of air to be passed through it. Stacks are arranged in series, in parallel, or in series-parallel, all depending on the desired increase in air temperature, the volume of air to be heated, and the pattern of the radiator itself. If the volume of air to be heated is small, as in house heating, the problem of designing and selecting indirect radiators is simplified, for often one radiator, one stack deep, is sufficient to take care of the volume of air. Fig. 18 shows a typical cast-iron indirect radiator arranged to take air from out of doors and discharge it by gravity through a flue into a room. These radiators are sometimes arranged so that all or part of the air is passed through or bypassed around the stack, with two dampers working in conjunction in such a manner that the air may be tempered and mixed at will. The dampers may be operated by hand or with a thermostat. If the volume of air to be heated is large, the problem is somewhat more involved, due to the large number of variables. All indirect heating problems, however, are solved in about the same general manner, whether the FIG. 18.-Indirect heating stack and hot air flue for steam or water. volume of delivered air is large or small. If the volume is large, the radiators may either be made larger or a number of them may be placed in parallel, or both, to increase the area through which the air passes, known as the free area. If the outside temperature is low, the amount of heating surface per square foot of free area must be increased, either by placing the stacks in series, or by selecting one stack whose particular design gives the necessary amount of heating surface per square foot of free area. Ns = The amount of heat to be delivered in the air in the rooms is the product of the volume, the rise in temperature, and the specific heat of the air, but since the volume and temperature rise depend on the square feet of heating surface N8 per square foot of free area, A = f, and the velocity of the air, the heat to be delivered also depends on the rates, f, and the velocity of the air, N being the number of stacks in series. If a velocity is assumed, and the temperature range, temperature conditions, and volume of air are known, the value of ƒ for those conditions may be computed from equation 6 or 7, Art. 166, and a stack of a certain pattern or a number of stacks in series of a certain pattern may be selected from Table 20. (See Art. 16a for method of computing the volume of air necessary.) After the pattern of stack and the number of stacks in series are selected, it is Q. necessary from the volume and the velocity to compute the free area from A = Following this, with values of the free area A, and the ratio f (Table 20), the total heating surface may be computed from Ns = Af. The B.t.u. per hr. required value of Ns thus computed should agree with the value of S computed from S = The (0.0m) X K transmission factor K is defined in Art. 16b. (0.0m) is the mean temperature difference between the steam and air and is computed from Eq. 3, Art. 166. With the total heating surface, the heating surface per section, and the number of stacks in series, the total number of sections and the number of sections per stack may be computed. (See illustrative problem in Art. 16c.) Where the radiators are in series, and N is used to denote the number of stacks, it is easily seen that the heating Na surface is increased while the free area remains constant, thus increasing the value of f. In such a case, f= 8 S Where one stack deep only is used, this reduces to f The above considerations assume that all the = = radiators are in series. Practically, where the volume is so great that a radiator is too long if it is built up of enough sections to satisfy the heating demand, it may be cut in two or more pieces making two or more radiators in parallel; in other words, two or more stacks in parallel. In that case, N = 1 the same as before, N only referring to the number of stacks deep or the number of stacks in series. The quantity & is the surface of all the radiators in any one row in parallel. 16a. Ventilation with Indirect Heating.-Care should be taken in assuming a temperature at which the air is to enter. The best temperature to assume depends on the desired number of air changes in the room per hour and varies between being too cool to be effective and too hot for comfort. The number of air changes per hour, neglecting infiltration, is the volume of air necessary to furnish the B.t.u. losses divided by the cubic contents of the room. If, by trial, certain temperature assumptions give too great or too little air change, new assumptions will have to be made before starting the design of appropriate stacks. Since the air leaves the room through foul air ducts at room temperature and is discharged into the atmosphere, which is often much below zero, it is easy to see that more heat must be furnished by the indirect radiators than is necessary to supply the heat losses; in fact, the exact amount to be furnished is that lost in the ducts, that used to supply the room losses, and that which is lost to the outside atmosphere through the vents. For example, if the outside atmosphere were at 10 deg. F., the air leaving the radiator at 125 deg. F., the temperature drop in the ducts at 5 deg. F., the temperature of the air entering the room at 120 deg. F., and the room temperature 70 deg. F., the TABLE 20.