1、51.1CHAPTER 51SNOW MELTING AND FREEZE PROTECTIONSnow-Melting Heat Flux Requirement 51.1Slab Design . 51.8Control 51.10Hydronic System Design . 51.10Electric System Design 51.13Freeze Protection Systems 51.18HE practicality of melting snow or ice by supplying heat to theTexposed surface has been demo
2、nstrated in many installations,including sidewalks, roadways, ramps, bridges, access ramps, andparking spaces for the handicapped, and runways. Melting elimi-nates the need for snow removal by chemical means, providesgreater safety for pedestrians and vehicles, and reduces the labor andcost of slush
3、 removal. Other advantages include eliminating piledsnow, reducing liability, and reducing health risks of manual andmechanized shoveling.This chapter covers three types of snow-melting and freeze pro-tection systems:1. Hot fluid circulated in slab-embedded pipes (hydronic)2. Embedded electric heate
4、r cables or wire3. Overhead high-intensity infrared radiant heatingDetailed information about slab heating can be found in Chapter6 of the 2012 ASHRAE HandbookHVAC Systems and Equipment.More information about infrared heating can be found in Chapter 16of the same volume.Components of the system desi
5、gn include (1) heat requirement,(2) slab design, (3) control, and (4) hydronic or electric systemdesign.1. SNOW-MELTING HEAT FLUX REQUIREMENTThe heat required for snow melting depends on five atmosphericfactors: (1) rate of snowfall, (2) snowfall-coincident air dry-bulbtemperature, (3) humidity, (4)
6、 wind speed near the heated surface,and (5) apparent sky temperature. The dimensions of the snow-melt-ing slab affect heat and mass transfer rates at the surface. Other fac-tors such as back and edge heat losses must be considered in thecomplete design.Heat BalanceThe processes that establish the he
7、at requirement at the snow-melting surface can be described by terms in the following equation,which is the steady-state energy balance for required total heat flux(heat flow rate per unit surface area) qoat the upper surface of asnow-melting slab during snowfall.qo= qs+ qm+ Ar(qh+ qe)(1)whereqo= he
8、at flux required at snow-melting surface, W/m2qs= sensible heat flux, W/m2qm= latent heat flux, W/m2Ar= snow-free area ratio, dimensionlessqh= convective and radiative heat flux from snow-free surface, W/m2qe= heat flux of evaporation, W/m2Sensible and Latent Heat Fluxes. The sensible heat flux qsis
9、 theheat flux required to raise the temperature of snow falling on the slabto the melting temperature plus, after the snow has melted, to raisethe temperature of the liquid to the assigned temperature tfof the liq-uid film. The snow is assumed to fall at air temperature ta. The latentheat flux qmis
10、the heat flux required to melt the snow. Under steady-state conditions, both qsand qmare directly proportional to the snow-fall rate s.Snow-Free Area Ratio. Sensible and latent (melting) heat fluxesoccur on the entire slab during snowfall. On the other hand, heat andmass transfer at the slab surface
11、 depend on whether there is a snowlayer on the surface. Any snow accumulation on the slab acts to par-tially insulate the surface from heat losses and evaporation. The in-sulating effect of partial snow cover can be large. Because snow maycover a portion of the slab area, it is convenient to think o
12、f the insu-lating effect in terms of an effective or equivalent snow-covered areaAs, which is perfectly insulated and from which no evaporation andheat transfer occurs. The balance is then considered to be the equiv-alent snow-free area Af. This area is assumed to be completely cov-ered with a thin
13、liquid film; therefore, both heat and mass transferoccur at the maximum rates for the existing environmental con-ditions. It is convenient to define a dimensionless snow-free arearatio Ar:Ar= (2)whereAf= equivalent snow-free area, m2As= equivalent snow-covered area, m2At= Af + As= total area, m2Ther
14、efore,0 Ar 1To satisfy Ar= 1, the system must melt snow rapidly enough thatno accumulation occurs. For Ar= 0, the surface is covered with snowof sufficient thickness to prevent heat and evaporation losses. Prac-tical snow-melting systems operate between these limits. Earlierstudies indicate that suf
15、ficient snow-melting system design informa-tion is obtained by considering three values of the free area ratio: 0,0.5, and 1.0 (Chapman 1952).Heat Flux because of Surface Convection, Radiation, andEvaporation. Using the snow-free area ratio, appropriate heat andmass transfer relations can be written
16、 for the snow-free fraction ofthe slab Ar. These appear as the third and fourth terms on the right-hand side of Equation (1). On the snow-free surface, maintained atfilm temperature tf, heat is transferred to the surroundings and massis transferred from the evaporating liquid film. Heat flux qhinclu
17、desconvective losses to the ambient air at temperature taand radiativelosses to the surroundings, which are at mean radiant temperatureTMR. The convection heat transfer coefficient is a function of windThe preparation of this chapter is assigned to TC 6.5, Radiant Heating andCooling.AfAt-51.2 2015 A
