September 4, 1914. THE COLLIERY GUARDIAN. 527 A COMBUSTION CHART FOR COAL AND OIL FUELS, AND A CHECK ON FLUE, EXHAUST AND PRODUCER CAS ANALYSES.* By Prof. J. D. Cormack, D.Sc. The object of this note is to describe a combustion chart which gives a simple and rapid means of checking flue and exhaust and producer gas analyses, so as to ascertain whether a given gas analysis is consistent with the combustion of a given coal or oil fuel. The chart gives also the air appropriated for the combustion and the excess air used. The composition of the gases is ascertained by a volumetric analysis, and it is unnecessary to reduce this to a weight analysis. Let the fuel, as burned, contain per lb.—C, lb. carbon, H1 lb. hydrogen, and O lb. oxygen, and let the volumetric analysis of the gases be :— Carbon dioxide.......... CO2, x Carbon monoxide ......... CO, y Oxygen.................... O, z Nitrogen ................. N, n Total.... 1 Any sulphur in the fuel is usually very small, requires little oxygen, and does not appear in the gas analysis. The nitrogen in the fuel may also be neglected, as its weight is very small compared with the total weight of the gases. Following the usual assumption, the available hydrogen per lb. of fuel as burned is H1 — g-, and the available hydrogen per lb. of carbon burned is H - ° H1 S TT1 oT...or H • Take the composition of the atmosphere as:— By volume: Oxygen, 21 per cent.; nitrogen, 79 per cent. By weight: Oxygen, 23’2 per cent.; nitrogen, 76*8 percent., the nitrogen being understood to include the inert gases. 11b. of carbon requires, for combustion to CO2, 32 t 8 i.e., g lb. oxygen, ~ and for combustion to CO, 16 4 jg, i.e., gib. oxygen. Therefore the air required for the combustion of lib. of carbon to CO2 is | X i-e*> 11’5 lb., and for combustion to CO, 5*75 lb. 1 lb. of hydrogen requires for combustion to H2O, 16 o', i.e., 81b. of oxygen, and the air required is, there- fore, 8 X i.e., 34*5 lb. In dealing with the combustion of carbon, two facts are notable:— I. When carbon combines with oxygen to form CO2, the volume of the CO2 produced is equal to the volume of the oxygen used. (In all cases volumes are compared at the same temperature and pressure.) When, however, the carbon forms CO, the volume of CO produced is twice the volume of the oxygen used. The following statement can, therefore, be made at once :—The ratio of the volume (or weight) of unused or excess air (or oxygen) to the volume (or weight) of the air (or oxygen) required for the combustion of carbon to C02 = — . x II. Equal volumes of CO2, CO (and CH4) contain the same weight of carbon. Hence it follows that if, of the total carbon C4 burned Cx (1 — S) is ourned to CO2, and CXS is burned to CO— Lr® = * , or 1 — s= —and S = (1) S y x+y x + y Consider the combustion of 1 lb. of carbon having H1 available hydrogen. Let S lb. be burned to CO, and 1 — S to CO2, and let there be m lb. excess air. Tne result of the combustion can be shown in tabular form. Take the volume of lib. of oxygen as V. When a volume V of oxygen is taken from the atmosphere, the 79 corresponding volume of nitrogen is V, or, say, rV. 21 Volume of oxygen required. (1-S) x For the combustion of (1 — S) lb. carbon to CO3 S lb. carbon to CO S x H1 lb. hydrogen toH2O Add mlb. excess air containing Volume Vol. of nitro- of gen corre- products. spending. |v...(l-S)fv...