ASHRAE FUNDAMENTALS SI CH 21-2017 Duct Design.pdf

1、21.1CHAPTER 21DUCT DESIGNBERNOULLI EQUATION . 21.1Head and Pressure. 21.2SYSTEM ANALYSIS. 21.2Pressure Changes in System 21.5FLUID RESISTANCE 21.6Friction Losses. 21.6Dynamic Losses . 21.8Ductwork Sectional Losses 21.13FAN/SYSTEM INTERFACE. 21.13MECHANICAL EQUIPMENT ROOMS. 21.15DUCT DESIGN 21.15Desi

2、gn Considerations. 21.15Design Recommendations 21.21Design Methods 21.22Industrial Exhaust Systems 21.28OMMERCIAL, industrial, and residential air duct system de-Csign must consider (1) space availability, (2) noise levels, (3) airleakage, (4) balancing, (5) fire and smoke control, (6) initial inves

3、t-ment cost, and (7) system operating cost. Deficiencies in duct design can result in systems that operateincorrectly or are expensive (increased energy) to own and operate.Poor design or lack of system sealing can produce inadequate air-flow rates at the terminals, leading to discomfort, loss of pr

4、oductiv-ity, and even adverse health effects. Lack of sound attenuation maylead to objectionable noise levels. Proper duct insulation eliminatesexcessive heat gain or loss.In this chapter, system design and calculation of a systems fric-tional and dynamic resistance (total pressure) to airflow are c

5、onsid-ered. Chapter 19 of the 2016 ASHRAE HandbookHVAC Systemsand Equipment examines duct construction and presents construc-tion standards for residential, commercial, and industrial HVAC andexhaust systems. For design guidance specific to residential sys-tems, refer to Manual D by ACCA (2014). 1.

6、BERNOULLI EQUATIONThe Bernoulli equation can be developed by equating the forceson an element of a stream tube in a frictionless fluid flow to the rateof momentum change. On integrating this relationship for steadyflow, the following expression (Osborne 1966) results:+ gz = constant, Nm/kg (1)wherev

7、 = streamline (local) velocity, m/sp = absolute pressure, Pa (N/m2) = density, kg/m3g = acceleration caused by gravity, m/s2z = elevation, mAssuming constant fluid density in the system, Equation (1) re-duces to+ gz = constant, Nm/kg (2)Although Equation (2) was derived for steady, ideal frictionles

8、sflow along a stream tube, it can be extended to analyze flow throughducts in real systems. In terms of pressure, the relationship for fluidresistance between two sections is+ p1+ g1z1= + p2+ g2z2+ pt, 12(3)whereV = average duct velocity, m/spt,12= total pressure loss caused by friction and dynamic

9、losses between sections 1 and 2, PaIn Equation (3), V (section average velocity) replaces v (streamlinevelocity) because experimentally determined loss coefficients allowfor errors in calculatingv2/2 (velocity pressure) across streamlines.On the left side of Equation (3), add and subtract pz1; on th

10、e rightside, add and subtract pz2, where pz1and pz2are the values of atmo-spheric air at heights z1andz2. Thus,(4)Atmospheric pressure at any elevation ( pz1and pz2) expressed interms of the atmospheric pressure paat the same datum elevation isgiven bypz1= pa gaz1(5)pz2= pa gaz2(6)Substituting Equat

11、ions (5) and (6) into Equation (4) and simpli-fying yields the total pressure change between sections 1 and 2.Assume no temperature change between sections 1 and 2 (no heatexchanger within the section); therefore, 1=2. When a heatexchanger is located in the section, the average of the inlet andoutle

12、t temperatures is generally used. Let = 1= 2, and ( p1 pz1)and ( p2 pz2) are gage pressures at elevations z1and z2.pt,12= + g(a )(z2 z1) (7a)pt,12= pt+ pse(7b)Rearranging Equation (7b) yieldspt= pt,1-2 pse(7c)whereps,1= static pressure, gage at elevation z1, Paps,2= static pressure, gage at elevatio

