Chapter 3 Flow Analysis

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Chapter 3Flow Analysis3.1IntroductionPipes and ducts are the veins and arteries of mechanical systems such as a powerplants, refineries, or HVAC systems. Without them, these systems could not exist.As in our own bodies, where the veins and arteries move blood through the pumpingaction of the heart, power plants require the circulation of a “working” fluid in orderto provide its functionality. In this chapter we will examine the basics of flow analysisand piping system design and develop the framework for the analysis of integratedmechanical systems such as powerplants, refineries, and airflow systems, to name justa few.3.2Flow in Pipes and ChannelsAnalysis of flow in piping systems always begins with some form of the mechanicalenergy equation. Initially, we may consider the Bernoulli equation:P1 V12P2 V22 z1 z2γ2gγ2g(3.1)where γ ρg.However, our intuition will tell us that losses throughout the system will limit application of Eq. (3.1). We may however use Eq. (3.1) to determine system behaviourunder ideal comditions, and use these results as a target for maximum performace. Inorder to apply Eq. (3.1) to a piping system, we must extend the Bernoulli equationto account for losses which result from pipe fittings, valves, and direct losses (friction)within the pipes themselves. The extended Bernoulli equation may be written as:XP1 V12P2 V22 z1 z2 hlosses(3.2)γ2gγ2gAdditionally, at various points along the piping system we may need to add energyto provide an adequate flow. This is generally achieved through the use of some sort33

34Mechanical Equipment and Systemsof prime mover, such as a pump. For a system containing a pump or pumps, we mustinclude an additional term to account for the energy supplied to the flowing stream.This yields the following form of the energy equation:X ẆpumpsXP1 V12P2 V22 z1 z2 hlossesγ2gṁgγ2g(3.3)Finally, if somewhere in the piping system a component extracts energy from thefluid stream, such as a turbine, the energy equation takes the form:X ẆpumpsX Ẇturbines XP1 V12P2 V22 z1 z2 hlossesγ2gṁgγ2gṁg(3.4)This statement of the mechanical energy equation from fluid dynamics says thatthe initial energy at point 1 plus any energy supplied by a pump or pumps mustbalance with the energy at point 2 plus any losses incurred due to fluid friction, pipefittings, valves etc., plus any energy output through mechanical conversion. In a fluidpiping system, whether we are dealing with a single pipe or a network of pipes, Eqs.(3.1-3.4) will be at the center of analysis. Note, that Eq. (3.4) is written to reflectthat all work terms have a positive sign convention. In the classic formulation of theenergy equation work done on a system and work done by a system have differentsigns. By isolating each component of shaft work separately we have included the signconvention by appropriately placing the work terms on the left side for work done ona system and the right side for work done by the system. This is best summarizedby Fig. 3.1.3.3Losses in Piping SystemsLosses in a piping system are typically catagorized as major and minor losses. Minorlosses in piping systems are generally characterized as any losses which are due topipe inlets and outlets, fittings and bends, valves, expansions and contractions, filtersand screens, etc. Essentially, everything within the system which is not a sectionof pipe or other major component. In fact, in many flow systems the minor lossescan account for more head loss, than the pipes themselves. The primary distinctioncomes in the manner in which major and minor losses are calculated.In the mechanical energy equation, head losses are computed from the followingexpression:X 4fi Li u2 X 4fi Li ρu2 hL iDi 2gor in terms of pressure loss, pL iDi2XXKiu2i X hcomp2g(3.5)Kiρu2i X pcomp2(3.6)

35Flow AnalysisCare must be taken, that the appropriate mean flow velocity is used in each termfor each individual length of pipe and each minor loss. The third group representsmajor losses due to components within a system that the fluid must flow through.Fig. 3.1 - Conservation of Mechanical Energy3.3.1Minor LossesMinor losses in systems are most often calculated using the concept of a loss coefficientor equivelent length method. In the loss coefficient method, a constant or variablefactor K is defined as:K Vhf2 /2g P1ρV 22(3.7)The associated head loss is related to the loss coefficient throughV2hf K2g(3.8)Values of K have been determined experimentally for many configurations of bends,fittings, valves etc. A comprehensive source of this loss coefficient data is found inthe excellent reference Handbook of Hydraulic Resistance by Idelchik (1986). Thisbook is now considered the leading source of design data for hydraulic systems. Someuseful expressions for the computing minor losses in pipe systems are supplied in thehandouts.Two useful expressions which are used often in flow modelling are the sudden contraction loss and teh sudden expansion loss. These may be predicted using the simplerelationships:

