ASHRAE REFRIGERATION IP CH 44-2010 ICE RINKS《溜冰场》.pdf

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1、44.1CHAPTER 44ICE RINKSApplications . 44.1Refrigeration Requirements . 44.1Ice Rink Conditions 44.4Equipment Selection 44.5Rink Floor Design 44.8Building, Maintaining, and Planing Ice Surfaces 44.10Imitation Ice-Skating Surfaces. 44.11NY level sheet of ice made by refrigeration (the term artificialA

2、 ice is sometimes used) is referred to in this chapter as an icerink regardless of use and whether it is located indoors or outdoors.Bobsled-luge tracks are not referred to as rinks but are referencedunder this chapter.An ice sheet is usually frozen by circulation of a heat transferfluid through a n

3、etwork of pipes or tubes located below the surfaceof the ice. The heat transfer fluid is predominantly a secondary cool-ant such as glycol, methanol, ethanol, or calcium chloride (seeChapter 31 of the 2009 ASHRAE HandbookFundamentals).R-22, R-404A, R-507, and R-717 are most frequently used forchilli

4、ng secondary coolants for ice rinks. R-12 and R-502 have alsobeen used; however, because of the phaseout of the CFC refrigerants,they should no longer be considered for use. Moreover, R-22 is alsobeing phased out, with North American production cuts starting in2010, so for new rink equipment selecti

5、on, R-22 and CFC replace-ments should be evaluated according to status and availability.In some rinks, R-22, and R-717 to a lesser degree, have beenapplied as a direct coolant for freezing ice. The direct-refrigerantrinks operate at higher compressor suction pressures and tempera-tures, thus achievi

6、ng an increased coefficient of performance(COP), compared to secondary coolants. The primary refrigerantcharge is greatly increased with this method of freezing. Because ofemissions regulations, the projected R-22 phaseout, building codes,and fire regulations, R-22 and R-717 should not be used to fr

7、eeze icedirectly in indoor rinks.APPLICATIONSMost ice surfaces are used for a variety of sports, although someare constructed for specific purposes and are of specific dimensions.Usual rink sizes include the following:Hockey. The accepted North American hockey rink size is 85 by200 ft. Radius corner

8、s of 28 ft are recommended by professional andamateur rules. The Olympic and international hockey rink size is100 by 200 ft, with 28 ft radius corners. Many rinks are consideredadequate with dimensions of 85 by 185 ft, 80 by 180 ft, and 70 by170 ft. In substandard size rinks, a corner radius of not

9、less than20 ft should be provided to allow use of mechanical resurfacingequipment.Curling. Regulation surface for this sport is 14 by 146 ft; how-ever, the width of the ice sheet is often increased to allow space forinstallation of dividers between the sheets, particularly at the cir-cles. Most curl

10、ing rinks are laid out with ice sheets measuring 15 by150 ft.Figure Skating. School or compulsory figures are generallydone on a “patch” measuring approximately 16 by 40 ft. Freestyleand dance routines generally require an area of 60 by 120 ft or more.Speed Skating. Indoor speed skating has traditio

11、nally beenperformed on hockey-sized rinks. The Olympic-sized outdoorspeed-skating track is a 1400 ft oval, 35 ft wide with 392 ft straight-aways and curves with an inner radius of 87.5 ft. Most speed-skatingovals were originally constructed outdoors, although some are nowconstructed indoors.Recreati

12、onal Skating. Recreational skating can be done on anysize or shape rink, as long as it can be efficiently resurfaced. Gen-erally, 25 to 30 ft2is allowed for each person actually skating. Thisratio may vary for large numbers of beginner skaters. An 85 by200 ft hockey rink with 28 ft radius corners ha

13、s an area of 16,327 ft2and will accommodate a mixed group of about 650 skaters.Public Arenas, Auditoriums, and Coliseums. Public arenas,auditoriums, field houses, etc., are designed primarily for spectatorevents. They are used for ice sports, ice shows, and recreationalskating, as well as for non-ic

14、e events, such as basketball, boxing,tennis, conventions, exhibits, circuses, rodeos, tractor events, andstock shows. The refrigeration system can be designed so that, withadequate personnel, the ice surface can be produced within 12 to16 h. However, general practice is to leave the ice sheet in pla

15、ce andto hold other events on an insulated floor placed on the ice. Thisapproach saves significant time, labor, and energy.Bobsled-Luge Tracks. The bobsled-luge track usually incorpo-rates steel piping embedded in the track and fed by an ammonia liq-uid recirculation system. Approximately 280,000 to

16、 315,000 ft ofpiping is required for an Olympic-sized track. The total refrigeratedsurface is 90,000 to 105,000 ft2. Refrigeration plant capacities in therange of 1100 to 1400 tons are required, depending on ambientdesign conditions, wind, and sun loads. The ammonia charge canexceed 200,000 lb. Beca