18、SHRAE HandbookHVAC Applications (SI)speed and a characteristic dimension of the snow-melting surface.This heat transfer coefficient is also a function of the thermody-namic properties of the air, which vary slightly over the temperaturerange for various snowfall events. The mean radiant temperatured
19、epends on air temperature, relative humidity, cloudiness, cloudaltitude, and whether snow is falling.The heat flux qefrom surface film evaporation is equal to theevaporation rate multiplied by the heat of vaporization. The evapo-ration rate is driven by the difference in vapor pressure between thewe
20、t surface of the snow-melting slab and the ambient air. The evap-oration rate is a function of wind speed, a characteristic dimensionof the slab, and the thermodynamic properties of the ambient air.Heat Flux EquationsSensible Heat Flux. The sensible heat flux qsis given by thefollowing equation:qs=
21、waterscp,ice(ts ta) + cp,water(tf ts)/c1(3)wherecp,ice= specific heat of ice, J/(kgK)cp,water= specific heat of water, J/(kgK)s = snowfall rate water equivalent, mm/hta= ambient temperature coincident with snowfall, Ctf= liquid film temperature, Cts= melting temperature, Cwater= density of water, kg
22、/m3c1= 1000 mm/m 3600 s/h = 3.6 106The density of water, specific heat of ice, and specific heat ofwater are approximately constant over the temperature range ofinterest and are evaluated at 0C. The ambient temperature andsnowfall rate are available from weather data. The liquid film tem-perature is
23、 usually taken as 0.56C.Melting Heat Flux. The heat flux qmrequired to melt the snowis given by the following equation:qm= watershif /c1(4)where hif= heat of fusion of snow, J/kg.Convective and Radiative Heat Flux from a Snow-FreeSurface. The corresponding heat flux qhis given by the followingequati
24、on:qh= hc(ts ta) + s(T4f T4MR)(5)wherehc= convection heat transfer coefficient for turbulent flow,W/(m2K)Tf= liquid film temperature, KTMR= mean radiant temperature of surroundings, K = Stefan-Boltzmann constant = 5.670 108W/(m2K4)s= emittance of surface, dimensionlessThe convection heat transfer co
25、efficient over the slab on a plane hor-izontal surface is given by the following equations (Incropera andDeWitt 1996):hc= 0.037 Re0.8LPr1/3(6)where kair= thermal conductivity of air at ta, W/(mK)L = characteristic length of slab in direction of wind, mPr = Prandtl number for air, taken as Pr = 0.7Re
26、L= Reynolds number based on characteristic length LandReL= c2(7)whereV = design wind speed near slab surface, km/hair= kinematic viscosity of air, m2/sc2= 1000 m/km 1 h/3600 s = 0.278Without specific wind data for winter, the extreme wind data inChapter 14 of the 2013 ASHRAE HandbookFundamentals may
27、be used; however, it should be noted that these wind speeds may notcorrespond to actual measured data. If the snow-melting surface isnot horizontal, the convection heat transfer coefficient might be dif-ferent, but in many applications, this difference is negligible.From Equations (6) and (7), it ca
28、n be seen that the turbulent con-vection heat transfer coefficient is a function of L0.2. Because ofthis relationship, shorter snow-melting slabs have higher convectiveheat transfer coefficients than longer slabs. For design, the shortestdimension should be used (e.g., for a long, narrow driveway or
29、 side-walk, use the width). A snow-melting slab length L = 6.1 m is usedin the heat transfer calculations that resulted in Tables 1, 2, and 3.The mean radiant temperature TMRin Equation (5) is theequivalent blackbody temperature of the surroundings of the snow-melting slab. Under snowfall conditions
30、, the entire surroundings areapproximately at the ambient air temperature (i.e., TMR= Ta). Whenthere is no snow precipitation (e.g., during idling and after snowfalloperations for Ar 1), the mean radiant temperature is approxi-mated by the following equation:TMR= T4cloud Fsc+ T4sky clear (1 Fsc)1/4(
31、8)whereFsc= fraction of radiation exchange that occurs between slab and cloudsTcloud= temperature of clouds, KTsky clear= temperature of clear sky, KThe equivalent blackbody temperature of a clear sky is primarilya function of the ambient air temperature and the water content ofthe atmosphere. An ap
32、proximation for the clear sky temperature isgiven by the following equation, which is a curve fit of data inRamsey et al. (1982):Tsky clear= Ta (1.1058 103 7.562Ta+ 1.333 102T2a 31.292 + 14.582)(9)whereTa= ambient temperature, K = relative humidity of air at elevation for which typical weather measu
33、rements are made, decimalThe cloud-covered portion of the sky is assumed to be at Tcloud.The height of the clouds may be assumed to be 3000 m. Thetemperature of the clouds at 3000 m is calculated by subtractingthe product of the average lapse rate (rate of decrease of atmospherictemperature with hei