13、n z2, PaV1= average velocity at section 1, m/sV2= average velocity at section 2, m/sa= density of ambient air, kg/m3 = density of air or gas in duct, kg/m3pse= thermal gravity effect, PaThe preparation of this chapter is assigned to TC 5.2, Duct Design.v22-p-+v22-p-+1V122-2V222-1V122- p1pz1pz1g1z1+

14、+2V222- p2+= pz2pz2g2z2pt 1-2,+ps 1,V122-+ps 2,V222-+21.2 2017 ASHRAE HandbookFundamentals (SI) pt= total pressure change between sections 1 and 2, Papt,1-2= total pressure loss caused by friction and dynamic losses between sections 1 and 2, Pa1.1 HEAD AND PRESSUREThe terms head and pressure are oft

15、en used interchangeably;however, head is the height of a fluid column supported by fluidflow, whereas pressure is the normal force per unit area. For liquids,it is convenient to measure head in terms of the flowing fluid. Witha gas or air, however, it is customary to measure pressure exerted bythe g

16、as on a column of liquid.Static PressureThe term p/g is static head; p is static pressure.Velocity PressureThe term V2/2g refers to velocity head, and V2/2gcrefers tovelocity pressure. Although velocity head is independent of fluiddensity, velocity pressure Equation (8) is not.pv= V2/2 (8)wherepv= v

17、elocity pressure, PaV = fluid mean velocity, m/sFor air at standard conditions (1.204 kg/m3), Equation (8) becomespv= 0.602 V2(9)Velocity is calculated byV = Q/1000A (10)whereQ = airflow rate, L/sA = cross-sectional area of duct, m2Total PressureTotal pressure is the sum of static pressure and veloc

18、ity pressure:pt= ps+ V2/2 (11)orpt= ps+ pv(12)wherept= total pressure, Paps= static pressure, PaPressure MeasurementThe range, precision, and limitations of instruments for measur-ing pressure and velocity are discussed in Chapter 36. The manom-eter is a simple and useful means for measuring partial

19、 vacuum andlow pressure. Static, velocity, and total pressures in a duct systemrelative to ambient space pressure can be measured with a pitot tubeconnected to a manometer. Pitot tube construction and locations fortraversing round and rectangular ducts are presented in Chapter 37.2. SYSTEM ANALYSIST

20、he total pressure change caused by friction, fittings, equipment,and net thermal gravity effect for each section of a duct system iscalculated by the following equation:(13)where= net total pressure change for i sections, Pa= pressure loss caused by friction for i sections, Papij= total pressure los

21、s caused by j fittings, including fan system effect (FSE), for i sections, Papik= pressure loss caused by k equipment for i sections, Pa= thermal gravity effect caused by r stacks for i sections, Pam = number of fittings within i sectionsn = number of equipment within i sections = number of stacks w

22、ithin i sectionsnup= number of duct sections upstream of fan (exhaust/return air subsystems)ndn= number of duct sections downstream of fan (supply air subsystems)From Equation (7), the thermal gravity effect for each nonhori-zontal duct with a density other than that of ambient air is deter-mined by

23、 the following equation: pse= g(a )(z2 z1) (14)wherepse= thermal gravity effect, Paz1and z2= elevation from datum in direction of airflow (Figure 1), ma= density of ambient air, kg/m3 = density of air or gas within duct, kg/m3g = 9.81 = gravitational acceleration, m/s2Example 1. For Figure 1, calcul

24、ate the thermal gravity effect for two cases:(a) air cooled to 34C, and (b) air heated to 540C. Density ofair at 34C is 1.477 kg/m3and at 540C is 0.434 kg/m3. Den-sity of ambient air is 1.204 kg/m3. Stack height is 15 m.Solution:pse= 9.81(a )z(a) For a(Figure 1A), pse= 9.81(1.204 1.477)15 = 40 Pa(b)

25、 For a(Figure 1B), pse= 9.81(1.204 0.434)15 = +113 Papti pfi pijj =1mpikk =1npseirr =1+=for i 12 nupndn+, ,=ptipfipseirFig. 1 Thermal Gravity Effect for Example 1Duct Design 21.3Example 2. Calculate the thermal gravity effect for the two-stack systemshown in Figure 2, where the air is 120C and stack

