Document pmNR6MM2dxj6mD0ezk52MVbbE

534 CHAPTER 28 1965 Guide And. Data Book Table 6 .... Heat Equivalent of ElectricMotors . For refrigerant temperatures below 30 F and storage tem peratures of 35 F and higher, off-cycle defrosting has been Motor hp H to H K to 3 3 to 20 Connected load ta refr (pace1 4,250 3,700 2,950 6to par (ftp) (hr) Motor losses outside refr tpoce* 2,545 2,545 2,545 . Connected load refr space* 1,700 1,150 400. r , generally used; defrosting by stopping compressor and allowing the relatively warm box air to pass through the coil until toe frost melts. Off-cycle defrosting for a 35 F storage tem perature requires the compressor to be off one hour for every two hours of compressor operation, and the load calculations are based on 16 hr operation. Where a positive defrost is used and the defrosting time is reduced, the compressor operating time can be increased; this averages about 20 hr.~ However, it is suggested that toe' equipment manufacturer's * For dm when both aaafttl output aad motor tones an dissipated within refrixereted space; motors driving fans (or forced circnUticn obit coolers. For oae when motor knees ere dissipated outside refrigerated space mad omdol work of motor ta expended within refrixersted space; pump os a eireu- tating triaa or water system, (aa motor outside refrigerated apaee driving fas circulating air withia refrigerated space. 1 Foruse when motor heat Iceeci arc diseipated within refrigerated apaee ead treefnl wprk opeaded outride of refrigerated spaoe; motor in refrigerated space driving pomp or fan located outside of apaee. ; recommendations on operating time be followed. For refrigerators maintaining temperatures below 34 F, off- cycle defrosting cannot be used and some positive method of defrosting must be provided.' ,\ : All of these methods add some heat to the refrigerator. The amounts of heat vary considerably with different methods. It - is suggested that the recommendations of the manufacturer be followed, both in the heat added due to.defrosting and the varying rates depending upon'the temperature, type of work, * ~ suggested operating time. clothing,.sice) etc. The average hourly load due to occupancy The calculated hourly load is used as a guide in we.Wting is shown in Table 7. When people go into the refrigerated space. equipment. Since the compressor sites are limited to com for short durations, they will carry with them a considerable; paratively large increments, toe compressor selection must be amount of heat over and above that listed in Table 7. There guided by the calculated load; and the actual selection of com fore, some allowance must be made if the traffic'load of this pressor may change the calculated time up or down. The type is heavy. : evaporator should be selected to.balance the selected com- SHORT METHOD / pressor capacity, and not the original calculated load: `Examples 2, 3, and 4 illustrate , typical load calculations There are many short or rule-of-thumb methods of est and compressor capacity selection. mating load. Some, based on external surface only. These are basically incorrect since the usage load will depend upon the- volume, and toe volume of a refrigerator is not proportional to the external surface. To illustrate, assume two refrigerators both with external areas equal to 1600 sq ft. If they are both 10 ft high, one may`have floor dimensions of 35 X 10 ft, another . 