reduce the pipe diameter to give spray or mist flow for a length of100 pipe diameters upstream of the distributor. Then proceed to Step 2.

2. Classify the fluid as pseudo-liquid, pseudo-vapor or mixed-phase, according to the following definitions:a. Call it pseudo-liquid if the vapor volume flow rate � 5% of the total for the mixture.b. Call it pseudo-vapor if the liquid volume flow rate � 5% of the total for the mixture.c. Call it mixed-phase if the vapor and liquid volume flow ratio lies between the above limits.

3. Determine the distributor diameter and the number and size of holes according to the procedures given in Subsections Band C, subject to the following rules.a. For pseudo-liquids, follow the procedure in Subsection B. Use mixture flow rates and physical properties, except use

liquid viscosity for calculating Re and looking up f.b. For pseudo-vapor, follow the procedure in Subsection C. Use mixture flow rates and physical properties.c. For mixed-phase, consult ExxonMobil Engineering.

4. Assuming uniform and proportionate flow of liquid and vapor through each of the outlet holes of the distributor as designedabove, check the flow regime (Figure 1) just upstream of the last hole. Sometimes in the case of large-diameterdistributors, the flow regime will change (because of reduced linear velocity) after part of the fluid has been distributed. Ifthe flow regime upstream of the last hole has changed to an undesirable pattern (see Step 1 above), locate the point in thedistributor where the transition took place by checking the flow regime upstream of other outlet holes, and taper thedistributor downstream of that point.

DISTRIBUTION MANIFOLDSDistribution of two-phase flow to parallel equipment should be done in a symmetrical fashion. Among the three manifold designsshown in Figure 7, manifold (a) is the best since it is the most symmetrical design; the flow is divided into two substreams by animpacting tee and then each substream again is divided into two substreams and so on. Manifold (b) is less symmetrical than(a) since it splits the flow with impacting as well as straight-through tees. Manifold (c) is the least symmetrical because all thesplitting is done by straight-through tees. As a result, Manifold (c) is expected to give the poorest distribution. The advantage of(c) is that it is simple to construct and requires relatively small plot space, which for large diameter pipes can be very significant.For applications requiring good flow distribution (< 15% maldistribution), a-type manifolds are recommended. For applications inwhich considerable maldistribution (15 to 50%) can be tolerated, b-type manifolds can be considered. C-type manifolds are notrecommended unless the two-phase mixture to be distributed is a pseudo-liquid or a pseudo-vapor for which up to 50%maldistribution is expected.There may be cases in which the design is limited by plot space and as a result, C-type manifolds are necessary to handlemixed-phase fluids. There may also be cases in which, because of changes in process requirement, C-type manifolds originallydesigned for single-phase fluid may be operating in the two-phase regime. ExxonMobil Engineering should be consulted in suchcases for ways to improve flow distribution.

INTEGRATED PRESSURE DROP CALCULATION FOR PIPING SYSTEMSUse the procedure below for calculating pressure drop in any flow system containing more than one single piping component(Method A):1. Break the system into sections of constant mass flow rates and constant nominal diameter. Then apply Steps 2 through 6,

below, to each of the sections.





2. For any section having a non-circular cross-section, calculate the equivalent hydraulic diameter, deq, from Eq. (9).3. Find the Reynolds Number, Rens, for each section from Eq. (14). For the first trial, use either the upstream or downstream

conditions to determine mixture density, �ns, (Eq. 10) and mixture viscosity, �2�, Eq. (12).4. Find the friction factor f from Figure 1 or 2 in Subsection B.5. If piping details are not available and cannot be estimated, assume for offsite lines an equivalent length of fittings of 20 to

80% of the actual pipe length and for onsite lines 200 to 500%. Estimate the pipe length from the plot plan, tower heights,pipe rack location, etc.When fittings are known or can be estimated, find their equivalent lengths from Eq. (37);

L x df

Keq ��

��

�

�� 2 08 10 2. (Customary) Eq. (37a)

L x df

Keq ��

��

�

�� 2 5 10 4. (Metric) Eq. (37b)

where:

Leq = Equivalent length of fitting, ft (m)

K = Sum of resistance coefficients of all fittings, dimensionless

The resistance coefficient K of bends, blanked-off tees, and valves is found from the appropriate Figures from Subsection B.For bends and blanked-off tees, Figures 5C and 5D should be used while Figures 5A and 5B should be used for valves. Donot add the K factors of contractions and expansions.For orifices, flow nozzles, and venturis, the resistance coefficient should be calculated from Eq. (38).

KC

ddo

��

��

�

��

��

�

��

12

14

Eq. (38)

where:

C = Flow coefficient, dimensionless [for orifices and nozzles, see Figures 7 and 8 ofSubsection B; for venturis, C is defined by Eq. (33).]

6. Add the equivalent lengths of the fittings in each section to the actual length of the section, and calculate the pressure dropin each section from Eq. (15), starting at the end of the system where the pressure is known. Calculate pressure drops inexpansions or contractions between sections by treating them as single components. Find the pressure drops in streamjunctions, such as tees and Y’s, from Eq. (8) of Subsection B, depending on the particular configuration.

7. Calculate pressure drop due to elevation changes, (P)e from Eq. (17).8. Check acceleration effects across the system using Step 9, Method A.9. Repeat Steps 3 through 8 with improved values of �ns and �2�, if necessary, until adequate convergence is obtained.