17、use elevation changes are significant, caremust be used in placing liquid recirculators, selecting ammoniapumps, and circuiting floor piping.REFRIGERATION REQUIREMENTSThe heat load factors considered in the following section includetype of service, length of season, use, type of enclosure, radiant l

18、oadfrom roof and lights, and geographic location of the rink with asso-ciated wet- and dry-bulb temperatures. For outdoor rinks, the suneffect and weather conditions (wind velocity and rain) must also beconsidered.Refrigeration requirements can be estimated fairly accuratelybased on data from a numb

19、er of rink installations with the pipes cov-ered by not more than 1 in. of sand or concrete and not more than1.5 in. of ice (a total of 2.5 in. sand or concrete and ice).Refrigeration load may be estimated by considering the larger of(1) the refrigeration necessary to freeze the ice to required cond

20、i-tions in a specified time, or (2) the refrigeration necessary to main-tain the ice surface and temperature during the most severe usageand operating conditions that coincide with the maximum ambientenvironmental conditions.In the time-to-freeze method, determine the (1) quantity of icerequired (ri

21、nk surface area multiplied by thickness); (2) heat load toreduce the water from application temperature to 32F, freeze thewater to ice, and reduce the ice to the required temperature; and(3) heat loads and system losses during the freezing period. TheThe preparation of this chapter is assigned to TC

22、 10.2, Automatic Ice-making Plants and Skating Rinks.44.2 2010 ASHRAE HandbookRefrigerationtotal requirement is divided by system efficiency and freezingperiod to determine the required refrigeration load or rate of heatremoval.Example 1. Calculate the refrigeration required to build 1 in. thick ice

23、 on a16,300 ft2rink in 24 h.Assume the following material properties and conditions:Then,qR= (Sys. losses)(qF+ qC+ qSR+ qHL)whereqR= refrigeration requirementqF= water chilling and freezingqC= concrete chilling loadqSR= refrigeration to cool secondary coolantqHL= building and pumping heat loadqF= 49

24、.6 tonsqC= = 10.3 tonsqSR= = 2.3 tonsqR= 1.15(49.6 + 10.3 + 2.3 + 50) = 129 tonsWhen no time restrictions for making ice apply, the estimatedrefrigeration load is the amount of heat removal needed to offset theusage loads plus the coincidental heat loads during the most severeoperating conditions. T

25、able 1 lists approximate refrigeration require-ments for various rinks with controlled and uncontrolled atmo-spheric conditions. Table 1 should only be used to check thecalculated refrigeration requirements. Table 2 shows the distributionof various load components for basic construction in two examp

26、lelocations.Heat LoadsEnergy and operating costs for ice rinks are very significant, andshould be analyzed during design. A good estimate of requiredrefrigeration can be calculated by summing the heat load compo-nents at design operating conditions. Heat loads for ice rinks consistof conductive, con

27、vective, and radiant components. Table 2, whichsummarizes load conditions for indoor rinks in different climates,is from ASHRAE research project RP-1289 (Suny et al. 2007).The study was performed on an ice test bench in a closed environ-ment, with the parametric results compared with a local commu-n

28、ity ice rink for accuracy. Note that curling rinks would havereduced ice surfacing loads, because resurfacing is done by peb-bling the surface, not by flooding via an ice resurfacer. Table 3sload data for outdoor rinks are from Connelly (1976). The amountof control over each load source is indicated

29、 as an approximate per-centage of the maximum reduction possible through effectivedesign and operation.Conductive Loads. If a rink is uninsulated, heat gain from theground below the rink and at the edges averages 2 to 3% of the totalheat load. Permafrost may accumulate and lead to frost heaving,whic

30、h is detrimental to both the rink and the piping. Heaving alsomakes it more difficult to maintain a usable ice surface, can affectthe buildings structural integrity, and is dangerous to users.Heat gain from the ground and perimeter is highest when the sys-tem is first placed in operation; however, i

31、t decreases as the temper-ature of the mass beneath the rink decreases and permafrostaccumulates. Ground heat gain is reduced substantially with insula-tion. Chapter 27 of the 2009 ASHRAE HandbookFundamentalsgives details on computing heat gain with insulation.Heat gain to the piping is normally abo

32、ut 2 to 4% of the totalrefrigeration load, depending on length of piping, surface area, andambient temperatures. The ice and frost that naturally accumulateon headers reduce the heat gain. Insulation can be applied to reduceheat gain to the piping and keep ice from accumulating. However,insulating h

33、eaders without obscuring visual inspection of joints(floor piping to the headers) is usually impractical. Headers may,with precautions and the use of steel headers and piping, be embed-ded in the rink floor. Embedded headers contribute to ice freezingand eliminate the trench-to-rink floor piping pen