36Mechanical Equipment and SystemsKSC 0.42(1 σ 2 )2(3.9)KSE (1 σ)2(3.10)A1 1A2(3.11)andwhere0 σ Here A1 and A2 denote the smaller and larger duct areas, respectively.3.3.2Major LossesMajor losses of head in a piping system are the direct result of fluid friction in pipesand ducting. The resulting head losses are usually computed through the use offriction factors. Friction factors for ducts have been compiled for both laminar andturbulent flows. Two widely adopted definitions of the friction factor are the Fanningfriction factor defined as:fF τ1ρV 22(3.12)and the Darcy friction factor defined as:dpDfD 1 dx 2ρV2 (3.13)Now for a circular pipe, the shear stress τ , is related to the pressure gradient dp/dx,by means of the following relation derived from a force balance on an element of fluidDh dpA dp (3.14)P dx4 dxThis leads to the definition of the hydraulic diameter and results in the followingrelationship between the two definitions:τ fD 4fF(3.15)In this course we will use the Fanning friction factor definition. Therefore, the headloss characteristic of a circular pipe may be written as:hf 4fFL V2Dh 2g(3.16)In future we will drop the subscript F and assume explicitly the definition of theFanning friction factor unless otherwise indicated. Note that Eqs. (3.5) and (3.6) arewritten in terms of the Fanning friction factor.

37Flow AnalysisFor a circular duct three different friction factor models are of interest. These are:fully developed laminar flow, developing laminar flow, and fully developed turbulentflow. The following design correlations for the friction factor are of interest for circularpipes.Fully Developed Laminar Flow, ReDh 2300In the fully developed laminar flow region in a circular tube, the Fanning frictionfactor which is oten written in terms of the friction factor-Reynolds number productisf ReDh 16orf (3.17)16ReDh(3.18)As we shall see later, for other non-circular ducts the friction factor is obtainedfromf CReDh(3.19)where the constant C is often tabulated for a specific duct shape.Developing Laminar Flow, ReDh 2300 and L LeIn the laminar entrance region, the pressure drop is much larger owing to thedevelopment of the hydrodynamic boundary layers, see Fig. 3.2. If a tube is short ornot much longer than the region where boundary layers develop an alternate approachis required to predict the pressure drop. The entrance length is defined such thatLe 0.05Dh ReDh(3.20)L Le Short DuctL Le Long Duct(3.21)whereA simple model developed by Muzychka (1999) for a duct of any length L is:fapp ReDh "µ3.44 L ¶2 (f ReDh )2#1/2(3.22)where the apparent friction factor fapp combines the effects of the wall shear stressand increase in pressure drop due to the accelerating core in the entrance region. Thedimensionless parameters L and ReDh are defined as:L LDh ReDhReDh ρV Dhµ(3.23)

38Mechanical Equipment and SystemsThe remaining parameter f ReDh is the fully developed friction factor-Reynoldsnumber product defined earlier. The pressure drop in terms of the mass flow rate andthe f Re group is obtained from the following equation: p 2(f ReDh )ṁνLDh2 Ac(3.24)Fig. 3.2 - Laminar Development of Velocity in a Circular Pipe,(From Fluid Mechanics, F.M. White, 2000).Fully Developed Turbulent Flow, ReDh 4000Fully developed turbulent flow in rough ducts may be characterized by the Colebrook expression which is the basis for the turbulent portion of the Moody diagram:µ¶12.51ǫ/Dh 2 log(3.25)3.7fDReDh fDThis expression, although highly accurate, is not very amenable to design due toits implicit nature. An alternate form proposed by Swamee and Jain (see Benedict,1980) which provides accuracy within 1.5 percent is given byf 1"16 logÃǫ/Dh5.74 9/103.7ReDh!#2(3.26)

39Flow AnalysisA correlation of the Moody diagram was developed by Churchill (1977). It spansthe entire range of laminar, transition, and turbulent flow in pipes. It consists of thefollowing expressions:f 2"µ8ReDh¶12 µ1(A B)3/2¶#1/12(3.27)where½·A 2.457 ln10.9(7/ReDh ) (0.27ǫ/Dh ) ¾16(3.28)andB 0.0000.000 0.200 0.400 0.600 0.8000.0000.000 0.200 0.400 0.600 0.800µ37530ReDhN 3N 4N 5TriangleSquarePentagon0Semi-CircleRectanglea b(3.29)N 6Hexagonyb¶16N Circleyybax0a0xEllipsea b2baxHyper-Ellipsea b2 x y 1 a b nn x y a b 10 n Fig. 3.3 - Typical Non-Circular Cross-sectionsLaminar Flow in Non-Circular DuctsIn many engineering systems, flow in a duct which is not circular in cross-sectionis quite common. The definitions above are still applicable through the definitionof the hydraulic diameter. The hydraulic diameter arises from Eq. (3.14). For anon-circular duct or channel the hydraulic diameter is defined as

