ASHRAE REFRIGERATION SI CH 24-2010 REFRIGERATED-FACILITY LOADS《冷藏设施负荷》.pdf

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1、24.1CHAPTER 24REFRIGERATED-FACILITY LOADSTransmission Load 24.1Product Load. 24.3Internal Load. 24.3Infiltration Air Load 24.4Equipment Related Load. 24.6Safety Factor . 24.7Load Diversity. 24.8OTAL refrigeration load includes (1) transmission load, whichTis heat transferred into the refrigerated sp

2、ace through its sur-face; (2) product load, which is heat removed from and produced byproducts brought into and kept in the refrigerated space; (3) inter-nal load, which is heat produced by internal sources (e.g., lights,electric motors, and people working in the space); (4) infiltrationair load, wh

3、ich is heat gain associated with air entering the refrig-erated space; and (5) equipment-related load.The first four segments of load constitute the net heat load forwhich a refrigeration system is to be provided; the fifth segmentconsists of all heat gains created by the refrigerating equipment.Thu

4、s, net heat load plus equipment heat load is the total refrigera-tion load for which a compressor must be selected.This chapter contains load calculating procedures and data forthe first four segments and load determination recommendations forthe fifth segment. Information needed for refrigeration o

5、f specificfoods can be found in Chapters 28 and 30 to 42.TRANSMISSION LOADSensible heat gain through walls, floor, and ceiling is calculatedat steady state as q = UA t (1)whereq = heat gain, WA = outside area of section, m2t = difference between outside air temperature and air temperature of the ref

6、rigerated space, KThe overall coefficient of heat transfer U of the wall, floor, orceiling can be calculated by the following equation:(2)whereU = overall heat transfer coefficient, W/(m2K)x = wall thickness, mk = thermal conductivity of wall material, W/(mK)hi= inside surface conductance, W/(m2K)ho

7、= outside surface conductance, W/(m2K)A value of 1.6 for hiand hois frequently used for still air. If theouter surface is exposed to 25 km/h wind, hois increased to 6.With thick walls and low conductivity, the resistance x/k makes Uso small that 1/hiand 1/hohave little effect and can be omitted from

8、the calculation. Walls are usually made of more than one material;therefore, the value x/k represents the composite resistance of thematerials. The U-factor for a wall with flat parallel surfaces of mate-rials 1, 2, and 3 is given by the following equation:(3)The thermal conductivities of several co

9、ld storage insulations arelisted in Table 1. These values increase with age because of factorsdiscussed in Chapter 25 of the 2009 ASHRAE HandbookFunda-mentals. Chapter 27 of that volume includes more complete tableslisting the thermal properties of various building and insulationmaterials.Table 2 li

10、sts minimum insulation thicknesses of expanded poly-isocyanurate board recommended by the refrigeration industry.These thicknesses may need to be increased to offset heat gaincaused by building components such as wood and metal studs, websin concrete masonry, and metal ties that bridge across the in

11、sulationand reduce the thermal resistance of the wall or roof. Chapter 27 ofthe 2009 ASHRAE HandbookFundamentals describes how to cal-culate heat gain through walls and roofs with thermal bridges. Metalsurfaces of prefabricated or insulated panels have a negligible effecton thermal performance and n

12、eed not be considered in calculatingthe U-factor.In most cases, the temperature difference t can be adjusted tocompensate for solar effect on heat load. Values in Table 3 applyover a 24 h period and are added to the ambient temperature whencalculating wall heat gain.The preparation of this chapter i

13、s assigned to TC 10.8, Refrigeration LoadCalculations.U11 hi xk 1 ho+-=Table 1 Thermal Conductivity of Cold Storage InsulationInsulationThermal Conductivityak, W/(m K)Polyurethane board (R-11 expanded) 0.023 to 0.026Polyisocyanurate, cellular (R-141b expanded) 0.027Polystyrene, extruded (R-142b) 0.0

14、35Polystyrene, expanded (R-142b) 0.037Corkboardb0.043Foam glassc0.044aValues are for a mean temperature of 24C, and insulation is aged 180 days.bSeldom-used insulation. Data are only for reference.cVirtually no effects from aging.Table 2 Minimum Insulation ThicknessStorageTemperature, CExpanded Poly

15、isocyanurate ThicknessNorthern U.S., mm Southern U.S., mm10 to 16 50 504 to 10 50 504 to 4 50 759 to 4 75 7518 to 9 75 10026 to 18 100 10040 to 26 125 125U1x1k1 x2k2 x3k3+-=24.2 2010 ASHRAE HandbookRefrigeration (SI)Latent heat gain from moisture transmission through walls,floors, and ceilings of mo

16、dern refrigerated facilities is negligible.Data in Chapter 27 of the 2009 ASHRAE HandbookFundamentalsmay be used to calculate this load if moisture permeable materialsare used.Chapter 14 of the 2009 ASHRAE HandbookFundamentalsgives outdoor design temperatures for major cities; values for 0.4%should