FLASHING CRITICAL FLOWThe flashing critical flow calculation method given below is for single-component systems when the stagnation conditions aregiven. It is based on the homogeneous equilibrium model, and as a result will underpredict the critical mass flux. Theunderprediction is more severe for saturated liquids and for two-phase streams of low quality (< 0.1). However, the accuracyimproves for streams at higher quality. Since this method underpredicts the mass flux, it provides a conservative estimate whencalculating the opening area required to vent a given mass flow rate. As a result, the method is suitable for sizing pressure reliefvalves. For more information on pressure relief valve sizing, please refer to EE.28E.90 and DP Section XV-C. It should benoted that if the stagnation conditions change, e.g., as a result of depletion of the fluid in the upstream device, the critical massflux will change. A new critical mass flux should be calculated based on the new stagnation conditions. ExxonMobil Engineeringshould be consulted for more sophisticated methods and for calculating the critical mass flux for subcooled liquids.1. Determine �LO, �GO, hLO, hGO, and CLO using EDL QUEST at the stagnation conditions. For the steam-water system, the

steam tables can be used.�LO, �GO = Specific volume of liquid and vapor at stagnation conditions, respectively, ft3/lbm (m3/kg)hLO, hGO = Enthalpy of liquid and vapor at stagnation conditions, respectively, Btu/lbm (kJ/kg)CLO = Heat capacity of liquid at stagnation conditions, Btu/lbm F (kJ/kg K)

2. Calculate �LGO and hLGO from the following equations:�LGO = �GO - �LO Eq. (39)

hLGO = hGO - hLO Eq. (40)

3. Calculate the two-phase specific volume �2�O from the following equation:�2�O = �LO + xO�LGO Eq. (41)

where:

xO = Stagnation quality

4. Calculate the correlation parameter � from the following equation:

��

� �

�

� �

� ��

��

�

x F C T Ph

O LGO

O

LO O O

O

LGO

LGO2 2

2

Eq. (42)

where:

TO = Stagnation temperature, R (K)

PO = Stagnation pressure, psi (kPa)

F = 0.185 (Customary)

= 1 (Metric)

5. Obtain � from Figure 8 at the calculated � and calculate the critical pressure, PC, from the following equation:PC = �PO Eq. (43)

6. Obtain GC* from Figure 8 at the calculated � and compute the critical mass flux GC in lbm/ft2-s (kg/m2-s) from the followingequation:

G F GP

C CO

O� 11

2*

� �

Eq. (44)

where:


= 31.62 (Metric)

SAMPLE PROBLEMSNote - In these problems, computed results in the two unit systems may differ because of roundoff in the calculations.





PROBLEM 1 - PRESSURE DROPGiven: Air and water are flowing through 200 ft (61 m) of a standard 2-inch (50 mm) steel line at a positive slope of 5 , followedby an elbow, a reducer, and 25 ft (7.5 m) of standard 1-1/2 inch (40 mm) line running straight up. Flow rates, conditions, andphysical properties are as follows:

CUSTOMARY METRIC

TemperatureInlet pressure

77�F40 psig

25�C280 kPa, gage

WaterFlow rateDensityViscositySurface tension

5.0 klbm/h62.2 lbm/ft30.894 cP72.0 dynes/cm

0.63 kg/s996 kg/m30.894 x 10-3 Pa�s72.0 mN/m

AirFlow rateDensityViscosity

7.88 x 10-3 klbm/h0.275 lbm/ft30.0184 cP

9.93 x 10-4 kg/s4.4 kg/m30.0184 x 10-3 Pa�s

Find: Total pressure drop.Solution: Use Method A to find the pressure drop.1. Divide the system into three sections:

� 200 ft (60 m) of 2 inch (50 mm) line and an elbow.� Sudden contraction from 2 inch (50 mm) to 1-1/2 inch (40 mm) line.� 25 ft (7.5 m) of 1-1/2 inch (40 mm) line.All pipes and fittings have circular cross-sections, so equivalent hydraulic diameter does not have to be calculated. Sincethe inlet pressure is given, calculate the pressure drops starting at the inlet and working toward the outlet.

2. Pressure drop in 200 ft (60 m) of 2 inch (50 mm) line and elbow:Customary Units Metric Units

Liquid flow rate

QL =5000

62 2 36003lb h

lb ft s hm

m

/( . / ) ( / )

QL = 0 63996 3

. //

kg skg m

= 0.0223 ft3/s = 6.33 x 10-4 m3/s

Vapor flow rate

QG =7 88

0 275 36003. /

( . / ) ( / )lb h

lb ft s hm

mQG = 9 93 10

4 4

4

3. /

. /x kg s

kg m

�

= 0.00796 ft3/s = 2.26 x 10-4 m3/s

Volume fraction liquid

� = QQ Q

L

L G�

� = QQ Q

L

L G�

= 0 02230 0223 0 00796

.. .�

= 6 33 106 33 10 2 26 10

4

4 4.

. .x

x x

�

� �

�

= 0.737 = 0.737

Density of inlet mixture [Eq. (10)]�ns = �L� + �G (1 - �) �ns = �L� + �G (1 - �)

= (62.2) (0.737) + (0.275) (1 - 0.737) = (996) (0.737) + (4.4) (1 - 0.737)






= 45.9 lbm/ft3 = 735 kg/m3

Viscosity of inlet mixture�2� = �L = 0.894 cP (0.894 x 10-3 Pa�s)

Inside diameter of standard 2 inch (50 mm) line (Table 1 of Subsection A)d = 2.067 in. (52.5 mm)

Average velocity of inlet mixture [Eq. (13)]

V2� =184

2( )Q Q

dL G�

V2� = 1273 106

2. ( )x Q Q

dL G

= ( ) ( . . )( . )

184 0 0223 0 007962 067 2

� = 1273 10 6 33 10 2 26 1052 5

6 4 4

2. ( . . )

( . )x x x� �

�

= 1.303 ft/s = 0.399 m/s

Reynolds Number of inlet mixture [Eq. (14)]

Rens =124 2

2

d V ns�

�

�

�Rens =

10 32

2

� d V ns�

�

�

�

= ( ) ( . ) ( . ) ( . ).

124 2 067 1303 45 90 894

= ( ) ( . ) ( . ) ( ).

10 52 5 0 399 7350 894 10

3

3

�

�x

= 17,150, call 17,000 = 17,220, call 17,000





Customary Units Metric Units

Inlet friction factor (Figure 2 of Subsection B)f = 0.072

Resistance coefficient of 2 inch (50 mm), 90 flanged elbow (Figure 5B of Subsection B)K = 0.37

Equivalent length of elbow [Eq. (37)]

Leq = 2 08 10 2. x df

K��

��

�

�� Leq = 2 5 10 4. x d

fK�

�

��

�

��

= ( . ) ..