26、 heights are 15 and30 m. Density of 120C air is 0.898 kg/m3; ambient air is 1.204 kg/m3.Solution: pse= 9.81(a )(z2 z1) = 9.81(1.204 0.898)(30 15) = 45 PaFor the system shown in Figure 3, the direction of air movementcreated by the thermal gravity effect depends on the initiating force(e.g., fans, wi

27、nd, opening and closing doors, turning equipment onand off). If for any reason air starts to enter the left stack (Figure3A), it creates a buoyancy effect in the right stack. On the otherhand, if flow starts to enter the right stack (Figure 3B), it creates abuoyancy effect in the left stack. In both

28、 cases, the produced thermalgravity effect is stable and depends on stack height and magnitudeof heating. The starting direction of flow is important when usingnatural convection for ventilation.To determine the fan total pressure requirement for a system, usethe following equation:Pt= for i = 1, 2,

29、 , nup+ ndn(15)whereFupand Fdn=sets of duct sections upstream and downstream of fanPt= fan total pressure, Pa = symbol that ties duct sections into system paths from exhaust/return air terminals to supply terminalsFigure 4 shows the use of Equation (15). This system has three sup-ply and two return

30、terminals consisting of nine sections connectedin six paths: 1-3-4-9-7-5, 1-3-4-9-7-6, 1-3-4-9-8, 2-4-9-7-5, 2-4-9-7-6, and 2-4-9-8. Sections 1 and 3 are unequal area; thus, they areassigned separate numbers in accordance with the rules foridentifying sections (see step 5 in the section on HVAC Duct

31、Design Procedures). To determine the fan pressure requirement,apply the following six equations, derived from Equation (15).These equations must be satisfied to attain pressure balancing fordesign airflow. Relying entirely on dampers is not economical andmay create objectionable flow-generated noise

32、.(16)Example 3. For Figures 5A and 5C, calculate the thermal gravity effect andfan total pressure required when the air is cooled to 34C. The heatexchanger and ductwork (section 1 to 2) total pressure losses are 170and 70 Pa respectively. The density of 34C air is 1.477 kg/m3; ambientair is 1.204 kg

33、/m3. Elevations are 21 and 3 m.Solution:(a) For Figure 5A (downward flow),Fig. 2 Multiple Stacks for Example 2Fig. 3 Multiple Stack AnalysisptiiFupptiiFdn+Ptp1 p3 p4 p9 p7 p5+=Ptp1 p3 p4 p9 p7 p6+=Ptp1 p3 p4 p9 p8+=Ptp2 p4 p9 p7 p5+=Ptp2 p4 p9 p7 p6+=Ptp2 p4 p9 p8+=Fig. 4 Illustrative 6-Path, 9-Sect

34、ion Systempse 9.81 az2z1=9.81 1.204 1.477 321=48 Pa=21.4 2017 ASHRAE HandbookFundamentals (SI)(b) For Figure 5C (upward flow),Example 4. For Figures 5B and 5D, calculate the thermal gravity effectand fan total pressure required when air is heated to 120C. Heatexchanger and ductwork (section 1 to 2)

35、total pressure losses are 170and 70 Pa respectively. Density of 120C air is 0.898 kg/m3; ambientair is 1.204 kg/m3. Elevations are 21 and 3 m.Solution:(a) For Figure 5B (downward flow),(b) For Figure 5D (upward flow),Fig. 5 Single Stack with Fan for Examples 3 and 4Ptpt,32 pse=170 70+48=192 Pa=pse 9

36、.81 az2z1=9.81 1.204 1.477 21 3=48 Pa=Ptpt,3-2 pse=170 70+48=288 Pa=pse 9.81 az2z1=9.81 1.204 0.898 321=54 Pa=Ptpt,32 pse=170 70+54=294 Pa=pse 9.81 az2z1=9.81 1.204 0.898 21 3=54 Pa=Ptpt,3-2 pse=170 70+54=186 Pa=Duct Design 21.5Example 5. Calculate the thermal gravity effect for each section of thes