34、etrations. Whenheaders are embedded in concrete, all joints from the steel floor pip-ing to the headers should, ideally, be welded. It may be difficult toremove air from this type of floor system.A circuit loop should be placed around the rink perimeter to pre-vent soft ice from developing at the ed

35、ges (see the section on RinkPiping and Pipe Supports). A circuit loop is especially important ifreturn bends are used and embedded in the concrete. If return bendsare embedded in the concrete, the pipe and the return bend should besteel with welded joints.Heat gain from coolant circulating pumps can

36、 represent upto 11% of the refrigeration load. Some facilities operate continu-ously. Energy consumption from pump operation can be reducedby using pump cycling, two-speed motors, multiple pumps, mul-tiple motors driving a single pump, or variable-speed motors withthe appropriate controls. High-effi

37、ciency pumps and motorsshould be used. Coolant flow should be sufficient at all times foracceptable chiller operation and to maintain a balanced flowthrough the piping grid.Equipment components should be selected for low energy con-sumption; they may be selected to operate at or feature low dis-char

38、ge pressure (oversized condenser), high suction pressure(oversized chiller), multiple compressors, and an intelligent controlsystem. Computer control of the refrigeration system is recom-mended.Ice resurfacing represents a significant operating heat load.Water is flooded onto the ice surface, normal

39、ly at temperaturesbetween 130 and 180F, to restore the ice surface condition. Theheat load resulting from the flood water application may be calcu-lated as follows:Qf= 8.33Vf1.0(tf 32) + 144 + 0.49(32 ti)whereQf= heat load per flood, Btu MaterialSpecific Heat, Btu/lbFTemperature,F Density or Weight

40、Initial Final6 in. concrete slab0.16 35 20 150 lb/ft3Supply water 1.0 52 32 62.5 lb/ft3 Ice 0.49 32 24 Ethylene glycol, 35%0.83 40 15 32,000 lb Latent heat of freezing water = 144 Btu/lb Building and pumping heat load = 50 tons of refrigerationSystem losses = 15%Mass of water = 16,300 ft21/12 ft62.5

41、 lb/ft3= 85,000 lb Mass of concrete = 16,300 ft21/2 ft150 lb/ft3= 1,230,000 lb Table 1 Range of Refrigeration Capacities for Ice RinksType of FacilityUp to 7 Months(Spring, Fall, Winter),ft2/ton 8 Months toYear-Round,ft2/ton Outdoors, unshaded 80 to 140 shaded 100 to 190 Sports arena 110 to 160 100

42、to 140 accelerated ice making 80 to 135 75 to 120 Ice recreation center 170 to 240 140 to 190 Curling rinks 200 to 380 150 to 200 Ice shows 80 to 160 75 to 130 85,000 lb 1 52 32144 Btu lb 0.49 32 24+24 h 12,000 Btu h ton-1,230,000 0.16 35 2024 12,000-32,000 0.83 40 1524 12,000-Ice Rinks 44.3Vf= floo

43、d water volume (typically 120 to 180 gal for a 100 by 200 ft rink), gal tf= flood water temperature, F ti= ice temperature, F The resurfacing water temperature affects the load and timerequired to freeze the flood water. Maintaining good water qualitythrough proper treatment may permit the use of lo

44、wer flood watertemperature and less volume.Convective Loads. Convective load from air to ice may be 28%or more of the total heat load to the ice (see Tables 2 and 3). The con-vective heat load is affected by air temperature, relative humidity,and air velocity near the ice surface. Design of the rink

45、 heating anddehumidification air distribution system should include precautionsto minimize the influence of air movement across the ice surface.The convection heat load may be estimated using the procedurefrom Appendix 5 in DOE (1980). The estimated convective heattransfer coefficient can be calcula

46、ted as follows:h = 0.6 + 0.00318Vwhereh = convective heat transfer coefficient, Btu/hft2FV = air velocity over the ice, ft/min The effective heat load (including the latent heat effect of con-vective mass transfer) is given by the following equation:Qcv= h(ta ti) + K(Xa Xi)(1226 Btu/lb)(18 lb /mol)w

47、hereQcv= convective heat load, Btu/hft2K = mass heat transfer coefficientta= air temperature, F ti= ice temperature, F Xa= mole fraction of water vapor in air, lb mol/lb mol Xi= mole fraction of water in saturated ice, lb mol/lb mol When the mole fraction of air is calculated using a relativehumidit

48、y of 80% and a dry bulb of 38F, Xais approximately 6.6103, and Xifor saturated ice at 100% and a temperature of 21Fis 3.6103. On the basis of the Chilton/Colburn analogy, K 0.17 lb/hft2(DOE 1980).In locations with high ambient wet-bulb temperatures, dehumid-ification of the building interior should be considered. This processlowers the load on the ice-making plant and reduces condensationand fog formation. Traditional air conditioners are inappropriatebecause the large ice slab tends to maintain a lower than normal dry-bulb temperature.Radiant Loads. Indoor ice rinks c