.2 08 10 2 0670 0072

0 372x ��

��

�

�� = ( . ) .

..2 5 10 52 5

0 00720 374x �

�

��

�

��

= 2.21 ft = 0.674 m

Total equivalent length of pipe and elbowL = 200 + 2.21 L = 60 + 0.674

= 202.21 ft, call 202 ft = 60.67 m, call 60.7 m

Frictional pressure drop [Eq. (15)]

(P)f =f V L

g dns

c

22

6� � (P)f =

2 22f V Ld

ns� �

= ( . ) ( . ) ( . ) ( )( ) ( . ) ( . )

0 0072 1303 45 9 2026 32174 2 067

2= ( ) ( . ) ( . ) ( ) ( . )

( . )2 0 0072 4 0 399 735 60 7

52 5

2

2

= 0.284 psi = 1.95 kPa

Superficial vapor velocity at the inlet [Eq. (16)]

Vsg =184

2Q

dG Vsg = 1273 106

2. x Q

dG

= ( ) ( . )( . )

184 0 007962 067 2 = ( . ) ( . )

( . )1273 10 2 26 10

52 5

6 4

2x x �

= 0.343 ft/s = 0.104 m/s

Elevation head factor (Figure 3)EH = 0.90

Elevation pressure drop [Eq. (17)]

(P)e = E HH L� �

144(P)e = 9.8 x 10-3 EH �L H

= ( . ) ( . ) ( sin )0 90 62 2 200 5144

o= (9.8 x 10-3) (0.90) (996) (60 sin 5 )

= 6.80 psi = 45.9 kPa







Pressure drop due to friction and elevationP = (P)f + (P)e P = (P)f + (P)e

= 0.284 + 6.80 = 1.95 + 45.9

= 7.08 psi = 47.9 kPa

Average pressure in the 2 inch (50 mm) line

P_

= P P1 22� P

_= P P1 2

2�

= 54 7 54 7 7 082

. ( . . )� � = 3813 3813 47 92

. ( . . )� �

= 51.16 psia = 357.4 kPa abs.

Vapor density at downstream conditions (assume ideal gas, isothermal)

�G2 = �G PP1 2

1�G2 = �G P

P1 2

1

= ( . ) ( . . ).

0 275 54 7 7 0854 7

� = ( . ) ( . . ).

4 4 3813 47 93813

�

= 0.239 lbm/ft3 = 3.847 kg/m3

Average vapor density

�

_

G = � �G G1 22�

�

_

G = � �G G1 22�

= 0 275 0 2392

. .� = 4 4 3 8472

. .�

= 0.257 lbm/ft3 = 4.12 kg/m3

Check the acceleration term [Eq. (18)]

J = 184

1 2

( )_

_W W W P

g d P P

L G G

c G

�

�

J = ( . ) ( )_

_162 109

41 2

x W W W P

d P P

L G G

G

�

�

= ( ) ( . . ) ( . ) ( . )( . ) ( . ) ( . ) ( . . ) ( . )

18 5 0 0 00788 0 00788 511632174 2 067 54 7 54 7 7 08 0 2574

�

�

= ( . )( . . )( . )( . )( . ) ( . ) ( . . ) ( . )

162 10 0 63 9 93 10 9 93 10 357 452 5 3813 3813 47 9 412

9 4 4

4x x x�

�

� �

= 9.24 x 10-5 = 9.11 x 10-5

Since J is much smaller than 0.1, the acceleration can be ignored. Also, since the pressure drop is a small fraction of theabsolute pressure (~ 14%), the effect of pressure drop on �G can be ignored for the purpose of calculating the pressure drop inthe 2 inch (50 mm) line.

3. Pressure drop in the reducer (use the procedure for sudden contraction):

Inside diameter of standard 1-1/2 inch (40 mm) line (Table 1 of Subsection A):d = 1.610 inch (40.89 mm)






Diameter ratiodd

1

2= 1610

2 067..

dd

1

2= 40 89

52 50..

= 0.779 = 0.779

Resistance coefficient (Figure 6 of Subsection B)K = 0.15

Density of vapor at the inlet to the reducer (calculated above)�G = 0.2394 lbm/ft3 (3.847 kg/m3)

Vapor flow rate

QG =7 88

0 2394 36003. /

( . / ) ( / )lb h

lb ft s hm

mQG = 9 93 10

3 847

4

3. /

. /x kg s

kg m

�

= 0.00914 ft3/s = 2.58 x 10-4 m3/s

Volume fraction liquid [Eq. (11)]

� = QQ Q

L

L G�

� = QQ Q

L

L G�

= 0 02230 0223 0 00914

.. .�

= 6 33 106 33 10 2 58 10

4

4 4.

. .x

x x

�

� �

�

= 0.709 = 0.71

Density of inlet mixture [Eq. (10)]�ns = �L� + �G (1 - �) �ns = �L� + �G (1 - �)

= (62.2) (0.709) + (0.2394) (1 - 0.709) = (996) (0.71) + (3.847) (1 - 0.71)

= 44.2 lbm/ft3 = 708.3 kg/m3

Friction pressure drop [Eq. (34)], based on the mixture density upstream of the reducer

(P)f = 0 282

4. KWdns s�

�

��

�

�� (P)f = 810 108

2

4. x KWdns s�

�

��

�

��

= ( . ) ( . ) ( )( . ) ( . )

0 28 015 54 42 161

2

4 = ( . ) ( . ) ( . )( . ) ( . )

81 10 015 0 63708 3 40 89

8 2

4x

= 0.00354 psi, call 0.004 psi = 0.02435 kPa, call 0.024 kPa

Pressure drop due to kinetic energy change [Eq. (35)], based on the mixture density upstream of the reducer

(P)K = 0 28 1 12

24

2 24

1. W

d dns ns� �

��

��

(P)K = 81 10 1 18 2

24

2 24

1. x W

d dns ns� ��

�

��

�

��

= ( . ) ( )( . ) ( . ) ( . ) ( . )

0 28 5 1161 44 2

12 067 44 2

24 4�

�

��

�

��

= ( . )( . )( . ) ( . ) ( . ) ( . )

8 1 10 0 631

40 89 708 3

1

52 5 708 38 2

4 4x ��

�

��

�

�

��

= 0.0149 psi, call 0.105 psi = 0.101 kPa







Total pressure drop in the reducer(P) = (P)f + (P)K (P) = (P)f + (P)K

= 0.004 + 0.015 = 0.024 + 0.101

= 0.019 psi, call 0.02 psi = 0.125 kPa

The change in mixture density across the reducer can be neglected.