37、ystem shown in Figure 6, and the systems net thermal gravity effect.Density of ambient air is 1.204 kg/m3, and the lengths are as follows:z1= 15 m, z2= 27 m, z4= 30 m, z5= 8 m, and z9= 60 m. Pressurerequired at section 3 is 25 Pa. Write the equation to determine the fantotal pressure requirement.Sol

38、ution: The following table summarizes the thermal gravity effectfor each section of the system as calculated by Equation (14). The netthermal gravity effect for the system is 118 Pa. To select a fan, use thefollowing equation:2.1 PRESSURE CHANGES IN SYSTEMFigure 7 shows total and static pressure cha

39、nges in a fan/duct sys-tem consisting of a fan with both supply and return air ductwork.Also shown are total and static pressure gradients referenced toatmospheric pressure.For all constant-area sections, total and static pressure losses areequal. At diverging transitions, velocity pressure decrease

40、s, abso-lute total pressure decreases, and absolute static pressure can in-crease. The static pressure increase at these sections is known asstatic regain.At converging transitions, velocity pressure increases in thedirection of airflow, and absolute total and absolute static pressuresdecrease.At th

41、e exit, total pressure loss depends on the shape of the fittingand the flow characteristics. Exit loss coefficients Cocan be greaterthan, less than, or equal to one. Total and static pressure grade linesfor the various coefficients are shown in Figure 7. Note that, for aloss coefficient less than on

42、e, static pressure upstream of the exit isless than atmospheric pressure (negative). Static pressure justupstream of the discharge fitting can be calculated by subtractingthe upstream velocity pressure from the upstream total pressure.At section 1, total pressure loss depends on the shape of the ent

43、ry.Total pressure immediately downstream of the entrance equals thedifference between the upstream pressure, which is zero (atmo-spheric pressure), and loss through the fitting. Static pressure ofambient air is zero; several diameters downstream, static pressure isnegative, equal to the sum of the t

44、otal pressure (negative) and thevelocity pressure (always positive).System resistance to airflow is noted by the total pressure gradeline in Figure 7. Sections 3 and 4 include fan system effect pressureFig. 6 Triple Stack System for Example 5Pt25 pt 1-7, pt 8-9, pse+ 25pt 1-7,+=pt 8-9, 118 pt 1-7, p

45、t 8-9, 93+=+Path(xx)Temp.,C,kg/m3z(zx zx),m(a xx), kg/m3pse,PaEq. (14)1-2 815 0.324 (27 15) +0.880 +1043-4 540 0.434 0 +0.770 04-5 540 0.434 (8 30) +0.770 1666-7 120 0.898 0 +0.306 08-9 120 0.898 (60 0) +0.306 +180Net Thermal Gravity Effect 118Fig. 7 Pressure Changes During Flow in Ducts21.6 2017 AS

46、HRAE HandbookFundamentals (SI)losses. To obtain the fan static pressure requirement for fan selectionwhere fan total pressure is known, usePs= Pt pv,o(17)wherePs= fan static pressure, PaPt= fan total pressure, Papv,o= fan outlet velocity pressure, Pa3. FLUID RESISTANCEDuct system losses are the irre

47、versible transformation ofmechanical energy into heat. The two types of losses are (1) frictionand (2) dynamic.3.1 FRICTION LOSSESFriction losses are caused by fluid viscosity and result from mo-mentum exchange between molecules (in laminar flow) or betweenindividual particles of adjacent fluid laye

48、rs moving at different ve-locities (in turbulent flow). Friction losses occur along the entireduct length.Darcy and Colebrook EquationsFor fluid flow in conduits, friction loss can be calculated by theDarcy equation:pf= (18)wherepf= friction losses in terms of total pressure, Paf = friction factor, dimensionlessL = duct length, mDh= hydraulic diameter Equation (24), mmV = velocity, m/s = density, kg/m3In the region of laminar flow (Reynolds numbers less