CHAPTER X CAPACITY OF REFRIGERATING MACHINES (Part 1)

CHAPTER X CAPACITY OF REFRIGERATING MACHINES In general there are two methods by which to determine the amomit of cooling effect produced by a refrigerating machine. Effect on Secondaky Refrigerating Mediums The first method is to measure the results produced as, for example, the cooling effect in degrees produced on a known quan- tity of brine or other mediums having known specific heats. The number of units of heat extracted per minute, hour or day being known, the capacity of the machine in pounds or tons of equivalent ice-melting capacity may be readily computed by dividing the number of units of heat extracted by the number of such units required in the same length of time to produce refrigeration at the rate of one ton per 24 hours. Effect on Primary Refrigerating Mediums The second method is to actually weigh or compute the number of pounds of the primary refrigerating fluid, assumed in this case to be ammonia, passing through the cycle of operations per unit of time and, by knowledge of the amount of refrigerating effect that a pound of the refrigerating fluid is capable of producing under the observed conditions of back pressure and liquid temperature,* finally arrive at a more or less accurate estimate of the amount of cooling effect being produced. Units The units involved in making such calculations are as follows: (1) British Thermal Unit (B.Lu,). — Equivalent to the specific heat of water, or the amount of heat required to raise the tempera- ture of a pound of water through V Fahrenheit at its temperature of maximum density, 39° Fahrenheit. (2) Pound of Refrigeration, — Equivalent to the expenditure of negative heat (absorption of heat) at the rate of 144 B.t.u. per * On account of the usual inaccuracy of pressure gauges, accurate determi- nations of liquid temperatures can best be made by means of thermometers set in mercury wells inserted in the liquid lines just before they enter the expan- sion valves. Digitized by V^OOQIC CAPACITY OF REFRIGERATING MACHINES 127 twenty-four hours; 144 B.t.u. being the latent heat of ice, or the amount of heat required to melt a pound of ice "from and at" 32° Fahrenheit (ice at 32*^ melting into water at the same tempera- ture). (3) Ton of Refrigeration. — Equivalent to the expenditure of negative heat (absorption of heat) at the rate of 2,000X144 B.t.u., or 288,000 B.t.u. per twenty-four hours; 288,000 B.t.u. being the amount of heat required to melt a ton of ice "from and a4i" 52*^ Fahrenheit. From the foregoing it is apparent that the capacity of a refrig- erant, which extracts heat at the rate of 288,000 B.t.u. per twenty- four hours, is one ton. The above rate is equivalent to 288,000-5- 24 = 12,000 B.t.u. per hour, or 288,000 -f- (24X60) =200 B.t.u. per minute. To absorb heat at this rate certain definite quantities of the primary refrigerating fluid must be evaporated per minute, hour or day; in addition to which if a secondary refrigerating medium, such as brine, is employed, a certain quantity must be cooled through a suflScient range of temperature to supply the amount of heat required to evaporate the primary fluid. Standard Conditions It has been proposed to employ as standard 185 pounds head pressure — under which pressure condensed anhydrous ammonia leaves the condenser at about 90° Fahrenheit — and 15.67 pounds back pressure, corresponding to an evaporating temperature in the cooler of 0° Fahrenheit. Under these standard conditions, where the anhydrous ammonia enters the refrigerator at 90° Fahrenheit and evaporates at 0° Fahrenheit, from 27 to 28 poimds of liquid must be evaporated per hour per ton of refrigerating effect produced. This means that the ammonia compressor must have an effective displacement capacity of about four cubic feet per minute per ton. Effect of Pressure The refrigerating capacity of evaporating refrigerants does not depend on the volume of gas evolved, except as volume depends on weight. The volume of gas varies widely with the pressure, but aside from the cooling effect that must be expended on the liquid to cool it from the condenser temperature down to the cooler tem- perature, the weight of refrigerant per unit of cooling capacity Digitized by V^OOQIC 128 ELEMENTARY MECHANICAL REFRIGERATION produced is practically constant under all conditions of tempera- ture and pressure.