20 X 20 ft. Calculation will show a 12.5 percent' : variation in volume. ' ... A short method based on volume as well as surface will give more dependable results than a method based on surface only. Of the four general sources of. heat in the refrigerator (wall losses, air change, product, and miscellaneous), the ex ternal surface will always .be proportional to the wall losses. The problem of simplification,is with the remaining three sources in terms of volume. Table 8 gives the probable values for the gains due to air changes, product, and miscellaneous. The values are based on experience, and judgment must be used in its application. Example B: Calculate the total load for a 20 ft X 10 ft X 10 ft (outside dimensions) refrigerated room, having 4 in. of cork insulation. The room is to be maintained at 30 P, while the out side air is at 90 F dry-bulb temperature and GO percent relative humidity. The product is 2000 lb of beef at 40 F to be cooled to 30 F in 24 hr. The electrical load in the room is 200 watts. SoUition A: The outride surface area of the room is 1000 sq ft (20 X 10 X 4 + 10 X 10 X 2). From Table 1, for a tempera ture difference of 60 F deg, the heat-gain factor for 4 in. cork insulation is found to be 108 Btu per (sq ft) (24 hr). The wall heat gain is 1000 X 108 - 108,000 Btu per 24 hr. Assuming that the walls are 1 ft thick, the internal volume of the room is 19 X 9 X 9 = 1540 cu ft.' From Table 4, the number of air changes per 24 hr is found to be 14. From Table 5, the heat required to cool the outride air is found to be 2.53 Btu per cu ft. The air change load is 1540 X 14 X 2.53 - 54,600 Btu per 24 hr. ' From Table 4, Chapter 69, the specific heat of beef is 0.77 Btu per (lb) (F deg). The product load is 2000 X 0.77 X (40 - 30) -i 15,400 Btu per 24 hr. The load due to electrical equipment in the room is 200 X 3.41 X 24 - 16,400 Btu per 24 hr. To find toe hourly load, toe total 24 hr load is divided by the desired compressor operating time!' Iran B-ro/24 He Refrigeration equipment is designed to operate continu ously without ill effect, and it is the defrost problem that de termines the compressor, operating time. . When toe.refrigerant temperature is 30 For higher, there is no frost and,the general practice has been to select equipment Wall gain Air change Product Electrical f. , ' - 108,000 54,600 15,400 16,400.. based on 20 or 22 hr operation. Calculated Load 194,400 -.Table 7 .. . Heat Equivalent of Occupancy Cooler temperature F 50; 40 30 20 10 0 rrlO :. - `} Heat equfxotont/persoo Bfo/fcr 720 ' 840 950 1,050 ' r,200" 1,300 - .v.i;400 Add 10 percent safety factor Total load 19,440 .213,840 For 16 hr compressor operation, the load becomes 213,840/16 13,360 Btu per hr. For 20 hr compressor operation, the load is 213,840/20 =* 10,690 Btu per hr. _ Solution B: (Short,method using Table 8). The wall heat gain is the'same as in Solution A -- 108,000 Btu per 24 hr. The usage, beat gain (Table 8) is 55.2 Btu'per (24' hr) (cu ft) or 1540 X 55.2. Refrigeration, Load - 535 Table 8.... -Usage Heat' Gain, Btu: per 74' hr for he Cu Ff Interior Capacity Vofeme-or ff Service* Heavy Heavy 400-. 500 Average Average: Heavy 600* Heavy 800- 1,000 Heavy - r,2do' 1,500 Average: Heavy Average 2,000 - Long storage 3,ooa.. 5,000 7,500- 10)000;- : ; Long storage Long storage Long storage 20,000 .50,000 75,000' 100,000' Long storage Long storage Long storage Long storage remperotore difference (ambrent toeip mmus itorage room terqp), F deg l 40 50 55 40 65- 70 75 B0 90 100 -.4.68 - 5.51. 3.30 4.56 2.28 3.55 187 234 220 276 132 165 182 228 91 - 114 142 .177 258 281 . 305 328 . 351 374 ` 421 468 303 331 358. . .386 413 441 496 551 182 198 215 231 248 264 297 330 261 -274 297 319 . 342 365 410 456 126 137 148 160,- 171 .182 205 228 196 213 231 249. ` 267 284 320 355 1.85 74 ' 93 102 ni 120 130 " 139 148 107* 185 2.88 115 , 144 158 173 188 202 216 230 259 288 1.61 64 81 84 97 105 113 121 129 145 161 2.52 101. 126 139 151 164 176. 189 202 227- 252 1.38 55 69 76 83 90 97 103 110 124 138 2.22 90 Hi 122 133 144 156 166 178 200 222 1.30 2.08 1.24 1.96 1.21 1.87 1.17 1.85 1.11 1.76 1.10 1.67 .995 1.58 .920 1.50 .835 .775 52.083.2 49;.6. 