(l-S)f Vr o o o 2S X iv... S X | Vr 8V ... * ( 0*232mV ■< of (. oxygen. * No water appears in the gas analysis. ... H1 x 8Vr and O ’232m V x r of ni rogen. The resulting gas analysis will be (s being the sum of the items in the two right-hand columns):— T = “ ® 2 8 For CO, y = 1 SX|V = 2,8 (3) * s For O, z = °'232toV 0232m V = zs (4) s For N, (1- S) IV + S x V + H1 x 8V + 0232m V n — r----2--------2---------------------- (5) * From a paper read before the Institution of Engineers and Shipbuilders in Scotland. Now from (2) and (3), | V = (» + ?/) s (6) o and substituting from (2), (3), (4), and (6) in (5), we have n — r £ a? 4- -| + z + 8H1x|(«?4-^)j also x + y 4- z + n = 1 n = r fl — — n 4- 3H1 (x + y) 1 ^■2 0'79 11 = 0-79^1- |+ 3H>(» + y)} or n = 0'79 1 + 3 H1 1 1 — 2 (7) When, therefore, the value of H1 and the amounts of CO2 and CO in the flue gases are known, the amount of nitrogen can be found, and the oxygen got by difference, and thus a check on the analysis is obtained. The equation shows that if the CO2 and CO are 791 Fig. 1. accurately ascertained, and if H1 is known, the oxygen and nitrogen can be obtained by a simple calculation. Take as an example the combustion of a coal of analysis :—Carbon, 81 per cent.; hydrogen, 4’5 per cent.; oxygen, 6'4 per cent.; other matters, 8T per cent. The ash per lb. of coal burned was 0'06 lb., and it contained 30 per cent, of carbon. Per lb. of fuel, C burned = 0'81 —0'06 X 0'3 — 0'792 lb. If any hydrogen was in the ashes owing to unburned coal, a correction could, if necessary, be made, but it would be extremely small. 8 0’037 H1 = ~<>792— = 0792 = 0’0467; The amounts of CO2 and CO found in the gases were 0'110 and 0'012 by volume. /. from (7) n = 079 {1-^— + 3 x 0-0467 (0'110 + 0’012) j =0 798 the oxygen must be 1 — 0'110 — 0'012 — 0'798 = 0'08, and the volumetric analysis should be:— CO2, 0'110; CO, 0'012; 0,0'080; N, 0'798. Construction of Chart. The result of the combustion of a fuel containing carbon and hydrogen can be readily shown by a com- bustion chart. (a) Combustion of Carbon to C02. If C is burned to CO2 in air, since the CO2 produced is equal to the O used, the nitrogen percentage remains unchanged. Take squared paper, height 100 units, horizontal and vertical units equal, and draw a line XXx at 79 (see fig. 1). At 100 draw a diagonal line at 45 degs. On XX15 from X, plot off the CO2, a?; then the vertical line ABDF gives the values AB. BD, and D F of the CO2, x; O, z; and N, n. For if XD = x, XD = AB, and since n — 79 = FD, then BD must = z, the oxygen. lib. of carbon, if completely consumed, requires 11'5 lb. of air. From statement I. the O measures the excess air, and if the O and CO2 are equal, the excess air is equal to the air required for the combustion of the carbon to CO2—i.e., in this case 11'5 lb. Taking any vertical line, measuring from XXj, the line X B, produced if necessary, will cut this line, and the inter- cept being proportional to BD, will, therefore, measure the excess air. When XD = D B—i.e., when CO2 and O are equal—XB is at 45 degs., and should cut the vertical line at a point in the scale which reads 11'5. We can use the same scale divisions if we place this vertical line at a horizontal distance of 11'5 from X, and the line at 45 degs. will then cut it at 11'5. This gives a convenient excess air scale per lb. of carbon burned. A number of scales may be placed at convenient distances from X, say, 2 x 11'5 ; A X 11'5, &c. These are shown on fig. 3, and the scales are convenient, as they involve no new horizontal lines. (b) Combustion of 1 lb. of carbon to C02 and CO, H1 available hydrogen also being burned. From (7) n = 0-79 — 079 g + 0 79 x 3H1 (as + y) For the value of H1 draw a line whose ordinates are 0'79 x 3 Hl (x + y), x + y being measured from X along XXn fig. 2. If the