CHAPTER XI COLD STORAGE DUTY (Part 1)

CHAPTER XI COLD STORAGE DUTY It is obviously impossible to detennine with any great degree of accuracy how much refrigeration it will take to cool a number of differently shaped cold storage boxes, built by imknown methods of construction, of several different insulating materials of un- known eflSciencies and various states of preservation, into which heat is admitted through the opening of doors and radiated from lights and workmen, and containing unknown quantities of dif- ferent kinds of products of varying heat absorbing capacities, stored for different lengths of time. When the above list of unknown quantities can be suflSciently reduced, however, calculations of the amoimt of cold storage capacity required to satisfy a certain set of conditions can readily be made. Since lights and workmen must necessarily be employed to a greater or less extent, and since no insulation can entirely prevent the inflow of heat, only a part of the refrigeration pro- duced by the refrigerating plant can be put to the useful work of cooling the stored products, the remainder being dissipated. Attempts are sometimes made to estimate cold storage duty by determining the number of cubic feet of space to be cooled and dividing that by the nmnber of cubic feet that a ton of refrigera- tion is supposed to cool imder average conditions. While it may be interesting to know the amount of space cooled by a ton of refrigeration under more or less similar conditions, such compari- sons are not only meaningless but are positively misleading, when the many varying conditions of operation are not definitely known. Since such calculations are at best inaccurate, they should be made only on the basis of the greatest number of known quanti- ties and carefully worked out assumptions regarding the remaining unknown quantities. These determinations may be simplified by the following brief method, which, together with the accompanying series of tables, will, when judiciously applied, be found accurate enough for all ordinary commercial requirements. The total amount of heat that the refrigerating machine must Digitized by V^OOQIC COLD STORAGE DUTY 149 remove from the cold storage compartment, is made up as fol- lows: a. Latent and sometimes specific heat of the products stored. b. Heat evolved by lights. c. Heat given ofif by workmen. d. Heat absorbed in the precipitation and freezing of moisture. e. Heat of air entering through open doors. f. Heat entering through the cold storage insulation. Cooling the Product The amount of refrigeration required to cool a given amount of food product through a given range in temperatiu'e is a prac- tically fixed quantity for a given product, but varies widely with different products. When cooling is not to be carried below the freezing point the amount of refrigeration required may be found by multiplying the specific heat of the product by the number of degrees through which it is to be cooled. If the product is also to be frozen, this amoimt of refrigeration must be increased by the amount of the latent heat of fusion, and if cooling is to be con- tinued below the freezing point, the refrigeration must be further increased by the specific heat of the product below 32° multi- plied by the number of degrees through which it is cooled below freezing point. The specific and latent heat of a number of products commonly preserved in cold storage are given in the following table: TABLE XVII—PROPERTIES OF FOOD PRODUCTS Product Specific Heat Above 32«F. Latent Heat of Freezing Specific Heat Below 32°F. Beef— lean .77 .60 .64 .68 .76 .82 .90 .67 .84 .80 .51 102 72 84 100 111 124 ii4 105 55 .41 Beef — fat .34 Butter .84 Cream .38 p}<rars . .40 F§^ ;;:::::::::::::::::::::: :::::: .43 Milk .47 Mutton .84 Oysters .44 Poultry .42 Pork — fat .30 Example.