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:= constant, ftlbf/lbm(1)wherev =
7、 streamline (local) velocity, fpsgc= dimensional constant, 32.2 lbmft/lbfs2p = absolute pressure, lbf/ft2 = density, lbm/ft3g = acceleration caused by gravity, ft/s2z = elevation, ftAssuming constant fluid density in the system, Equation (1) re-duces to= constant, ftlbf/lbm(2)Although Equation (2) w
8、as derived for steady, ideal frictionlessflow 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+ 1z1= + p2+ 2z2+ pt, 12(3)whereV = average duct velocity, fpspt,12= total pressur
9、e loss caused by friction and dynamic losses between sections 1 and 2, lbf/ft2In Equation (3), V (section average velocity) replaces v (streamlinevelocity) because experimentally determined loss coefficients allowfor errors in calculatingv2/2gc(velocity pressure) across stream-lines.On the left side
10、 of Equation (3), add and subtract pz1; on the 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
11、= pa az1(5)pz2= pa az2(6)Substituting Equations (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
12、 section, the average of the inlet andoutlet temperatures is generally used. Let = 1= 2, and ( p1 pz1)and ( p2 pz2) are gage pressures at elevations z1and z2.pt,12= (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, l
13、bf/ft2ps,2= static pressure, gage at elevation z2, lbf/ft2The preparation of this chapter is assigned to TC 5.2, Duct Design.v22gc-p-gzgc-+v22gc-p-gzgc-+1V122gc-ggc-2V222gc-ggc-1V122gc-p1pz1pz1ggc-1z1+ +2V222gc-p2+= pz2pz2ggc-2z2pt 12,+ggc-ggc-ps 1,V122gc-+ps 2,V222gc-+ggc-+21.2 2017 ASHRAE Handbook
14、Fundamentals V1= average velocity at section 1, fpsV2= average velocity at section 2, fpsa= density of ambient air, lbm/ft3 = density of air or gas in duct, lbm/ft3pse= thermal gravity effect, lbf/ft2 pt= total pressure change between sections 1 and 2, lbf/ft2pt,1-2= total pressure loss caused by fr
15、iction and dynamic losses between sections 1 and 2, lbf/ft21.1 HEAD AND PRESSUREThe terms head and pressure are often 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
16、 head in terms of the flowing fluid. Witha gas or air, however, it is customary to measure pressure exerted bythe gas on a column of liquid.Static PressureThe term pgc/g is static head; p is static pressure.Velocity PressureThe term V2/2g refers to velocity head, and V2/2gcrefers tovelocity pressure
17、. Although velocity head is independent of fluiddensity, velocity pressure Equation (8) is not.pv= (V/1097)2(8)wherepv= velocity pressure, in. of waterV = fluid mean velocity, fpm1097 = conversion factor to in. of waterFor air at standard conditions (0.075 lbm/ft3), Equation (8) becomespv= (V/4005)2
18、(9)where 4005 = (10972/0.075)1/2. Velocity is calculated byV = Q/A (10)whereQ = airflow rate, cfmA = cross-sectional area of duct, ft2Total PressureTotal pressure is the sum of static pressure and velocity pressure:pt= ps+ (V/1097)2(11)orpt= ps+ pv(12)wherept= total pressure, in. of waterps= static
19、pressure, in. of waterPressure 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 vacuum andlow pressure. Static, velocity, and total pressures in a duct s
20、ystemrelative 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 ANALYSISThe total pressure change caused by friction, fittings, equipment,and net t
21、hermal gravity effect for each section of a duct system iscalculated by the following equation:(13)where= net total pressure change for i sections, in. of water= pressure loss caused by friction for i sections, in. of waterpij= total pressure loss caused by j fittings, including fan system effect (F
22、SE), for i sections, in. of waterpik= pressure loss caused by k equipment for i sections, in. of water= thermal gravity effect caused by r stacks for i sections, in. of waterm = number of fittings within i sectionsn = number of equipment within i sections = number of stacks within i sectionsnup= num
23、ber 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 the following equation:
24、pse= 0.192(a )(z2 z1) (14)wherepse= thermal gravity effect, in. of waterz1and z2= elevation from datum in direction of airflow (Figure 1), fta= density of ambient air, lbm/ft3 = density of air or gas within duct, lbm/ft30.192 = conversion factor to in. of waterExample 1. For Figure 1, calculate the
25、thermal gravity effect for two cases:(a) air cooled to 30F, and (b) air heated to 1000F. Densityof air at 30F is 0.0924 lbm/ft3and at 1000F is 0.0271 lbm/ft3.Density of ambient air is 0.075 lbm/ft3. Stack height is 40 ft.Solution: pse= 0.192(a )z(a) For a(Figure 1A), pse= 0.192(0.075 0.0924)40 = 0.1
26、3 in. of water(b) For a(Figure 1B), pse= 0.192 (0.075 0.0271)40 = +0.37 in. of waterExample 2. Calculate the thermal gravity effect for the two-stack systemshown in Figure 2, where the air is 250F and stack heights are 50 and100 ft. Density of 250F air is 0.0558 lbm/ft3; ambient air is 0.075 lbm/ft3