4. Calculate the pressure drop in the 1-1/2 inch (40 mm) line, using inlet conditions to reducer (as being close enough to thoseat the reducer outlet):

Mixture viscosity at the inlet to the 1-1/2 inch (40 mm) line�2� = �L = 0.894 cP (0.894 x 10-3 Pa�s)

Average velocity of inlet mixture [Eq. (13)]

V2� =184

2( )Q Q

dL G�

V2� = 1273 106

2. ( )x Q Q

dL G�

= ( ) ( . . )( . )

184 0 0223 0 009141610 2

� = 1273 10 6 33 10 2 58 1040 89

6 4 4

2. ( . . )

( . )x x x� ��

= 2.23 ft/s = 0.682 m/s

Reynolds Number of inlet mixture [Eq. (14)]

Rens =124 2

2

d V ns�

�

�

�Rens =

10 32

2

� d V ns�

�

�

�

= ( ) ( . ) ( . ) ( . ).

124 161 2 23 44 20 894

= ( ) ( . ) ( . ) ( ).

10 40 89 0 682 7080 894 10

3

3

�

�x

= 22,000 = 22,094, call 22,000

Inlet friction factor (Figure 2 of Subsection B)f = 0.0070

Length of 1-1/2 inch (40 mm) lineL = 25 ft (7.5 m)

Friction pressure drop [Eq. (15)]

(P)f =f V L

g dns

c

22

6� � (P)f =

2 22f V Ld

ns� �

= ( . ) ( . ) ( . ) ( )( ) ( . ) ( . )

0 007 2 23 44 2 256 32174 1610

2= ( ) ( . ) ( . ) ( . ) ( . )

.2 0 007 0 682 708 3 7 5

40 89

2

= 0.124 psi = 0.846 kPa






Superficial vapor velocity at the inlet [Eq. (16)]

Vsg =184

2Q

dG Vsg = 1273 106

2. x Q

dG

= ( ) ( . )( . )

184 0 00914161 2 = ( . ) ( . )

( . )1273 10 2 58 10

40 89

6 4

2x x �

= 0.649 ft/s = 0.197 m/s

Elevation head factor (Figure 3)EH = 0.825

Elevation pressure drop [Eq. (17)]

(P)e = E HH L� �

144(P)e = 9.8 x 10-3 EH �L H

= ( . ) ( . ) ( )0 825 62 2 25144

= (9.8 x 10-3) (0.825) (996) (7.5)

= 8.91 psi = 58.6 kPa

Pressure drop due to friction and elevationP = (P)f + (P)e P = (P)f + (P)e

= 0.124 + 8.91 = 0.846 + 58.6

= 9.03 psi = 59.4 kPa

The acceleration term [Eq. (18)] will again be neglected. Check the effect of pressure drop on the elevation term.P1 = 54.7 - 7.08 - 0.02 P1 = 381.3 - 47.9 - 0.125

= 47.6 psia = 333 kPa, abs.

P2 = 47.6 - 9.03 P2 = 333 - 59.4

= 38.57 psia = 273.6 kPa, abs.

P_

= P P1 22� P

_= P P1 2

2�

= 47 6 38 572

. .� = 333 273 62� .

= 43.08 psia, call 43.1 psia = 303.3 kPa, call 303 kPa

Vsg = 0 649 47 6431

. ..

�

��

�

�� Vsg = 0197 333

303. �

��

�

��

= 0.717 ft/s = 0.216 m/s

EH = 0.82 EH = 0.82







(P)e = ( . ) ( . ) ( )0 82 62 2 25144

(P)e = (9.8 x 10-3) (0.82) (996) (7.5)

= 8.85 psi = 58.22 kPa

P = 0.124 + 8.85 P = 0.846 + 58.22

= 8.97 psi (vs. 9.03 psi) = 59.1 (vs. 59.4)

5. Pressure drop over the entire systemP = 7.084 + 0.02 + 8.97 P = 479.46 + 0.125 + 59.1

= 16.07 psi, call 16.1 psi = 107.1 kPa, call 107 kPa

PROBLEM 2 - FLOW PATTERN IN HORIZONTAL PIPEGiven: Same as in Problem 1.Find: The two-phase flow regime at the average conditions in the 2 inch (50 mm) line.Solution: For the 2 inch (50 mm) line, which is nearly horizontal, use Figure 1 and the procedure for horizontal flow under“Determining the Flow Regime.”Average vapor density in the 2 inch (50 mm) line (from Problem 1)

�

_

G = 0.257 lbm/ft3 (4.12 kg/m3)

Weight fraction vapor in the mixture

x = WW W

G

L G�

x = WW W

G

L G�

= 0 007885 0 0 00788

.. .�

= 9 93 100 63 9 93 10

4

4.

. .x

x

�

��

= 0.001574 = 0.001574

Parameter X [Eq. (1)]

X = 530 7 1 1 2 1 3

1 6./ /

/��

��

�

��

��

�

��

xx

G L

L

�X = 2100 1 1 2 1 3

1 6��

��

�

��

��

�

��

xx

G L

L

�

�

/ /

/

= ( . ) ..