40Mechanical Equipment and Systems4Ac(3.30)Pwhere Ac is the cross-sectional area and P is the wetted perimeter of the non-circularduct.Dh Table 1Typical values of f ReDh forNon-Circular DuctsShapef ReDhEquilateral Triangle .41Circle16Elliptic 2:116.82Elliptic 4:118.24Elliptic 8:119.15Rectangular 2:115.55Rectangular 4:118.23Rectangular 8:120.58Parallel Plates24Equation (3.30) has been adopted by virtually all hydraulics engineers. As we shallsee, the concept of the hydraulic diameter is quite useful in applications where the flowis turbulent, since the turbulent friction factor is only weakly dependent upon ductshape. This is not the case when the flow is laminar. Laminar flow friction factors area strong function of duct shape. As a result, we must be careful when approximatingfriction factors for non-circular ducts. Table 1 gives some typical values of the f ReDhproduct for some common ducts. It is evident that the hydraulic diameter conceptdoes not allow for a single value of the friction factor to be used for non-circular ducts.For developing laminar flows Eq. (3.22) is still applicable since the term whichaccounts for the entrance region pressure drop, is independent of duct shape. On theother hand as we have seen the second term which accounts for the fully developedflow component, is a strong function of duct shape. Therefore, one merely has tosubstitute the appropriate value of the f ReDh group for the duct of interest. Equation(3.22) provides accuracy within 5 percent. All of the results of Table 1 can easily bepredicted from the following expression which is derived from the analytical solutionfor the rectangular ductµ ¶12β 3/24 Ac· f ReDh (3.31)³ 192πP(1 β) 1 5 tanhβπ β2

Flow Analysis41where 1 β is the duct aspect ratio, A is the cross-sectional area, and P isthe perimeter of the non-circular duct. Note, that β 100 is a fair approximation tothe parallel plate channel. This expression may be used to predict the f ReDh valuefor many other non-circular cross-sections within 10 percent, provided that thesmallest re-entrant corner angle is greater than 15 degrees. The aspect ratio β formost non-circular ducts is simply a measure of the slenderness of the duct.Flow Through Porous MediaIn many engineering systems, filtration beds or packed columns are used in variousprocesses. These components consist of a channel or duct which contains some formof porous material or a collection of randomly packed spheres or other non-sphericalparticle. A simple model for predicting pressure drop through packed columns wasdeveloped by Ergun (1952). The model which is now commonly referred to as theErgun equation is a composite solution containing two asymptotic results. One is forslow viscous flow and the other for highly turbulent flow. The resulting model takesthe following form:¶µ( p/L) Dp75 (1 ǫ)2 1.75 1 ǫf 1 2(3.32) 334Reǫ2ǫρVDpo2where Dp is the diameter of the spherical or equivalent particle, ǫ Vf ree /Vtotal isthe porosity of the bed or column which is defined as the volume of voids divided bythe volume of the bed or column, and Vo ṁ/(ρAc ) is the superficial velocity whichwould result in the bed or column if no packing were present. The Ergun equationpredicts the pressure drop (or flow) through porous media or packed columns quitewell. It is widely used in the chemical process industries for modelling filtration beds,columns containing catalyst pellets, and percolation systems. Additional informationon the flow of fluid through porous media may be found in Churchill (1987), Bear(1972), and McCabe et al. (1985).Example 3.1Examine the airflow through a heat sink which is composed of a number of channelsformed by a series of plates. The system parameters are: fin height H 118 mm, finthickness t 1.25 mm, fin spacing b 2.5 mm, and fin length in direction of flowL 30 cm. The air properties may be taken to be ρ 1.2 kg/m3 , µ 1.8 10 5P a · s. Calculate the pressure drop as a function of approach velocity V . Includeminor losses at the entrance and at the exit. What is the pressure drop when V 2m/s?Example 3.2A filtration component consists of a rectangular channel having dimensions H 10 cm, W 50 cm, and L 100 cm, containing carbon pellets of diameter 5mm. Develop the pressure drop versus mass flowrate characteristic for the followingporosities (due to packing arrangement): ǫ 0.5, 0.7, 0.9. The fluid properties maybe taken to be ρ 1000 kg/m3 , µ 1.0 10 3 P a · s. What is the pressure drop