27、.Solution: pse= 0.192(0.075 0.0558)(100 50) = 0.18 in. of waterFor the system shown in Figure 3, the direction of air movementcreated by the thermal gravity effect depends on the initiating force(e.g., fans, wind, opening and closing doors, turning equipment onand off). If for any reason air starts
28、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 cases, the produced thermalgravity effect is stable and depends on stack height and magni
29、tudepti pfi pijj =1mpikk =1npseirr =1+=for i 12 nupndn+, ,=ptipfipseirDuct Design 21.3of 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, , nup+
30、ndn(15)whereFupand Fdn= sets of duct sections upstream and downstream of fanPt= fan total pressure, in. of water = 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 retu
31、rn 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 D
32、uctDesign 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 no
33、ise.(16)Example 3. For Figures 5A and 5C, calculate the thermal gravity effect andfan total pressure required when the air is cooled to 30F. The heat ex-changer and ductwork (section 1 to 2) total pressure losses are 0.70 and0.28 in. of water respectively. Density of 30F air is 0.0924 lbm/ft3;ambien
34、t air is 0.075 lbm/ft3. Elevations are 70 and 10 ft.Solution:(a) For Figure 5A (downward flow),Fig. 1 Thermal Gravity Effect for Example 1ptiiFupptiiFdn+Fig. 2 Multiple Stacks for Example 2Fig. 3 Multiple Stack AnalysisPtp1 p3 p4 p9 p7 p5+=Ptp1 p3 p4 p9 p7 p6+=Ptp1 p3 p4 p9 p8+=Ptp2 p4 p9 p7 p5+=Ptp
35、2 p4 p9 p7 p6+=Ptp2 p4 p9 p8+=Fig. 4 Illustrative 6-Path, 9-Section System21.4 2017 ASHRAE HandbookFundamentals (b) For Figure 5C (upward flow),Example 4. For Figures 5B and 5D, calculate the thermal gravity effect andfan total pressure required when air is heated to 250F. Heat exchangerand ductwork
36、 (section 1 to 2) total pressure losses are 0.70 and 0.28 in.of water, respectively. Density of 250F air is 0.0558 lbm/ft3; ambient airis 0.075 lbm/ft3. Elevations are 70 and 10 ft.Solution:(a) For Figure 5B (downward flow),(b) For Figure 5D (upward flow),pse 0.192 az2z1=0.192 0.075 0.0924 10 70=0.2
37、0 in. of water=Ptpt,32 pse=0.70 0.28+0.20=0.78 in. of water=Fig. 5 Single Stack with Fan for Examples 3 and 4pse 0.192 az2z1=0.192 0.075 0.0924 70 10=0.20 in. of water=Ptpt,3-2 pse=0.70 0.28+0.20=1.18 in. of water=pse 0.192 az2z1=0.192 0.075 0.0558 10 70=0.22 in. of water=Ptpt,32 pse=0.70 0.28+0.22=
38、1.20 in. of water=pse 0.192 az2z1=0.192 0.075 0.0558 70 10=0.22 in. of water=Ptpt,3-2 pse=0.70 0.28+0.22=0.76 in. of water=Duct Design 21.5Example 5. Calculate the thermal gravity effect for each section of thesystem in Figure 6 and the systems net thermal gravity effect. Densityof ambient air is 0.
39、075 lbm/ft3, and the lengths are as follows: z1= 50 ft,z2= 90 ft, z4= 95 ft, z5= 25 ft, and z9= 200 ft. Pressure required at sec-tion 3 is 0.1 in. of water. Write the equation to determine the fan totalpressure requirement.Solution: The following table summarizes the thermal gravity effect foreach s
40、ection of the system as calculated by Equation (14). The netthermal gravity effect for the system is 0.52 in. of water. To select a fan,use the following equation:2.1 PRESSURE CHANGES IN SYSTEMFigure 7 shows total and static pressure changes in a fan/duct sys-tem consisting of a fan with both supply
41、 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 decreases, abso-lute total pressure decreases, and absolute static pres
42、sure 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 the exit, total pressure loss depends on the shape of the fitting
43、and 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 one, static pressure upstream of the exit isless than atmospheric
44、 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 entry.Total pressure immediately downstream of the entrance equals
45、 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 total pressure (negative) and thevelocity pressure (always posit
46、ive).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 5Pt0.1 pt 1-7, pt 8-9, pse+ 0.1pt 1-7,+=pt 8-9, 0.52 pt 1-7, pt 8-9, 0.42+=+Path(xx)Temp.,F,lbm/ft3z(zx zx),ft(a xx), lbm/
47、ft3pse,in. of water Eq. (14)1-2 1500 0.0202 (90 50) +0.0548 +0.423-4 1000 0.0271 0 +0.0479 04-5 1000 0.0271 (25 95) +0.0479 0.646-7 250 0.0558 0 +0.0192 08-9 250 0.0558 (200 0) +0.0192 +0.74Net Thermal Gravity Effect 0.52Fig. 7 Pressure Changes During Flow in Ducts21.6 2017 ASHRAE HandbookFundamenta
48、ls 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, in. of waterPt= fan total pressure, in. of waterpv,o= fan outlet velocity pressure, in. of water3. FLUID RESISTANCEDuct system losses are the irreversible transformation ofmechanical energy into heat. The two types of losses are (1) frictionand (2) dynamic.3.1
