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

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 25.91by 60.96 m. Radius co

8、rners of 8.53 m are recommended by profes-sional and amateur rules. The Olympic and international hockeyrink size is 30.48 by 60.96 m, with 8.53 m radius corners. Manyrinks are considered adequate with dimensions of 25.9 by 56.4 m,24.4 by 54.9 m, and 21.3 by 51.8 m. In substandard size rinks, a cor-

9、ner radius of not less than 4.5 m should be provided to allow use ofmechanical resurfacing equipment.Curling. Regulation surface for this sport is 4.3 by 45 m; how-ever, the width of the ice sheet is often increased to allow space forinstallation of dividers between the sheets, particularly at the c

10、ir-cles. Most curling rinks are laid out with ice sheets measuring 4.5 by46 m.Figure Skating. School or compulsory figures are generallydone on a “patch” measuring approximately 5 by 12 m. Freestyleand dance routines generally require an area of 18 by 36 m or more.Speed Skating. Indoor speed skating

11、 has traditionally beenperformed on hockey-sized rinks. The Olympic-sized outdoorspeed-skating track is a 400 m oval, 10 m wide with 112 m straight-aways and curves with an inner radius of 25 m. Most speed-skatingovals were originally constructed outdoors, although some are nowconstructed indoors.Re

12、creational Skating. Recreational skating can be done on anysize or shape rink, as long as it can be efficiently resurfaced. Gen-erally, 2.3 to 2.8 m2is allowed for each person actually skating. Thisratio may vary for large numbers of beginner skaters. A 26 by 61 mhockey rink with 8.5 m radius corner

13、s has an area of 1517 m2andwill 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 85 000 to

16、96 000 m ofpiping is required for an Olympic-sized track. The total refrigeratedsurface is 8300 to 9800 m2. Refrigeration plant capacities in therange of 4000 to 5000 kW are required, depending on ambientdesign conditions, wind, and sun loads. The ammonia charge canexceed 90 000 kg. Because elevatio

17、n 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 loadfrom roof

18、 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 number of rink i

19、nstallations with the pipes cov-ered by not more than 25 mm of sand or concrete and not more than38 mm of ice (a total of 63 mm 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 condi-tions in a spe

20、cified 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 (rink surface area

21、multiplied by thickness); (2) heat loadto reduce the water from application temperature to 0C, 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 10.2, Automatic

22、Ice-making Plants and Skating Rinks.44.2 2010 ASHRAE HandbookRefrigeration (SI)total 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 25 mm thick iceon a 1500 m2

23、rink 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= 172.8 kWqC= = 33.

24、5 kWqSR= = 7.9 kWqR= 1.15 (172.8 + 33.5 + 7.9 + 170) = 442 kWWhen 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. Table 1 lists approx

25、imate 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 examplelocations. Heat L

26、oadsEnergy 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, convective, and radia

27、nt 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-nity ice rink for a

28、ccuracy. 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 as an approximate

29、 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,which is detrimental t

30、o 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, it decreases as the

31、 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 about 2 to 4% of the

32、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 headers without obs

33、curing 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 penetrations. Whenhea

34、ders 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 edges (see the secti

35、on 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 represent upto 11

36、% 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-efficiency pumps and m

37、otorsshould 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-charge pressure (overs

38、ized 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, normally at temperatures

39、between 55 and 80C, to restore the ice surface condition. The heatload resulting from the flood water application may be calculated asfollows:Qf= 1000Vf4.2(tf 0) + 334 + 2.0(0 ti)whereQf= heat load per flood, kJMaterialSpecific Heat, kJ/(kgK)Temperature,CDensity or MassInitial Final150 mm concrete s

40、lab 0.67 2 6 2400 kg/m3Supply water 4.18 11 0 1000 kg/m3Ice 2.04 0 4 Ethylene glycol, 35% 3.5 5 9 14 000 kgLatent heat of freezing water = 334 kJ/kgBuilding and pumping heat load = 170 kW of refrigerationSystem losses = 15%Mass of water = 1500 m20.025 m1000 kg/m3= 37 500 kgMass of concrete = 1500 m2

41、0.150 m2400 kg/m3= 540 000 kgTable 1 Range of Refrigeration Capacities for Ice RinksType of FacilityUp to 7 Months(Spring, Fall, Winter),m2/kW8 Months toYear-Round,m2/kWOutdoors, unshaded 2.1 to 3.7 shaded 2.6 to 5.0 Sports arena 2.9 to 4.2 2.6 to 3.7accelerated ice making 2.1 to 3.6 2.0 to 3.2Ice r

42、ecreation center 4.5 to 6.3 3.7 to 5.0Curling rinks 5.3 to 10.0 4.0 to 5.3Ice shows 2.1 to 4.0 2.0 to 3.437 500 kg 4.18 11 0344 kJ kg 2.04 0 4+24 h 3600 s/h-540 000 0.67 2 624 3600-14 000 3.5 5 924 3600-Ice Rinks 44.3Vf= flood water volume (typically 0.4 to 0.7 m3for a 30 by 60 m rink), m3tf= flood

43、water temperature, Cti= ice temperature, CThe resurfacing water temperature affects the load and timerequired to freeze the flood water. Maintaining good water qualitythrough proper treatment may permit the use of lower flood watertemperature and less volume.Convective Loads. Convective load from ai

44、r 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 heating anddehumidification air distribution system should include precautionsto mi

45、nimize 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 calculated as follows:h = 3.41 + 3.55Vwhereh = convective heat transfer coefficient, W/(m2K

46、)V = air velocity over the ice, m/sThe 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)(2852 kJ/kg)(18 kg/mol)whereQcv= convective heat load, W/m2K = mass heat transfer coefficientta= air temperature, Cti=

47、 ice temperature, CXa= mole fraction of water vapor in air, kg mol/kg molXi= mole fraction of water in saturated ice, kg mol/kg molWhen the mole fraction of air is calculated using a relativehumidity of 80% and a dry bulb of 3.3C, Xais approximately 6.6103, and Xifor saturated ice at 100% and a temp

48、erature of 6.1Cis 3.6103. On the basis of the Chilton/Colburn analogy, K 0.23 g/(sm2) (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 create a unique condition wherea large, relatively cold plane (the ice sheet) is maintain