42Mechanical Equipment and Systemswhen ṁ 50 kg/s? Is it better to operate three filters in series or three in parallel.Why?Example 3.3Examine the electronics packaging enclosure described below. Nine circuit boardsare placed in an enclosure with dimensions of W 50 cm, H 25 cm, and L 45cm in the flow direction. If the airflow required to adequately cool the circuit boardarray is 3 m/s over each board, determine the fan pressure required to overcome thelosses within the system. Assume each board has an effective thickness of 5 mm,which accounts for the effects of the circuit board and components. You may furtherassume that the roughness of the boards is 2.5 mm. The air exhausts to atmosphericpressure. What effect would adding a louvered grill to the back of the cabinet haveon the required fan pressure? How will your flow equations change? In your analysisinclude the effect of entrance and exit effects due the reduction in area. The densityof air at 20 C is ρ 1.2 kg/m3 and the viscosity is µ 1.81 10 5 P a · s.3.4Pipe NetworksMost engineering systems are comprised of more than one section of pipe. In factin most systems a complex network of piping is required to circulate the workingfluid of a particular thermal system. These networks consist of series, parallel, andseries-parallel configurations. We will examine each of these separately. However,before we proceed, a brief summary of the classification of piping system problemsis necessary. This classification determines the type of solution which is obtainablewhen any two of the following principal design parameters are specified: head loss or p, volumetric flowrate Q, and pipe diameter D.Pipe flow problems fall into three categories. In Category I problems the solutionvariable is the head loss or pressure drop p. The problem is specified such that thevolumetric flow Q, the length of pipe L, the size or diameter D, are all known alongwith other parameters such as the pipe roughness and fluid properties. These typesof problems yield a direct solution for the unknown variable p. In a Category IIproblem, the head loss (h or p) is specified and the volumetric flow Q is sought.Finally in a Category III problem, both the head loss and volumetric flow are specified,but the size or diameter of the pipe D is sought. Category I and Category II problemsare considered analysis problems since the system is specified and only the flow iscalculated. Whereas Category III problems are considered design problems, as theoperating characteristics are known, but the size of the pipe is to be determined. BothCategory II and Category III problems require an iterative approach in solution. TYPE I - p Q or ṁ, L, D TYPE II - Q or ṁ p, L, D TYPE III - L or D p, Q or ṁ, D or L

43Flow AnalysisDepending upon the nature of the flow (and solution process), it may be requiredto recompute other parameters such as the relative roughness at each iterative pass,since the ǫ/D ratio will change as the pipe diameter changes. However, with mostmodern computational software, we may solve “iterative” problems rather efficientlyand need not resort to classic methods such as Gaussian elimination.The simplest pipe networks are those that are either entirely composed of the seriestype of arrangement or the parallel type of arrangement as shown in Fig. 3.4. Wewill axamine these two arrangements first and then proceed to the more complexseries-parallel network.Fig. 3.4 - Series and Parallel Arrangements,(From Fluid Mechanics, F.M. White, 2000).3.4.1Pipes in SeriesIn a series arrangement of pipes, the volumetric flow at any point in the systemremains constant. Thus, for an arrangement of N pipes, the volumetric flow is givenbyQ1 Q2 Q3 · · · QN constant(3.33)V1 A1 V2 A2 V3 A3 · · · VN AN constant(3.34)orThe head loss in the system is the sum of the individual losses in each section ofpipe. That is pA B p1 p2 p3 · · · pN(3.35)

44Mechanical Equipment and SystemsExample 3.4Examine the series piping system consisting of three pipes each having a lengthL 1 m. The diameter of the first pipe is D1 0.02664 m, the diameter of thesecond pipe is D2 0.07792 m, and the diameter of the third pipe is D3 0.05252m. Assume that the roughness ǫ 0.0005 m for each pipe. The fluid properties maybe taken to be ρ 1000 kg/m3 , µ 1.0 10 3 P a · s. Develop the pressure dropversus mass flowrate characteristic for the system assuming at first no minor lossesand then include minor losses. What is the pressure drop when the mass flowrate isṁ 20 kg/s? How significant are the minor losses relative to the piping losses?3.4.2Pipes in ParallelIn a parallel arrangement the total head loss or pressure drop across the system isconstant. That is p1 p2 p3 · · · pN constant(3.36)On the other hand, the volumetric flow through the system is the sum total of theindividual flow in each pipe. That isQA B Q1 Q2 Q3 · · · QN(3.37)Example 3.5Examine the parallel piping system consisting of three pipes each having a lengthL 1 m. The diameter of the first pipe is D1 0.02664 m, the diameter of thesecond pipe is D2 0.07792 m, and the diameter of the third pipe is D3 0.05252m. Assume that the roughness ǫ 0.0005 m for each pipe. The fluid properties maybe taken to be ρ 1000 kg/m3 , µ 1.0 10 3 P a · s. Develop the pressure dropversus mass flowrate characteristic for each of pipes in the system and the pressuredrop versus total mass flowrate characteristic, assuming no minor losses. What is the

Chapter 3 Flow Analysis 3.1 Introduction . Initially, we may consider the Bernoulli equation: P1 . energy equation work done on a system and work done by a system have different signs. By isolating each component of shaft work separately we have included the sign

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