( . ) ( . )( . ) ( . )

/ /

/530 7 1 0 0015740 001574

0 257 0 89462 2 72 0

1 2 1 3

1 6��

��

�

��

��

�

��

= 2100 1 0 0015740 001574

412 0 894 10966 72 0

1 2 3 1 3

1 6��

��

�

��

��

�

��

�..

( . ) ( . )( ) ( . )

/ /

/x

= 1147 = 1154

Parameter Y [Eq. (2)]

Y =3962 1 2 1 2

x Wd G L� �

/ / = 3 24 107

2 1 2 1 2.

/ /x x W

d G L� �

= ( ) ( . ) ( . )( . ) ( . ) ( . )/ /396 0 001574 5 007882 067 0 257 62 22 1 2 1 2 = ( . ) ( . ) ( . )

( . ) ( . ) ( )/ /3 24 10 0 001574 0 631

52 5 412 966

7

2 1 2 1 2x

= 0.183 = 0.182





From Figure 1 and using the above values of X and Y, one finds that the flow regime is plug flow, not far from the boundary withstratified flow. However, since the line is slightly inclined at a positive slope of 5 , which increases the tendency toward plugflow, one would conclude that the flow regime is plug flow.

PROBLEM 3 - FLOW PATTERN IN VERTICAL RISERGiven: A mixture of oil and gas is transported through a 11.81 inch (300 mm) I.D. vertical riser. The flow rates up the riser are:

Oil: 570.3 klbm/h (71.9 kg/s)Gas: 87.8 klbm/h (11.1 kg/s)

Fluid properties at pipeline conditions are as follows (assume constant at all positions):Density: �G = 3.13 lbm/ft3 (50.18 kg/m3)

�L = 40.63 lbm/ft3 (651.3 kg/m3)Viscosity: �G = 0.015 cP (0.15 x 10-4 Pa�s)

�L = 0.5 cP (0. x 10-3 Pa�s)Surface tension: � = 25 dynes/cm (25 mN/m)

Find: Flow pattern in vertical riser.Solution:


Parameter � [Eq. (4)]

� = 1041 4

./�

�L�

��

�� = 0 52

1 4.

/

�L�

�

��

= 104 40 6325

1 4. . /

�

��

�

�� = 0 52 6513

25

1 4. . /

� �

��

= 1.17 = 1.17

Mass fraction of gas

x = WW W

G

G L�

x = WW W

G

G L�

= 87 887 8 570 3

.. .�

= 111111 719

.. .�

= 0.133 = 0.134

Liquid superficial velocity [Eq. (5)]

Vsl = 50 931

2.( )� x W

dL�Vsl = 1273 10

162.

( )x

x WdL

�

�

= ( . ) ( . ) ( . . )( . ) ( . )

50 93 1 0133 87 8 570 340 63 1181 2� � = ( . ) ( . ) ( . . )

( . ) ( )1273 10 1 0134 719 111

6513 300

6

2x � �

= 5.13 ft/s = 1.56 m/s

Calculating the ordinate of Figure 2�Vsl = (1.17) (5.13) �Vsl = (1.17) (1.56)

= 6.0 ft/s = 1.83 m/s


Parameter � [Eq. (6)]






� = � �2 451 4

1 3./

/�

��L

G��

�� = � �0 486

1 41 3.

//�

��L

G��

��

= � �2 45 40 6325

3131 4

1 3. . ./

/�

��

�

�� = � �0 486 6513

255018

1 41 3. . .

//�

��

�

��

= 4.04 = 4.05

Gas superficial velocity [Eq. (7)]

Vsg = 50 93 2.x W

dG�Vsg = 1273 106

2. xx W

dG�

= ( . ) ( . ) ( . . )( . ) ( . )

50 93 0133 87 8 570 3313 1181 2

� = ( . ) ( . ) ( . . )( . ) ( )

1273 10 0134 719 1115018 300

6

2x �

= 10.21 ft/s = 3.11 m/s

Calculating the abscissa of Figure 2�Vsg = (4.04) (10.21) �Vsg = (4.05) (3.11)

= 41.2 ft/s = 12.6 m/s

From Figure 2 the flow pattern is identified as Annular/MIST.

PROBLEM 4 - HOLDUP AND PRESSURE DROP IN VERTICAL RISERGiven: Air and water flow up a vertical riser of 1 inch (25.4 mm) I.D. at the following rates:

Water: 7.92 klbm/h (1.0 kg/s)Air: 0.09 klbm/h (1.135 x 10-2 kg/s)

The pipe is 50 ft (15.24 m) long and the inlet pressure is 100 psia (689.5 kPa abs.). The mixture temperature is 80 F (300 K).Under these conditions, the inlet air density is calculated to be 0.5 lbm/ft3 (8.02 kg/m3). The water density is 62.5 lbm/ft3(1002 kg/m3). Other properties are:

Air viscosity: 0.02 cP (2 x 10-5 Pa�s)Water viscosity: 1.0 cP (1.0 x 10-3 Pa�s)Surface tension: 72 dynes/cm (72 mN/m)

Find: Liquid holdup in the pipe and the total pressure drop.Solution: Assume outlet pressure = 60 psia (413.7 kPa abs).


Outlet air density

�G2 = �G PP1 21

�G2 = �G PP1 21

= ( . ) ( )0 5 60100

= ( . ) ( . ).

8 02 413 7689 5

= 0.3 lbm/ft3 = 4.81 kg/m3






Average air density

�

_

G = � �G G1 22�

�

_

G = � �G G1 22�

= 0 5 0 32

. .� = 8 02 4 812

. .�

= 0.4 lbm/ft3 = 6.42 kg/m3

Cross-section area of pipe

A = �d2

41

144A = �d2

641

10

= ( . ) ( )( ) ( )

31416 14 144

2= ( . ) ( . )

( ) ( )31416 25 4

4 10

2

6x

= 0.005454 ft2 = 5.06 x 10-4 m2

Total mass velocity

Gt =( . . )

.

7 92 0 09 10003600

0 005454 2

��

��

�

��

��

�

��

klbh

lbklb

hs

ft

m m

m Gt = ( . . ) /.