-DATA ON TYPES OF CAST-IRON INDIRECT RADIATORS For friction in lb. per sq. ft. for any velocity, divide by 150, square the result, and multiply by quantity in last column. For inches of water, multiply this result by 0.1925. Multiply by number of stacks deep for total friction. ratio of total heat to the useful heat is the same as the ratio of the temperature differences; thus, using the above temperature assumptions, or Total heat (125 deg. 10 deg.) 135 70 deg.) 50 B. t. u. losses X 135 50 Again, using these same temperature conditions, the average weight of air per cu. ft. specific heat 0.24, the required volume of air in cubic feet is = 166. Heat Given Up by Indirect Radiators.-The heat transmitted by indirect radiators depends on two things: the transmission factor K, and the mean temperature difference between the fluids, in this case either steam and air or hot water and air. The transmission factor K is the number of B.t.u. per square foot of heating surface per hour per degree of mean temperature difference between the fluids. It varies only with the velocity DIAGRAM 5.-Condensation chart to be used in connection with Diagram 6. Diagram 5 gives values of K for various velocities. The writer has assumed that a velocity was known and that the relation would hold true whether the air was circulated by gravity or by fan. The mean temperature difference of fluids transmitting heat through surfaces is, in any case, 1 Formulas given were deduced from experiments by F. L. Busey and W. H. Carrier and are considered classic contributions to the science of Heating Engineering. They are used by all blower and heater maufac turers for their capacity tables. DIAGRAM 6.-Air temperature increments for cast iron and pipe coil indirect radiation under various conditions of steam and water pressures and temperatures. Diagram 6 gives values of ƒ for various temperature conditions. 16c. Illustrative Problem.-Design a heater to furnish heat to room 1, second floor, Fig. 7, p. 1094. 96,430 B.t.u. per hr., cubic contents = 34,272 cu. ft., and air changes due to From Table 13, total heat loss infiltration = 0.375 per hr. Assume = This is good ventilation, therefore the temperature conditions were well chosen for outside air at -10 deg. F. By inspection, it will be seen that if the temperature of the atmosphere rises, it will reduce the number of air changes, unless tempering dampers and a by-pass for cold air are employed. Diagram 6 gives a value of (0, 260,360 B. t. u. per hr. 2.20 X 152.0 0m)= = The next step in the problem is to decide whether pipe coils or cast-iron heaters will be used. First we will choose to use "Vento" cast-iron heaters. A velocity of 150 ft. per min. from Diagram 5 gives a value of K = 2.20; 107,150 152.0; and the free area is = 12 sq. ft. If the heating surface is 60 X 150 779 = 65. From Eq. 6, Art. 16b, 12 220 10 = = 65.2 - - 125 (0.1175 X 150 † 152.5) log 220 Values of ƒ may also be read directly from Diagram 6. Start with a value of 0. 01- 230, follow dotted line in direction of arrow up to curve of 0, 02 95, follow dotted line to the right to the cast-iron radiator diagram to where the velocity 150 ft. per min., and on the scale directly above, read the value of ƒ = = = 65.1. N = 65.2. Let f' 1 =sq. ft. of surface per sq. ft. 21.7; N = 4, f' 16.3; and These two values of ƒ agree very well, so we can assume f free area per stack. Then, when N 1, f' = = N5, f' 13.0. The number of stacks may be varied at will by selecting a radiator of appropriate design from Table 20. Values of ƒ are given for each type of radiator. In this case, we will assume 4 stacks deep. Then ƒ'= 65.1 = 16.3. From Table 20, a regular section "Vento" 9 in. wide, 72 in. high, 5 in. c. to c., and with 19 sq. ft. of heating surface per section, has a value of ƒ' = 17.2, making f 17.2 X N = 68.8, or a little above the requirements. The total number of sections in the first stack, = or the first row of stacks as the case may be, is 12 sq. ft. free area = 10.9, say 11 sections. The total number of sections then is 11 X 4 44, and the total heating surface is 44 X 19 =836 sq. ft., or somewhat more than the 780 sq. ft. required. The width of the air spaces should be a governing factor in selecting a type of section for indirect gravity heating, as the air is liable to pass without being heated if the sections are too far apart. The previous selection of |