* At 10 pounds back pressure, for example, corresponding to an evaporating temperature of about 8° below zero, the volume of am- monia gas is 10.8 cubic feet per pound. At 32 pounds back pres- sure, corresponding to an evaporating temperature of about 19**, the volume is about 6 cubic feet. This means that operating at 10 pounds back pressure a compressor must pass approximately 66 per cent more volume of gas per unit of capacity than when operating at 32 pounds pressure. Computed Capacity — Example The method of arriving at the size of the compressor required for the performance of a given refrigerating duty per twenty-four hours is as follows: The latent heat, or the amount of heat ex- pressed in B.t.u., required to evaporate a pound of anhydrous ammonia at 10 pounds back pressure is about 560. If, for example, 150 pounds condenser pressure is carried, the liquid ammonia will pass to the refrigerator at about 84 degrees. The evaporating temperature corresponding to 10 pounds back pressure is —8°, making a difference of temperature of 92° through which the liquid must be cooled before it can produce any useful cooling effect in the refrigerator. For approximations,- the specific heat of the ammonia liquid may be taken to be the same as that of water, i.e., unity, in which case the expenditure of 92 B.t.u. will be neces- sary to cool the liquid from the temperature of the condenser to that of the refrigerator. Subtracting this amount from the latent heat — 560 B.t.u. — we have 468 B.t.u. remaining for the perform- ance of useful work in the refrigerator. Pounds Refrigeration The latent heat of ice, taken as the unit pound capacity of refrigeration, is 144. The evaporation of one pound of anhydrous ammonia under the foregoing conditions will produce 468 -r- 144 = 3.25 pounds of refrigeration; this regardless of the time in which the evaporation takes place. Pounds of Ammonia The evaporation of one pound of anhydrous ammonia per mih- This presupposes only limited superheating when the refrigerant in question is in the gaseous state. Digitized by V^OOQIC CAPACITY OF REFRIGERATING MACHINES 129 ute under these conditions would produce refrigeration at the rate of 468^200 = 2.34 tons per twenty-four hours. At the rate of one pound per hour the tonnage rate per twenty-four hours would be 468-^12,000 = 0.039. Cubic Feet of Ammonia At 10 pounds back pressure each pound of anhydrous ammonia evaporated has a volume of about 10.8 cubic feet. A compressor displacing this volume of gas per minute will have a capacity of 2.34 tons, or a ton capacity for every 10.8 -^2.34 = 4.6 cubic feet of piston displacement per minute. Capacity of Compressor The method of computing the capacity of a refrigerating ma- chine of a given size when operating at a given number of revolu- tions, from the apparent displacement in cubic feet per minute, involves two other very important factors: First, the back pres- sure at which the compressor is operated; second, the displacement efficiency of the particular compressor in question when operating at the back pressure in question. The amount of refrigeration produced is directly dependent on the number of pounds of the refrigerating fluid evaporated, due allowance being made for the range of temperature through which the liquid must be cooled before it can do useful work in the refrig- erator, and the cooling effect available from superheating of the gas. Displacement Efficiency of Compressor When the compressor itself is employed as a meter, i.e., when the amount of the refrigerating medium is computed from the number of cubic feet of volume swept out per minute by the piston, it is necessary to assume or determine the compressor displacement efficiency in order to arrive at the actual displacement from the apparent displacement indicated by the volume swept out. The weight of a refrigerating medium vapor is directly proportional to its absolute pressure. For a given back pressure the weight of the gas per cubic foot can be determined directly from published tables of the properties of the refrigerating medium in question. * According to the standards adopted by the Ice Machine Builders in 1903, the evaporation of 27.7 pounds of ammonia under the "standard" conditions of 185 pounds head pressure (90** Fahrenheit) and 15.67 pounds back pressure (0® Fahrenheit) constituted a ton capacity. On this basis, approximately 5 cubic feet of displacement would be allowed per ton per twenty-four hours, in a compressor of 80 per cent, displacement efl&ciency. Digitized