10 0 011355 06 4 2�

�

kg sx m

= 408 lbm/ft2-s = 1998.7 kg/m2-s

Volumetric flow rate of liquid

QL =7 92 1000

360062 5 3

.

. /

klbh

lbklb

hs

lb ft

m m

m

m

�

��

�

��

��

�

��

QL = 101002 3

. //

kg skg m

= 0.0352 ft3/s = 0.001 m3/s

Volumetric flow rate of gas

QG =0 09 1000

36000 4 3

.

. /

klbh

lbklb

hs

lb ft

m m

m

m

�

��

�

��

��

�

��

QG = 0 01356 42 3. /. /

kg skg m

= 0.0625 ft3/s = 0.00177 m3/s

Volume fraction liquid [Eq. (11)]

� = QQ Q

L

L G�

� = QQ Q

L

L G�

= 0 03520 0352 0 0625

.. .�

= 0 0010 001 0 00177

.. .�

= 0.36 = 0.36







Two-phase viscosity [Eq. 21)]�2� = �L� + �G (1 - �) �2� = �L� + �G (1 - �)

= (1) (0.36) + (0.02) (1 - 0.36) = (1.0 x 10-3) 0.36 + (2 x 10-5) (1 - 0.36)

= 0.373 cP = 0.373 x 10-3 Pa�s

Average mixture [Eq. (13)]

V2� =184

2( )Q Q

dL G�

V2� = 1273 106

2. ( )x Q Q

dL G�

=( ) ( . . )184 0 0352 0 0625

12�

= 1273 10 0 001 0 0017725 4

6

2. ( . . )

( . )x �

= 17.98 ft/s = 5.47 m/s

Determining two-phase Reynolds Number:

1. Parameter � [Eq. (28)]

� = 6 28 1040 9 0 19 0 205 0 7 2 75

0 435 0 72.. . . . .

. .x WW G

L

G

L G G

t L

�

��

�

��

� �

(Customary)

Since Gt > 50 lbm/ft2-s, set Gt = 50 lbm/ft2-s

� = 6 28 10 7 920 09

1 72 0 4 0 0250 62 5

40 9 0 19 0 205 0 7 2 75

0 435 0 72. ..

( ) ( ) ( . ) ( . )( ) ( . )

. . . . .

. .x �

��

�

��

= 0.88

� = 8 75 10130 9 0 19 0 205 0 7 2 75

0 435 0 72.. . . . .

. .x WW G

L

G

L G G

t L

�

��

�

��

� �

(Metric)

Since Gt > 244.35 kg/m2-s, set Gt = 244.35 kg/m2-s

� = 8 75 10 100 0135

10 72 8 42 2 10244 35 1002

130 9 3 0 19 0 205 0 7 5 2 75

0 435 0 72. ..

( ) ( ) ( . ) ( )( . ) ( )

. . . . .

. .x x�

��

�

��

� �

= 0.88

2. From Figure 6, RL = 0.52 at � = 0.88.

3. Two-phase density [Eq. (23)]

�2� = � � � �L

L

G

LR R

2 211

��

�

( )�2� = � � � �L

L

G

LR R

2 211

��

�

( )

= ( . ) ( . ).

( . ) ( . ).

62 5 0 360 52

0 4 1 0 361 0 52

2 2�

�

�

= ( ) ( . ).

( . ) ( . ).

1002 0 360 52

6 42 1 0 361 0 52

2 2�

�

�

= 15.92 lbm/ft3 = 255.21 kg/m3






4. Reynolds Number [Eq. (22)]

Re2� =124 2 2

2

d V � �

�

�Re2� =

10 32 2

2

� d V � �

�

�

�

= ( ) ( ) ( . ) ( . ).

124 1 17 98 15 920 373

= ( ) ( . ) ( . ) ( . ).

10 25 4 5 47 255 210 373 10

3

3

�

�x

= 95,158, call 95,000 = 95,062, call 95,000

Single-phase Fanning friction factor [Eq. (24)]

f1� = 0 0014 0125

20 32. .

(Re ) .�

�

= 0 0014 012595000 0 32. .

( ) .�

= 0.0046

At � = 0.36, f2�/f1� = 1.9 from Figure 5.Two-phase friction factor

f2� = (1.9) f2� = (1.9) (0.0046) = 0.00874

Frictional pressure drop [Eq. (25)]

(P)f =f V L

g dc

2 22

2

6� � ��

(P)f =2 2 2

22f V L

d� � ��

= ( . ) ( . ) ( . ) ( )( ) ( . ) ( )

0 00847 17 98 15 92 506 32174 1

2= ( ) ( . ) ( . ) ( . ) ( . )

.2 0 00847 5 47 255 21 15 2

25 4

2

= 11.65 psi = 80.1 kPa

Pressure drop due to gravity [Eq. (29)]

(P)e = � �� L L G LR R L� �( )1144

(P)e = 9.8 x 10-3 [�L RL + �G (1 - RL)] L

= � �( . ) ( . ) ( . ) ( . )62 5 0 52 0 4 1 0 52 50144� �

= 9.8 x 10-3 [(1002)(0.52) + (6.42) (1 - 0.52)](15.24)

= 11.35 psi = 78.28 kPa

Pressure drop due to acceleration:1. Upstream properties

�G = 0.5 lbm/ft3 �G = 8.02 kg/m3

QG =0 09 1000

36000 5 3

.

. /

klbh

lbklb

hs

lb ft

m m

m

m

�

��

�

��

��

�

��

��

�

��

QG = 0 011358 02 3. /. /

kg skg m

= 0.05 ft3/s = 0.00142 m3/s







� = QQ Q

L

L G�

� = QQ Q

L

L G�

= 0 03520 0352 0 05

.. .�

= 0 0010 001 0 00142

.. .�

= 0.41 = 0.41

� = 1.03 (from [Eq. (28)]

RL = 0.54 (from Figure 6)

2. Downstream properties (assume ideal gas, isothermal, downstream pressure 60 psia)�G = 0.3 lbm/ft3 �G = 4.81 kg/m3

QG =� � � �0 09 1000 1

36000 3

.