17、be used.Additional information on thermal insulation may be found inChapters 25 and 26 of the 2009 ASHRAE HandbookFundamentals.Chapter 18 of that volume discusses load calculation procedures ingreater detail.Heat Gain from Cooler FloorsHeat gain through cooler concrete slab floors is predicted using

18、procedures developed by Chuangchid and Krarti (2000), who devel-oped a simplified correlation of the total slab heat gain for coolersbased on analytical results reported earlier by Chuangchid andKrarti (1999). Parameters in the solution include slab size and ther-mal resistance, insulation thermal r

19、esistance, soil thermal conduc-tivity, water table depth and temperature, and indoor and outdoorair temperatures. The design procedure accommodates four slabinsulation configurations: no insulation, uniform horizontal insu-lation, partial-horizontal-perimeter insulation, and partial-vertical-perimet

20、er insulation. The slab size characteristic is expressed as theratio of slab area A to exposed slab perimeter P. The result is an esti-mate of the annual mean and amplitude cooler floor heat gain that,when combined, gives the instantaneous floor heat at a specifictime of year. The time-variation of

21、the ground-coupled heat gainq(t) for cooler slabs isq(t) = qm qacos (t ) (4)whereqm= annual mean slab floor heat gain, kWqa= amplitude of annual variation slab floor heat gain, kW = phase lag between cooler floor heat gain and outdoor air temperature variation, dayst = time, days = constant for annu

22、al angular frequency, 0.0172 radians/day These quantities are functions of input parameters such as buildingdimensions, soil properties, and insulation thermal resistance. Notethat the time of origin (t = 0) in Equation (4) corresponds to the timewhen the soil surface temperature is minimum (typical

23、ly January 15in most U.S. locations). Details of the calculation method are givenby Chuangchid and Krarti (2000).For the conditions in Table 4, results for cooler slab floor heatgain are computed based on the method. As shown in Figure 1, heatgain through the cooler floor significantly varies and ma

24、y even benegative at certain times of the year. The influence of partial-horizontal and partial-vertical-perimeter insulation on cooler floorheat gain for Table 4 conditions can also be seen in Figure 1.Figure 2 shows the results for maximum heat transfer per unit areaqmax/A by calculation where typ

25、ical cold and warm climate temper-atures are applied. These outdoor temperatures are given in Table 5.qmax/A is the sum of the mean and amplitude of heat gain per unitarea and is, therefore, the maximum heat transfer rate that occurs atsome time during the year. The influence of floor slab width and

26、Table 3 Allowance for Sun EffectTypical Surface TypesEastWall, KSouth Wall, KWestWall, KFlatRoof, KDark-colored surfacesSlate roofing 5 3 5 11Tar roofingBlack paintMedium-colored surfacesUnpainted wood 4 3 4 9BrickRed tileDark cementRed, gray, or green paintLight-colored surfacesWhite stone 3 2 3 5L

27、ight colored cementWhite paintNote: Add to the normal temperature difference for heat leakage calculations to compensate for sun effect. Do not use for air-conditioning design.Table 4 Example Input Data Required to Estimate Cooler Floor Heat GainRequired Information Example ValuesSoil thermal conduc

28、tivity ks1.51 W/(mK)Soil thermal diffusivity s7.12 107m2/sTotal floor area A 10 m 15 m = 150 m2Exposed perimeter P 50 mSlab thickness y 100 mmSlab thermal resistance Rf0.587 (m2K)/WInsulation thermal resistance Ri(partial uniform, partial, vertical)3.52 (m2K)/WPartial insulation length c 1 mCooler i

29、nside temperature Tr1.67CAnnual mean Tm4.4CAnnual amplitude Ta33.3 KWater table depth b Ignored (if depth is greater than 2 m, water table effect can be ignored)Table 5 Typical Annual and Annual Amplitude Outdoor Temperatures for Warm and Cold ClimatesTemperatureTypical ColdClimate, CTypical WarmCli

30、mate, CAnnual mean Tm4.4 21.1Annual amplitude Ta33.3 16.7Fig. 1 Variation of Cooler Floor Heat Gain over One Year forConditions in Table 1Fig. 1 Variation of Cooler Floor Heat Gain over One Year for Conditions in Table 4Refrigerated-Facility Loads 24.3length is captured by the A/P parameter. Widths

31、of both 10 and 30 mwere used to generate Figure 2. The fact that the qmax/A curves inFigure 2 coalesce for different slab areas while still having the sameA/P ratio and the same indoor, annual mean outdoor, and annualamplitude outdoor temperatures suggests that slab area does notaffect the cooler fl

32、oor slab heat gain rate per unit area with floor areasgreater than about 100 m2.PRODUCT LOADThe primary refrigeration loads from products brought into andkept in the refrigerated space are (1) heat that must be removed tobring products to storage temperature and (2) heat generated byproducts (mainly