78.4 48.4 74.8 46.8 74.0 44.4 70.4 44.0 66.8. 65 104 62 : 98 : 60.5 93.5 i 58.5 92.5 : 55.5 - 88.0 55.0 : 83.5 71.5 > 78 114 i 125 68.2 74.4 108 ;ns 66.6 - t 72.6 103 112 i 64 - 70 102 . 1U 61.1 66.6 96.8 106 60.5 66 91.9 , 100 39-8, 63.2, 36.8 60.0 33:4 31.0 : 49.8 : 79.0 ' 46.0 r 75.0 :'41.8 ; 38.8 54.7 86.9 50.6 82.5 45.9 42.6 59.7 94.8 55.2 90.0 50.1 46.5 84.5 135 80.6 128 78.7 122 76 120 72.2 115 .71.5 108 04.7 103 59.8 97.5 54.3 50.4 91. 97.5 104 146: , 156 `166 86.8 . 93 99.2 137 147 157 84.7' - 90.7 96.8 131 ' 140 150 1L7-. 187 H2 176 . 109 1'68 82- ; 83 : -94 130 , 139 148 77.r . 83.3 88.8 123: ` 132 141 77 ' 82.5 : 88 1L7 1 125 134 105 167' 100. 158 - 99 150. 69\7' 74.7 111. 119 64.4' 69 105 . 113 58-.5' ' 62.7 54.0 58.1 ! 79.6 126 73.6 120 66.8 62 .j89.6 142 82.8 135 ,'75.2 . 69.8 130 208 124 196 121 187 117 185 111 176 110 167 99.5 158 92 150 83.5 77.5 . .750 . 30.0 .576 23.0 .403 16.1 .305- 12.2 .240 1 9.6 .187 . .178 .176' .173 7*. 487.12. 7.04 6.92 37.5 41.3 28.8 31.7 20.2 - ' 22.2 15.3 10.8> 12.0 . 13-.2 45.0 48.8 52\5' 56.2 60.0 34.6 37.3 40.3 43.2 46.1 24.2 26.2 28.2 30.2 32.2 I8i.3! : 19.$ 1 21.4 : 22:0 - 24:4. : 14.4= i -15.6' .. 181& 18.0 19-.2- 67.5 75.0 51.8 " 5T.0 36.3 40.3 27*.5- > ,30.0 21-6- .-24.0 9.35 8.90 8.80 8.65 10.3 11.2 i2.2 ! .13.1' i 14.0 1 10.6 16U- nsiz! 9-79' . 10.7 ; 11.6 [ 12.5- . 13.4 14.2. 16.0. . 17.8 9.68 . 10*.61 1L.5 12.3 13.2 . 14a 15.8- , 17.6- 9.52 10.4- 11.2' 12. ! 13.0 13.8 ..15.6 : 17 . * Fw tad heavy terriec. producttoedis.b**ed.oa.product eotering-at 10 deg abovetherc&iseratCEtcapcratgreifbftlcog-stnrege. the catering* temperstore " egwarimstdy wpl to the refrigerator temperature. Where the product load ts unusual, do.not use this table. - 85,200 Btu per 24 hr. The total' Toad- is 108,000 + 85,200 193,2fX)-BtU'per 24 hr. ~'Fxampfe~3."Fmdthciotallosdfur a refrigerated-storage-room; 15 ft X 11 ft X 9 ft (outride dimensions) with 4 in. cork insulation, two triple-glass windows, each 6 ft X 5 ft. The ambient air tem perature is 90 F and the room temperature is to be 40 F. Two men work in the room from 8 A.M. to 4 P.M., and 800 watts of light ing and motors are in operation during the same period. The product, 2000 lb of beef, is to be cooled from 60 F to 40 F be tween 6 P.M. and 6 A.M. Softdtoa; The gross surface area of the outride walla is (15 X 9 + If X 9 + 15 X 11) 2 - 798 sq ft. The glass area is 6 X 5 X 2 ** i ft. The net insulated surface area is 798 -- GO -- 738 sq ft. From Table 1, the heat gain factor for the insulated wall is louad to be 90 Btu per (sq ft) (24 hr), while the factor for the glass ?rea 18 per (sq ft) (214 hr), for a 50 F deg temperature dif- terence. The heat gain through the walls is 739 X 90 = 66,400 *'P" 24 hr, and the heat gain through the is 60 X 350 " 21.000 Btu per 24 hr. _ The internal volume of the room is 14 X 10 X 8= 1120 cu ft. from Table 4, the number of air changes per 24 hr is found to be 16.6. The heat required to cool air from 90 F to 40 F is 2.26 Btu Per cu ft (from Table 5). The air change load is 1120 X 16.6 X 2.26-42,000 Btu per 24 hr._ Tne product load is .(2000 X 0.77 X 20) (24 + 12 - 61,600 Btu per 24 hr. ' The-load-due. to people is-2'X 750-X-24' = 30,000 Blupper 24 hr.The electrical load is 800 X 3.4i X 24 -- 65,500 Btu-per.24hr. In-order.tordetennine thehouriyfoad requirements,.a compari son of the day and'night' loads is necessary.' Item Btc m 24 Hn Htoar Dat Wall Load Air Change Load Product People Electrical Calculated Load 87,400 42,000 61,600 -- -- 191,100 87,000 42,000 -- 36,000 65,500 230,500 10% Safety Factor 19,100 23,100 Total 210,200 253,600 The day load is greater and should be used in selecting equip ment. Example 4: Determine the load for a banana storage room,