.

�

��

�

��

QG = 0 011354 81

..

= 0.83 ft3/s = 0.00236 m3/s

� = 0 03520 0352 0 083

.. .�

� = 0 0010 001 0 00236

.. .�

= 0.30 = 0.30

� = 0.72

RL = 0.48

3. Pressure drop(P)a in Customary Units

(P)a = 7 25 11

114

2 2

24

2 2

1

.d

QR

QR d

QR

QR

G G

L

G G

L

G G

L

G G

L

��

��

��

�

��

�

��

��

�

��

��

��

��

�

��

�

��

��

�

��

��

��

��

��

� � � �

= 7 25 11

0 3 0 0831 0 48

62 5 0 03520 48

11

0 5 0 051 0 54

62 5 0 03520 544

2 2

4

2 2.

( )( . ) ( . )

.( . ) ( . )

. ( )( . ) ( . )

.( . ) ( . )

.�

��

�

��

��

��

�

��

��

�

��

��

��

�

��

��

��

�

��

��

�

��

�

��

��

�

��

��

= 0.138 psi

(P)a in Metric Units

(P)a = 162 10 11

11

94

2 2

24

2 2

1

. xd

QR

QR d

QR

QR

G G

L

G G

L

G G

L

G G

L

��

��

��

�

��

�

��

��

�

��

��

��

��

�

��

�

��

��

�

��

��

��

��

��

� � � �

=

162 1025 4

4 81 0 002361 0 48

1002 0 0010 48

162 1025 4

8 02 0 001421 0 54

1002 0 0010 54

9

4

2 2 9

4

2 2.( . )

( . ) ( . ).

( ) ( . ).

.( . )

( . ) ( . ).

( ) ( . ).

x x�

��

�

��

�

��

�

��

�

�

��

�

�

��

��

�

��

�

�

��

�

�

��

�

��

�

��

�

�

��

�

�

��

= 0.96 kPa


(P)a is small compared to (P)f + (P)e. As a result, the total pressure drop is:





(P)t = (P)f + (P)e (P)t = (P)f + (P)e

= 11.65 + 11.35 = 80.1 + 78.28

= 23 psi = 158.4 kPa

Outlet pressure P2

P2 = P1 - (P)t P2 = P1 - (P)t

= 100 - 23 = 689.5 - 158.4

= 77 psi = 531.1 kPa

The calculated outlet pressure is substantially larger than the assumed pressure of 60 psi (413.7 kPa). The procedure isrepeated and now the assumed outlet pressure is 77 psi (531.1 kPa). Without going into details, the results of the seconditeration are given below:

RL = 0.53

(P)f = 11.02 psi (76 kPa)

(P)e = 11.58 psi (79.84 kPa)

(P)a can again be neglected

(P)t = 22.6 psi (155.84 kPa)

Outlet pressureP2 = P1 - (P)t P2 = P1 - (P)t

= 100 - 22.6 = 689.5 - 155.84

= 77.4 psi = 533.7 kPa

The calculated outlet pressure is very close to the assumed pressure of 77 psi (531.1 kPa)Final Solution:

Holdup in pipe = 0.53Total pressure drop = 22.6 psi (155.84 kPa)

PROBLEM 5 - CRITICAL MASS FLUXGiven: The vent of a vessel containing a steam/water mixture is opened. The vessel is at 514.7 psia (3548.9 kPa) and 470 F(516.7 K) and the mass fraction of steam is 0.5.Find: Initial critical mass flux.Solution: At the stagnation condition, 514.7 psia (3548.9 kPa) and 470 F (516.7 K).

�LO = 0.0198 ft3/lbm (0.00124 m3/kg)�� from the steam tables

�GO = 0.9006 ft3/lbm (0.0562 m3/kg) �

�LGO = �GO - �LO = 0.9006 - 0.0198 = 0.8808 ft3/lbm (Customary)

= 0.0562 - 0.00124 = 0.055 m3/kg (Metric)







Two-phase specific volume�2�O = �LO + xo�LGO �2�O = �LO + xo�LGO

= 0.0198 + (0.5) (0.8808) = 0.00124 + (0.5) (0.0549)

= 0.46 ft3/lbm = 0.0287 m3/kg

Also from the steam tableshGO = 1204.7 Btu/lbm (2802.1 kJ/kg)

hLO = 452.84 Btu/lbm (1053.2 kJ/kg)

hLGO = hGO - hLO = 1204.7 - 452.8 = 751.9 Btu/lbm (Customary)

= 2802.1 - 1053.2 = 1748.9 kJ/kg (Metric)

CLO = 1.1 Btu/lbm- F (4.6 kJ/kg-K) (from International Critical Tables)

Correlation parameter [Eq. (42)]Customary Units:

� =2

0202185.0 �

��

��

��

LGO

LGOOOLOLGOOh

PTCx �

��

�

��

= ( . ) ( . ).

( . ) ( . ) ( ) ..

..

0 5 0 88080 46

0185 11 470 460 514 70 46

0 88027519

2�

� �

��

�

��

= 1.25

Metric Units:

� =2

0202��

��

��

LGO

LGOOOLOLGOOh

PTCx �

��

�

��

= ( . ) ( . ).

( . ) ( . ) ( . ).

..

0 5 0 05490 0287

4 6 516 7 3548 90 0287

0 05491748 9

2�

��

��

= 1.25

From Figure 8 GC* = 0.6

Critical mass flux [Eq. (44)]

GC = 68 032

. *G PC

O

O� �

GC = 31622

. *G PC

O

O� �

= ( . ) ( . ) ..

68 03 0 6 514 70 46

= ( . ) ( . ) ..