by V^OOQIC 130 ELEMENTARY MECHANICAL REFRIGERATION The displacement efficiency of compressors, or the ratio of their apparent to their actual cubical displacement, is not easy to determine, and it is the exception rather than the rule that the builders of compressors can themselves give the efficiency of their various machines when operated under different heads and back pressures. As a matter of fact, the only exact means of determining such efficiencies is by the laborious method of checking the amount of refrigerating fluid apparently passed through the compressor, with the amount, determined by weight, actually passed through the expansion valve. Even this method has a serious shortcoming in the case of "wet" compression machines in that a considerable amount of unevaporated liquid may pass through the compressor and this liquid appearing in the weights of liquid passed through the expansion valve indicates a higher displacement efficiency than the compressor deserves'. Apparent Cubical Displacement To compute the refrigerating capacity of an ammonia com- pressor of the assumed cylinder dimensions of 19x38 operating at 45 revolutions per minute under the above suggested standard conditions of 185 pounds head and 15.67 pounds back pressure: If r = radius, or one-half cylinder diameter, Z= length of stroke, both in inches, n= number of revolutions per minute. Then il = area of a 19-inch piston = [1] A = 7tr^ or 3.1416X9.5X9.5=283.53 sq. in; F= volume of 19x38-inch cylinder. „,^ Ttr'l 283.53X38 ^ ^^^ [2] V^—- = -^-- — =6.235 cu. ft. ^ 1728 1728 Z)= apparent displacement per minute double-acting compressor. [3] Z)=2 ^^1^ =2X6.235X45 = 561.15 cu. ft. Somewhat simplified this expression becomes [4] I)=rf2 i n 0.00090903; or, since I n 2 I n — =— — = piston speed P S in feet per minute. [5] Z)=d2 PS 0.00545418 and 19X19x885X0.00545418 = 561.15 cu. ft. Digitized by VjOOQIC CAPACITY OF REFRIGERATING MACHINES 131 Effect of Working Pressures The amount of heat absorbed in the evaporation of any liquid depends on the temperature of the liquid supplied and the pressure under which it evaporated as well as on the latent heat of vaporization of the liquid. Water, for example, fed to a boiler at a temperature below that of the steam generated must be heated up to the boiling point before it can be evaporated. Liquid ammonia fed into expansion coils at a temperature above that of the ammonia vapors must similarly be cooled down to the boiling point corresponding to the pressure. Since the only means of cooling the latter is by its own evaporation, it is evident that just so much liquid as evaporates in cooling itself can do no useful refrigerating work on other products. The greater the range in temperature through which it must be cooled, whether on account of the abnormally high liquid temperature or low evaporating temperature, the less will be the useful cooling effect available per pound of refrigerant. The total heat absorbing capacity, Rbp oi a pound of liquid refrigerant, is known as its latent heat of vaporization R. And its value depends on the back pressure 6 p at which it evaporates. The amount of heat Q that must be extracted to cool a pound of liquid refrigerant from the temperature corresponding to the head pressure h p down to that of the back pressure 6 p is the difference between the values of the "heat of the liquid" Q imder the two conditions. The available cooling effect C per pound of refrigerant is accordingly [6] c Rbp ~ Qhp Qbv Available Latent heat Heat of the Heat of the Cooling ef- ■ = ■ of vaporiza- ■- liquid under — ■ liquid under fect per tion at back head pres- back pres- pound' . pressure b p. - sure sure As tables of the properties of refrigerating mediums do not always give values for the "heat of the liquid" at different temperatures, the number of heat units required to cool the liquid may be arrived at by multiplying the number of degrees (<i-fe) through which the liquid is cooled by S, the specific heat of the liquid, and expression [6] becomes [7], in which have also been substituted values corresponding to standard conditions. Digitized by V^OOQIC 132 ELEMENTARY MECHANICAL REFRIGERATION Since the displacement of ammonia compressors is expressed in cubic feet, the available cooling effect per cubic foot of gaseous [ Rbp 1 f «i ^ [ <« 1 f ^ ^ Latent r s 1 Temper- temper- Available heat of Specific ature ature cooling effect per pound— evapo- heat corre- corre- m ■ = ■ ration at taken as ■ sponding — sponding back unity to head to back 465.5 pressure (15.67) 555.5 1 pressure hp 90** pressure bp 0^ B.T.u. PER Cubic Foot refrigerant passing through the compressor is most frequently em- ployed. This may be readily detennined by dividing the expres- sion [7] by Vy the volume occupied by a pound of ammonia under back pressure b p and for standard conditions becomes [8]C= RbpXS {ti-h) 555.5- 1 (90-0) = 51.56 B.t.u. V 9.028 If the apparent displacement in cubic feet per minute as deter- mined by expression [5] be multiplied by the cooling effect per cubic foot as determined by expression [8], the result will be the apparent capacity of the compressor expressed in B.t.u. per minute, which value divided by 200 gives the apparent tonnage capacity of the compressor per twenty-four hours. Expressed as an equation: f D \ f C I Apparent 1 v i ^^^li^^g effect I displacement / I in B. t. u. per \ per min. / • \ ^"* ^^' Cooling effect in tons per 24 hrs. 561.15X51.56 200 / B. t. u. per min. I equivalent to one \ ton per 24 hrs. = 144.66 tons. Cubic Feet Displacement per Ton Since 200 B.t.u. per minute is the equivalent of a ton per twenty-four hours, 200 divided by 51.56, the number of B.t.u. of cooling effect available per cubic foot under standard conditions ♦ For various values of the specific heat of anhydrous ammonia determined by a number of authorities see Transactions of A. S. M. £., pp. 522-3, Vol. 26,1905. Digitized by V^OOQIC CAPACITY OF REFRIGERATING MACHINES 133 gives 3.88, the number of cubic feet of gas that must be actually displaced per ton. Assuming that the compressor has a displace- ment efficiency of 80 pet/cent the apparent displacement per ton will be 3.88 ^.80 = 4.85 cubic feet.* Approximate Nominal Capacity of Compressors Dividing the constants of equations [4] and [5] by 5 (see foot- note) gives equations [9] and [10], which will be foimd convenient in arriving at the approximate nominal capacity of a compressor when its dimensions and speed of operation are known. [9] Tons = d' in 0.000181806 [10] Tons =d' PS 0.00109081 Substituting values as above, = 19X19X38X45X0.000181806 = 112.23 tons. = 19X19X285X0.00109081 = 112.23 tons. The above equations give approximate results only, and only in the special case of standard conditions and approximately 80 per cent compressor displacement efficiency. To find the capacity of a compressor in the general case in which it is operated at other back and head pressures, the cooling effect per cubic foot actual displacement must be determined in each case by equation [8]. Furthermore, since the efficiency of ammonia compressors is subect to wide variations, both through diversity of design and diversity of operating conditions to which the same machine is often subjected, the efficiency of the individual compressor should be determined under its own operating conditions. This can be accomplished most accurately by determining the quantity of refrigerating fluid actually passing through the system and com- paring this amount with the apparent amount computed from the displacement of the compressor. Capacity Determined by Test — ^Weighing Primary Refrig- erant The best way to determine the amoimt of liquid refrigerant is to weigh it. Fig. 41 represents a condenser, a pair of weighing tanks and their connections. For testing, crosses are inserted in * The above is arrived at on the basis of unity taken as the latent heat of anhydrous ammonia. The somewhat higher value of 1.1 sometimes employed would make the required actual displacement about 4 cubic feet and the apparent about 5. Digitized by V^OOQIC 134 ELEMENTARY MECHANICAL REFRIGERATION the inlet and outlet lines, and valves and additional pipes are attached as indicated. When usiug the weighing tanks, the outlet valve D is closed or blanked off and, as the valves in the new con- nections are closed, the liquid refrigerant collects in the receiver. from Compressor^ ^frmtmrm I To Refrigerator Fig. 41. — Diagram of System for Weighing Liquid Refrigerant The weighing tank A is filled by opening valves E and G, after which valve G is closed and the gross weight of the tank and its contents is determined. The weight of the refrigerant is then found by subtracting tlae net weight of the tank and the liquid in the bottom connections. While the liquid in tank A is being weighed, tank B is supplying the cooler through valve J. Alter- nate filling and emptying of the two tanks