34、ght) and the altitude from the atmospherictemperature Ta. The average lapse rate, determined from the tables ofU.S. Standard Atmospheres (COESA 1976), is 6.4 K per 1000 m ofelevation (Ramsey et al. 1982). Therefore, for clouds at 3000 m,Tcloud= Ta 19.2 (10)Under most conditions, this method of appro
35、ximating the tem-perature of the clouds provides an acceptable estimate. However,when the atmosphere contains a very high water content, the tem-perature calculated for a clear sky using Equation (9) may bewarmer than the cloud temperature estimated using Equation (10).When that condition exists, Tc
36、loudis set equal to the calculated clearsky temperature Tsky clear.Evaporation Heat Flux. The heat flux qerequired to evaporatewater from a wet surface is given bykairL-VLair-Snow Melting and Freeze Protection 51.3qe= dry airhm(Wf Wa)hfg(11)wherehm= mass transfer coefficient, m/sWa= humidity ratio o
37、f ambient air, kgvapor/kgairWf= humidity ratio of saturated air at film surface temperature, kgvapor/kgairhfg= heat of vaporization (enthalpy difference between saturated water vapor and saturated liquid water), J/kgdry air= density of dry air, kg/m3Determination of the mass transfer coefficient is
38、based on the anal-ogy between heat transfer and mass transfer. Details of the analogyare given in Chapter 5 of the 2013 ASHRAE HandbookFunda-mentals. For external flow where mass transfer occurs at the con-vective surface and the water vapor component is dilute, thefollowing equation relates the mas
39、s transfer coefficient hmto theheat transfer coefficient hcEquation (6):hm= (12)where Sc = Schmidt number. In applying Equation (11), the val-ues Pr = 0.7 and Sc = 0.6 were used to generate the values inTables 1 to 4.The humidity ratios both in the atmosphere and at the surface ofthe water film are
40、calculated using the standard psychrometric rela-tion given in the following equation (from Chapter 1 of the 2013ASHRAE HandbookFundamentals):W = 0.622 (13)where p = atmospheric pressure, kPapv= partial pressure of water vapor, kPaThe atmospheric pressure in Equation (13) is corrected for alti-tude
41、using the following equation (Kuehn et al. 1998):p = pstd(14)wherepstd= standard atmospheric pressure, kPaA = 0.0065 K/mz = altitude of the location above sea level, mTo= 288.2 KAltitudes of specific locations are found in Chapter 14 of the 2013ASHRAE HandbookFundamentals.The vapor pressure pvfor th
42、e calculation of Wais equal to thesaturation vapor pressure psat the dew-point temperature of the air.Saturated conditions exist at the water film surface. Therefore, thevapor pressure used in calculating Wfis the saturation pressure atthe film temperature tf. The saturation partial pressures of wat
43、ervapor for temperatures above and below freezing are found intables of the thermodynamic properties of water at saturation or canbe calculated using appropriate equations. Both are presented inChapter 1 of the 2013 ASHRAE HandbookFundamentals.Heat Flux Calculations. Equations (1) to (14) can be use
44、d to de-termine the required heat fluxes of a snow-melting system. However,calculations must be made for coincident values of snowfall rate,wind speed, ambient temperature, and dew-point temperature (or an-other measure of humidity). By computing the heat flux for eachsnowfall hour over a period of
45、several years, a frequency distributionof hourly heat fluxes can be developed. Annual averages or maxi-mums for climatic factors should never be used in sizing a systembecause they are unlikely to coexist. Finally, it is critical to note thatthe preceding analysis only describes what is happening at
46、 the uppersurface of the snow-melting surface. Edge losses and back losseshave not been taken into account.Example 1. During the snowfall that occurred during the 8 PM hour onDecember 26, 1985, in the Detroit metropolitan area, the followingsimultaneous conditions existed: air dry-bulb temperature =
47、 8.3C,dew-point temperature = 10C, wind speed = 31.7 km/h, and snowfallrate = 2.54 mm of liquid water equivalent per hour. Assuming L =6.1 m, Pr = 0.7, and Sc = 0.6, calculate the surface heat flux qofor asnow-free area ratio of Ar= 1.0. The thermodynamic and transportproperties used in the calculat
48、ion are taken from Chapters 1 and 33 ofthe 2013 ASHRAE HandbookFundamentals. The emittance of thewet surface of the heated slab is 0.9.Solution:By Equation (3),qs= 1000 2100(0 + 8.3) + 4290(0.56 0) = 14.0 W/m2By Equation (4),qm= 1000 334 000 = 235.6 W/m2By Equation (7),ReL= = 4.13 106By Equation (6)
49、,hc= 0.037 (4.13 106)0.8(0.7)1/3= 24.8 W/(m2K)By Equation (5),qh= 24.8(0.56 + 8.3) + (5.670 108)(0.9)(273.74 264.94)= 258.8 W/m2By Equation (12),hm= = 0.0206 m/sObtain the values of the saturation vapor pressures at dew-point tem-perature 10C and film temperature 0.56C from Table 3 in Chapter 1of the 2013 ASHRAE HandbookFundamentals. Then, use Equation(13) to obtain Wa= 0.00160 kgvapor/kgairand Wf= 0.00393 kgvapor/kgair.By Equation (11),qe= 1.33 0.0206(0.00393 0.00160)