— It is required to cool 10,000 pounds of freshly killed poultry through 68° Fahr. The specific heat as given above is 0.80. The number of B. t. u. to be removed will be 0.80X10,000X68 = 544,000. Dividing this result by 144 (number of B. t. u. per pound of refrigeration), the amount of cooling duty is found to be 3,777.7 pounds. If the poultry is frozen, the addi- Digitized by V^OOQIC 150 ELEMENTARY MECHANICAL REFRIGERATION tional refrigeration required will be 10,000 X 105 = 1,050,000 B. t. u. or ( -5- 144) 7,292 pounds, and if additional cooling to 0^ Fahr. is required, the additional cold necessary will be 10,000X0.42X32 = 134,000 B. t. u. or 933.3 pounds. The total refrigeration duty required to cool the products through 68® Fahr., freeze it at 32® Fahr., and then chill it to 0® Fahr., would be 3,777.7+7,292-1- 933.3 = 12,003 pounds or dividing by 2,000 (pounds per ton), 6 tons. The following table may be found convenient in estimating the amoimt of refrigeration required to chill beef, pork, and sausage through 64"^ Fahr., or from 104° Fahr. to 40° Fahr.: TABLE XVIII— REFRIGERATION REQUIRED TO COOL MEATS Sausage Products — Poultry Fat Medium Lean Fat (16% Water) Specific Heat B.t.u. to cool 1,000 pounds VFahi B.t.u. to cool 1,000 pounds 64'Fahr Pounds Refrigeration per 1,000 pounds (64°Fahr.) . Pounds of Meat cooled 64° per ton Refrigeration Average wt. Carcass Carcasses cooled per ton . . Poultry Beef Fat Beef Medium Beef Lean Pork Fat .80 .60 .68 .77 .51 800 600 680 770 510 51200 38400 43520 49280 32640 355.55 266.66 302.22 333.66 226.66 5,625 7,500 750 lbs. 10 6.615 750 lbs. 8.82 5,844 750 7.78 8,765 250 35.3 .65 650 41600 228.88 6,923 It will be noted that ten 750 pound fat beeves, and thirty-five 250 pound hogs require one ton of refrigeration for the cooling of the meat alone. In estimating the cooling capacity of a medium for packing house work, a ton of refrigeration is allowed for from five to seven beeves weighing from 700 to 750 pounds, and for from fifteen to twenty-four hogs weighing 250 pounds. Still another rough rule sometimes employed is to allow a ton of re- frigeration for from 3 to 4,000 pounds of meats cooled. These larger figures are intended to give ample reserve capacity to pro- vide for ordinary insulation and other losses encountered packing house practice. m Water Cooling Since the specific heat of water is unity, the number of heat imits to be extracted in order to produce a given drop in temper- ature of a given quantity of water is found by simply multiplying the weight in pounds by the range cooled through in degrees. If, for example, 20,000 pounds of water is to be cooled one degree, 1,000 pounds 20 degrees, 400 pounds 50 degrees, or in fact any number of pounds through a range of temperature that Digitized by V^OOQIC COLD STORAGE DUTY 151 will give a product 20,000 pound degrees, 20,000 B. t. u. will be required for the cooling. One U. S. gallon of water at 62° F. weighs 8.336 pounds. The cooling of 20,000 gallon degrees will accordingly require 8.336 X 20,000 or 166,720 B. t. u. If the cooling is accomplished in 24 hours the amount of refrigeration required will be 166, 720 -^ 288,000 = .5789 tons. If done in one hom* the equivalent rate per 24 hours will be 24 times as great or .5789X24 = 13.893 tons. Table XIX shows the amoimt of refrigeration required to cool 1,000 gallons of water per minute, hour and 24 hours through different ranges of temperature. If 1,000 gallons of water be cooled 50 degrees in one hour, the equivalent cooling efifect per 24 hours will be ten times the value given in the table for 5 degrees, or 34.733 tons; if 53 degrees, ten times the value for 5 degrees, or 34.733 plus that given for 3 degrees or 2.0840, making 36.817 tons. TABLE XIX— REFRIGERATION REQUIRED TO COOL WATER — GALS. — DUTY IN TONS PER 24 HOURS 1,000 Gals. Cooled per | 1,000 Gals. Cooled per Degrees Degrees Cooled Cooled Minute Hour 24 Hours Minute Hour 24 Hours 1 41.68 0.6946 .02894 21 875.28 14.5879 .60778 2 83.36 1.3893 .05789 22 916.96 15.2824 .61672 3 125.04 2.0840 .08682 23 958.64 15.9770 .65564 4 166.72 2.7786 .11577 24 1000.32 16.6716 .69456 6 208.40 3.4733 .14471 25 1042.00 17.3664 .72352 6 250.08 4.1679 .17364 26 1083.68 18.0612 .75248 7 291.76 4.8646 .20259 27 1125.36 18.7598 .78142 8 333.44 5.5590 .23154 28 1167.04 19.4584 .81036 9 375.12 6.2519 .26048 29 1208.72 20.1491 .83931 10 416.80 6.9466 .28942 30 1250.40 20.8399 .86826 11 458.48 7.6412 .30836 31 1292.08 21.5379 .89721 12 500.16 8.3358 .34728 32 1333.76 22.2360 .92616 13 641.84 9.0306 .37624 33 1375.44 22.9289 .95510 14 583.52 9.7292 .40518 34 1417.12 23.6218 .98404 15 625.20 10.4199 .43413 35 1458.80 24.3147 1.01298 16 666.88 11.1180 .46308 36 1500.48 25.0076 1.04192 17 708.56 11.8109 .49202 37 1542.16 25.7023 1.07086 18 750.24 12.5038 .52096 38 1583.84 26.3970 1.09980 19 791.92 13.1985 .54990 39 1625.52 27.0918 1 . 12874 20 833.60 13.8933 .57884 40 1667.20 27.7866 1.15768 Wort Cooling Table XX shows the amount of refrigeration expressed in tons per 24 hours required to cool 100 bbls. of water per hour and one bbl. per minute through different ranges of temperature. To apply this table to wort cooling multiply the number of bbls. of 31 gallons by the specific gravity of the wort and this product by the specific heat of the wort corresponding to the specific gravity as shown in Fig. 43. Digitized by V^OOQIC 152 ELEMENTARY MECHANICAL REFRIGERATION TABLE XX— REFRIGERATION REQUIRED TO COOL WATER (BBLS. OF 31 GALS.) DUTY IN TONS OF REFRIGERATION PER 24 HOURS Degrees 100 Bbla. of 31 1 Bbl. of 31 Degrees 100Bbls.of31 1 Bbl. of 31 Cooled Gals., per Hour Gals., per Min. Cooled Gals., per Hour Gals., per Min. 1 2.1645 1.2927 21 45.2445 27.1467 2 4.3090 2.5854 22 47.3990 28.4394 3 6.4635 3.8781 23 49.5535 29.7331 4 8.6180 5.1708 24 51.6988 31.0248 6 10.7725 6.1635 25 53.8625 32.3175 6 12.9270 7.7562 26 66.1370 33.6102 7 15.0815 9.0489 27 68.1715 34.9029 8 17.2360 10.3416 28 60.3260 36.1956 9 19.3905 11.6343 29 62.4806 37.4883 10 21.5950 12.9270 30 64.6350 38.7810 11 23.6995 14.2197 31 66.7895 39.0737 12 25.8494 15.5124 32 68.9440 41.3664 13 28.0685 16.8051 33 71.0985 42.6591 14 30.1630 18.0978 34 73.0530 43.9518 15 32.3175 19.3905 35 75.4075 45.2445 16 34.4720 20.6832 36 77.5620 46.5372 17 36.5265 21.9759 37 79.7165 47.8299 18 38.7810 23.2686 38 81.8710 44.1226 19 40.9355 24.5613 39 83.9255 60.4153 20 43.0900 25.8540 40 86.1800 51.6581 TABLE XXI— PRODUCTS OF SPECIFIC GRAVITY AND SPECIFIC HEAT OF WORT OF DIFFERENT PER CENT STRENGTH Strength % Product Strength Product 8 9 10 11 12 13 14 0.9742 0.9741 0.9665 0.9631 0.9609 0.9571 0.9536 15 16 17 18 19 20 666666 • Refrigeration required to cool wort »that required to cool equal quantity of water, multi- plied by the above "product" corresponding to strength of wort. Fig. 43 Digitized byCjOOQlC COLD STORAGE DUTY 153 Suppose it is desired to find the amount of refrigeration neces- sary to cool 100 bbls. of wort per hour having a strength of 12% through a range of 40'' F. It is found from table 20, the refrigeration required to cool a like amount of water is 86.18 tons. Since the specific gravity of the wort, as determined from Fig. 43, is 1.049, the weight of the wort cooled will be 1.049 times as great as for water; but since the specific heat is only .916 the refrigeration per pound will be only .916 as great. The product of these two factors .9609 — given for different strengths of wort in Table XXI — shows that the amount of refrige- ration required to cool a given quantity of wort of 12% strength is .9609 as great as for the same quantity of water or in the above example of 100 bbls. per hour, .9609X86.18 or 82.80 tons. The same result might have been obtained from Table XIX by first reducing the quantity in bbls. to gallons, or direct from the following equation in which for clearness the above values have been substituted. / Bbls. of\ / Gala. \ /Wt.water\ / No. de- \ / Specific \ / Specific \ ( wortper \ I V^t bbl. W per gal., \ / grees F. \ / gravity A [ heat of A / !>« of V hour M wort M lbs. ){ cooled M of wort M wort If ^i^ \B=100/ \ G=31 / \ L=8.34 / (ti-t«)=40/ \««--i ^o/ ^a— 01 «/ ' Jjoolmg. d y V of wort / V wort I / coolimt =40/ \Sq=1.049/ \8p=.916 /_ t^"^, ;eration per 24 hour8= 12000) \ 24 hre. (B.t.u. per hour equivalent to a ton of refrigeration per 24 hour8= 12000) ^ [16] ^T = 82.80^ This expression when applied to wort cooling expressed in bbls. of 31 gallons cooled per hour, becomes: Tons = .021545 X {t - h) Xsg. Xsh, in which (t - ti) = range of temperature cooled through. (40® F.) /Sgf.