3162 0 6 3548 90 0287

= 1365 lbm/ft2-s = 6671 kg/m2-s





NOMENCLATURE

C = Flow coefficient for orifices, nozzles and venturis, dimensionlessCLO = Heat capacity of liquid at stagnation condition, Btu/lbm- F (kJ/kg-K)d = Inside pipe diameter, in. (mm)EH = Elevation head factor, dimensionlessf = Fanning friction factor, dimensionlessG = Mass velocity, lbm/ft2-s (kg/m2-s)

GC* = Normalized critical mass flux defined by Eq. (44), dimensionless

gc = Gravitational conversion factor 32.174 ft-lbm/lbfs2

h = Enthalpy, Btu/lbm (kJ/kg)H = Summation of vertical uplegs, ft (m)J = Acceleration group [Eq. (18)], dimensionlessK = Resistance coefficient, dimensionlessL = Length of pipe, ft (m); actual length of pipe plus equivalent lengths of fittings, ft (m)P = Pressure, psi (kPa)

p_

= Average pressure = 0.5 (P1 + P2), psi (kPa)

P = Pressure drop, psi (kPa)Q = Volumetric flow rate, ft3/s (m3/s)RL = Liquid holdup, dimensionlessRe = Reynolds Number, dimensionlessV = Linear fluid velocity, averaged overflow cross-section, ft/s (m/s)� = Specific volume of fluid, ft3/lbm (m3/kg)W = Mass flow rate, klbm/h (kg/s)x = Mass fraction gas or vapor in mixture with liquid also called quality, dimensionlessX = Parameter in Eq. (1) and Abscissa of Figure 1Y = Parameter in Eq. (2) and Ordinate of Figure 1� = Critical pressure ratio defined by Eq. (43), dimensionless� = Volume fraction liquid in mixture with gas or vapor, dimensionless� = Viscosity, cP (Pa�s)� = Density of fluid, lbm/ft3 (kg/m3)

�

_= Average fluid density, lbm/ft3 (kg/m3)

� = Surface tension of liquid, dynes/cm (mN/m)� = Parameter in Eq. (6)� = Parameter in Eq. (4)� = Parameter in Eq. (28)� = Parameter in Eq. (42)






Subscripts (unless indicated otherwise)A = Aira = Due to accelerationc = Criticale = Due to elevationeq = Equivalentf = Due to frictionG = Gas or vaporK = KineticL = LiquidLG = Difference between vapor and liquid propertiesns = No velocity slipO = Stagnation conditionsg = Superficial gassl = Superficial liquidt = TotalWA = Water-air system1 = Upstream conditions or location2 = Downstream conditions or location1� = Single-phase2� = Two-phase





Figure 1Two-Phase Flow Regimes In Horizontal Pipe

Bubble or Froth FLow

Plug Flow

Slug Flow

Spray Flow

Annular Flow

Stratified Flow

Wavy Flow

F 2 x

W

d2 (�

G)1/

2 (�

L)1/

2Y

=

100

10

1

0.1

2

3456789

2

3456789

2

3456789

12 3 4 5 6 7 8911

2 3 4 5 6 7 89 2 3 4 5 6 7 89 2 3 4 5 6 7 89 2 3 4 5 6 7 890.1 10410

1 1102

1103

1

* From Reference (1). Meaning of Symbols differ from those in the original article. See Nomenclature. DP14DF1

F1 = 530.7 (Customary) = 2100 (Metric)

F 2 = 3

96

(C

usto

mar

y)

= 3

.24

x 10

7 (

Met

ric)

Where:��

��

��

�

� �

��

��

6/1

3/12/1

11

L

LG

xxFX






Figure 2ATwo-Phase Flow Regimes In Vertical Pipe (Customary)






10

1

0.1

0.010.1 1.0 10 100 500

�Vsg , ft/s

Annular-MistFrothSlugBubble

�V sl

, ft/

s





Figure 2BTwo-Phase Flow Regimes In Vertical Pipe (Metric)

3

1.0

0.1

0.01

0.003

�V sl

, m

/s

0.03 0.1 1 10

�Vsg , m/s

100 300

Annular-MistFrothSlugBubble

DP14Df2b






Figure 3AElevation Head Factor For Two-Phase Pressure Drop (Customary)

1.0

.9

.7

.8

.6

.5

.4

.3

0.2

0.1

9

8

7

6

5

4

3

0.022010 30 40 50 60 70 80

Superficial Vapor Velocity, Vsg , ft/s

Elev

atio

n H

ead

Fact

or, E

H

0 2 4 6 8 10

Superficial Vapor Velocity, ft/s

1.0

.9

.7

.8

.6

.5

.4

.3

.2

DP14Df3a





Figure 3BElevation Head Factor For Two-Phase Pressure Drop (Metric)

1.0

.9

.7

.8

.6

.5

.4

.3

.2

0.1

9

8

7

6

5

4

3

0.022 4 10 14 16 20 24

Superficial Vapor Velocity, m/s

Elev

atio

n H

ead

Fact

or, E

H

Superficial Vapor Velocity, m/s

6 8 12 18 22

0 1 2 3

DP14Df3b






Figure 4Liquid Holdup Correlation For Horizontal Pipe

0.1

1

0.0110-3 10-2 0.1 1

Volume Fraction Liquid, �

Frac

tion

Liqu

id H

oldu

p, R

L

Re2� x 10-3

0.1

0.5

1

2.5

5

10

25

50

100

200

DP14Df4





Figure 5Friction Factors In Two-Phase Flow

3.0

2.5

2.0

1.5

1.0

0.510-4 10-3 10-2 0.1 1

Volume Fraction Liquid at Inlet, �

f2�f1�

DP14Df05






Figure 6Liquid Holdup For Vertical Upflow Pipe

1.0

0.8

0.6

0.4

0.2

00.001 0.01 0.1 1 10 100

�

RL =

Liq

uid

Hol

dup

DP14Df06





Figure 7Manifold Designs For Distributing Two-Phase Flow

(b)

(a)

(c) DP14Df07






Figure 8Critical Flow Of Vapor - Liquid Mixtures

1.0

.5

01 10 100

�

�

G*c

Normalized Critical Mass Flux orCritical Pressure Ratio

DP14Df08

flujo fluido exxon

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