33、 fruits and vegetables) in storage. The quantity ofheat to be removed can be calculated as follows:1. Heat removed to cool from initial temperature to some lowertemperature above freezing:Q1= mc1(t1 t2)(5)2. Heat removed to cool from initial temperature to freezing pointof product:Q2= mc1(t1 tf)(6)3

34、. Heat removed to freeze product:Q3= mhif(7)4. Heat removed to cool from freezing point to final temperaturebelow freezing point:Q4= mc2(tf t3)(8)whereQ1, Q2, Q3, Q4= heat removed, kJm = mass of product, kgc1= specific heat of product above freezing, kJ/(kgK)t1= initial temperature of product above

35、freezing, Ct2= lower temperature of product above freezing, Ctf= freezing temperature of product, Chif= latent heat of fusion of product, kJ/kgc2= specific heat of product below freezing, kJ/(kgK)t3= final temperature of product below freezing, CThe refrigeration capacity required for products broug

36、ht intostorage is determined from the time allotted for heat removal andassumes that product is properly exposed to remove the heat in thattime. The calculation isq = (9)whereq = average cooling load, kWn = allotted time, hEquation (9) only applies to uniform entry of product into stor-age. The refr

37、igeration load created by nonuniform loading of awarm product may be much greater over a short period. See Chapter28 for information on calculating the cooling load of warm product.Specific heats above and below freezing for many products aregiven in Table 3 of Chapter 19. A products latent heat of

38、fusion maybe estimated by multiplying the water content of the product (ex-pressed as a decimal) by the latent heat of fusion of water, which is334 kJ/kg. Most foods freeze between 3 and 0.5C. When the ex-act freezing temperature is not known, assume that it is 2C.Example 1. 100 kg of lean beef is t

39、o be cooled from 18 to 4C, then frozenand cooled to 18C. The moisture content is 69.5%, so the latent heatis estimated as 233 kJ/kg. Estimate the cooling load.Solution:Specific heat of beef before freezing is listed in Table 3, Chapter 19,as 3.52 kJ/(kgK); after freezing, 2.12 kJ/(kgK).To cool from

40、18 to 4C in a chilled room:1003.52(18 4) = 4928 kJTo cool from 4C to freezing point in freezer:1003.524 (2) = 2112 kJTo freeze: 100233 = 23 300 kJTo cool from freezing to storage temperature:1002.12 (2) (18) = 3392 kJTotal: 4928 + 2112 + 23 300 + 3392 = 33 732 kJ(Example 3 in Chapter 19 shows an alt

41、ernative calculation method.)Fresh fruits and vegetables respire and release heat during storage.This respiration heat varies by product and its temperature; the colderthe product, the less heat of respiration. Table 9 in Chapter 19 givesheat of respiration rates for various products.Calculations in

42、 Example 1 do not cover heat gained from productcontainers brought into the refrigerated space. When pallets, boxes,or other packing materials are a significant portion of the total massintroduced, this heat load should be calculated.Equations (5) to (9) are used to calculate the total heat gain. An

43、ymoisture removed appears as latent heat gain. The amount of mois-ture involved is usually provided by the end user as a percentage ofproduct mass; so, with such information, the latent heat componentof the total heat gain may be determined. Subtracting the latent heatcomponent from the total heat g

44、ain determines the sensible heatcomponent.INTERNAL LOADElectrical Equipment. All electrical energy dissipated in therefrigerated space (from lights, motors, heaters, and other equip-ment) must be included in the internal heat load. Heat equivalents ofelectric motors are listed in Table 6.Forklifts.

45、Forklifts in some facilities can be a large and variablecontributor to the load. Although many forklifts may be in a spaceat one time, they do not all operate at the same energy level. Forexample, the energy used by a forklift while it is elevating or low-ering forks is different than when it is mov

46、ing.Processing Equipment. Grinding, mixing, or cooking equip-ment may be in refrigerated areas of food processing plants. OtherFig. 2 Variation of Qmax/A with A/P Using Conditions fromTables 1 and 2Fig. 2 Variation of qmax/A with A/P Using Conditions from Tables 4 and 5Q2Q3Q4+3600n-24.4 2010 ASHRAE

47、HandbookRefrigeration (SI)heat sources include equipment for packaging, glue melting, orshrink wrapping. Another possible load is the makeup air for equip-ment that exhausts air from a refrigerated space.People. People add to the heat load, in amounts depending onfactors such as room temperature, ty

48、pe of work being done, type ofclothing worn, and size of the person. Heat load from a person qpmay be estimated asqp= 272 6t (10)where t is the temperature of the refrigerated space in C. Table 7shows the average load from people in a refrigerated space as cal-culated from Equation (10).When people

49、first enter a storage they bring in additional surfaceheat. Thus, when many people enter and leave every few minutes,the load is greater than that listed in Table 7 and must be adjusted.A conservative adjustment is to multiply the values calculated inEquation (10) by 1.25.Latent Load. The latent heat component of the internal load isusually very small compare