= specific gravity of the wort. (1.049) /SA. = specific heat of the wort. (.916) which values substituted in the above equation give tons per 24 hours =82.80 as above. Lights The heat generated by artificial lights in cold storage com- partments often becomes of considerable importance. The amount of refrigeration required to neutralize the heat radiated by electric lights depends on the candle power of the lights and their efficiency, expressed in watts per candle power. If, for example, an ordinary low efficiency lamp consuming 3.5 watts per candle power is employed, the heat per 16 c. p. will be 16X3.5X0.05685=3.1836 B.t.u. per minute or 191.0 B.t.u. per hour, which quantities divided respectively by 0.1 and 6.0, the number of B.t.u. per minute and hour equivalent to a pound of Digitized by V^OOQIC 154 ELEMENTARY MECHANICAL REFRIGERATION refrigerating duty per twenty-four hours, gives 31.83 pounds. One ton of refrigeration equivalent to 2,000 pounds of ice melting capacity will accordingly be required for every sixty-three lights. The heat generated by an ordinary gas light is 4,800 B. t. u. per hour, to absorb which requires 800 pounds of refrigeration per twenty-four hours (24 X 4800 -r 144 = 800) . On this basis two and one-half gas lights will absorb a ton of refrigeration per twenty- four hours (2000^800 = 23^). Similarly each workman employed in cold storage compart- ments radiates about 500 B. t. u. per hour, equivalent to 83J pounds of refrigeration (24X500-7-144 = 835). On which basis a ton of refrigeration will have to be supplied for every 24 work- men per twenty-four hours (20004-83i = 24). In commercial practice some authorities allow a ton of refrigeration for every 30 workmen, presupposing a radiation of 400 B. t. u. per man. Air Cooling and Moisture Precipitation Before proceeding to illustrate the method of calculating the amoimt of refrigeration required to cool a mixture of air and water vapor it may be advisable to define terms. Air is a mechanical mixture of nitrogen and oxygen in the prac- tically constant proportion of 80 parts of the former to 20 parts of the latter, a very small per cent., about 3 or 4 hundredths of one per cent, of which is replaced by carbon dioxide. Into this uniform mechanical mixture water vapor enters in widely varying proportions. When the air contains all the moisture that it can hold it is said to be saturated. The higher the temperature of the air the more water vapor it is capable of absorbing before be- coming saturated. At a given temperature saturated air always contains a certain fixed quantity of water vapor. It must be remembered, however, that the temperature of the air does not fix the amount of moisture that it contains except in the limiting case of saturation. In the general case the air is not saturated, and may contain different amounts of water at the same temperature as it varies in degree of saturation; or it may contain different amounts of moisture at the same degree of satura- tion at different temperatures. Since the amount of water that it is possible for air to hold in suspension increases with increasing temperature, and decreases with decreasing temperature, it is evident that the air may or may not contain less moisture after Digitized by VjOOQIC COLD STORAGE DUTY 155 H ^ O Bo OH O ^ § W P3 ^ 6§ 6? 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Table XXII shows the amount of vapor in poimds per thou- sand cubic feet of air at diflEerent degrees of saturation at different temperatures. At 100® Fahr., for example, 1,000 cubic feet of saturated air will contain 2.82 pounds of water vapor, while at 75"^ Fahr. the amoimt is only 1.33 poimds, or less than one-half that quantity, and at 15° Fahr. it is still further reduced to about one-tenth of what it is at 75® Fahr. In the general