ordinates are calculated for x 4- y — 0'21, they will be found to agree very nearly with the scale divisions already adopted. Thus, for x 4- y — 0'21, H1 = 0'1, 0'79 x 3 H1 (x 4- y) — 0'04977, very nearly five divisions. The scale shown on fig. 3 should, therefore, be a little to the right, say, at 21'1. Draw a line whose ordinates are 0'79 X j, the CO line 2 When y = 0'20 the ordinate is 0'079. For any value of x 4- y draw the vertical line A B D G F, the line X D being the H1 line for the fuel as burned. Mark off DE = KL =0’79g. Also GD = 0 79 x 3H1 (as + y). Then GE = G D — D E = 0 79 x 3 H1 (x + y) - 0 79g. FE = 0-79 + GE = 0-79 + 0-79 x3H1 (x + y)—0;79 g . F E = n, and A B = x -I- y. :. B E = 1 — (x 4- y) — n — z. As before, tbe oxygen or B E measures the excess air. Through X draw lines X B and X E, and read off the excess air, per lb. of carbon burned, as the intercept between these lines on any convenient excess air scale. If the air required or appropriated for combustion of the fuel per lb. of carbon burned is IT, then Ti _ 100 .< 8 n qv , 4q . wi^S being burned U — 23 2 ^3 I1 — o) + 3k + H > 0Q = 11-5 (1 - 2 + 3 H1 J = x+y b + — I+3H1 (x+y')' = U-5 + ^A<3H1(a: + 2/)_H = ll-5 + ^A x from (7) ‘ GE = 11'5 4-14'6 x^-q. n — 0 79 = GE andXG= x 4- y To show this on a suitable scale for 1 lb. of carbon burned to CO2, when CO is absent and H1 = O, the ment II., or equations (2) & (3). 8 = TZTZ? ioo£ Fig. 2. scale should read 11'5 lb., the weight of air required for the combustion of 11b. of carbon to CO2. The G E additional air appropriated is 14'6 X and is pro- portional to the ordinates of the line XE. If X G is made 14'6, the additional air is the ordinate at 14'6, and if X G is made 29'2, the additional air is half the ordinate at 29'2. To save confusion the scale is put at 29'2, the line XXx corresponds to 11'5 lb, and the scale shown in fig. 3 is thus determined. The reading on this scale of XE produced gives the appropriated air or the air required, per lb. of carbon burned, for the combustion as it actually took place. In using the chart a line can be drawn for any given value of H1. Some H1 lines are shown for coal and oil fuels. Taking the example already given, the intercept between X B and X E on the excess air scale = 7 45 lb. excess air per lb. of fuel = 7'45 x 0'792 = 5'9 lb. The line X E gives on scale for the air required for combustion 12 6lb. appropriated air, or air for combustion per lb. of fuel = 12'6 x 0'792 = 9'95 lb. The chart may be used in many ways. For example:— (1.) It may be used, as explained above, to check the analysis of the gases, and obtain the appropriated and excess air in any case where the analyses of the fuel as burned, and of the resultant gases, are given. ^2.) It shows the analysis resulting from the com- bustion, complete or incomplete, of a fuel with any given amount of excess air. (3.) It may be used to obtain the air leakage into, say, an economiser. From the analysis at inlet and outlet find the excess air at inlet and outlet, and the difference is the leakage. (4.) It may be used to check the analysis of producer gas, and to determine the air used, &c. An example of (4) may be useful. A series of tests in a producer, air and steam blown, are given by Prof. Bone in the Journal of the West of Scotland Iron and Steel Institute. 1910-11, page 161. The fuel used was Lancashire nut coal, and its analysis and the gas analysis for one test were:— „ . Per Producer gas Per H ilOl ~ cent. (by volume). cent. C! 78-0 CO.. 5-10 H 5'4 CO . 27'3 0 .. 10'0 H 15’5 S 1'0 N 49'05 N 1'4 CH4 3’05 Ash .. 4'2 100 100